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Question: <p>I'm trying to look for a table for magnetic moments for all elements of the periodic table for my research project in computational chemistry, can anyone direct me to a suitable one?</p>
Answer: <p>Unpaired electrons will give rise to a magnetic moment equal to the 'spin only' value <span class="math-container">$\mu_s=g\sqrt{S(S+1)}$</span> in units of Bohr Magnetons, <span class="math-container">$e\hbar/2m_e$</span>, for example H atoms with <span class="math-container">$S=1/2$</span>. <span class="math-container">$g$</span> is the magnetogyric ratio <span class="math-container">$\approx 2$</span>. </p>
<p>There may also be an orbital contribution (from orbital angular momentum) depending on what levels are filled then the equation becomes <span class="math-container">$\mu_{s+L}=\sqrt{4S(S+1)+L(L+1)}$</span> where <span class="math-container">$L$</span> is the orbital angular quantum number, e.g. <span class="math-container">$V^{3+},\,S=1,\,L=3$</span> gives <span class="math-container">$\mu_{s+L}=4.47$</span>. </p>
<p>(Many transition metal complexes have values differing from this formula because of bonding between metal and ligand restricts the orbital angular momentum. You will need to consult texts on Ligand Field Theory and come to grips with the Van Vleck equation to understand this topic in more detail.)</p>
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https://chemistry.stackexchange.com/questions/112948/table-for-magnetic-moments-for-all-elements-in-periodic-table
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Question: <p>Can someone explain me how can I use the periodic table in order to find the specific heat capacities of some elements? (for example, of Aluminium )</p>
<p>The question I have encountered is as follows:</p>
<blockquote>
<p>Given the specific heat capacities of the following liquids, use the periodic table in order to find the specific heat capacities of the metals <span class="math-container">$Al, Fe, Cu, Au $</span> .</p>
<p><span class="math-container">$ H_2 O - 4.18 J /g \cdot C $</span>, <span class="math-container">$ C_2 H_5 OH - 2.46 $</span> , <span class="math-container">$CCl_4 - 0.861 $</span> , <span class="math-container">$CCl_2 F_2 - 0.598 $</span> .</p>
</blockquote>
Answer: <p>Either you have a periodic table that features the specific heat capacity of elements in their standard state at STP (standard temperature and pressure), in which case you merely have to look it up, or you don't, and there's nothing you can do.</p>
<p>Specific heat capacity of materials, even elemental solids, are physical properties resulting from complex behavior of the systems, and cannot be determined by simple laws or equations. They have to be determined experimentally, or by <a href="http://en.wikipedia.org/wiki/Ab_initio_quantum_chemistry_methods" rel="nofollow">first principles calculations</a>.</p>
<hr>
<p>PS: water, methanol, carbon tetrachloride and dichlorodifluoromethane have got nothing to do with it.</p>
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https://chemistry.stackexchange.com/questions/1258/specific-heat-capacity-and-the-periodic-table
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Question: <p>Why do some people say that hydrogen should be above lithium in the periodic table and others argue it should be above fluorine?</p>
Answer: <p>This comes down to the fact that hydrogen has a $1s$ valence shell, capable of holding only two electrons.</p>
<p>In some sense, hydrogen is like the halogens, in that it can achieve the electronic configuration of a noble gas (namely, helium, which has a full $1s$ level) by gaining an electron. This reduction yields the hydride ion, $\ce{H-}$. It should be noted that hydride does not have the same stability as the halogen anions (e.g. hydrogen has an <a href="https://en.wikipedia.org/wiki/Electron_affinity_%28data_page%29">electron affinity</a> of 0.75 eV vs. 3.40 eV for fluorine).</p>
<p>In another sense, hydrogen is like the alkali metals, as it only contains a single electron in its valence shell. Whilst the alkali metals can attain noble gas configurations by losing their valence electrons, it's a bit abstract to talk about hydrogen doing the same by losing its single electron to generate a proton ($\ce{H^{+}}$)(1). Nevertheless there is strong analogy between the electronic structure of hydrogen and the alkali metals which motivates hydrogen being often placed in this group.</p>
<p>Periodic table makers sometimes get away with duplicating hydrogen above the alkali metals <em>and</em> the halogens, or putting it in no group in particular.</p>
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https://chemistry.stackexchange.com/questions/4164/where-hydrogen-belongs-in-the-periodic-table
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Question: <p>In searching online, I've noticed there are a lot of different ways to group the elements of the periodic table.</p>
<p>Take mercury in the two tables linked below, for example:</p>
<ul>
<li><a href="http://i.telegraph.co.uk/multimedia/archive/02101/periodic_2101916b.jpg" rel="nofollow">http://i.telegraph.co.uk/multimedia/archive/02101/periodic_2101916b.jpg</a></li>
<li><a href="http://www.sciencegeek.net/tables/PToE_basic.pdf" rel="nofollow">http://www.sciencegeek.net/tables/PToE_basic.pdf</a></li>
</ul>
<p>On the first table, it's explicitly outside of the transition metals. On the second table, it's included in the "transition metals" group.</p>
<p>Mercury is an example of this inconsistent classification, but in general, why is there no canonical grouping for all elements, across the periodic table? I see some tables mentioning "metalloids," but others not doing so. Or, sometimes elements are classified in entirely different groups altogether (like zinc, cadmium and mercury in the above-linked tables). Or, the lanthanides and actinides being included sometimes in the transition metals group, and sometimes not.</p>
<p>So which classification is right? Or rather, which one is the "most right"? Really: which one should I focus my efforts on remembering?</p>
Answer: <p>There is no "most right classification scheme" for the elements of the periodic table, worth memorizing above all others. This is because there is no single, 'universal' set of criteria that effectively, unarguably, and neatly divides the entire periodic table into groups. Depending on the properties of interest, different classifications/groupings are more or less appropriate.</p>
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https://chemistry.stackexchange.com/questions/33453/periodic-table-groups-which-grouping-is-right
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Question: <p>I am developing a website and later on an app with the periodic table. The reason I started was that I thought almost all the other periodic tables was awfully ugly.
I do not only want to make it a well-looking periodic table, but I want to make it easy and smart to use combined with the design. </p>
<p><strong>My question to you:</strong>
Which features do you require and want from an online periodic table?
I am also asking for the smallest features you can come up with. The more the better!</p>
<hr>
<p>Features I already have got:</p>
<ul>
<li>The periodic table view with number, symbol, name, and type of element (like noble gas)</li>
<li>Detailed information on an element on a specific page for each element including:
<ul>
<li>Number, name, and symbol</li>
<li>Discovery, etymology, and meaning</li>
<li>Electron configuration</li>
<li>Electronegativity</li>
<li>Density</li>
<li>Melting and boiling point</li>
<li>Phase at room temperature</li>
</ul></li>
</ul>
<p>I hope you chemists can help me. By the way, the service will be free of charge to use when it is developed.</p>
<p>Right now it looks like this:
<a href="https://i.sstatic.net/5lWbP.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/5lWbP.png" alt="The Periodic table"></a></p>
<p>and when you click on one of the elements:
<a href="https://i.sstatic.net/lnbTG.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/lnbTG.png" alt="Detailed description of Oxygen"></a></p>
Answer:
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https://chemistry.stackexchange.com/questions/85676/what-do-you-chemists-need-from-a-periodic-table
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Question: <p>In “periodic table”, the adjective is related to the noun <em>period</em>, and <a href="http://www.etymonline.com/index.php?term=periodic&allowed_in_frame=0" rel="noreferrer">comes from</a> Ancient Greek <em>περίοδος</em> through French <em>périodique</em>. In “periodic acid”, it is formed from the prefix <em>per-</em> and <em>iodic</em> (like <em>peroxide</em> and <em>permanganate</em>).</p>
<p><a href="http://en.wiktionary.org/wiki/periodic#Etymology_1" rel="noreferrer">Wiktionary</a> lists their respective UK pronunciations of as <code>/pɪə(ɹ).iˈɒdɪk/</code> and <code>/ˌpɜːraɪˈɒdɪk/</code>, markedly different: <code>pɪə</code> (as in <em>piece</em>) vs. <code>pɜː</code> (as in <em>perfume</em>); then <code>i</code> (as in <em>it</em>) vs. <code>aɪ</code> (as in <em>eye</em>).</p>
<p>However, is that distinction really made in practice? Would a native US/UK/Aussie speaker make the difference when talking in the lab?</p>
Answer: <p>I am an Australian English speaker and yes, this distinction is made in practice.</p>
<p>Possibly the only reasonable opportunity to use the <code>pɪə</code> pronunciation in the name of a compound is in the case of the entertaining molecule <em>periodane</em>, which is actually named after the periodic table. This <a href="http://onlinelibrary.wiley.com/doi/10.1002/qua.20948/abstract" rel="noreferrer">molecule</a> (and <a href="http://onlinelibrary.wiley.com/doi/10.1002/qua.21322/abstract" rel="noreferrer">later</a> a number of different plausible isomers) was identified computationally by a methodology called 'mindless chemistry' which optimises randomly generated molecular graphs. <em>Periodane</em> is a stable configuration of each atom on the second row of the periodic table, with the exception of neon (although some people are <a href="http://www.sciencedirect.com/science/article/pii/S0166128008007677" rel="noreferrer">working</a> on that).</p>
<p><img src="https://i.sstatic.net/qpeqX.gif" alt="enter image description here"></p>
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https://chemistry.stackexchange.com/questions/306/should-one-pronounce-periodic-the-same-in-periodic-acid-and-periodic-table
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Question: <p>I'm familiar with the periodic table / periodic system, but I wonder why it's called "periodic" since there seems not much periodic about (there seems to be little or no predictability of which elements are stable) and from physics I know Z-value of elements and that is not periodic either. So what is a "period" in this context?</p>
Answer: <p>I agree that the "periods" in the periodic table are not mathematically regular. The simplest definition is that a <em>period</em> begins when a new <em>s</em>-subshell starts to fill. Recall that in the <em>s, p, d, f</em> subshells there are 2, 8, 10, and 14 electrons, respectively, so the periods have to get bigger over time and thus cannot be regular. As to the lack of predictability of stability, that may be true, but the periodic table is pretty good at predicting the <em>chemistry</em> of elements by collecting together similar elements into <em>groups</em> (the columns). This similarity is mostly expressed by the <em>s</em> and <em>p</em> subshells, which is where a lot of the reactivity occurs.</p>
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https://chemistry.stackexchange.com/questions/4529/what-is-a-period-of-the-periodic-table
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Question: <p>My physics lecturer was introducing the idea of what a Theory of Everything was, and said something like this:</p>
<blockquote>
<p>Other sciences have their own theories of everything too, fundamental axioms from which everything else can be derived from. In mathematics, we have ZFC, and well, the entire framework of abstract mathematics and axiomatic systems. Even in Chemistry, all of modern chemistry can be derived from the periodic table and the laws of atomic structure. <strong>Well ... except for two words: Organic Chem.</strong></p>
</blockquote>
<p>Is this true? Can organic chemistry somehow not be derived within the framework of the periodic table?</p>
Answer: <p>While all chemistry is unified because all chemical compounds are made from things in the periodic table, that isn't like mathematics and much of physics. In these sciences many important complexities are derivable from a simple set of axioms.</p>
<p>Some chemistry is simple to derive or deduce from the periodic table but not all of it. The electronic properties of iron atoms tell you a lot about many iron compounds; the location of alkali metals in the table tells us a lot about their common properties.</p>
<p>But carbon is odd. The properties of simple compounds follow simple periodic patterns. But there are more organic compounds (consisting of carbon skeletons, usually with C-C bonds) than all other chemical put together. The complexity and variety of carbon chemistry is almost unfathomably large. None of this is obvious or predictable from the periodic table. You can't deduce the enormous variety from the simple properties of carbon atoms or anything given in the periodic table.</p>
<p>That is probably the sort of distinction your lecturer was aiming at. </p>
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https://chemistry.stackexchange.com/questions/61680/does-organic-chemistry-somehow-defy-the-periodic-table
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Question: <p>The periodic table has 7 periods and they have 2,8,8,18,18... elements respectively from 1 to 7. But from what I understand, the periods each state the number of electron shells that the elements in that period has. So if that is the case, shouldn't period 3 have more elements, since it can hold up to 18 electrons, and therefore it can have up to 18 more protons from the largest atomic number element in period 2? (Since period 3 has a M electron shell, it can also have a p orbital, and in total therefore can have 18 elements.)</p>
<p>The same could be said for elements in group 4, group 5, group 6 and so on since they can have f,g, and h orbitals as well. Am I missing something here?</p>
Answer: <blockquote>
<p>So if that is the case, shouldn't period 3 have more elements, since
it can hold up to 18 electrons, and therefore it can have up to 18
more protons from the largest atomic number element in period 2?</p>
</blockquote>
<p>Indeed, elements of the same period have the same number of electron shells, but the "problem" is that in accordance with the <a href="https://en.wikipedia.org/wiki/Aufbau_principle#Madelung_energy_ordering_rule"><em>Madelung/Janet/Klechkowski rule</em></a>, the $\mathrm{4s}$ orbital is occupied <em>before</em> the $\mathrm{3d}$ orbital. As a result, the $\mathrm{3d}$ orbital could be filled only when the $\mathrm{4s}$ orbital is already filled, i.e. only for elements of the 4th period.</p>
<p>Similarly, the $\mathrm{5s}$, the $\mathrm{5p}$, and the $\mathrm{6s}$ orbitals are occupied <em>before</em> the $\mathrm{4f}$ orbital. As a result, the $\mathrm{4f}$ orbital could be filled only when the $\mathrm{5s}$, the $\mathrm{5p}$, and the $\mathrm{6s}$ orbitals are already filled, i.e. only for elements of the 6th period.</p>
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https://chemistry.stackexchange.com/questions/34549/confusion-with-the-periodic-table
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Question: <p>On the periodic table, period 2 and 3, 4 and 5, 6 and 7 and so on have similar blocks, identical length and groups. What are such pairs of analogous periods called?</p>
Answer: <p>In a bit of a web search, there does not seem to be any particular names for families of periods in the Periodic Table. However, according to <a href="http://www.webelements.com/nexus/chemistry_page/periods-periodic-table" rel="nofollow noreferrer">WebElements</a>, </p>
<blockquote>
<p>The f-block elements are assigned to Periods 6 (lanthanoids) and 7 (actinoids) since that is where they are located in the full, or extended, version of the periodic table.</p>
</blockquote>
<p>However, these, according to the image below, only apply to the Lanthanides and Actinides, not all of periods 6 and 7.</p>
<p><img src="https://i.sstatic.net/P5tMr.png" alt="enter image description here"></p>
<p><a href="http://en.wikipedia.org/wiki/Periodic_table" rel="nofollow noreferrer">Image source</a></p>
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https://chemistry.stackexchange.com/questions/7022/pair-of-analogous-periods-on-the-periodic-table
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Question: <p>Why there are three separate columns for 8B group in old version of the periodic table?</p>
Answer:
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https://chemistry.stackexchange.com/questions/152244/my-question-is-regarding-periodic-table-why-there-are-three-columns-for-8b-grou
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Question: <p>What is the most rare element in periodic table? I have a lot of confusion about this.</p>
Answer: <p>Well, the first question to answer would be: where? The rarest element in the universe? the sun? in meteorites? on earth? in the ocean? in humans? We can tap in to the curated datasets provided by Mathematica to get the answer:</p>
<p><img src="https://i.sstatic.net/2wSWX.png" alt="Mathematica graphics" /></p>
<p>Elements up to atomic number 98 have been found in trace quantities; however, the dataset does not contain any information about the trans uranium elements which were <a href="http://en.wikipedia.org/wiki/Synthetic_element" rel="noreferrer">once thought to be synthetic</a>.</p>
<p>Here's a plot of abundance as a function of atomic number:</p>
<p><img src="https://i.sstatic.net/N4Hrv.png" alt="Mathematica graphics" /></p>
<p>We can see that the elements beyond about Z = 82 (lead) have a marked decrease in abundance, and this is consistent with all of the elements with atomic number above 82 being radioactive.</p>
<h1>Where do we come from?</h1>
<p>We can compare the elemental abundance of these individual "spaces" to speculate on where humans come from. Below are plots showing the correlation of the elemental abundances. The plot labels give the best fit equation:</p>
<p><img src="https://i.sstatic.net/EUPeq.png" alt="Mathematica graphics" /></p>
<p>The best slope is observed when comparing our abundance to that of the Earth's crust; however it does appear as if the scatter is smallest for the Us vs. the Ocean plot.</p>
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https://chemistry.stackexchange.com/questions/7462/what-is-the-most-rare-element-in-periodic-table
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Question: <p>We know that atomic radius decreases along a period and increases along a group. (Same goes for metallic character)</p>
<p>But if we take 2 elements A & B(A is at somewhere in the top and left in periodic table and B is somewhere is below and right of B) then which among A and B will have greater atomic radius and more metallic character?</p>
Answer: <p>If the period effect for B overrules the group effect, B will have the greater atomic radius than A.</p>
<p>If the group effect for B overrules the period effect, B will have the smaller atomic radius than A.</p>
<p>The same for the element metallic character.</p>
<p>For the p elements, you can see the metalicity boundary goes somewhat diagonally.</p>
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https://chemistry.stackexchange.com/questions/163493/confusion-regarding-trends-in-modern-periodic-table
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Question: <p>I came across the following question:</p>
<blockquote>
<p>If each orbital can hold a max. of 3 e– what is the number of elements in the 4th period of the periodic table? </p>
</blockquote>
<p>I was unable to even start the thought process of answering the question because I'm rather horrible at discerning the period number using the electronic configuration, the number of elements that can be accommodated in a period etc. If someone could please explain it to me, that would be so great, especially since I've googled this too many times already and haven't been able to wrap my head around it yet! </p>
<p>Thanks in advance :) Cheers! </p>
Answer: <p>In the 4th period of the periodic table there is 1 s orbital, 3 p orbitals, and 5 d orbitals. That's a total of 9 orbitals. $9 \times 3=27$</p>
<p>That's 27 elements.</p>
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https://chemistry.stackexchange.com/questions/57572/electronic-configuration-and-the-periodic-table
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Question: <p>My teacher said that on the periodic table there is a "nose" formed by Al, Zn, Ag, and Cd. She said that they are all fixed charged (+3, +2, +1, and +2 respectively), and said that if I write them in ionic equations, I just say Silver Nitrate instead of Silver (I) Nitrate. She also said to put all Al as +3 charge in all cases, etc. But I did some research and found out that you do say Silver (I) Nitrate (actually both are fine, but my teacher specifically said <strong>never</strong> put the (I) in such cases). So is my teacher wrong? And are there cases where the charges of the elements are not what my teacher says?</p>
Answer: <p>All of these elements can form compounds in other oxidation states. Aluminium forms some compounds in the +1 state (e.g. see the section in <a href="https://en.wikipedia.org/wiki/Aluminium_iodide" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Aluminium_iodide</a>), as does Zinc (see the section in <a href="https://en.wikipedia.org/wiki/Compounds_of_zinc" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Compounds_of_zinc</a>) and Cadmium (e.g. <a href="https://en.wikipedia.org/wiki/Cadmium(I)_tetrachloroaluminate" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Cadmium(I)_tetrachloroaluminate</a>). But Silver is the element that shows this most often in the list given - for instance it forms fluorides in the +1, +2 and +3 oxidation states (<a href="https://en.wikipedia.org/wiki/Silver_fluoride" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Silver_fluoride</a>). As such I wouldn't say including the oxidation states in the formula is wrong, but on the other hand if you omit it everybody who knows there is an ambiguity will understand what is implied.</p>
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https://chemistry.stackexchange.com/questions/123729/the-nose-of-the-periodic-table
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Question: <p>Reading about heavy water, I have found that the chemical formula for it can be written 2H<sub>2</sub>O or D<sub>2</sub>O.</p>
<blockquote>
<p>Deuterium is an isotope of hydrogen that is found in large quantities in water, more than one atom per ten thousand hydrogen atoms has a deuterium nucleus. The isotope is denoted “2H” or “D”, and is normally known as “heavy hydrogen”. Deuterium is used in a number of conventional nuclear reactors in the form of heavy water (D<sub>2</sub>O), and it will probably also be used as fuel in fusion reactors in the future (Source: <a href="https://www.sciencedaily.com/releases/2009/05/090511181356.htm" rel="nofollow noreferrer">ScienceDaily</a>).</p>
</blockquote>
<p>Encyclopædia Britannica <a href="https://www.britannica.com/science/deuterium" rel="nofollow noreferrer">says the formula can be</a> <sup>2</sup>H<sub>2</sub>O. Is this correct?</p>
<p>2H (or <sup>2</sup>H) = D is interesting to me as D is not in the periodic table of elements.</p>
<p>I thought that formulae should be formed using the symbols within the periodic table. What <strong>are</strong> the rules surrounding these chemical formulae?</p>
Answer: <p>According to this entry in the IUPAC Gold book</p>
<p><a href="https://goldbook.iupac.org/html/D/D01648.html" rel="noreferrer">https://goldbook.iupac.org/html/D/D01648.html</a></p>
<p>the symbol for Deuterium is H together with the isotopes notation. </p>
<p>However deuterium and tritium are routinely indicated by D and T respectively. Their use is dictated by the framework. Besides nuclear reactions, when only straightforward chemical properties* are considered, it is convenient to use a single letter rather than H and super/sub-scripts.</p>
<p>In organic nomenclature, while writing names for molecules, the presence of deuterium is signalled by the symbol (d).</p>
<p>*note that isotopic effect might however affect chemical properties at fine levels. The toxicity of heavy water is a notable striking example. </p>
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https://chemistry.stackexchange.com/questions/112352/chemical-formulae-and-the-periodic-table-of-elements
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Question: <p>I was wondering why the elements in the periodic table were disposed the way they are. I understand, of course, that they are put in increasing atomic number fashion, but I'd like to know more about the topic. Could you link some reference that examine in detail the structure of the periodic table?</p>
Answer: <p>I do not address the question of history, of Mendeleyev's investigations, that is answered in <a href="https://chemistry.stackexchange.com/questions/15883/how-was-mendeleev-able-to-develop-his-table">How was Mendeleev able to develop his table?</a></p>
<p>Modern understanding of basic principles follows. Let us have a piece of periodic table</p>
<pre><code> ... A0 B0 ...
... A1 B1 ...
.............
</code></pre>
<p>Element "B0" must have the next atomic number after "A0", <em>i.e.</em> one more electron. It is different chemically, although may have (albeit not necessarily) similar physical properties.</p>
<p>Element "A1" must be <em>chemically</em> similar to "A0", although physical properties may differ significantly (as a general rule, ".1" elements are denser than ".0"). It happens that for each element there exist its "analog" in the next period, that has similar chemical properties (based on similar outer-shell electron configuration). This pattern, once noticed, permitted for good predictions (search for Mendeleyev's "eka" elements in Internet for more information).</p>
<p>Periods have different lengths. Analog's atomic number may be 8, 18, or 32 ahead off the prototype's atomic number (the number of electrons expended to "raise" all the configuration one shell up). This difference is due to quantummechanical effects.</p>
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https://chemistry.stackexchange.com/questions/15926/according-to-what-criteria-was-the-periodic-table-of-elements-ordered
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Question: <p>Which is the most reactive element in the periodic table? Is it francium, caesium, lithium or fluorine?</p>
Answer: <p>Fluorine is the most reactive non radioactive element on periodic table.</p>
<p>It exists in gaseous form at room temperature and even reacts with glass. So it is almost impossible to store it in pure form.and caesium is the most reactive metal</p>
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https://chemistry.stackexchange.com/questions/53831/which-is-the-most-reactive-element-in-the-periodic-table
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Question: <p>I wish to know the trends in the periodic table and why that trend exists, for the following parameters:</p>
<ol>
<li>Atomic size</li>
<li>Ionization energy</li>
<li>Metallic character</li>
<li>Positive ion size</li>
<li>Negative ion size</li>
<li>Electronegativity</li>
</ol>
<hr>
<p>My progress (if my reasoning is wrong for any of these please correct me):</p>
<ol>
<li><p>Atomic size - I memorized from my class lecture that the greatest atomic size is towards the bottom-left of the periodic table. I think the reason for this is because as you go down the groups (IA, IIA, etc.) you add more electrons. Adding more electrons means the atom will logically grow bigger. I am not sure why the answer is not bottom-right, though. </p></li>
<li><p>Ionization energy - This is the energy required to remove an electron from an atom. If there are more electrons then the electron will be further from the nucleus and by Coulomb's law, since the distance increases, the force attraction that keeps the electron in the nucleus's orbit is weak. Thus, the ionization energy is the highest toward the top-left.</p></li>
<li><p>I have no clue what to think for metallic character.</p></li>
<li><p>Positive ion size I know the answer is that it increases towards the bottom left, but I don't know why.</p></li>
<li><p>Negative ion size I know the answer is that it increases towards the bottom left, but I don't know why.</p></li>
<li><p>Electronegativity I am pretty sure is highest at F which is 4.0 and is lowest towards the bottom left. So the trend increases towards the top-right.</p></li>
</ol>
Answer: <p><ol>
<li>You are correct on atomic size being due to number of electrons
and their <em>shells</em> and the reason why atomic size decreases from
left to right is due to the number of the protons being greater
on the right than on the left of the periodic table with same number of shells due to shell theory. <a href="http://en.wikipedia.org/wiki/Atomic_radius">(Reference)</a>.</li>
<li>The concept you are forgetting is
<a href="http://en.wikipedia.org/wiki/Valence_electron">Valence Electrons</a>.
Atoms are stabilized most often when there outer most shell is
filled with 8 valence electrons (except for hydrogen and helium with
0 or 2). The elements on the left side of the periodic table lose
electrons easily since they have a very large atomic radius allowing
for lower ionization energy. on the right side of the periodic
table(nitrogen, oxygen , and halogens) have a smaller atomic radius
creating a stronger pull making it harder to lose an electron thus
they will steal an electron to become stable. </li>
<li>The best way to summarize this is to quote Chemistry.about.com </p>
<blockquote>
<p>"There are trends in
metallic character as you move across and down the periodic table.
Metallic character decreases as you move across a period in the
periodic table from left to right. This occurs as atoms more readily
accept electrons to fill a valence shell than lose them to remove
the unfilled shell. Metallic character increases as you move down an
element group in the periodic table. This is because electrons
become easier to lose as the atomic radius increases, where there is
less attraction between the nucleus and the valence electrons
because of the increased distance between them."<br>
<a href="http://chemistry.about.com/od/periodicitytrends/a/Metallic-Character.htm">(Reference)</a></li>
<li>Positive ion size decreases due to the lack of electrons which allows for the protons to pull stronger on fewer electrons.</li>
<li>The inverse applies for negative ion size. Due to the fact that there are more electrons, the size will increase since they will occupy more space.</li>
<li>Electronegativity is the pull a elements has on electrons. This determines how it bonds with other elements i.e.(whether it shares or looses electrons). If a element has over a 1.7 Electronegativity difference then these the element with the smaller electronegativity will give away several electrons 1-4. Elements on the left side of the periodic table have the smallest Electronegativity due to their atomic size and inability to gain enough electrons to be stable ions. Elements from left to right in a period show increasing elctronegativity, and electrons form up to down in a group will have a lower electronegativity. Also fluorine has a 3.98 elctronegativity. <a href="http://answers.yahoo.com/question/index?qid=20080924041039AAyrZDW">(Reference)</a>.</li>
</ol></p>
</blockquote>
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https://chemistry.stackexchange.com/questions/2843/what-trends-exist-in-the-periodic-table
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Question: <p>I have often wondered about diagonal relationships between elements on the periodic table, and the most often cited explanations revolve around charge-density considerations.</p>
<p>But other than that, what other factors could possibly contribute to this phenomenon?</p>
<p><strong>EDIT:</strong> What I mean by "diagonal relationships" is that there are certain similarities in chemical properties that have been observed in diagonally adjacent neighbors in the 2nd and 3rd period example: Li-Mg, Be-Al, and B-Si.</p>
<p><a href="http://en.wikipedia.org/wiki/Diagonal_relationship">Link to a wikipedia article (not much to be found here though)</a></p>
<p>The explanation that I personally have most often encountered is that the charge-to volume ratios (charge density), say for Li-Mg cations is roughly the same, and hence could account for some of the similarities in behaviors.</p>
<p>What I am looking for is what can be some <em>other</em> contributing factors (if any) to this phenomenon. I am not expecting very concrete answers because I believe this phenomenon is not very well understood.</p>
Answer: <p>What I remember from studying C-P 'diagonal relationships' (for multiple bonds e.g. carbene vs phosphinidene, alkyne vs phosphaalkyne) is that similar electronegativity also played a role, also by affecting the valence orbitals.</p>
<p>However, perhaps it is better viewed from a different vantage point:
the main group elements behave rather similarly from 3rd row onwards, but there is a distinct difference between the 2nd and 3rd row elements and the explanation is pretty straightforward:
The 2s and 2p orbitals are roughly similar in 'size' since the 2p orbitals are the first p-orbitals and are pretty compact. Therefore 2s and 2p orbitals mix happily.
Going to the third row, the 3p orbitals need to be orthogonal to the 2p orbitals and as a consequence they are much more diffuse than the 3s orbitals, resulting in less efficient s-p hybridization. This also manifests itself in the so-called 'inert pair' effect whereby e.g. Si will resist hybridization in favor of a lone pair. There is some beautiful work by Knutzelnigg from the 80s (?) describing these effects.
This 2nd-3rd row discrepancy is somewhat counterbalanced if you step to right, increasing Z and thereby contracting the orbitals giving rise to this apparent diagonal relationship.</p>
<p>Similar arguments make that for the 3d TMs the hybridization is more efficient for 4s-3d than for the 4d/5d TMs. Consequently, 3d TMs tend to behave differently from their heavier sisters, which in turn are closer together (although for the 5d's the relativistic effects start to kick in more heavily...).</p>
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https://chemistry.stackexchange.com/questions/32591/diagonal-relationships-in-the-periodic-table
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Question: <p>Hydrogen is not in the first group as it was before and it is now placed above the periodic table? So why is it still categorized as a metal in some books?</p>
Answer: <p>There is a lot that could be said about this question, but a simple explanation goes like this:</p>
<ul>
<li><p>Some periodic tables categorize hydrogen as a metal (or at least an alkali metal) because it has one electron in its outermost orbital - just like Lithium, Sodium, Potassium, etc., which are obviously metals.</p></li>
<li><p>Other authors of chemistry textbooks realize that hydrogen seems to love the gas-phase so much over the solid phase that it just doesn't make sense to call it a metal, irrespective of what atomic orbital pattern may tell us.</p></li>
</ul>
<p>Hence the differences between authors of texts and where they put Hydrogen on their periodic table.</p>
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https://chemistry.stackexchange.com/questions/55464/what-is-the-position-of-hydrogen-in-the-periodic-table
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Question: <p>I am studying Periodic Table and have doubts in the group 3 (III-B) periods 6 and 7.</p>
<ol>
<li>If f-block is not a part of d-block, then why so they place inner transition metals in the boxes of group 3 periods 6, 7?</li>
<li>Since they place so, why are sometimes La and Ac placed while sometimes Lu and Lr are placed? (are these 2 sets of elements part of Group 3, even if the other inner transition elements in between are not!?)</li>
<li>Are all the other f-block elements in between the extreme element La and Lu, or between Ac and Lr, part of the Group 3? (regardless of whether this is anomaly because they are not part of d-block)</li>
</ol>
Answer: <p>I can at least answer your first question, 1). The periodic table developed by Mendeleev is an incredibly useful tool for predicting the physical and chemical properties of elements, but you must understand that it has limitations. I linked the wikipedia page on the history of the periodic table. The page under "Mendeleev's predictions and inability to incorporate the rare-earth metals" is a quick summary of why the periodic table arranges lanthanoides and actinoides the way it does. It also has plenty of links for further reading. <a href="https://en.wikipedia.org/wiki/History_of_the_periodic_table#Mendeleev%27s_predictions_and_inability_to_incorporate_the_rare-earth_metals" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/History_of_the_periodic_table#Mendeleev's_predictions_and_inability_to_incorporate_the_rare-earth_metals</a></p>
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https://chemistry.stackexchange.com/questions/185425/periodic-table-group-3-period-6-7
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Question: <p>Normally with the periodic table the lanthanide series is separated out because it's long and would make the table wide. I looked for an expanded version and found this:</p>
<p><a href="https://i.sstatic.net/VH9ax.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/VH9ax.png" alt="enter image description here"></a></p>
<p>I found it kind of strange that scandium and yttrium are above the lanthanide and actinide series on the far left.</p>
<p>Does it make sense for those two elements to be there? I always assumed each column had certain characteristics like same number of valence electrons, that there were certain characteristics that held true for the whole table, but in this format I don't know how to understand the patterns or why the table ends up taking this shape. </p>
Answer: <p>La and Ac have <span class="math-container">$d^1$</span> electrons in their valence shells, rather than <span class="math-container">$f^1$</span> electrons.</p>
<p>The long table you found looks like that for a several reasons.</p>
<p>The trends going down Sc-Y-La are like those seen in groups 1 and 2. The trend going down Sc-Y-Lu is like that of groups 4 to about 10. Since lanthanide chemistry is basically that of trivalent alkali or alkaline earth metals, this tips the balance in favour of group 3 as Sc-Y-La.</p>
<p>Also helping is that group 4 is the first in which the really characteristic properties of transition metals (variable oxidation states; colour; paramagnetism) are seen. Ditto, same thing happens with group 12, which is why they're shown as post-transition metals.</p>
<p>And there's the lanthanide contraction, which starts at Ce and finishes at Lu. If you show Lu as being a transition metal it mushes the end of the Ln contraction into the d-block.</p>
<p>And there's the periodic law, which says that the chemical elements, if arranged according to their atomic numbers, show an approximate repetition of properties after certain regular but varying intervals. Here, La is the first element after Y that shows the approximate repetition in properties.</p>
<p>This research supports the periodic law outcome:</p>
<ul>
<li>Glawe H, Sanna A, Gross EKU & Marques MAL 2016, “The optimal one dimensional periodic table: a modified Pettifor chemical scale from data mining”, <em>New Journal of Physics,</em> vol. 18, 093011, <a href="https://iopscience.iop.org/article/10.1088/1367-2630/18/9/093011/pdf" rel="nofollow noreferrer">https://iopscience.iop.org/article/10.1088/1367-2630/18/9/093011/pdf</a></li>
<li>Restrepo G 2017, "Building classes of similar chemical elements from binary compounds and their stoichiometries", in MA Benvenuto (ed.), <em>Elements old and new: Discoveries, developments, challenges, and environmental implications,</em> American Chemical Society, Washington DC, pp. 95-110</li>
</ul>
<p>Both sources point to La being more distinct from the Ln than is the case for Lu.</p>
<p>Finally, if you count the differentiating electron discrepancies in each block of the periodic table, noting the above table has a split-d block, you'll find that an Sc-Y-La table has 12 such discrepancies whereas an Sc-Y-Lu table has 13. The current <em>agreed-within-IUPAC</em> table has 14 differentiating electron discrepancies!</p>
<p>As many people have a hard time accepting the idea of a split-d block, I personally feel that a better solution would be a Sc-Y-Lu table with group 3 being shown as Sc-Y-Lu-Lr <em>and</em> La-Ac. That would have the extra benefit of resolving long-standing discussions about the composition of group 3, which have been going on since about 1982, when Jensen published his paper in JChemEd, arguing for Sc-Y-Lu:</p>
<ul>
<li><a href="http://www.che.uc.edu/jensen/w.%20b.%20jensen/reprints/018.%20la%20vs%20lu.pdf" rel="nofollow noreferrer">http://www.che.uc.edu/jensen/w.%20b.%20jensen/reprints/018.%20la%20vs%20lu.pdf</a></li>
</ul>
<p>Nothing much has to change. Just add a 3 above La-Ac in the f-block, in addition to the one over Sc-Y-Lu-Lr in the d-block.</p>
<p>A chemistry book chapter on this group 3 would make fascinating reading (a good thing given, to date, that group 3 is supposed to be the least studied group).</p>
<p>Hope that helps.</p>
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https://chemistry.stackexchange.com/questions/99223/what-is-the-true-depiction-of-the-periodic-table
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Question: <p>What are the different diagrams/tables used to organize the elements other than the Periodic Table? What are the advantages/disadvantages of each?</p>
Answer: <p>One periodic table that apparently physicists find useful is ADOMAH periodic table, for its usefulness is finding electron configuration of a particular element. More information at this <a href="http://www.perfectperiodictable.com/userguide" rel="nofollow noreferrer">guide</a>. A disadvantage to this table, along with any new table configuration you have not experienced, will be the lack of knowledge in the trends on your new table. There are many spiral looking tables, but the trends such as the change in atomic radii, or electronegative characteristics, might not be as obvious, and you might have to find a guide or relearn old tricks to methods you would learn in your first year of chemistry. At least the ADOMAH Table is built in such a way that these trends should still be recognizable, although clearly not the same as the normal table you find in any high-school science classroom.</p>
<p>Of particular interest to me are dynamic periodic tables. These are tables that are basically computer applications which will change according to what you are looking for. One outstanding example is the widely popular <a href="http://www.ptable.com/" rel="nofollow noreferrer">ptable.com</a> which will maintain all the common trends you probably know already, and dynamically show you various states of the elements when changing the temperature. I learned from this table that rhenium has the highest known boiling temperature, while I thought it was tungsten (although tungsten has the highest melting temperature of the transition metals). And better yet, carbon has the highest melting temperature of all elements, very cool table! The disadvantage is obviously that you cannot carry it around in your notebook without some flavor of a mobile computer and/or internet connection.</p>
<p>I am sure there are many other great tables still to be found, but I hope you have learned something new from my post.</p>
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https://chemistry.stackexchange.com/questions/901/what-are-the-alternatives-to-the-periodic-table-of-the-elements
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Question: <p>I was studying about the periodic table recently, and was reading a topic associated with oxides of halogens, and came across the following line</p>
<blockquote>
<p>The bromine oxides, $\ce{Br2O}$, $\ce{BrO2}$, $\ce{BrO3}$ are the least stable halogen oxides (<strong>Middle row anomaly</strong>) and exist only at low temperatures. They are very powerful oxidizing agents.</p>
</blockquote>
<p>So, I went to search this on Google, and found these lines</p>
<blockquote>
<p>It refers to the instability of oxides of bromine as compared to relative stability of oxides of chlorine and Iodine at room temperature, the former being stable only at low temperatures.</p>
</blockquote>
<p>But the line above is simply restating what the book had already told.
So, is their any specific reason why this happens, or is this only due to the experimental data we have gathered?</p>
Answer: <p>This is due to the transition metal contraction. </p>
<p>Bromine has the electron configuration $\ce{[Ar] 4s^{2} 3d^{10} 4p^{5}}. $The 3d orbital has no radial nodes and is therefore quite contracted (close to the nucleus), so there is relatively little repulsion between the 3d electrons and the 4p electrons. This makes it much harder to acheive high oxidation states of bromine (I, IV, VI for the examples you give) because the ionization energies are higher than you might expect from a simplistic approach to periodic trends. </p>
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https://chemistry.stackexchange.com/questions/31113/middle-row-anomaly-of-the-periodic-table
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Question: <p>How can one read from the periodic table the number of outer shell electrons that an atom has, to predict how these atoms will make bonds with other atoms? For example to see that hydrogen ($\ce{H}$) has 1 electron free, while carbon ($\ce{C}$) has 4 and oxygen ($\ce{O}$) has 2? This allows us to infer that carbon can make four bonds with 4 different hydrogen atoms. It's not clear what in the periodic table would tell us this; it is neither in the group nor in the period of an atom, it seems.</p>
<p>Related to this: can someone summarize concisely all the properties (like number of free outer shell electrons) that one can read off about each element from its position in the periodic table? I was having a hard time finding a concise summary of these.</p>
Answer: <p>Start with a periodic table that shows the electron configurations. You might try the <a href="http://www.ptable.com/" rel="nofollow">Dynamic Periodic Table</a>; as you mouse over an element, its electron configuration is presented. </p>
<p>Then go to <a href="http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/intro2.htm" rel="nofollow">Electron Configurations in the Periodic Table</a>, <a href="http://www.mikeblaber.org/oldwine/chm1045/notes/Struct/EPeriod/Struct09.htm" rel="nofollow">Electronic Structure of Atoms</a> or <a href="http://en.wikipedia.org/wiki/Block_%28periodic_table%29" rel="nofollow">Block (periodic table)</a> to see how shells and orbitals are added. It's pretty straightforward until the transition metals, and things get much more complicated for the lanthanides and actinides, where a new shell may start before an inner shell is complete.</p>
<p>The general rule is: an orbital is more "stable" when filled, or half-filled, so fluorine, for example, can readily accept one more electron to finish its outer <strong>2s2 2p5</strong> shell and it will then have the configuration of neon, <strong>2s2 2p6</strong>. Many periodic tables don't present the inner shells, using instead the shorthand of <strong>[Ne] 3s2 3p5</strong> for chlorine, rather than the full <strong>1s2 2s2 2p6 3s2 3p5</strong>.</p>
<p>In fact, you're not alone in trying to understand the electron structure; see the current debate on the position of Lw in the table: <a href="https://www.sciencenews.org/article/new-data-synthetic-element-trigger-rethink-periodic-table" rel="nofollow">New data on synthetic element trigger rethink of periodic table</a>.</p>
<p>BTW, note there are different ways a shell can be completed. Carbon will <strong>share</strong> its four outer electrons with four hydrogen atoms; this is called a <em>covalent</em> bond. Chlorine outright "steals" an electron from sodium; this is an ionic bond. See <a href="http://www.kentchemistry.com/links/bonding/covalentlewisdot.htm" rel="nofollow">Covalent Lewis Dot Structures</a> and <a href="https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/basic-concepts-of-chemical-bonding-9/the-ionic-bond-72/ionic-bonding-and-electron-transfer-335-1484/" rel="nofollow">Ionic Bonding and Electron Transfer</a> for more info.</p>
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https://chemistry.stackexchange.com/questions/31615/reading-number-of-outer-shell-electrons-and-other-properties-from-periodic-table
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Question: <p>there are 118 currently existing elements on the periodic table, and recently I've found like 77 elements that quote on quote "may exist" on our periodic table. Is this true?</p>
Answer:
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https://chemistry.stackexchange.com/questions/89409/is-there-any-evidence-that-there-are-75-extra-elements-on-the-periodic-table
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Question: <p>I am searching about some basics in chemistry. I was looking for the molecular structure of all periodic table element molecules.
eg: Hydrogen molecule: as <span class="math-container">$\ce{H2}$</span>; structure:<img src="https://i.sstatic.net/lwUhA.jpg" alt="enter image description here" /></p>
<p>I was able to find out several others too. But when I reach to certain molecules like <span class="math-container">$\ce{He}$</span> I was not able to find out the structure. <br/>
Does anyone know from where I can get the molecular structure of all periodic table elements. And if not, could you please explain the case of Helium and other noble gases. I am very poor in chemistry. Anyone showing me a reference on how to find the answer is welcome. Thanks in advance, and Happy New Year to all... :)</p>
Answer: <p><img src="https://i.sstatic.net/4R93I.jpg" alt="" /></p>
<p><img src="https://i.sstatic.net/tA50z.png" alt="" /></p>
<p>The noble gases (group <span class="math-container">$18$</span>) are rather inert, so their molecule is just a single atom.</p>
<p>The following elements <strong>usually</strong> consist of molecules made up of two atoms: <span class="math-container">$\ce{H2}$</span>, <span class="math-container">$\ce{N2}$</span>, <span class="math-container">$\ce{O2}$</span>, <span class="math-container">$\ce{F2}$</span>, <span class="math-container">$\ce{Cl2}$</span>, <span class="math-container">$\ce{Br2}$</span>, <span class="math-container">$\ce{I2}$</span>.</p>
<p>Astatine (<span class="math-container">$\ce{At}$</span>) should also belong to that list but it is not confirmed because of its short half-life.</p>
<p>Sulfur exists as (<span class="math-container">$\ce{S8}$</span>); phosphorus exists as (<span class="math-container">$\ce{P4}$</span>).</p>
<p>The structure of boron (<span class="math-container">$\ce{B}$</span>), arsenic (<span class="math-container">$\ce{As}$</span>), selenium (<span class="math-container">$\ce{Se}$</span>), antimony (<span class="math-container">$\ce{Sb}$</span>), and tellurium (<span class="math-container">$\ce{Te}$</span>) are rather complicated.</p>
<p>Carbon (<span class="math-container">$\ce{C}$</span>) has too many forms, but its most common forms are graphite and diamond, which is demonstrated below.</p>
<p>Silicon (<span class="math-container">$\ce{Si}$</span>) and germanium (<span class="math-container">$\ce{Ge}$</span>) both assume the structure of diamond.</p>
<p>Metals all adopt a metallic structure. Read more <a href="http://www.chemguide.co.uk/atoms/structures/metals.html" rel="nofollow noreferrer">here</a>.</p>
<hr />
<h2>Further</h2>
<p>You might go to <a href="https://en.wikipedia.org/wiki/Allotropy" rel="nofollow noreferrer">here</a> to learn about the elements I skipped.</p>
<hr />
<h2>Gallery</h2>
<p>Sulfur:</p>
<p><img src="https://i.sstatic.net/g4mUCm.png" alt="" /></p>
<p>Phosphorus:</p>
<p><img src="https://i.sstatic.net/AFR5Om.jpg" alt="" /></p>
<p>Carbon (as graphite):</p>
<p><img src="https://i.sstatic.net/WfeFGm.png" alt="" /></p>
<p>The common structure of diamond, silicon, and germanium (left diamond, right graphite):</p>
<p><img src="https://i.sstatic.net/umglz.jpg" alt="" /></p>
<hr />
<p>Credits:</p>
<p>All images are from Wikipedia.</p>
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https://chemistry.stackexchange.com/questions/22228/molecular-structure-of-all-periodic-table-element-molecules-exceptional-cases-et
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Question: <p>According to the binding energy per nucleon vs mass number graph, it is observed that iron-56 has the maximum value of binding energy per nucleon ($\pu{8.75 MeV}$). It means that iron-56 is the most efficiently bound nucleus meaning that it has the least average mass per nucleon. This is the approximate basic reason why iron and nickel are very common metals in planetary cores since they are produced profusely as end products in supernovae and in the final stages of silicon burning in the stars. So, in one word, iron is quite stable. </p>
<p><a href="https://i.sstatic.net/PG3WB.gif" rel="noreferrer"><img src="https://i.sstatic.net/PG3WB.gif" alt="Nucleon binding energy by number of nucleons in the nucleus of a given atom"></a></p>
<p>But, what about helium and other noble gases? They are considered the most stable elements in the whole periodic table. But their binding energy per nucleon value is less than iron-56. So, they are not stable as iron-56. Is that true? <strong>Is iron the most stable element in the periodic table both structurally and chemically?</strong></p>
Answer: <p>Yes, $^{56}\ce{Fe}$ has the most stable nucleus, and $\ce{He}$ is the most chemically inert element. These are different and unrelated qualities, pretty much like physical fitness and intelligence in a man. As for structural stability, there is no such thing in chemistry (there is one in architecture and another in mathematics, but those are out of scope of this question).</p>
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https://chemistry.stackexchange.com/questions/40407/is-iron-the-most-stable-element-in-the-periodic-table
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Question: <p>Why is there a diagonal line cutting through some of the nonmetals and metalloids on the periodic table in groups IIIA, IVA, VA, VIA, and VIIA?</p>
<p><img src="https://i.sstatic.net/3zpgH.jpg" alt="Periodic table of elements"></p>
<p>Is there any historical background about it?</p>
Answer: <p><strong>The red line is the <a href="http://en.wikipedia.org/wiki/Dividing_line_between_metals_and_nonmetals" rel="noreferrer">dividing line between metals and non-metals</a></strong>. It is rather arbitrary; as Mendeleev himself wrote: <em>“It is...impossible to draw a strict line of demarcation between metals and nonmetals, there being many intermediate substances.”</em> Most modern periodic tables do not feature it.</p>
<p><strong>The elements with a blue background are the <a href="http://en.wikipedia.org/wiki/Metalloid" rel="noreferrer">metalloids</a></strong>. Against, it's a rather arbitrary choice, but it corresponds to elements that have some properties of metals. Wikipedia describes them in the following way:</p>
<blockquote>
<p>Physically, metalloids usually have a metallic appearance but they are brittle and only fair conductors of electricity; chemically, they mostly behave as (weak) nonmetals. They can, however, form alloys with metals. Ordinarily, most of the other physical and chemical properties of metalloids are intermediate in nature</p>
</blockquote>
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https://chemistry.stackexchange.com/questions/6116/diagonal-line-going-through-groups-iiia-via-on-the-periodic-table
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Question: <p>I am looking for a full set of data on every element on the periodic table. (and if possible those only speculated about)</p>
<p>When I say full I mean everything. Electron configurations, Valence electrons, melting point, Atomic mass, Isotopes and what they decay into and all the rest.</p>
<p>Is there such a database anywhere?</p>
Answer: <p>You can find all these information on <a href="http://www.wolframalpha.com" rel="nofollow">Wolfram Alpha</a>. You can use the website directly to obtain these information, and you can call their <a href="http://products.wolframalpha.com/api/" rel="nofollow">API</a> to directly fetch the data online from your own program. Of course, you can also use Mathematica to write your program and query the database with <a href="http://reference.wolfram.com/mathematica/ref/WolframAlpha.html" rel="nofollow">WolframAlpha</a> function.</p>
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https://chemistry.stackexchange.com/questions/7189/a-full-verion-of-the-periodic-table-for-computer-game
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Question: <p>Why exactly does hydride acidity increase across period and down group in periodic table? What is the explanation with respect to electrons? I can't figure this out because for 1st period etc H is an anion while for the right and middle of the periodic table H is a cation in compounds formed. I feel confused. Does this have something to do with electronegativity of element that forms hydride?</p>
Answer: <p>Yes, but that is somewhat circular reasoning, as electronegativity is largely just putting a numerical value to the trends in properties you are observing. A better answer is to compare the effective nuclear charge holding on to the valence electrons as you go across or down the Table. Going across the table, each electron gets added to the same shell, which means it shields the other electrons from the added proton poorly. Consequently, Z* (the positive charge effectively felt by the outermost electrons) increases, and these electrons are more tightly held. At some point the atom holds onto not only its own valence electrons, but those of the H atom when the H atom leaves, and so the H atom leaves as H+. Conversely, at the far left of the table, the Group 1A atoms hold on so loosely to their lone electron that the H atom is able to leave with the metal's electrons, as H-.</p>
<p>Going down the table, you add a whole electron shell between the nucleus and outermost electrons with each period. That is counterbalanced a bit by the fact that the shells become more closely spaced, but nevertheless for main group elements the extra shielding of the extra shells usually matters more, and so effective nuclear charge falls going down a group.</p>
<p>Finally, bear in mind the "acidity" of a hydride is the tendency of the compound to break up with the H atom leaving as H+. So the greater the Z* of the partner, the more likely the H atom leaves without its electron.</p>
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https://chemistry.stackexchange.com/questions/97667/why-does-hydride-acidity-increase-across-period-and-down-group-in-periodic-table
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Question: <p>Why are the groups of elements on the periodic table organized into areas represented by the letters s,p,d,f,g, and h?
What does this mean?</p>
Answer: <p>The letters are related to the electron orbitals, which were originally observed through spectroscopy. <a href="https://en.wikipedia.org/wiki/Electron_configuration" rel="noreferrer">The lines shown in the spectroscope were named <em>sharp</em>, <em>principal</em>, <em>diffuse</em> and <em>fine</em> (or <em>fundamental</em>).</a></p>
<p>With a strong magnetic or electrostatic field, these separate into one, three, five or seven lines, or energy levels. There can be up to two electrons (with opposite spin) with the same energy level, according to the <a href="https://en.wikipedia.org/wiki/Pauli_exclusion_principle" rel="noreferrer">Pauli exclusion principle</a>, so this implies there can be double the numbers above, or 2 <em>s</em> electrons, 6 <em>p</em> electrons, 10 <em>d</em> electrons and 14 <em>f</em> electrons. Thus, the Periodic Table is based on the number of electrons in each elements orbitals.</p>
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https://chemistry.stackexchange.com/questions/78557/on-the-periodic-table-why-are-groups-of-elements-organized-by-letter
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Question: <p>I'm currently reading in-depth about the layout of the Periodic Table, and I wondered why the table has 3 columns in its Group VIII:</p>
<p><a href="https://i.sstatic.net/ThqdHm.png" rel="noreferrer"><img src="https://i.sstatic.net/ThqdHm.png" alt="enter image description here"></a></p>
<p>As I understand, this is an old notation of the table, now deprecated. But why was Group VIII a triple-sized group? </p>
<p>The answer might be trivial, but it's interesting. </p>
Answer: <p>In the beginning of the <span class="math-container">$20$</span>th century, the periodic table had <span class="math-container">$8$</span> columns, and not <span class="math-container">$18$</span> as today. Electrons and of course <span class="math-container">$spdf$</span> electrons were unknown. The numerous transition elements were included in this table by doubling the line between Argon and Krypton, then between Krypton and Xenon. It was going this way.</p>
<p>The <span class="math-container">$4$</span>th period was made of two lines. So each square (each box) contained two elements written on two lines. The first seven atoms <span class="math-container">$\ce{_{19}K - _{25}Mn}$</span> occupied the columns <span class="math-container">$\ce{I - VII}$</span>, and their oxidation state may always have been equal to the number of the column. Then the three atoms <span class="math-container">$\ce{_{26}Fe}$</span>, <span class="math-container">$\ce{_{27}Co}$</span> and <span class="math-container">$\ce{_{28}Ni}$</span> did not belong to any column. We will come back to them later on. Then the seven next atoms <span class="math-container">$\ce{_{29}Cu - _{35}Br}$</span> had to be put in columns <span class="math-container">$1$</span> to <span class="math-container">$\ce{VII}$</span>, under the corresponding elements : <span class="math-container">$\ce{_{29}Cu}$</span> in the same box and under <span class="math-container">$\ce{_{19}K}$</span>, <span class="math-container">$\ce{_{30}Zn}$</span> in the same box and under <span class="math-container">$\ce{_{20}Ca}$</span>, etc. up to <span class="math-container">$\ce{_{25}Mn}$</span> above <span class="math-container">$\ce{_{35}Br}$</span>. With this system all atoms had at least one oxidation state equal to the number of their column. The same procedure was adopted for <span class="math-container">$5$</span>th period, with <span class="math-container">$\ce{_{37}Rb}$</span> and <span class="math-container">$\ce{_{47}Ag}$</span> in the first column,<span class="math-container">$\ce{_{38}Sr}$</span> above <span class="math-container">$\ce{_{48}Cd}$</span>, etc.</p>
<p>This table was useful, except nobody knew what to do with the three atoms <span class="math-container">$\ce{_{26}Fe}$</span>, <span class="math-container">$\ce{_{27}Co}$</span> and <span class="math-container">$\ce{_{28}Ni}$</span> having all valence <span class="math-container">$2$</span>, and of course of the six corresponding atoms of the platinum group, immediately under them : <span class="math-container">$\ce{_{44}Ru}$</span> to <span class="math-container">$\ce{_{46}Pd}$</span> and <span class="math-container">$\ce{_{76}Os}$</span> to <span class="math-container">$\ce{_{78}Pt}$</span>. There was plenty of discussion at that time about what to do with them. Usually these elements were "rejected" in the 8th column, with the noble gases. or they were put in a separate column made specially for them. This problem was finally solved with the discovery of the <span class="math-container">$spdf$</span> electrons.</p>
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https://chemistry.stackexchange.com/questions/50413/why-are-there-3-columns-in-group-viii-of-the-periodic-table
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Question: <p>In the standard Periodic Table layout , all the elements up to 56 are in order i.e are in the same layout table. However, lanthanides and actinides are always shown separately from the layout like in this layout:
<a href="https://i.sstatic.net/C4q1r.png" rel="noreferrer"><img src="https://i.sstatic.net/C4q1r.png" alt="enter image description here"></a></p>
<p>What is the reason behind this structure? </p>
<p>Is this standard layout or can I represent it like this too?
<a href="https://i.sstatic.net/jqntf.png" rel="noreferrer"><img src="https://i.sstatic.net/jqntf.png" alt="enter image description here"></a></p>
Answer: <p>The Periodic Table arranges elements in blocks as each type of orbital fills with electrons - $s,p,d,f,g,h$. Alkali metals and alkaline earths are $s$-block filling (but could be one $s$-block slot). $p$-block six electrons to fill are trelides, tetralides, pnticides, chalcogenides, halides, inert gases (but could be one $p$-block slot). Transition metal $d$-block is ten elements (but could be one $d$-block slot). Filling the $f$-block are 14 elements, lanthanoids and actinoids. That gets sloppy to print and the elements are (or at least were) overall obscure. They get condensed.</p>
<p>Representing the Periodic Table has become an an art form. The plain vanilla variety is terse and useful.</p>
<p><a href="http://en.wikipedia.org/wiki/Alternative_periodic_tables" rel="noreferrer">http://en.wikipedia.org/wiki/Alternative_periodic_tables</a></p>
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https://chemistry.stackexchange.com/questions/10478/why-lanthanides-and-actinides-are-shown-separate-from-standard-periodic-table-la
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Question: <p>I am confused about the zig-zag pattern of metalloids in the periodic table. Why are metalloids arranged in a zig-zag line? Can anyone answer my question?</p>
Answer: <p>In addition to Pritt Balagopal's answer, in which he has clarified the trends in metallic character across the Periodic Table, I would also like to explain why the metalloids occur in a zig-zag fashion. </p>
<p>Firstly, I would like to again clarify that it is possibly a coincidence that the metalloids start from boron, not carbon or any other non-metal in that period. My guess could be the start of electron occupation of the p subshells. Note that boron has one electron in its p subshell. Thus, its chemical properties start to change to become more non-metallic in nature. </p>
<p>Diagonal relationships is a common occurrence in the Periodic Table. However, it is often used to describe the relationship between two elements which are diagonal to each other, not a whole line of them. Thus, I would like to clarify again that this is just my conjecture, not what is the true reason. But I believe this should be part of the correct explanation as it is logically sound. </p>
<p><a href="https://i.sstatic.net/XMui3.jpg" rel="noreferrer"><img src="https://i.sstatic.net/XMui3.jpg" alt="enter image description here"></a></p>
<p>This site offers a more in-depth explanation as to why diagonal relationships occur:
<a href="https://chem.libretexts.org/LibreTexts/University_of_Denver/Chem_2132%3A_Chemistry_of_the_Elements/Unit_3%3A_Descriptive_Chemistry/09%3A_Building_a_Network_of_Ideas_to_Make_Sense_of_the_Periodic_Table/9.3%3A_The_Diagonal_Effect" rel="noreferrer">https://chem.libretexts.org/LibreTexts/University_of_Denver/Chem_2132%3A_Chemistry_of_the_Elements/Unit_3%3A_Descriptive_Chemistry/09%3A_Building_a_Network_of_Ideas_to_Make_Sense_of_the_Periodic_Table/9.3%3A_The_Diagonal_Effect</a></p>
<p>In short, the diagonal relationships are based on two effects: the increasing effective nuclear charge across a period and the decreasing effective nuclear charge down a group. Effective nuclear charge increases across a period because the number of shielding electrons in the inner shells is constant while there are additional protons in the nucleus of the atoms of the elements across the period. Effective nuclear charge decreases down the group as there is an addition shell of shielding electrons from one element to the next in the group. (These are similar to the trends in metallic character pointed out in Pritt Balagopal's answer.)</p>
<p>These two competing effects cancel each other out and thus, chemists believe that the chemical properties of elements in a diagonal should be similar. The image shows some examples. </p>
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https://chemistry.stackexchange.com/questions/76568/why-are-metalloids-found-in-a-zig-zag-line-on-the-periodic-table
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Question: <p><a href="https://i.sstatic.net/Kysf7.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Kysf7.jpg" alt="enter image description here"></a></p>
<p>Row 1: 2 elements</p>
<p>Row 2: 8 elements</p>
<p>Row 3: 8 elements</p>
<p>Row 4: 18 elements</p>
<p>Row 5: 18 elements</p>
<p>Row 6: 32 elements</p>
<p>Row 7: 32 elements</p>
<p>In other words: 2, 8, 8, 18, 18, 32, 32</p>
<p>Why does the first row have only 2 elements, but all the next rows have a pattern? There's a pattern of two rows having the same number of elements <strong>that only starts after the first row.</strong></p>
<p><strong>EDIT:</strong> I just noticed <a href="https://en.wikipedia.org/wiki/Alternative_periodic_tables#Left_step_periodic_table_.28Janet.2C_1928.29" rel="nofollow noreferrer">this alternate periodic table!</a> It's first two rows have two elements each! Why does it work there? (I can't interpret the s, p, d, f part on the left). Which one is the right periodic table?</p>
Answer: <p>The pattern is better expressed this way:</p>
<pre><code>Row 1: 2 elements
Row 2: 2+6 elements
Row 3: 2+6 elements
Row 4: 2+6+10 elements
Row 5: 2+6+10 elements
Row 6: 2+6+10+14 elements
Row 7: 2+6+10+14 elements
</code></pre>
<p>The reason comes down to how the electrons fill the available energy levels. The thing that differentiates one element from another is the <em>proton number</em>, and each time a proton is added and a new element defined it requires one more electron to neutralise the charge. That electron naturally occupies the lowest available energy level in the atom.</p>
<p>The energy levels available are defined by the quantum state, including the quantum numbers $n, l,$ and $m_l$. $n$ is the principle quantum number and relates to the period the element is in, or the shell. $l$ is the angular momentum quantum number which defines the sub shell <em>s</em>, <em>p</em>, <em>d</em>, <em>f</em>, of which there are $n$ subshells whose values are $l=0,\dots {n-1}$. The magnetic quantum number $m_l$ further subdivides the subshell into orbitals, of which there are $2l+1$ orbitals whose values are $m_l=-l,\dots {+l}$.</p>
<pre><code>subshell number of orbitals subshell
label l value (number of ml values) electron capacity
s 0 1 2
p 1 3 6
d 2 5 10
f 3 7 14
</code></pre>
<p>Each orbital (i.e. each value of $m_l$) may contain 2 electrons The available quantum numbers are:</p>
<pre><code>n l values (subshells) ml values total shell electron capacity
1 0 (s) 0 2
2 0,1 (s,p) -1,0,1 2+6
3 0,1,2 (s,p,d) -2,-1,0,1,2 2+6+10
4 0,1,2,3 (s,p,d,f) -3,-2,-1,0,1,2,3 2+6+10+14
</code></pre>
<p>The pattern is there, but it doesn't seem to match up because by row 7 you would have 98 electrons in the shell and might expect the row to contain 98 elements. This is not the case! The energy levels for the $l$ values with large deviations from $0$ (i.e. orbitals with high angular momentum) become increasingly far apart, so even though <em>3d</em> orbitals exist in the third row, they are so much higher in energy than the <em>3p</em> orbitals that they are higher even than the <em>4s</em> orbitals. This happens again in row 4 where <em>4f</em> orbitals are so much higher in energy than <em>4d</em> orbitals, they are even higher than <em>5s</em>.</p>
<p>The different in energy is not always so large in specific cases, so there are examples where bonding or ligands or symmetry cause the energy levels to switch around, but otherwise the trend is true. If they can produce (or discover) some elements in the next row, we might expect to see the <em>5g</em> shell finally start to get filled and have a new block in the table, but realistically it doesn't look like they will be stable atoms.</p>
<p>I think I'm missing the reason why we arrange the elements into this table in the first place. The periodic table is arranged so that all elements in the same column have the same number of outer electrons. This is useful because elements with the same outer shell configuration of electrons react in similar ways, so they are grouped together in columns. This has the effect of creating 'periods' that begin with a reactive metal and end with an inert gas. Row 1 has two elements, and it's special in a way because it's the first in the series. Hydrogen has 1 electron, and a maximum capacity of 2 electrons. This means it behaves in a similar way as other elements that only have 1 electron in their outer shell (group 1), and also in a similar way to elements that only need 1 more electron to complete their outer shell (group 7). Where do we position it then? Often it is placed somewhere in the middle. Helium then has a full outer shell, which means it has many similarities to other elements with a full outer shell and goes in group 8 (or group 0) on the right.</p>
<p>Edit:
The alternative period table that you mention, the left-hand version, is laid out so as you read it from left to right you fill up the orbitals. This takes in to account that situation I describe where <em>3d</em> orbitals are higher than <em>4s</em>, and using this left-hand table you clearly see that. However this alternative table does not allow you to easily read the valance shell configuration like the Mendelev one does. Other types of table highlight different kinds of properties or relationships, which can be more useful in certain situations. The labels on the left (1s; 2s; 2p 3s; etc) relate to the outer subshell in that row, so Carbon is in the "2p" section of that chart while Magnesium is in the "3s" section on the same row.</p>
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https://chemistry.stackexchange.com/questions/44640/in-the-periodic-table-why-doesnt-the-2nd-row-have-exactly-2-elements
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Question: <p>Period 1 of the periodic table contains 2 elements (<span class="math-container">$1s^1$</span> and <span class="math-container">$1s^2$</span>).</p>
<p>Period 2 contains 8 elements (<span class="math-container">$2s^1$</span>, <span class="math-container">$2s^2$</span>, <span class="math-container">$2p^1$</span>, <span class="math-container">$2p^3$</span>, ..., <span class="math-container">$2p^6$</span>).</p>
<p>By the same argument, period 3 might contain 18 elements (<span class="math-container">$3s^1$</span>, <span class="math-container">$3s^2$</span>, <span class="math-container">$3p^1$</span>, <span class="math-container">$3p^3$</span>, ..., <span class="math-container">$3p^6$</span>, <span class="math-container">$3d^1$</span>, <span class="math-container">$3d^2$</span>, <span class="math-container">$3d^{10}$</span>). Why does period 3 of the periodic table contain 8 elements instead of 18?</p>
<p>I think that it could be explained by means of atomic stability, but I need help to understand what happens.</p>
<p>Thank you in advance.</p>
Answer: <p>You could say that the period number tells you about the largest value of principal quantum number in which a electron is present.</p>
<p>Electrons are filled according to <span class="math-container">$n+l$</span> rule. It states as <span class="math-container">$n+l$</span> increases, the energy of the orbital increases. If <span class="math-container">$n+l$</span> is the same, then the bigger <span class="math-container">$n$</span> has larger energy.</p>
<p>So <span class="math-container">$\mathrm{4s}$</span> would be filled before <span class="math-container">$\mathrm{3d}$</span>. Therefore elements containing <span class="math-container">$\mathrm{3d}$</span> electrons would have their <span class="math-container">$\mathrm{4s}$</span> orbital filled, and therefore, come in 4th period. There 3rd period contains elements having <span class="math-container">$\mathrm{3s}$</span> and <span class="math-container">$\mathrm{3p}$</span> electron as their last electron which are equal to 8.</p>
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https://chemistry.stackexchange.com/questions/119812/why-does-period-3-of-the-periodic-table-contain-8-elements-instead-of-18
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Question: <p>I understand thermodynamics in a physicsy way - at the level of Callen's <em>Thermodynamics</em>. However, my chemical thermodynamics is quite rusty: concepts like fugacity and activity coefficients are only vaguely familiar from undergrad courses.</p>
<p>I'm looking for a <em>well written, rigorous and succinct</em> guide to advanced chemical thermodynamics - at the level where I can read papers on ENRTL equations of state, or understand Pitzer models fully. If it is geared towards chemical engineers (like myself) that would be a bonus.</p>
<p>Basically, I consider Callen's book a classic, and wonder if there's anything that good within chemical thermodynamics.</p>
<p>Does such a book exist? Thanks</p>
Answer: <p>I would suggest:</p>
<p>1) <a href="http://rads.stackoverflow.com/amzn/click/0071247084" rel="nofollow">Introduction to Chemical Engineering Thermodynamics</a> by Smith, Van Ness, and Abbott;</p>
<p>2) <a href="http://books.unbeatablesale.com/chemical_biochemical_engineering_thermodynamics_by_sandler_stanley_i_1152064025.php" rel="nofollow">Chemical, Biochemical, and Engineering Thermodynamics</a> by Sandler; and</p>
<p>3) This is more suitable for advanced reading or for a graduate student: <a href="http://web.mit.edu/testerel/thermo/" rel="nofollow">Thermodynamics and Its Applications</a> by Tester and Modell.</p>
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https://chemistry.stackexchange.com/questions/34724/best-guide-to-chemical-thermodynamics
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Question: <p>I just started learning chemical thermodynamics and have come upon the definitions for extensive and intensive properties. I had a great deal of confusion over the exact meaning of intensive properties, but I believe I have come to a proper understanding of it after doing some extra reading. I will present my understanding below, and I just need to clarify that this is exactly what intensive properties are.
<br><br>
Intensive properties are defined differently for homogeneous and heterogeneous systems. For homogeneous systems, intensive properties are those properties that possess the same value at different points within the system and for the system as a whole. For heterogeneous systems, it no longer has anything to do with the value of the property for the system as a whole, but instead transforms into a local property, concerning only small volume elements within the system.<br><br>
Is this correct, or is there some brushing up needed? As of now I am only concerned with the most general understanding of intensive properties. Hope you could shed some light on this. Thank you.</p>
Answer:
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https://chemistry.stackexchange.com/questions/39499/intensive-properties-clarification-chemical-thermodynamics
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Question: <p>Can one be understood on the basis of the other or are they not interrelated at all?</p>
<p>The first thing my kinetics textbook demonstrated was how thermodynamics ignores time taken for a process whereas kinetics considers the time and rate of the process and hence kinetics is a more effective way to view change.</p>
<p>Also, I saw examples of highly thermodynamically feasible reactions which were not kinetically favourable in the sense that they were so slow that the entire process could be neglected as not happening.</p>
<p>Can one understand the thermodynamics of a reaction by studying it kinetics?</p>
<p>Or can one figure out the kinetics of a reaction by thermodynamical consideration?</p>
<p>Also, would it be a wise observation to say that thermodynamics is a largely theoretical subject while kinetics is a more experimental field?</p>
Answer: <p>No. Both I think are equally theoritical or experimental. You must study both seperately. Thermodynamics doesn't speak about rates. While kinetics doesn't tell you stability.
If a reaction produces more than one product, thermodynamics will tell you which is more stable, while that doesn't necessarily mean it is the major product. It may happen that rate of formation of other product is more under certain conditionds. Read about thermodynamic vs kinetic product and you will get it.</p>
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https://chemistry.stackexchange.com/questions/44155/what-is-the-relation-between-chemical-thermodynamics-and-chemical-kinetics
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Question: <blockquote>
<p>For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant <span class="math-container">$K$</span> in terms of change in entropy is described by:</p>
<p>(A) With the increase in temperature, the value of <span class="math-container">$K$</span> for exothermic reaction decreases because the entropy change of the system is positive</p>
<p>(B) With the increase in temperature, the value of <span class="math-container">$K$</span> for endothermic reaction increases because of an unfavorable change in entropy of the surroundings decreases</p>
<p>(C) With the increase in temperature, the value of <span class="math-container">$K$</span> for endothermic reaction increases because the entropy change of the system is negative</p>
<p>(D) With the increase in temperature, the value of <span class="math-container">$K$</span> for exothermic reaction decreases because the favorable change in entropy of the surroundings decreases</p>
</blockquote>
<p>Now, I am having trouble understanding option (B) (which in fact, is one of the correct options).</p>
<p>Since the process is endothermic and the temperature is rising, the value of <span class="math-container">$K$</span> increases, so heat must flow from surroundings to the system to favor the reaction which implies <strong>increase</strong> in an unfavorable change in entropy of surroundings rather than <strong>decrease</strong> as given in the option.</p>
<p>Kindly clarify my doubt. Also, any insights (other than the ones already given in the links attached below) into other options are very appreciated.</p>
<p>This question might look Duplicate of the questions already posted here:</p>
<ul>
<li><p><a href="https://chemistry.stackexchange.com/questions/115842/relation-between-equilibrium-constant-and-entropy-change">relation between equilibrium constant and entropy change</a></p>
</li>
<li><p><a href="https://chemistry.stackexchange.com/questions/109907/effect-of-temperature-on-equilibrium-constant-in-terms-of-entropy-change">Effect of temperature on equilibrium constant in terms of entropy change
</a></p>
</li>
</ul>
<p>But since this question is not focusing on the same problem, so I hope, it will not be marked as duplicate.</p>
Answer: <p>I agree it's not a duplicate because, while you are asking about the same homework problem discussed in the other question you linked, the issue you are having is different. Hence, because the last answer didn't address your specific confusion, you are asking a different question about that same homework problem.</p>
<p>Here's where you've got things mixed up: Whether a reaction is exothermic or endothermic depends on the inherent nature of the reactants and products, not the equilibrium constant. I.e., <span class="math-container">$\Delta H^{o}_{rxn}$</span> is the enthalpy change for a <em>complete</em> conversion of reactants to products at standard state, irrespective of the equilibrium constant. It's <em>not</em>, as I believe you were thinking, the enthalpy change to go from pure reactants to the equilibrium mixture. Hence if you could, say, change <span class="math-container">$K_{eq}$</span> purely by tweaking <span class="math-container">$\Delta S^{o}_{rxn}$</span>, then <span class="math-container">$\Delta H^{o}_{rxn}$</span> wouldn't change just because you've changed <span class="math-container">$K_{eq}$</span>. </p>
<p>So B is correct. If you raise T (and here we're assuming <span class="math-container">$\Delta H^{o}_{rxn}$</span> is independent of T), and <span class="math-container">$\Delta H^{o}_{rxn} > 0$</span>, the reason the reaction shifts to the right is this: In calculating <span class="math-container">$\Delta S^{o}_{surr}$</span>, we have to divide <span class="math-container">$q_{rev, surr} = q_{surr}$</span> by T, and T is now higher. Hence, for the same heat flow from the surroundings (which is given by <span class="math-container">$-\Delta H^{o}_{rxn}$</span>), the resulting unfavorable entropy change in the surroundings is less.</p>
<p>[I've used standard state here, which means it's carried out at a constant external pressure of 1 bar, but what I've written applies at any constant external pressure.]</p>
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https://chemistry.stackexchange.com/questions/132737/entropy-and-equilibrium-constant-chemical-thermodynamics
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Question: <p>We know that in thermodynamics, reversible process never occur since entropy should be conserved. However, in chemistry we do have a lot of reversed chemical reactions that forms reactants from products, previously formed by a direct reaction. So, what is the misunderstanding here?</p>
Answer: <p>There is no misunderstanding at all.</p>
<p>The second law of thermodynamics can be roughly expressed as: in a given process the entropy of the universe will always increase (defining universe as system + surroundings). </p>
<p><strong>Reversible processes</strong></p>
<p>For reversible processes the entropy of the universe remains constant. Most natural occurring processes are irreversibles.</p>
<p>A reversible process is a process where you can change its direction by inducing infinitesimal changes to the system. Throughout the entire process the system is at equilibrium, so from one state A to a state B you will not observe macroscopic changes.</p>
<p>The system goes from state A to state B in the following way: A + dA = B where B is infinitesimally different from A. These changes are applied over properties of the system like temperature and pressure by modifying its surroundings.</p>
<p><strong>Reversible chemical reaction</strong></p>
<p>In the case of a reversible chemical reaction it means that if you combine the products of that reaction you’ll obtain at least some of the reactants again. Usually these reactions require the same amount of energy that was consumed or released to obtain the products (This means that you must take into account the enthalpy of the forward reaction). </p>
<p>A typical example are lead batteries where lead reacts with sulfuric acid to form lead sulfate and <span class="math-container">$\ce{H+}$</span> ions while the battery is discharging and then you apply a direct current to revert the reaction and obtain lead, sulfuric acid and lead oxide again.</p>
<p>Even when a reaction is thermodynamically possible you must take into account the kinetics parameters. For example: it is thermodynamically possible to turn graphite into diamond but kinetically at standard conditions (1 atm, 25<span class="math-container">$^\circ$</span>C) it will happen in millions of years.</p>
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https://chemistry.stackexchange.com/questions/108623/whats-the-difference-between-reversible-process-in-thermodynamics-and-reversible
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Question: <p>I took organic chemistry a few years ago at my university, and although I went through the motions to pass the exams, I didn't understand the thermodynamic reasoning behind <em>why</em> the reactions actually occurred. For instance, I memorized that Claisen Condensation reactions occur between two esters in the presence of a strong base, but I don't fully grasp WHY the reaction happens in the first place. <a href="https://i.sstatic.net/GKnnl.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/GKnnl.png" alt="enter image description here"></a> Why are the products more stable than the reactants in cases like these? Does this reaction always occur whenever these components are in a mixture, or is there a sort of catalyst to trigger it? I feel similarly confused for many other organic reactions (e.g., Aldol condensations, Fischer esterification, Sn2, etc). I suppose I'm wondering if there is a general thermodynamic reasoning I'm missing in organic chemistry.</p>
Answer: <p>Whether or not the product is "more stable" would really depend on the sense in which you consider its reactivity. The <span class="math-container">$\mathrm{p}K_\mathrm{a}$</span> of a <span class="math-container">$\beta$</span>-keto ester is about 11, so in that sense, it is much more reactive than either of the reactants. Once the <span class="math-container">$\beta$</span>-keto ester is formed it is quickly deprotonated. It remains predominantly in this form because the anion is so stable, and the reaction is incapable of reversing. Thermodynamics does not always determine the result of an organic reaction. Many times, kinetics is the deciding factor.</p>
<p>The Claisen condensation is catalyzed by the base, as it is recovered in the end, and though I am sure it occurs spontaneously without it, I doubt it happens very often. I don't think ethyl acetate would be as popular a solvent as it is if it quickly and spontaneously converted to ethyl acetoacetate.</p>
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https://chemistry.stackexchange.com/questions/115221/thermodynamics-behind-organic-chemical-reactions
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Question: <blockquote>
<p>Which of the following is/are correct for spontaneous isothermal
chemical reaction?</p>
<p>(A) <span class="math-container">$\Delta H = 0$</span>, because <span class="math-container">$\Delta T = 0$</span></p>
<p>(B) <span class="math-container">$\Delta S = 0$</span></p>
<p>(C) <span class="math-container">$\Delta U = 0$</span>, because <span class="math-container">$\Delta T = 0$</span></p>
<p>(D) <span class="math-container">$\Delta G < 0$</span></p>
</blockquote>
<p>My answer: <span class="math-container">$\mathrm{A, C, D}$</span></p>
<p>Reasoning: </p>
<p><span class="math-container">$U$</span> is a state function and only depends on the initial and final states and their temperature. Since <span class="math-container">$\Delta T = 0$</span>, <span class="math-container">$\Delta U = 0$</span></p>
<p><span class="math-container">$\Delta H = \Delta U + \Delta(PV)$</span>, but <span class="math-container">$PV = \text{constant}$</span>, so <span class="math-container">$\Delta H = 0$</span></p>
<p><span class="math-container">$\Delta G <0$</span> - the requirement for any reaction to be spontaneous. </p>
<p>But answer given is only option D. </p>
<p>I think the flaw in my <span class="math-container">$\Delta H = 0$</span> argument is that reactants can be solids and liquids as well and we cannot apply <span class="math-container">$PV = \text{constant}$</span> to them. Am I right? </p>
<p>But I still can't figure out why <span class="math-container">$\mathrm C$</span> is wrong. </p>
Answer: <p>The internal energy and enthalpy of a <strong><em>pure</em></strong> ideal gas depends only on temperature. But, the internal energy and enthalpy of an ideal gas <strong><em>mixture</em></strong> depends both on temperature and on the amount of each chemical species that is present. To be more precise, it is equal to the internal energy and enthalpy per mole of each component as if it were a pure species times the number of moles of that species present, summed over all the species in the mixture: <span class="math-container">$$U=\sum{n_iu_i}$$</span>and<span class="math-container">$$H=\sum{n_ih_i}$$</span>So, even at constant temperature, <span class="math-container">$$\Delta U=\sum{(\Delta n_i)u_i}$$</span>and<span class="math-container">$$\Delta H=\sum{(\Delta n_i)h_i}$$</span>You should have been taught this when they were covering the thermodynamics of mixtures and chemical reactions. </p>
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https://chemistry.stackexchange.com/questions/106828/thermodynamics-of-a-spontaneous-isothermal-chemical-reaction
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Question: <blockquote>
<p>Chemical reactions occur at constant temperature and pressure.</p>
</blockquote>
<p>Consider a gaseous, equilibrium reaction: <span class="math-container">$\ce{2NO2(g) <=> N2O4(g)}$</span>. Most questions/textbooks formulate such questions by stating: <em>The reaction happens at (<span class="math-container">$T$</span>) temperature and (<span class="math-container">$P$</span>) pressure.</em></p>
<p>This gives the impression that the surrounding Temperature and pressure are constant.</p>
<ul>
<li>The first confusion that then arises is: For thermodynamic quantities such as <em>enthalpy</em> that involve pressure, which pressure should we use, internal or external?</li>
</ul>
<p>After reading some chem.SE answers, it seems to me that <span class="math-container">$P_{sys}=P_{surr}$</span> is a basic assumption. However this doesn't seem to go well with the example I mentioned: Since the pressure of the system will be given by <span class="math-container">$(n_1)RT/V + (n_2)RT/v$</span>, where <span class="math-container">$n_1$</span> and <span class="math-container">$n_2$</span> are the moles of <span class="math-container">$\ce{NO2}$</span> and <span class="math-container">$\ce{N2O4}$</span>. Clearly, as the reaction progresses, (<span class="math-container">$n_1+n_2$</span>) changes, and thus,the pressure of the system changes. If the surrounding pressure is kept constant, this beaks the "assumption": <span class="math-container">$P_{sys}=P_{surr}$</span>.</p>
<p>Am I missing something?</p>
Answer: <p>You have correctly stated the pressure of the system is <span class="math-container">$$p = \frac {(n_1 + n_2)RT}{V}$$</span></p>
<p>But at the end of your question you incorrectly imply the volume is constant and the pressure changes. That is not true. Constant pressure means the external constant pressure ( like the atmospheric one ) keeps the system pressure constant. Imagine there is a massless, frictionless piston, ensuring <span class="math-container">$p_\mathrm{ext} = p_\mathrm{sys}$</span>. It is similar as a thermostatic water bath keeps the system temperature constant.</p>
<p>So less confusing is to rewrite the ideal gas state equation :</p>
<p><span class="math-container">$$V = \frac {(n_1 + n_2)RT}{p}$$</span></p>
<p>There are 4 variables: <span class="math-container">$p, V, T, n_\mathrm{tot}=n_1 + n_2$</span>.
But there are just 3 degress of freedom = 3 independent variables.</p>
<p>Either <span class="math-container">$V, n_\mathrm{tot}, T$</span> are given, and <span class="math-container">$p=f(V, n_\mathrm{tot},T)$</span>,<br />
either <span class="math-container">$p, V, T$</span> are given, and <span class="math-container">$n_\mathrm{tot}=f(p,V,T)$</span>,<br />
either <span class="math-container">$p, n_\mathrm{tot}, T$</span> are given, and <span class="math-container">$V=f(p,n_\mathrm{tot},T)$</span>,<br />
either <span class="math-container">$p, n_\mathrm{tot}, V$</span> are given, and <span class="math-container">$T=f(p,n_\mathrm{tot},V)$</span>.</p>
<p>We have already given 2 variables: <span class="math-container">$p, T$</span>, with <span class="math-container">$n_\mathrm{tot},V$</span> remaining free.<br />
But there is the equilibrium relation</p>
<p><span class="math-container">$$K_p = \frac {p_{\ce{N2O4}}} {({p_{\ce{NO2}} )}^2}=\frac {p \cdot ( 1 - x_{\ce{NO2}}) } {({p \cdot x_{\ce{NO2}} )}^2}$$</span></p>
<p><span class="math-container">$$K_p \cdot p \cdot {({x_{\ce{NO2}} )}^2} + x_{\ce{NO2}} - 1 = 0$$</span></p>
<p>Solving the quadratic equation we would finally get <span class="math-container">$n_\mathrm{tot}$</span>, so all 3 degrees of freedom are saturated with given values for <span class="math-container">$p, T, n_\mathrm{tot}$</span>.</p>
<p>So in our case, the following applies:</p>
<p><span class="math-container">$p, n_\mathrm{tot}, T$</span> are given, and <span class="math-container">$V=f(p,n_\mathrm{tot},T)=\dfrac {n_\mathrm{tot}RT}{p}$</span></p>
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https://chemistry.stackexchange.com/questions/138417/thermodynamics-of-chemical-reactions-at-constant-pressure
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Question: <p>In chemical thermodynamics, the condition "constant temperature and pressure" appears almost everywhere. However, I felt a small ambiguity in the phrase so I wanted to clear my doubts. When we say "constant <span class="math-container">$T$</span> and <span class="math-container">$p$</span>" in chemical thermodynamics, which of these two statements do we actually mean?</p>
<ul>
<li><span class="math-container">$T$</span> and <span class="math-container">$p$</span> are the same at the beginning and end of the reaction</li>
<li><span class="math-container">$T$</span> and <span class="math-container">$p$</span> are the same throughout the entire reaction</li>
</ul>
<p>If we care about path-dependent variables, then there is a major difference between the two meanings.</p>
Answer: <p>As a chemical engineer, my perspective on this differs from those of the other responders. I regard a process occurring at constant temperature <em>and</em> pressure as one in which</p>
<ol>
<li><p>The system is held in contact with a constant temperature over the portion of its interface with the surrounding through which heat transfer is occurring, not necessarily the initial temperature of the system, throughout the process</p>
</li>
<li><p>The external pressure (or more precisely, the external compressive stress) at the portion of the system interface with its surroundings at which displacement work is being done is held constant (not necessarily at the initial pressure of the system) throughout the process.</p>
</li>
</ol>
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https://chemistry.stackexchange.com/questions/141929/meaning-of-constant-t-and-p
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Question: <p>Wikipedia article on a <a href="https://en.wikipedia.org/wiki/Component_(thermodynamics)" rel="nofollow">component (in thermodynamics)</a> first mentions that:</p>
<blockquote>
<p>In thermodynamics, a <strong>component</strong> is a chemically-independent constituent
of a system. The number of components represents the minimum number of
independent species necessary to define the composition of all phases
of the system.</p>
</blockquote>
<p>A bit later it is said in the same article that:</p>
<blockquote>
<p>The number of components is equal to the number of distinct chemical
species (constituents), minus the number of chemical reactions between
them, minus the number of any constraints (like charge neutrality or
balance of molar quantities).</p>
</blockquote>
<p>How does the second statement follows from the first one?</p>
Answer: <p>Well, a reaction comes with an equilibrium constant, i.e., mathematically speaking, some equation on constituents. Now we have N unknowns (constituents) and M equations (constraints and equilibrium constants). Supposedly N>M. This means we can't determine all unknowns. To do so, we must fix N-M free variables. These are our components.</p>
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https://chemistry.stackexchange.com/questions/38896/on-the-notion-of-a-component-in-thermodynamics
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Question: <p>I have a book at home called <a href="https://www.wiley.com/en-us/Chemical+Thermodynamics+for+Process+Simulation,+2nd,+Completely+Revised+and+Enlarged+Edition-p-9783527343256" rel="nofollow noreferrer">Chemical Thermodynamics for Process Simulation</a> (I have the first edition) and at the very beginning of chapter 2 of the book one may read the following:</p>
<p><a href="https://i.sstatic.net/sK6nN.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/sK6nN.jpg" alt="CO2-water system at high pressure"></a></p>
<p>No reference is given. I have never found anything about it so far and I am curious to know whether this statement is true.</p>
Answer: <p>Yes! Here's a video of Cody floating NaK (liquid) or ethanol/water (liquid) on compressed xenon (gas):
<a href="https://youtu.be/AsP4yMY-a6U" rel="nofollow noreferrer">https://youtu.be/AsP4yMY-a6U</a></p>
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https://chemistry.stackexchange.com/questions/122344/is-it-really-possible-for-a-liquid-to-stay-afloat-a-gas-vapor
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Question: <p>The way kinetics is taught at the undergraduate level (Arrhenius and collision theory) chemical equilibrium is determined governed immensely by activation energy of the reaction. According to thermodynamics, however equilibrium is a function of free energy change. In a way, thermodynamics and kinetics seem to contradict each other. What insight am I missing. Is the energy in energy vs reaction coordinates supposed to be Gibbs free energy instead of enthalpy? Would this imply activation energy changes as a reaction proceeds? Is the way kinetics is taught wrong?</p>
Answer: <p>A great deal of chemistry is determined by the interaction of kinetics and thermodynamics. The world would be a dull place if we didn't have both.</p>
<p>While thermodynamics determines what directions a reaction can go, the kinetics often determines whether the reaction can happen. Take a simple example: diamond is not the stable from of the element carbon at room temperature; graphite is. If the world were purely determined by thermodynamics all carbon would be graphite and men would be unable to impress women with the expensive shiny jewellery they buy them. But this reaction doesn't happen.</p>
<p>The reason it doesn't is because to convert diamond into graphite we would have to have to radically restructure the chemical bonding in diamond from a tetrahedral network breaking at least one bond for every carbon atom. Breaking all those bonds requires a huge input of energy to kick off the process (thermodynamics says you will get that energy plus a little more back if you can make this happen). You can think of diamond with a whole bunch of broken carbon-carbon bonds as an <em>intermediate</em> structure on the path from diamond to graphite. But that intermediate structure needs a great deal of energy to create and there simply isn't enough energy at room temperature to get there. The reaction will proceed at very high temperatures (and, if the pressure is high enough, diamond will be the thermodynamically favoured form). This is how the earth (or industry) creates diamonds: at high temperatures there is enough energy to break the carbon-carbon bonds and the equilibrium can be established (and, at high pressures, will favour diamond and at lower pressures graphite). </p>
<p>The point of all this is that, in chemistry, what happens involves the <em>interplay</em> of thermodynamics and kinetics. If it takes a lot of energy to get to an intermediate, the thermodynamic products cannot be reached. So being able to achieve an equilibrium depends not just on the start and end products of a reaction but the intermediate structures or compounds that you have to pass though to interconvert them. If those structures are hard to get to because you don't have enough energy around then thermodynamics is irrelevant to what will happen. you always have to think through the <em>mechanism</em> behind the reaction and understand the energy required to create the intermediate states along the reaction path.</p>
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https://chemistry.stackexchange.com/questions/25077/kinetics-vs-thermodynamics
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Question: <p>I'm a high school student with basic knowledge about chemical thermodynamics. I have three questions regarding the thermodynamic state functions $H, S, G$. I have searched a lot in textbooks but they don't give the complete information as to whether the properties taken are that of the system or surroundings.</p>
<p>Q1):-Enthalpy is given by the formula</p>
<p>$H = U +PV$ .
What I want to know is that whether $P$ mentioned above is the internal pressure of the system or the externally applied pressure?</p>
<p>Q2):- The change in entropy is given by the formula
$\Delta S= $$\int_1^2$${\frac{dq_{reversible}}{T}}$. Is the temperature mentioned of the system or the surroundings. </p>
<p>Q3)-On the same lines what temperature is taken to define the Gibbs free energy function:-$G=H-TS$.</p>
<p>Thanks in advance.</p>
Answer: <p>The three functions you mentioned are applicable only to thermodynamic equilibrium states of the system, for which the pressures and temperatures of the system and its surroundings match one another to within insignificantly small differences. In the integral to get the entropy change (Q2), the same matching applies; that is, a reversible path consists of a continuous sequence of thermodynamic equilibrium states.</p>
<p>Note that, at the interface between the system and the surroundings, the temperature and pressure of the system always matches the temperature and pressure of the surroundings (irrespective of whether the system is at equilibrium). It's just that, along an irreversible path of the system (for which the system passes through a sequence of non-equilibrium states), the pressure and temperature within the system vary with spatial position, and their average values (averaged over the volume of the system) do not match those of the surroundings at the interface.</p>
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https://chemistry.stackexchange.com/questions/82322/chemical-thermodynamic-functions
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Question: <p>I'm not a chemist, but I need some insight into chemical thermodynamics. </p>
<p>In the book that I'm reading <a href="http://www.amazon.co.uk/Physical-Chemical-Equilibrium-Engineers/dp/0470927100/ref=sr_1_1?ie=UTF8&qid=1374225908&sr=8-1&keywords=physical%20and%20chemical%20equilibrium%20for%20chemical%20engineers" rel="nofollow">(Nevers, 2012)</a>, equation (4.28) relates the chemical potential of a <strong>pure species</strong> to it's Gibbs energy per mol:</p>
<p>$\qquad\mu_a^{(1)} = \bigg( \dfrac{\partial G}{\partial n_a} \bigg)_{T,P,n_b...} = g_a^{(1)}$</p>
<p>where the subscript stands for the species $a$, the superscript stands for it's phase, $\mu$ is the chemical potential and $g$ is the Gibbs energy per mol.</p>
<p>I understand the first equal sign. Though, I don't understand the second equal sign. What am I missing?</p>
Answer: <p>The key is that the Gibbs free energy is an extensive quantity. This means that for a single-component system it can be written as
$$ G\left(T,P,N\right) = N\cdot g\left(T,P\right) $$
where <em>N</em> is the number of particles and <em>g</em> is the Gibbs free energy per particle. In other words, if you double the number of particles and double the volume (at fixed temperature and pressure), then <em>G</em> will also double. </p>
<p>From the above expression you can see that the chemical potential is just the Gibb's free energy per particle:
$$\mu\left(T,P\right) = g\left(T,P\right)$$</p>
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https://chemistry.stackexchange.com/questions/5625/why-are-partial-molar-gibbs-energy-and-gibbs-energy-per-mol-equal
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Question: <p>In the book of Chemical, biochemical and engineering thermodynamics 4th edition from Stanley I. Sandler (Wiley, 2006), there is a problem about the ammonia producing reaction at page 770.</p>
<p>13.17 The simple statement of the LeChatelier-Braun principle
given in Sec. 13.1 leads one to expect that if
the concentration of a reactant were increased, the
reaction would proceed so as to consume the added
reactant. This, however, is not always true. Consider
the gas-phase reaction
Show that if the mole fraction of nitrogen is less than
0.5, the addition of a small amount of nitrogen to the
system <strong>at constant temperature and pressure</strong> results
in the reaction of nitrogen and hydrogen to form ammonia
whereas if the mole fraction of nitrogen is
greater than 0.5, the addition of a small amount of
nitrogen leads to the dissociation of some ammonia
to form more nitrogen and hydrogen. Why does this
occur?</p>
<p>So I approached this problem the following way:</p>
<p>\begin{align}
\ce{N_2 + 3H_2<=>2NH_3}
\end{align}</p>
<p>The equilibrium constant:</p>
<p>$$K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}$$</p>
<p>Now when we write this in terms of reactant consumption X:</p>
<p>$$K_c=\frac{(2x)^2}{([N_2]-x)([H_2]-3x)^3}$$</p>
<p>Now, when I use some values to try to prove this. I never get a negative X value, so I am not getting ammonia dissociation? Mind that I am not an expert on chemical thermodynamics and this question was also proposed to us in an industrial chemical engineering class where we discussed the Haber-Bosch process.</p>
<p>Can someone help me out here? Why is this not working? Or what is the right strategy to approach this problem.</p>
<p>Thank you in advance</p>
Answer:
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https://chemistry.stackexchange.com/questions/53324/chemical-equilibrium-and-le-chatelier-braun-for-ammonia-production
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Question: <p>Is Light/Radiant Energy considered Kinetic Energy or Potential Energy?</p>
<p>I have started studying Thermodynamics, and the concept of internal energy was introduced, and defined as the sum of kinetic and potential energies of a system.</p>
<p>Another important definition was that:</p>
<p>ΔU = q + w , where</p>
<p>q = heat absorbed/released by the system</p>
<p>w = work done on the system by surroundings/work done on surroundings by system</p>
<p>U = Internal Energy of a system</p>
<p>There are lots of chemical reactions, such as combustion reactions, where light is produced, but there aren't many examples in thermodynamics concepts that deal with light energy. Where does light fit into this concept of internal energy?</p>
<p>If light is produced by a chemical reaction, is that a form of work done by a system on the surroundings?</p>
<p>If light is absorbed, is that a form of work done on the system by the surroundings?</p>
Answer: <p>Absorbing a photon will increase the potential energy of a molecule, and possibly a little kinetic energy change if rotational and vibrational energy is changed in the process. The translational kinetic energy is not changed immediately after absorption (there if no external force to do this) but can be imparted into products after reaction when bonds break. Green photons have <span class="math-container">$\approx 25000\,\mathrm{cm^{-1}}$</span> vs thermal energy of <span class="math-container">$\approx 210\,\mathrm{cm^{-1}}$</span> so a lot of energy can be released on reaction. If light is emitted then this can simply heat up the surroundings by being absorbed by unreactive molecules or some work could be extracted if these photons are absorbed by, say, a photodiode/solar panel. (Excited molecules can also release energy by non-radiative means but this just heats up the surrounding molecules which energy then diffuses away into other molecules))</p>
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https://chemistry.stackexchange.com/questions/180479/is-light-energy-defined-as-work-in-thermodynamics
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Question: <p>Thermodynamics has always been a tough thing for me. There are lots of assumptions in this subject (those assumptions, I know, are necessary, I know the science of thermodynamics is a very practical science).<br>
First Law of Thermodynamics states mathematically:
<span class="math-container">$$\Delta U=Q+W$$</span>
(with proper sign conventions must be used). This is just a law of conservation of energy and a very straightforward equation, but when we come to chemical thermodynamics this equation changes its form and becomes:
<span class="math-container">$$\Delta U=Q+p\,\Delta V$$</span>
My intuition says as soon as pressure and volume comes in any equation it becomes specifically for <em>gases</em>. So, my first question is: </p>
<p><strong>Why thermodynamical equations are just for gases?</strong> </p>
<p>Let's imagine an isothermal expansion of a gas (that simple piston and gas experiment) under a constant pressure, now work <span class="math-container">$W$</span> is
<span class="math-container">$$W=p\,\Delta V$$</span>
but if use ideal gas Law equation i.e.
<span class="math-container">$$pV=nRT$$</span>
<span class="math-container">$$p\,\Delta V = \Delta nRT + nR\,\Delta T\tag1$$</span> </p>
<p>since the expansion is isothermal therefore <span class="math-container">$\Delta T = 0$</span> and I can think that during expansion no atom or molecule has been annihilated therefore <span class="math-container">$\Delta n = 0$</span>, so after all we get
<span class="math-container">$$p\,\Delta V = 0$$</span>
<span class="math-container">$$W=0$$</span> </p>
<p><strong>I want to know my mistakes in above consideration.</strong></p>
<p>There is a question in my book:</p>
<blockquote>
<p>A swimmer coming out of a pool is covered with a film of water weighing <span class="math-container">$18\ mathrm g$</span>. How much heat must be supplied to evaporate this water at <span class="math-container">$298\ \mathrm K$</span>? Calculate the internal energy change of vaporization at <span class="math-container">$100\ \mathrm{^\circ C}$</span>. <span class="math-container">$\Delta_\mathrm{vap}H^\circ = 40.66\ \mathrm{kJ\ mol^{-1}}$</span> for water at <span class="math-container">$373\ \mathrm K$</span> </p>
</blockquote>
<p>My book give its solution like this
<span class="math-container">$$\ce{H2O(l) -> H2O(g)}$$</span>
Amount of substance of <span class="math-container">$18\ \mathrm g$</span> of <span class="math-container">$\ce{H2O(l)}$</span> is just <span class="math-container">$1\ \mathrm{mol}$</span>. Since, <span class="math-container">$\Delta U=Q-p\,\Delta V$</span>, therefore,<br>
<span class="math-container">$$\Delta U=\Delta H-p\,\Delta V $$</span>.
<span class="math-container">$$\Delta U=\Delta H-\Delta nRT$$</span>
<span class="math-container">$$\Delta U=40.66 \times 10^3\ \mathrm{J\ mol^{-1}}-1\ \mathrm{mol}\times8.314\ \mathrm{J\ K^{-1}mol^{-1}}\times373\ \mathrm K$$</span>
<span class="math-container">$$ \Delta U=37.56\ \mathrm{kJ\ mol^{-1}}$$</span> </p>
<p>I have a lot of problems with this solution which goes directly to the foundations of science of thermodynamics. (I must say it's because of these books that science becomes a rotten subject, these books destroy the real essence of science).<br>
<strong>How is <span class="math-container">$\Delta n=1\ \mathrm{mol}$</span>?</strong><br>
<strong>Why temperature is taken as <span class="math-container">$373\ \mathrm K$</span> and not <span class="math-container">$298\ \mathrm K$</span>, since the process starts at <span class="math-container">$298\ \mathrm K$</span> we should use it?</strong><br>
<strong>At <span class="math-container">$373\ \mathrm K$</span> the process becomes an isothermal one (latent heat) so <span class="math-container">$\Delta U$</span> ought to be zero, if we think the process of vaporization starts from <span class="math-container">$373\ \mathrm K$</span>.</strong> </p>
<p>Any help will be much appreciated. Thank you. </p>
Answer: <blockquote>
<p>Why thermodynamical equations are just for gases?</p>
</blockquote>
<p>They are not. The equation <span class="math-container">$\Delta U = q + P\Delta V$</span> applies to any phase (gas, liquid, solid...) when <em>only</em> pV work is done. In the particular form of the equation you present, the pressure is in addition constant during the work. </p>
<p>Gases are (1) an easy way to introduce thermodynamics concepts because they allow for a simplified analysis and may exhibit dramatic behavior, and (2) they provide a link to understanding the behavior of other phases. They also happen to be inherently important for practical and historical reasons during the development in the science.</p>
<blockquote>
<p>I want to know my mistakes in above consideration.</p>
</blockquote>
<p>There are no mistakes. Consider the ideal gas law <span class="math-container">$$V=\frac{nRT}{p}$$</span> If you assume that <span class="math-container">$p$</span>, <span class="math-container">$n$</span> and <span class="math-container">$T$</span> are constant, then the dependent variable <span class="math-container">$V$</span> will also be constant.</p>
<p>What's probably causing the confusion is that the water is undergoing a phase change. We assume that the liquid does no work, that the change in volume is only due to the formation of vapor. In practice <span class="math-container">$\pu{\Delta n=+ 1 mol}$</span> <em>for the gas</em>, <span class="math-container">$\pu{\Delta n= - 1 mol}$</span> for the liquid, and <span class="math-container">$\pu{\pu{\Delta n= 0}}$</span> for all of the water. We ignore the work done when reducing the amount of liquid because it is small compared to that done when the gas is formed (the change in volume of gas is much greater).</p>
<blockquote>
<p>How is Δn=1 ?</p>
</blockquote>
<p>See the answer to the previous question. You are converting <span class="math-container">$\pu{18 g}$</span> of liquid water into vapor. Since the molecular weight of water is <span class="math-container">$\pu{18 g/mol}$</span>, you are converting <span class="math-container">$\pu{1 mol}$</span> of water.</p>
<blockquote>
<p>Why temperature is taken as 373 K and not 298 K, since the process starts at 298 K we should use it?</p>
</blockquote>
<p>I agree, if the work is actually performed at a lower temperature than <span class="math-container">$\pu{373 K}$</span>, so this is an estimate. It is assumed that water is "boiling off" the skin at <span class="math-container">$\pu{373 K}$</span> and <span class="math-container">$\pu{1 atm}$</span> vapor pressure (the boiling point of water at <span class="math-container">$\pu{1 atm}$</span> of pressure is <span class="math-container">$\pu{373 K}$</span>). Maybe not an accurate portrayal of what is going on, but it gets you to practice the theory. </p>
<hr>
<blockquote>
<p>I have lot of problems with this solution which goes directly to the foundations of science of thermodynamics. (I must say it's because of these books that science becomes a rotten subject, these books destroy the real essence of science).</p>
</blockquote>
<p>The problems more generally are that (1) we have an intuition about the way the world works, based on everyday observations, and this intuition sometimes misinforms us; some of the effort of education is to develop a more accurate intuition; and (2) when teaching science, practice problems are sometimes abstract and don't reflect real life situations except approximately. For instance, when you dry yourself after swimming, you are far from a thermodynamic equilibrium, with air currents, sun heating your skin, and a low water vapor pressure all playing a potential role. Modeling all this is beyond the scope of an introductory course. Probably the most important approximations here (assuming a closed system free of mass flow) are that the enthalpy of vaporization is constant over a broad temperature range and/or (as you rightly pointed out) that the water evaporates at <span class="math-container">$373 K$</span>. </p>
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https://chemistry.stackexchange.com/questions/119640/relation-between-first-law-of-thermodynamics-and-ideal-gas-law
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Question: <p>I'm using the following handbook <em>"Engineering and Chemical Thermodynamics 2nd edition"</em> by Koretsky. In the book they have an alternate formulation for <em>h(P,T)</em> though it was never explained how they got it (<a href="https://i.sstatic.net/R3HrJ.jpg" rel="nofollow noreferrer">source</a>, equation 5.43).</p>
<p>I'm trying to prove:
<span class="math-container">$$dh=(c_p-{\beta}vP)dT + vdP$$</span>
with <span class="math-container">$\beta$</span> the volumetric thermal expansion coefficient. I keep ending up with:
<span class="math-container">$$dh=Tds+vdP$$</span>
<span class="math-container">$$dh=T(\frac{c_p}{T}dT-{\beta}vdP)+vdP$$</span>
<span class="math-container">$$dh=c_pdT + (v-{\beta}vT)dP$$</span></p>
<p>How do you get the <span class="math-container">$-{\beta}vPdT$</span> term in the first equation?</p>
Answer:
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https://chemistry.stackexchange.com/questions/166677/alternate-formulation-of-hp-t
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Question: <p>There are 5 most common <a href="https://en.wikipedia.org/wiki/Thermodynamic_potential#:%7E:text=It%20is%20the%20energy%20of,from%20an%20expression%20for%20U." rel="nofollow noreferrer">thermodynamics potential</a> -</p>
<p><a href="https://i.sstatic.net/mfkxV.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/mfkxV.png" alt="enter image description here" /></a></p>
<p>But we will discuss about only three of them -</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Internal_energy" rel="nofollow noreferrer">Internal energy</a> (∆U)</li>
<li><a href="https://en.wikipedia.org/wiki/Enthalpy" rel="nofollow noreferrer">Enthalpy</a>(∆H)</li>
<li><a href="https://en.wikipedia.org/wiki/Gibbs_free_energy" rel="nofollow noreferrer">Gibbs free energy</a> (∆G)</li>
</ul>
<p>As Gibbs free energy is also a thermodynamics potential that means it is not related to only energy of bonds and <a href="https://en.wikipedia.org/wiki/Interatomic_potential" rel="nofollow noreferrer">interatomic interactions</a> but also to <a href="https://en.wikipedia.org/wiki/Free_entropy" rel="nofollow noreferrer">entropy potential</a>.</p>
<hr />
<p><strong>1. So now my first question is that</strong> <strong>what thermodynamics potential actually represent physically not mathmatically?</strong></p>
<hr />
<p>Now talk about those three thermodynamics potential -</p>
<p>I find it most confusing that how internal energy , enthalpy and Gibbs free energy are different form of energies or potential although looking at the mathmatical expressions we can easily say Gibbs free energy is different from enthalpy by a additional term -TS or enthalpy is different from internal energy as it has a additional term pV. But how they are different form of energies theoretically?</p>
<p>Let's consider a example of chemical reaction to understand what I am actually asking.</p>
<p><a href="https://i.sstatic.net/lXGL2.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/lXGL2.jpg" alt="potential energy vs reaction coordinate " /></a></p>
<p>This is most common diagram used in our textbooks which shows potential energy of system when reactant is converting into product. but the difference of potential energy between reactant and product is ∆U or ∆H or ∆G ? And how these 3 terms (∆U,∆H,∆G) <em><strong>physically</strong></em> represent 3 different kind of potential energies between reactant and product ?</p>
<p><strong>Physically</strong> means in terms of bond energies, interatomic forces and interactions , molecular kinetic energy , vibrational energy, rotational energy etc.</p>
<p>Ok finally I address my question in more clarified language - internal energy of a system (here reactants and products) as per it's defination includes all kinds of kinetic and potential energies of system so when we say there is difference between reactant's and product's internal energy by ∆U , we have included all kind of energy differences between reactant and product. but when we say there is a difference ∆H between enthalpy ( or ∆G between Gibbs free energy ) of reactant and product , does it represent any <strong>specific type of potential energy in internal energy</strong> difference between reactant and product means <em><strong>does ∆H ( or ∆G) is a specific part of internal energy?</strong></em></p>
Answer: <p>Note that the term state function is more common that the thermodynamic potential, while both are correct.</p>
<p>Considering <span class="math-container">$U$</span>, <span class="math-container">$H$</span>, <span class="math-container">$F$</span> and <span class="math-container">$G$</span> as different forms of energy may be misleading.</p>
<p>The state function <strong>internal energy</strong> <span class="math-container">$\ce{U}$</span> is intuitively well understood as the total system kinetic and potential energy. Formally, the energy E=mc^2 should be included to, but as the interest is usually in <span class="math-container">$U$</span> changes, it is not involved, being many orders higher than the rest.</p>
<p>The state function <strong>enthalpy</strong> <span class="math-container">$H = U + pV$</span> (as pV is a state function too) is not other form of energy, but made up state function. At isobaric conditions, enthalpy change is energy added to the system that is used</p>
<ul>
<li>to increse the system internal energy</li>
<li>to do mechanical volume work on the neighbourhood, due volume changes.</li>
</ul>
<p>For thermally induced system expansion:</p>
<p><span class="math-container">$$\mathrm{d}H = \delta Q - \delta W = C_V \mathrm{d}T + p\mathrm{d}V = C_V \mathrm{d}T + p \left(\frac{\partial V}{\partial T}\right)_p \mathrm{d}T \\
= \left(C_V + p \left(\frac{\partial V}{\partial T}\right)_p \right)\mathrm{d}T = C_p \mathrm{d}T$$</span></p>
<p><span class="math-container">$\Delta H - \Delta U$</span> may be formally considered as energy, that system does not store in form its internal energy, but lends/borrows it to/from its neighbourhood via mechanical work.</p>
<p><strong>(Free) Gibbs energy (or "free enthalpy)</strong> <span class="math-container">$G = H -TS = U + pV - TS$</span> is not other form of energy either, but yet another made up state function. It is the maximal part of enthalpy, that is, at least theoretically, available to do work, allowed by the 2nd law of thermodynamic.</p>
<p>For reversible isothermal and isobaric process, <span class="math-container">$-\frac{\Delta G_\text{sys}}{T} = \Delta S_\text{sys} + \Delta S_\text{surr}$</span></p>
<p>Imagine 2 almost identical, isolated systems:</p>
<ul>
<li>The system A contains 2 thermally connected identical metal blocks at equilibrium, at <span class="math-container">$\pu{T = 300 K}$</span> and temperature independent heat capacity.</li>
<li>The system B is identical to the system A, but the blocks have the temperature respectively <span class="math-container">$\pu{290 K}$</span> and <span class="math-container">$\pu{310 K}$</span>.</li>
</ul>
<p>Both systems have the same internal energy and enthalpy. But the system A has higher entropy <span class="math-container">$S$</span> and lower Gibbs energy <span class="math-container">$G$</span>. The difference is the part of the thermal energy of the blocks that can be used to do some work, e.g. via thermal engines or thermocouples generating electricity.</p>
<p>Note that the part of the thermal energy available to do work has the same form as the part of the thermal energy not available to do work. The difference is in quantity, not quality.</p>
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https://chemistry.stackexchange.com/questions/175333/thermodynamics-potential-and-differences-between-them
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Question: <p>In my thermodynamics course, we introduced the chemical potential as a modification of the first + second law of thermodynamics in the case of a system that can exchange particles with its surroundings (consider only PV work):</p>
<p><span class="math-container">\begin{equation}
dU = TdS - pdV + \sum_{j=1}\mu_j dN_j
\end{equation}</span>
where the sum is over all the components of the system. This makes sense to me - even if we keep everything else constant, just by adding more matter to the system the energy should increase. Now, this should be true for a closed system undergoing a chemical reaction, so the 3rd term on the RHS could be non-zero. If we consider a quasistatic process, we can identify the first and second terms on the RHS as the heat and work. So the chemical potential term appears to represent some sort of energy transfer separate from heat and work, but how is this possible? Aren't heat and work the only ways to transfer energy?</p>
<p>And if the chemical potential term represents some sort of work in this case, how come the term which we defined to just explain the increase of the internal energy with the increase of the amount of matter in the system also coincidentally can now be interpreted as a work term?</p>
<p>In other words, my question is how do I physically interpret the chemical potential term?</p>
<p>I asked this question in Physics Stack Exchange (I am a physics student), but I wasn't able to obtain an answer that made complete sense to me (I was told that internal energy doesn't include bond energy, but I couldn't find a source that confirms this).</p>
Answer:
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https://chemistry.stackexchange.com/questions/160446/meaning-of-chemical-potential-in-chemical-reactions
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Question: <p>I wonder if there is a way to measure the chemical potential of a substance in a two-component mixture and also to account for its dependence on the number of moles. Explaining it further, suppose there is a mixture of a light component (like methane, C1) and a heavier one (like C20). Given temperature and pressure, one could add some amount of the light component to the mixture and that would rise the chemical potential of that first component. Thermodynamics tell us that the chemical potential can be expressed by</p>
<p>$$d \mu_i = - \bar S_i dT + \bar V_i dP + \sum_{m=1}^{n} \left(\frac{\partial \mu_m}{\partial N_i}\right)_{T,P,N_j[i]} d N_m $$</p>
<p>Being pressure and temperature constant, the increase in the chemical potential by adding some amount of light component depends entirely on the summation of $({\partial \mu_m}/{\partial N_i})_{T,P,N_j[i]}$. That is precisely the point: how could I measure or calculate that derivative or, alternatively, variations in the chemical potential by adding mass to the mixture?</p>
<p>Of course, one could use a cubic equation of state or the like and calculate fugacity and chemical potential etc. But I am interested in the more fundamental aspects of thermodynamics. One of them is that, given P-V-T data and specific heats, in addition to the composition, one is supposedly able to calculate all the thermodynamic properties of a mixture or of a component. </p>
<p>Now, the practical side of this question is this: if there is a two-phase mixture of C1 and C20, under ordinary T and P, the vapour phase is almost pure C1, but the liquid phase is a mixture of both. For given P and T, composition can be measured somehow. Given an increase in pressure, say $\Delta P$, with coexisting phases, there must be an increase in the concentration of C1 in the mixture, $\Delta x_1$ (or $\Delta N_1$, if the number of moles is to be used instead of concentration). The change in composition can obviously be calculated by equating the chemical potential of C1 in the vapour and in the liquid phases. While calculating the change in chemical potential in the vapour phase is quite easy because it is pure C1 and
$d \mu_1 = \bar V_1 dP$,
the chemical potential in the liquid phase depends on the above mentioned derivative:</p>
<p>$$d \mu_1 = \bar V_1 dP + \sum_{m=1}^2 \left(\frac{\partial \mu_m}{\partial N_1}\right)_{T,P,N_2} d N_m $$</p>
<p>Finally, by "measuring" chemical potential I mean any way of obtaining it from other variables that can be measured by ordinary instruments in a lab. And, just to mention, I have used the notation of Modell and Reid, "Thermodynamics and its applications".</p>
Answer:
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https://chemistry.stackexchange.com/questions/75115/how-to-measure-chemical-potential
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Question: <blockquote>
<p><strong>This question has been re-asked on <a href="https://hsm.stackexchange.com/">History of Science and Mathematics</a>:</strong><br>
<a href="https://hsm.stackexchange.com/q/3581">Who is Hræðraford, the “learned clerk” “writing in modern chemical Latin”?</a></p>
</blockquote>
<hr>
<p>Truesdell, C. <em>Rational Thermodynamics</em>. New York, NY: Springer New York, 1984. <a href="http://dx.doi.org/10.1007/978-1-4612-5206-1" rel="nofollow noreferrer">http://dx.doi.org/10.1007/978-1-4612-5206-1</a>, p. 41 says:</p>
<blockquote>
<p>… in the numerous papers and books whose titles and topics include "irreversible"—the learned clerk HRÆðRAFORD, writing in modern chemical Latin, has recently called their field "thermodynamice inreparabilis"—…</p>
</blockquote>
<p>Who is this "learned clerk" Hræðraford?</p>
Answer:
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https://chemistry.stackexchange.com/questions/14040/who-is-hr%c3%a6%c3%b0raford-the-learned-clerk-writing-in-modern-chemical-latin
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Question: <p>suppose a system in chemical equilibrium like ice melting to form water and said water freezing into ice again. it seems to me that a state in chemical equilibrium would have less entropy that a state that isn't in equilibrium since (at least intuitively), the number of microstates that yield a macrostate that isn't in equilibrium is higher than the number of microstates that would yield a state in equilibrium.</p>
<p>however, LeChatelier's principle states that a system in chemical equilibrium, when perturbated, tends to return to equilibrium. this implies that if i move a system in equilibrium (fewer microstates/low entropy) to a state that isn't in equilibrium (more microstates/higher entropy) then it will tend to return to the state of equilibrium(low entropy) which seems to break the 2nd law of thermodynamics which states that a system will always evolve from a macrostate of low entropy to a macrostate of higher entropy.</p>
<p>i believe there's something wrong with my reasoning, but i can't exactly find where</p>
Answer: <p>We can test you assumption that entropy is not maximised at equilibrium by finding the chance that it will be in a non-equilibrium state. We observe that a gas always fills the volume available to reach its equilibrium state. As a test we can find the chance that a gas would fill just <span class="math-container">$99$</span>% of the volume available i.e. a vacuum of size 1% suddenly appears in the gas, or equivalently that if a gas fills just 1% of a room what is the chance that it will remain there.</p>
<p>Imagine that there are <span class="math-container">$10^{24}$</span> molecules, just over a mole, and that the other <span class="math-container">$1$</span>% of the volume is empty. Is the gas now in a lower entropy non-equilibrium state or not? We can find the chance of this using the result <span class="math-container">$m_2/m_1 = (V_2/V_1)^n$</span>,(shown below) which is the number of microstates <span class="math-container">$m_2$</span> in volume 2 (<span class="math-container">$V_2 = 0.99$</span>), divided by those in volume 1 (<span class="math-container">$V_1 = 1$</span>), the probability is</p>
<p><span class="math-container">$$ \displaystyle prob= (0.99)^{10^{24}} \approx 10^{-4.36\cdot 10^{21} }$$</span></p>
<p>which is utterly minute and hard to comprehend. This is the chance that the gas will remain where it is or alternatively that the gas in a room will spontaneously move so as to fill just 1% .</p>
<p>Suppose that <span class="math-container">$10^{12}$</span> observations could be made each second to see if a void appeared, this would mean taking <span class="math-container">$3\cdot 10^{19}$</span> /year and if this continued for the age of the universe (<span class="math-container">$10^{10})$</span> years would make <span class="math-container">$\approx 10^{29}$</span> observations. However, this would not be of any help at all as it represents only <span class="math-container">$4\cdot 10^{\large{-4\cdot 10^{21} -29}}$</span> of the total, an utterly insignificant amount. Just to be clear <span class="math-container">$10^{10^{21}}$</span> is <span class="math-container">$1$</span> followed by</p>
<p><span class="math-container">$$\displaystyle \mathrm{1000 \times a\; million \times a\; million\times a\; million\; zeros} !$$</span></p>
<p>The minute chance of observing even a small void illustrates that the equilibrium state is overwhelmingly probable, in fact far greater than all other possible states combined and so at equilibrium entropy is maximised.</p>
<p>(As an aside. Nowadays certain types of single molecule experiments are possible. If the number of molecules is small say 100 then fluctuations about equilibrium can be observed from time to time but measured over long a time time, relative to period of the fluctuations the similar conclusion is reached)</p>
<p>Derivation</p>
<p>In a gas the atoms/molecules (particles) themselves occupy virtually no volume compared to the volume occupied by the gas, thus the number of spaces available is vastly more than the number of particles. We divide the space occupied up into <span class="math-container">$w$</span> minute cells which is a number far greater than the number of particles <span class="math-container">$n$</span>. The total number of configurations is found using the familiar (Binomial) equation,</p>
<p><span class="math-container">$$\displaystyle m=\frac{w!}{n!(w-n)!}$$</span></p>
<p>Using the Sterling formula for factorials and simplifying gives</p>
<p><span class="math-container">$$\displaystyle \ln(m)= w\ln(w)-n\ln(n) -(w-n)\ln(w-n)$$</span></p>
<p>Because <span class="math-container">$n/w \ll 1$</span> the log can be rearranged to</p>
<p><span class="math-container">$$\ln(w-n)=\ln(1-n/w)-\ln(w)$$</span></p>
<p>and then approximating as <span class="math-container">$\ln(1-n/w) = -n/w$</span> can be done without any significant error to give</p>
<p><span class="math-container">$$\displaystyle \ln(m)=n\ln\left(\frac{w}{n}\right)+n$$</span></p>
<p>Suppose that the volume the gas is irreversibly expanded from <span class="math-container">$V_1 \to V_2$</span> and define <span class="math-container">$r = V_2/V_1$</span>, therefore the number of spaces/cells <span class="math-container">$w$</span> changes as <span class="math-container">$w \to rw$</span>. The ratio of the number of configurations in the final to initial volume is found using the last equation again and is</p>
<p><span class="math-container">$$\displaystyle \ln\left(\frac{m_2}{m_1}\right)=n\ln\left(\frac{rw}{n}\right)-n\ln\left(\frac{w}{n}\right)=n\ln(r)$$</span></p>
<p>meaning that</p>
<p><span class="math-container">$$\displaystyle \frac{m_2}{m_1}=\left(\frac{V_2}{V_1}\right)^n\qquad $$</span></p>
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https://chemistry.stackexchange.com/questions/186796/does-lechateliers-principle-contradict-the-second-law-of-thermodynamics
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Question: <p>I just started reading <em>Thermodynamics</em> by Enrico Fermi; he started defining <em>state of a system</em> for <em>homogeneous mixture of several chemical compounds</em> & then <em>non-homogeneous system</em> as: </p>
<p><img src="https://lh3.googleusercontent.com/-h94NSuXpbfc/VfwufkLkYfI/AAAAAAAAAFE/pn6cJBq1C-M/s419-Ic42/screenshot-www.amazon.com%2525202015-09-18%25252021-00-57.jpg" alt="image"></p>
<p>I thought non-homogeneous system to be a mixture of several homogeneous chemical compounds separated by boundaries. But Fermi seems to have done it otherwise; he has defined the state separately for <code>homogeneous mixture of several chemical compounds</code> & non-homogeneous system that is, <code>mixture of several homogeneous chemical compounds</code>. I am not getting how they are different at all. Can anyone please explain how they are different? What's the difference between <em>homogeneous mixture of several chemical compounds</em> & <em>non-homogeneous system</em>?</p>
Answer: <p>You are right that normally a non-homogeneous system would be one where there is boundary between two internally homogeneous phases (like milk which is basically oil and water mixed at a very fine scale).</p>
<p>But Fermi as also allowing for the possibility that there may be only one phase where the concentration of its components varies <em>continuously</em> in space (if there is more than one phase the variation is <em>discontinuous</em> by definition). </p>
<p>It certainly isn't a common situation if the phase in in equilibrium where we would expect things like diffusion to create a uniform concentration of things over time, at least in liquids and gases). But it is very common in the real world where equilibrium is rarely reached (imagine the state of a poorly stirred cup of tea where the concentrations of tea-stuff will be larger near the tea bag than far away from the tea bag).</p>
<p>It is also very common in places where it is hard to get to equilibrium which is often true in solids (I don't know whether this is what Fermi was thinking of). Pyrex glass, for example, is apparently a single phase but the structure is deliberately made different near the surface than far away from the surface to create tension in the surface layers that protect the glass from thermal shock. But the properties vary continuously. </p>
<p>The examples I can think of may not be relevant to what Fermi was saying as they mostly are metastable and won't be at equilibrium which is often what thermodynamics restricts itself to worrying about.</p>
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https://chemistry.stackexchange.com/questions/37544/difference-between-homogeneous-mixture-of-several-chemical-compounds-non-ho
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Question: <p>In the book, it is mentioned the formula for <span class="math-container">$\Delta U$</span> in a bomb calorimeter without any derivation:</p>
<blockquote>
<p><span class="math-container">$$\Delta U = q_v = \frac{Q\times M\times \Delta T}{m}$$</span>
where
<span class="math-container">$$Q=\textrm{heat capacity of calorimeter,}$$</span>
<span class="math-container">$$M=\textrm{molecular mass of sample,}$$</span>
<span class="math-container">$$m=\textrm{mass of sample used, and}$$</span>
<span class="math-container">$$\Delta T=\textrm{change in temperature of water in the bath}$$</span></p>
</blockquote>
<p>I am confused regarding this formula. Can anyone give me the derivation of this formula (or a corrected formula)?</p>
<p>[I am 11-grader and am studying chemical thermodynamics. I can distinguish between <span class="math-container">$C$</span> as an extensive property and <span class="math-container">$c$</span> and <span class="math-container">$C_m$</span> as intensive properties.]</p>
<p>Any help would be appreciated :)</p>
<p><strong>NOTE</strong>: I know that a formula is <span class="math-container">$q_v=cm\Delta T$</span>, I want to know how the book got to the formula mentioned previously.</p>
Answer: <p>The formula in the book is correct. They are trying to get the change in internal energy per mole of sample. From the first law, for this constant volume system (no work), <span class="math-container">$$\Delta U_{\textrm{total}}=q=C\Delta T$$</span>where C is the heat capacity of the calorimeter. This equation assumes that the heat capacity of the water in the bath is lumped into C, and that the temperature change of other parts of the calorimeter is the same as that of the water.</p>
<p>The number of moles of sample is m/M. So, <span class="math-container">$$\Delta U_{\textrm{per mole}}=\Delta U_{\textrm{total}}\frac{M}{m}=C\Delta T\frac{M}{m}$$</span>In their notation, they use the symbol Q to represent the heat capacity of the calorimeter C.</p>
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https://chemistry.stackexchange.com/questions/144442/confusion-in-calculating-delta-u-from-a-bomb-calorimeter
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Question: <p>I know about the <a href="https://en.wikipedia.org/wiki/Belousov%E2%80%93Zhabotinsky_reaction" rel="nofollow noreferrer">Belousov-Zhabotinskii reaction</a> as an example of "chemical oscillator", e.g. a reaction that can produce repeating patterns over time as it exhausts its reagents. I also know about the <a href="https://en.wikipedia.org/wiki/Briggs%E2%80%93Rauscher_reaction" rel="nofollow noreferrer">Briggs-Rauscher reaction</a> but I wonder how many such chemical oscillators are possible? Can somebody provide me with a good review paper and/or book reference written specifically to survey the range of all possible chemical oscillator systems?</p>
<p>To note, I know there are a lot of people outside of chemistry (particularly people interested in chaos theory and non-equilibrium thermodynamics) who do detailed research on modeling the detailed behavior of such oscillating systems, and I've found several such papers/reviews. But, what I'm looking for is a resource focusing on the <em>chemical properties</em> of oscillator systems, not the mathematics or dynamics modeling.</p>
Answer:
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https://chemistry.stackexchange.com/questions/70586/reference-request-review-paper-overview-of-chemical-oscillators
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Question: <p>This is not a homework question. I am not in any course of study. The following question arises from my personal curiosity. If 1 mole of sodium sulfate and 2 moles of potassium chloride are dissolved in water, and the solution is gradually evaporated, which salts precipitate, and in which order?</p>
<p>There are four possible single salts that could be precipitated, namely <span class="math-container">$\ce {Na2SO4, K2SO4, NaCl, KCl}$</span>, as well as mixed salts with the general formula <span class="math-container">$\ce{((Na,K)Cl)_x.((Na,K)2SO4)_{1-x}}$</span>, but I don't know how to work out which ones, or what the sequence would be, since I am a complete ignoramus regarding the chemical thermodynamics of these salts, their solutions, and their precipitation from solution. I do understand the basic concept of Gibbs free energy, which I guess would be involved here.</p>
Answer:
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https://chemistry.stackexchange.com/questions/134484/which-salts-precipitate-from-a-potassium-sodium-chloride-sulfate-solution
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Question: <p>I'm studying equilibria and thermodynamics and came across these two terms.</p>
<p>My problem is, unlike other thermodynamic properties that I can understand physically like volume, pressure, enthalpy etc. <em>do these two quantities have physical significance</em> or, is it just that we define fugacity (without any physical meaning) just to make sure the equation</p>
<p><span class="math-container">$$\mu - \mu_0 = RT\ln\frac{p}{p_0}$$</span></p>
<p>looks the same for real gases? If yes, then what is the motivation for the name “fugacity” meaning literally “escaping tendency”?</p>
<p>Similarly, do we just define chemical potential randomly, since it's extremely <em>useful in various calculations</em> when the composition varies, or does it have some physical significance?</p>
<p>Also, why do we call it chemical potential?</p>
Answer: <p>The chemical potential is the partial derivative of the Gibbs free energy with respect to the number of moles of the specified species at constant T, P, and numbers of moles of all other species. The equation you wrote is the what this partial derivative reduces to for a species in an ideal gas mixture, where P is the partial pressure of the species in the mixture. For a real gas, the fugacity of the species replaces the partial pressure of the species in the same equation for the chemical potential.</p>
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https://chemistry.stackexchange.com/questions/145224/physical-significance-of-chemical-potential-and-fugacity
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Question: <p>I am studying solutions for my thermodynamics course and I am a little confused about the chemical potential of a solute. I found the following from Atkins:</p>
<blockquote>
<p>For a solute $\ce B$,obeying Henry's Law $p_\ce B=K_\ce B\, x_\ce B$, where $x_\ce B$ is the molar fraction</p>
</blockquote>
<p>$$\mu =\mu^*+RT\ln\left(\frac{p}{p^*}\right)=μ^*+RT\ln\left(\frac{K}{p^*}\right)+RT\ln(x)$$</p>
<p>It then states:</p>
<p>$$\mu(\textrm{standard})=\mu^*+RT\ln\left(\frac{K}{p^*}\right)$$</p>
<p>Therefore, the chemical potential of B in an ideal-dilute sol. is:</p>
<p>$$\mu=\mu(\textrm{standard})+RT\ln(x)$$</p>
<p>The text often states that anything with the * means it is in its <strong>pure form</strong>, so $\mu^*$ would be chemical potential of $\ce B$ not mixed in with a solvent...but how does this physically differ from $\mu(\textrm{standard})\,?$</p>
Answer:
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https://chemistry.stackexchange.com/questions/49197/what-is-the-difference-between-standard-chemical-potential-%ce%bc-standard-and-the
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Question: <p>I'm taking chemical thermodynamics this semester, and currently seeing things like the Maxwell's Relations.</p>
<p>My professor never uses explicitly the variable "n" and always works with molar quantities like $\bar{V}$ and stuff like that.</p>
<p>I think this simplifies the algebra a great deal, and can be useful if you have the temptation to get a "molar quantity". However, I don't like it right now.</p>
<p>I think theres nothing that can assure me that if I were to work with the variable $n$ in my derivations, I won't somehow get a quadratic dependence of my thermodynamic equation with $n$, or logarithmic, who knows.</p>
<p>But it seems that there is always a linear dependence with $n$. This makes physical sense to me, however, how can I prove this to myself? Maybe this is more of a question of mathematics, but it has scientific implications as well.</p>
<p>Thanks.</p>
Answer: <p>Extensive properties actually have a linear dependence with the common magnitudes used for amount of substance.</p>
<p>Mathematical speaking, you work in some representation in which one magnitude is represented with a function of some variables, and the later represent other physical variables. For example $S(U,V,N)$. When a magnitude represents a extensive property, the function is a homogeneous function of of order 1 of the extensive variables, that is:
$S(\lambda E, \lambda V, \lambda N) =\lambda \,S(E,V,N)$</p>
<p>We choose $\lambda = 1/N$, so
$S(\lambda E, \lambda V, \lambda N) =\lambda \,S(E,V,N)$, using lower case symbols for (say molar) properties,</p>
<p>$ N \,S(e, v, 1) = S(E,V,N)$</p>
<p>So, taking in mind that your equations are formulated for finding values for your thermodynamic functions, they must be stated in the form (for the previous example):</p>
<p>$S(E,V,N) =$ $whatever$</p>
<p>If you work with molar properties you will find just:</p>
<p>$S(e, v, 1) =$ $whatever$ $for$ $a$ $mol$, but using the third equation:</p>
<p>$S(E,V,N) = N\times$ $whatever$ $for$ $a$ $mol$</p>
<p>So the homogeneous (order 1) property will assure you that there is linear dependence with $N$.</p>
<p>Note: Thermodynamic functions that also depend on intensive variables have a zero order dependence in the homogeneous sense with these variables.</p>
<p><strong>EDITED:</strong></p>
<p>After reading the @Curt F. answer, to avoid confusions I decided made a comment: The explaining above works in the scope of 'traditional' thermodynamics, where no surface effects are involved. There are another reasons that can leads to draw wrong conclusions about this linearity, for example, if external fields are involved. But these cases are not in the scope of any course on thermodynamics that I aware of. </p>
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https://chemistry.stackexchange.com/questions/26785/will-all-thermodynamic-equations-have-a-linear-dependence-with-moles
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Question: <p>Can anyone help me to understand what the "Chemical Potential" is? I can see that in batteries the electrical energy is stored as "chemical potential", according to <a href="http://www.sciencedirect.com/science/article/pii/S1369702115003181" rel="nofollow noreferrer">this</a>:</p>
<blockquote>
<p>Li-ion batteries, Na-ion batteries, and Mg-ion batteries, reversibly
convert between electrical and chemical energy via redox reactions,
thus storing the energy as chemical potential in their electrodes.</p>
</blockquote>
<p>So what is chemical potential? According to Wikipedia, it is defined as:</p>
<blockquote>
<p>In thermodynamics, chemical potential, also known as partial molar
free energy, is a form of potential energy that can be absorbed or
released during a chemical reaction or phase transition</p>
</blockquote>
<p>Which is quite straight forward, in a way. However, it doesn't really make me understanding it entirely, therefore I would like to ask if there is anyone that can help me understand this. I'm always trying to use analogies and therefore I would like to ask the following question:</p>
<blockquote>
<p>Can the chemical potential be seen like the latent heat, where the
energy is stored as bonds and, in order to break the bonds, one needs
to add extra energy.</p>
</blockquote>
<p>Or is it more complex than this? Please see below a sketch with what I imagine happens inside. Did I understand it right?</p>
<p><a href="https://i.sstatic.net/pt5le.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/pt5le.png" alt="enter image description here" /></a></p>
<p>It's hard for me to understand what keeps the energy there. Any help counts!</p>
<p>Thank you!</p>
Answer:
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https://chemistry.stackexchange.com/questions/66863/understanding-chemical-potential-li-ion
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Question: <p>I want to ask a question about chemical potential, <span class="math-container">$\mu$</span>.</p>
<p>I dislike the use of integrals to describe quantities in thermodynamics. If we observe the definition of chemical potential for constant pressure:</p>
<p><span class="math-container">$$\mu_i = \left(\frac{\partial G}{\partial N_i} \right)_{T, P, N_i \neq N_j}$$</span></p>
<p>which I can understand to mean <em>the change in Gibbs free energy G for a small change caused by the number of molecules of species <strong>i</strong></em>.</p>
<p>However, using some high school mathematics, I wondered if the following interpretation was valid.</p>
<p>If we consider that if <span class="math-container">$N_i = 0$</span>, <span class="math-container">$G = 0$</span> and logically if <span class="math-container">$N_i = N_i$</span>, <span class="math-container">$G = G_i$</span>, integration by separation of variables will yield:</p>
<p><span class="math-container">$$\partial G = \mu_i \times \partial N_i$$</span>
<span class="math-container">$$\int_{0}^{G_i} \partial G = \mu_i\int_{0}^{N_i}\partial N_i$$</span></p>
<p>such that</p>
<p><span class="math-container">$$G = \mu_i \times N_i$$</span></p>
<p>and hence</p>
<p><span class="math-container">$$\mu_i = \frac{G}{N_i}$$</span>
where the chemical potential of a species <span class="math-container">$\mu_i$</span> can better be expressed as the Gibbs free energy of a species <span class="math-container">$G$</span> divided by the amount of substance <span class="math-container">$N_i$</span>.</p>
<p>Is this a more reasonable yet valid understanding of the chemical potential?</p>
<p><strong>EDIT</strong> a similar equation is derived in [1].</p>
<p>[1] Chen, L. (2019). Chemical potential and Gibbs free energy. MRS Bulletin, 44(7), 520-523. doi:10.1557/mrs.2019.162</p>
Answer:
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https://chemistry.stackexchange.com/questions/136948/integration-of-chemical-potential
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Question: <p>We're learning about chemical reactions in class and we recently did an experiment where we combined copper and nitric acid. Both were at relatively room temperature and when they were combined the temperature rose very quickly. Isn't this contradicting the first law of thermodynamics? If the thermal energy is neither created nor destroyed but only transferred than where did all that extra thermal energy come from? please help!</p>
Answer: <p>The energy was in the chemicals, in the form of bonds. Energy is liberated when there is bond formation. As the reaction between $\ce{Cu}$ and $\ce{HNO3}$ is feasible at room temperature so it happened. </p>
<p>And another thing, room temperature is a nice temperature for that reaction because first we need to break bonds of $\ce{HNO3}$ to make new bond, which is easily done at this temperature.</p>
<p>Now coming to your question, the answer is 'no'. The first law is still safe. You know it wrong. The truth is (from wikipedia) : "the total energy of an isolated system is constant; energy can be transformed from one form to another, but cannot be created or destroyed". </p>
<p>Give special attention to the word 'total energy', you said the 'thermal energy' in your question. And the total energy (atleast in this case) is conserved in the form of summation of thermal energy and bond energy.</p>
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https://chemistry.stackexchange.com/questions/41664/where-does-the-thermal-energy-come-from-in-a-chemical-change-when-two-substances
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Question: <p>(This question is taken from Problem 1.1(b) of the book Chemical Thermodynamics: Principles and Applications.)<span class="math-container">$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$</span></p>
<blockquote>
<p>Prove <span class="math-container">$$\pd HTp = \pd UTV + \left[ V - \pd HpT \right] \pd pTV. \tag{1}$$</span></p>
</blockquote>
<p>First, I expressed <span class="math-container">$H$</span> as <span class="math-container">$U + PV$</span>, and took the partial derivative against <span class="math-container">$T$</span>, keeping pressure constant:</p>
<p><span class="math-container">$$\pd HTp = \pd UTp + p\pd VTp \tag{2}$$</span></p>
<p>Expressing <span class="math-container">$U$</span> as a function of <span class="math-container">$(T, V)$</span>:</p>
<p><span class="math-container">$$\pd UTp = \pd UTV + \pd UVT \pd VTp \tag{3}$$</span></p>
<p>Using the well-known identity that <span class="math-container">$$\pd pVT \pd TpV \pd VTp = -1, \tag{4}$$</span></p>
<p><span class="math-container">$$\pd UVT \pd VTp = -\pd UpT \pd pTV = \left[ V - \pd UpT - V\pd ppT \right]\pd pTV \tag{5}$$</span></p>
<p>This, of course, means that</p>
<p><span class="math-container">$$\pd HTp = \pd UTV + \left[ V - \pd HpT \right]\pd pTV + p\pd VTp \tag{6}$$</span></p>
<p>Based on my working, there is an additional <span class="math-container">$$p\pd VTp$$</span> term. Is my working incorrect? I can’t find my mistake.<span class="math-container">$\endgroup$</span></p>
Answer: <p><span class="math-container">$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$</span></p>
<p>I would start from</p>
<p><span class="math-container">$$ dH = \pd HTp dT + \pd HpT dp$$</span></p>
<p>This gives rise to </p>
<p><span class="math-container">$$ \pd HTV = \pd HTp + \pd HpT \pd pTV$$</span></p>
<p>in which you can recognize some components of the solution. Since</p>
<p><span class="math-container">$$dH = dU +PdV + VdP $$</span> </p>
<p>so that </p>
<p><span class="math-container">$$ \pd HTV = \pd UTV + V \pd pTV$$</span></p>
<p>it follows that </p>
<p><span class="math-container">$$ \pd UTV + V \pd pTV = \pd HTp + \pd HpT \pd pTV$$</span></p>
<p>which is readily rearranged to obtain the desired equation.
<span class="math-container">$\endgroup$</span></p>
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https://chemistry.stackexchange.com/questions/126955/proving-a-thermodynamic-relation-between-partial-h-partial-t-p-and-part
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Question: <p>The fundamental equation of thermodynamics, as us chemists (and chemical engineers!) are used to seeing it, is </p>
<p>$$ dG = - S~dT + V~dP + \sum_{i}\mu_i~dN_i$$</p>
<p>This gives the Gibbs free energy as a function of temperature, pressure, and composition, <em>assuming there are no other relevant forces other than mechanical pressure</em>. </p>
<p>The other day I watched <a href="https://www.youtube.com/watch?v=uJ_NpWnCXzM" rel="nofollow">a video</a> on the <a href="https://en.wikipedia.org/wiki/Magnetic_refrigeration" rel="nofollow">magnetocaloric effect</a>. Obviously, there are non-pressure magnetic forces acting in such systems. <strong>What's the proper form of the fundamental equation for magnetocaloric materials?</strong> </p>
<p>Suppose that the magnetocaloric material used is chemically pure and non-reactive during the magnetization process. Then we could get rid of the $\sum_{i}\mu_i~dN_i$ term. I suppose its also reasonable to assume that pressure is constant during magnetization / demagnetization process, and that the volume of the material is unchanged by magnetization so probably we could dispense with the $V~dP$ term as well (is that true?). </p>
<p>That leaves us with $dG = -S~dT + \rm{MAGNETIC~STUFF}$. The $\rm{MAGNETIC ~STUFF}$ term probably has a $B$ or $H$ or something like that in it to represent the imposed magnetic field, but what else goes in there?</p>
Answer: <p>The fundamental equation for paramagnetic system is as follows<br>
$$dU=TdS+BdM+\mu dN$$<br>
Where B is the external magnetic field intensity and M is the magnetic moment (Here I have neglected the P-V work of the system). For your chemically pure and non-reactive system the above equation simplifies to<br>
$$dU=TdS+BdM$$<br>
Using Legendre Transformations:<br>
$$y(0)=U(S,M)$$
$$dy(0)=dU=TdS+BdM$$
$$dy(1)=dA(T,M)=-SdT+BdM$$
$$dy(2)=dG(T,B)=-SdT-MdB$$
Now changing the order of the equation, we can find enthalpy,
$$y(0)=U(M,S)$$
$$dy(0)=dU=BdM+TdS$$
$$dy(1)=dH(S,B)=-MdB+TdS$$</p>
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https://chemistry.stackexchange.com/questions/33329/fundamental-equation-of-thermodynamics-for-magnetocaloric-materials
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Question: <p>After doing some research and reading I found some problems which I will try to state as clearly as possible. </p>
<p>The definition of Gibbs Free Energy says "<em>the greatest amount of mechanical work which can be obtained from a given quantity of a certain substance in a given initial state, without increasing its total volume or allowing heat to pass to or from external bodies, except such as at the close of the processes are left in their initial condition.</em> (I have taken the definition from the wikipedia). The equation of the Gibbs energy <span class="math-container">$$ G= H -TS$$</span> is also clear.
<strong>But the way I understand it is this: the maximum non-expansion work , like transporting an electron, breaking a chemical bond, moving real life things etc. which can be obtained from a system</strong> . These things made sense during the study of Second Law of thermodynamics but as I moved to Chemical Equilibrium these concepts began to tremble (for me).<br>
In elementary classes it is said that Equilibrium is a state where the composition of reactants and products do not change over time, but in higher chemistry classes it is said that equilibrium is a condition corresponds to <span class="math-container">$$ \Delta G = 0$$</span> So, my first question is how do these two concepts mean same thing? The next thing which causes problem is chemical potentials. If we adhere to the formal meaning of <em>potential</em> i.e. something which is stored and can be used when proper conditions are met, so <em>chemcial potential</em> would mean <em>the potential of substance to react</em> and again this is related to Gibbs energy whose definition I gave above. So, how do chemical potential and Gibbs energy can have any relation?<br>
A question which is off topic over here but I want to mention it, why do we bother so much about standard things like <span class="math-container">$$ \mu_\mathrm A = \mu_\mathrm A^\circ + RT \ln(p_\mathrm A)$$</span> why we wanted to express it in that standard (that little circle) form ?<br>
I want to make myself clear that the conception of Gibbs energy was quite clear to me in the context of thermodynamics, we simply meant it to be the work which can be extracted from a substance, but it all-pervading use has made me to doubt myself, just like in mathematics the number <span class="math-container">$\mathrm e$</span> appears in odd places. Even if Gibbs energy (according to the understanding that gave above in bold) is appearing mathematically then also it would be having some physical meaning because Thermodynamics and Equilibrium are natural sciences and not the mathematics. </p>
<p>Thank you, any help will be much appreciated. </p>
Answer: <p>Your first question:</p>
<p>Before the second law was understood it used to be thought that the maximum amount of work that could be extracted from a reaction was <span class="math-container">$-\Delta H$</span> but many experiments showed that this was not the case. It is true that, according to the first law, (the law of conservation of energy), that external work done must be equal to the loss of energy of the system, unless some heat is given to or taken from the surroundings. This was first understood by Gibbs. In an isothermal reaction working under reversible conditions, the heat absorbed from the surroundings is <span class="math-container">$T\Delta S$</span>, if this is positive then the work done will be even greater than the heat of reaction and so heat will be taken from the thermostat.</p>
<p>Second question:</p>
<p>The <span class="math-container">$\Delta G$</span> you mention is actually the gradient of the free energy with extent of reaction <span class="math-container">$(\partial G/\partial \xi)_{T,p}$</span> at constant temperature and pressure. This differential is sometimes written as <span class="math-container">$\Delta G'$</span>. The equation in general is <span class="math-container">$\Delta G'=\Delta G^\text{o}+RT\ln(Q)$</span> where <span class="math-container">$Q$</span> is the ratio of partial pressures, for example, in a gas phase reaction. At equilibrium a plot of <span class="math-container">$G$</span> vs <span class="math-container">$\xi$</span> reaches a minimum and the gradient is zero, i.e. <span class="math-container">$\Delta G'=0$</span> then <span class="math-container">$\Delta G^\text{o}=-RT\ln(K_p)$</span> where <span class="math-container">$K_p$</span> is the equilibrium constant and is used instead of <span class="math-container">$Q$</span> at equilibrium. This reaction tells us how the position of equilibrium can be determined in terms of standard state free energies of reactants and products at 1 atm pressure.</p>
<p>(The extent of reaction is zero when only reactants are present and is one when one mole of reactants have changed to product) </p>
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https://chemistry.stackexchange.com/questions/122162/what-is-the-true-meaning-of-gibbs-energy-and-chemical-potential
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Question: <p>Engel and Reid's <em>Physical Chemistry</em> [1, p. 308] defines the electrochemical potential in terms of the charge in units of the electron charge <span class="math-container">$(z),$</span> Faraday constant <span class="math-container">$(F),$</span> and electrical potential <span class="math-container">$(\phi)$</span> as follows:</p>
<blockquote>
<p>The electrochemical potential is a generalization of the chemical potential to include the effect of an electrical potential on a charged particle.
It is the sum of the normal chemical potential <span class="math-container">$\mu$</span> and a term that results from the nonzero value of the electrical potential:</p>
<p><span class="math-container">$$\tilde{\mu} = \mu + zF\phi \tag{11.3}$$</span></p>
</blockquote>
<ol>
<li><p>Is the electrical potential <span class="math-container">$\phi$</span> due to electric field generated from charge separation between electrode and solution, or is it an externally applied electric field?</p>
</li>
<li><p>Isn't <span class="math-container">$zF\phi$</span> electrical potential energy? How can we simply add it to chemical potential?</p>
</li>
</ol>
<h3>Reference</h3>
<ol>
<li>Engel, T.; Reid, P. <em>Thermodynamics, Statistical Thermodynamics, and Kinetics: Physical Chemistry</em>, 4th ed.; Pearson: New York, <strong>2019</strong>. ISBN 978-0-13-480458-3.</li>
</ol>
Answer: <p>The chemical potential of a substance <span class="math-container">$i$</span> in a system is generally defined in relation to the Gibbs energy differential:</p>
<p><span class="math-container">$$\mathrm{d}G = \left( \frac{\partial G}{\partial n_i} \right)_{T,p}\mathrm{d}n_i = \mu_i\ \mathrm{d}n_i \tag{1}$$</span></p>
<p>The related Gibbs energy differential for a charge is:</p>
<p><span class="math-container">$$\mathrm{d}G = \left( \frac{\partial G}{\partial q} \right)_{T,p}\mathrm{d}q = \phi \mathrm{d}q = \phi (zF \mathrm{d}n_i) = (\phi zF) \mathrm{d}n_i = \mu_{i, E}\cdot \mathrm{d}n_i \tag{2}$$</span></p>
<p>Then, it can be summed up:</p>
<p><span class="math-container">$$\mathrm{d}G = \tilde{\mu_i}\cdot \mathrm{d}n_i = \left(\mu_i + \mu_{i, E}\right)\cdot \mathrm{d}n_i =\left(\mu_i + \phi zF\right)\cdot \mathrm{d}n_i \tag{3}$$</span></p>
<p>By other words, the electrochemical potential of a charged molecular entity is its chemical potential dependent on local electrostatic potential:</p>
<p><span class="math-container">$$\tilde{\mu_i}=\tilde{\mu_i}(\phi)\tag{4}$$</span></p>
<p>This allows us to modify chemical potentials of molecular entities at the electrode surface by externally forced potential. This is used during electrolysis, when reactions that would not be otherwise spontaneous become spontaneous.</p>
<hr />
<p>Commenting the feedback:</p>
<p>I have incorrectly evaluated your level from your question. :-) I think I have not written nothing extraordinary.</p>
<p>The contribution of electrostatic potential can be written as dG/dn the same way as is for the chemical potential. It formally contributes to the summary electrochemical potential.</p>
<hr />
<p>Regarding the 1st question, it is potential of any origin. It may be forced externally, it may be caused local charge disbalance, it may be caused by the electrode surface dual layer, it may be the bulk solution volume potential, there may be Ohmic potential gradient....</p>
<p>But the potential at electrode/solution boundary is the most important, as that is the place where redox reactions occur.</p>
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https://chemistry.stackexchange.com/questions/174677/why-is-electrochemical-potential-defined-as-sum-of-electrical-potential-and-appa
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Question: <p><span class="math-container">$ΔG=-nFe$</span> (<a href="https://chemistry.stackexchange.com/questions/72457/why-delta-g-nfe-and-not-f-integraldne/176835#176835">example</a>) and <span class="math-container">$μ = μ° + RT\ln a$</span> are used very frequently in chemical thermodynamics.</p>
<p>However, if the Gibbs energy and pressure <span class="math-container">$P^\circ,G^\circ$</span> in the standard state of an ideal gas are known, when an isothermal reaction <span class="math-container">$(P^\circ,G^\circ) \to (P,G)$</span> occurs, if the pressure P at the end of this reaction is known, the Gibbs energy G at the end state can be expressed by the following formula.</p>
<p><span class="math-container">$$G\ =G^\circ+\ nRT\ln\left(\frac{P}{P^\circ}\right)$$</span></p>
<p>Using the extensivity of G and partially differentiating with respect to n, the "Gibbs energy per mole", that is, the chemical potential μ of this system, is</p>
<p><span class="math-container">$$\mu=\mu^\circ+\ RT\ln\left(\frac{P}{P^\circ}\right)$$</span></p>
<p>So, If we further assume that in an ideal solution where solute A is volatile, and the mole fraction of substance A is <span class="math-container">$\chi_A$</span> and the vapor pressure of the pure substance A is <span class="math-container">$P_A^\bullet$</span>, then the vapor pressure <span class="math-container">$P_A$</span> of A satisfies the following equation.</p>
<p><span class="math-container">$$P_A=\ \chi_A\ P_A^\bullet$$</span></p>
<p>Therefore, if substance A is volatile and the vapor is an ideal gas, the chemical potential of the vapor of substance A is:</p>
<p><span class="math-container">$$\mu_A=\mu_A^\circ+RT\ln\left(\frac{\chi_A\ P_A^\bullet}{P^\circ}\right)\ \\ =\mu_A^\circ+RT\ln\left(\frac{\ P_A^\bullet}{P^\circ}\right)\ +RT\ln\left(\chi_A\right)$$</span></p>
<p>When we define<span class="math-container">$\mu_A^\bullet$</span> as follows</p>
<p><span class="math-container">$$\mu_A^\bullet := \mu_A^\circ+RT\ln\left(\frac{\ P_A^\bullet}{P^\circ}\right)\ $$</span></p>
<p>we get
<span class="math-container">$$\mu_A = \mu_A^\bullet +RT\ln\left(\chi_A\right)$$</span></p>
<p>However, if A is non-volatile (for example NaCl),</p>
<p><span class="math-container">$$P_A^\bullet =0$$</span></p>
<p>and so</p>
<p><span class="math-container">$$\ln\left(\frac{\ P_A^\bullet}{P^\circ}\right) =\ln (0) = -\infty$$</span></p>
<p>Of course, for volatile non-ideal solvents an empirical correction using activity values could be made. The chemical potential μ is expressed as a function of the standard chemical potential μ° and the activity a as follows:</p>
<p><span class="math-container">$μ = μ° + RT \ln a$</span></p>
<p>However, in the case of non-volatile solvents such as NaCl, there is the major problem of ln(0)=-∞ mentioned above, so derivation using vapor pressure seems inherently difficult, and even if activity is used, it seems necessary to use a logic that at least avoids "ln(0)=-∞".</p>
<p>Despite this, many textbooks use the same formula for the chemical potential of an ideal solution of a non-volatile solute as for a volatile solute.</p>
<p>My question: <em>Why does the following equation hold even when the A is non-volatile? How does the following equation derive when A is non-volatile?</em></p>
<p><span class="math-container">$μ = μ° + RT \ln a$</span></p>
Answer: <p>I think I was able to solve the problem myself using the following method:</p>
<p><strong>●The chemical potential of an ideal gas</strong> is:
<span class="math-container">$$\mu=\mu_0+RT\ln\left(\frac{P}{P_0}\right) \tag{0-1}$$</span></p>
<p>Where,</p>
<ul>
<li><span class="math-container">$\mu$</span>: Chemical potential</li>
<li><span class="math-container">$\mu_0$</span>: Standard state chemical potential</li>
<li><span class="math-container">$R$</span>: Gas constant</li>
<li><span class="math-container">$T$</span>: Temperature</li>
<li><span class="math-container">$P$</span>: System pressure</li>
<li><span class="math-container">$P_0$</span>: Standard state pressure</li>
</ul>
<p><strong>●In the case of volatile components</strong> (volatile solutes and solvents)<br></p>
<p>When the mole fraction of a volatile component is <span class="math-container">$x$</span>, applying Henry's law under the condition that "the volatile components of the dilute solution are in equilibrium with the gas phase," the saturated vapor pressure P of the volatile component is given as follows:</p>
<p><span class="math-container">$$P=K_{H}x \tag{1-1}$$</span></p>
<p>Here, <span class="math-container">$K_\mathrm{H}$</span> is Henry's constant.</p>
<p>Substituting Henry's law into the formula (0-1) for the chemical potential of an ideal gas,</p>
<p><span class="math-container">$$\mu = \mu_0 + RT \ln \left( \frac{K_H x}{P^0} \right)\tag{1-2}$$</span></p>
<p>Rearranging this,</p>
<p><span class="math-container">$$\mu = \mu_* + RT \ln x\tag{1-3}$$</span></p>
<p>Here,
<span class="math-container">$$\mu_* = \mu_0 + RT \ln \left( \frac{K_H}{P^0} \right)\tag{1-4}$$</span></p>
<p>Expression using activity: To take into account the difference between ideal and actual behavior, we introduce activity a instead of concentration <span class="math-container">$x$</span>,</p>
<p><span class="math-container">$$\mu = \mu_* + RT \ln a\tag{1-5}$$</span></p>
<p><strong>●In the case of non-volatile solutes</strong><br>
For non-volatile solutes, the vapor pressure is <span class="math-container">$P=0$</span>, and direct substitution into the basic formula is mathematically inappropriate. For simplicity, consider the case where one non-volatile solute is dissolved in a volatile solvent.</p>
<p>Therefore, consider the application of the Gibbs-Duhem equation. When the total amount of substance is constant and the temperature and pressure are constant, the Gibbs-Duhem equation can be written as follows;</p>
<p><span class="math-container">$$n_1d\mu_1 + n_2 d\mu_2 = 0\tag{2-1}$$</span></p>
<p>Here, <span class="math-container">$n_1$</span> and <span class="math-container">$n_2$</span> are the amounts of solute and solvent, respectively, and <span class="math-container">$\mu_1$</span> and <span class="math-container">$\mu_2$</span> are the chemical potentials of the solute and solvent, respectively. If this is rewritten using the mole fractions <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span> of the solute and solvent, it becomes as follows.</p>
<p><span class="math-container">$$x_1 d\mu_1 + x_2 d\mu_2 =0\tag{2-2}$$</span></p>
<p>here,the mole fraction <span class="math-container">$x_1$</span> of solvent 1 and the mole fraction <span class="math-container">$x_2$</span> of solute 2 are defined as follows:</p>
<p><span class="math-container">$$x_1 = \frac{n_1}{n_1 + n_2}\tag{2-3}$$</span>
<span class="math-container">$$x_2 = \frac{n_2}{n_1 + n_2}\tag{2-3a}$$</span></p>
<p>From (2-2), we get
<span class="math-container">$$d\mu_2 = - \frac{x_1}{x_2} d\mu_1 \tag{2-4}$$</span></p>
<p>Assuming that Raoult's law holds true, the chemical potential of the solvent can be written as follows using the same discussion as in the previous section.
<span class="math-container">$$\mu_1 = \mu_1^* + RT \ln x_1\tag{2-5}$$</span></p>
<p>Differentiating both sides gives us the following:</p>
<p><span class="math-container">$$d\mu_1 = RT \frac{dx_1}{x_1}\tag{2-6}$$</span></p>
<p>From (2-4)＆(2-6), we get:
<span class="math-container">$$d\mu_2 = - \frac{x_1}{x_2} d\mu_1
= - \frac{x_1}{x_2} RT \frac{dx_1}{x_1}
= - RT \frac{dx_1}{x_2}\tag{2-7}$$</span></p>
<p>On the other hand, (2-3) shows the fact that the ratio of solvent to solute is always 100% (or 1) overall.
<span class="math-container">$$x_1 + x_2 =1\tag{2-8}$$</span></p>
<p>Differentiating both sides of (2-8) gives us the following:</p>
<p><span class="math-container">$$d{x}_1=−d{x}_2\tag{2-9}$$</span></p>
<p>From the above (2-6),(2-9), the following equation holds for the chemical potential of the solute:
<span class="math-container">$$d\mu_2 = RT \frac{dx_2}{x_2}\tag{2-10}$$</span></p>
<p>Integrating both side of (2-19) gives us:
<span class="math-container">$$\mu_2 = \mu_2^* + RT \ln x_2\tag{2-11}$$</span></p>
<p>This method allows us to properly describe the chemical potential even for non-volatile solutes.</p>
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https://chemistry.stackexchange.com/questions/184433/how-to-derive-the-chemical-potential-of-an-ideal-solution-of-a-non-volatile-solu
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Question: <p>In thermodynamics, to find the expression for the chemical potential of a pure gas, we can find this expression:</p>
<p><span class="math-container">$$\frac{\mathrm d\mu}{\mathrm dP} = \frac{\mathrm dV}{\mathrm dn} = \frac{RT}{P} \quad \text{(from $\frac{\mathrm dg}{\mathrm dP} = V$)}.$$</span></p>
<p>But why does this formula not work for all gases, even in a mixture? Why does this formula only work for pure substances?</p>
<p>Because even in a mixture, we have
<span class="math-container">$\dfrac{dV}{\mathrm dn} = \dfrac{RT}{P}$</span>
for a gas, and <span class="math-container">$\dfrac{dV}{\mathrm dn}$</span> for a liquid, so it does not change whether it is in a mixture or not. So why doesn't the formula work in all cases? Where is the error, since all the steps seem correct?</p>
<p>Thank you.</p>
Answer: <p>Gibbs' theorem tells us that the partial molar free energy (chemical potential) of a species in an ideal gas mixture is the same as that of the pure species at the same temperature and at a pressure equal to the partial pressure of the species in the gas mixture. So, <span class="math-container">$$\mu=\mu^0(T)+RT\ln(Px)$$</span> where <span class="math-container">$P$</span> is the total pressure in bars, and <span class="math-container">$x$</span> is the mole fraction of the species in the mixture. This equation applies both to a pure species and to a species in an ideal gas mixture.</p>
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https://chemistry.stackexchange.com/questions/188654/why-does-the-formula-for-chemical-potential-change
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Question: <p>I have a system of a pure ideal gas <span class="math-container">$a$</span>, which I call 'the reference', with a known <span class="math-container">$P^{ref}, V^{ref}, T^{ref}$</span> and <span class="math-container">$n^{ref}_a$</span>. I know the chemical potential formula for an ideal gas, so can calculate how it changes as its variables change.</p>
<p>I place it next to an unknown system with unknown chemical potential <span class="math-container">$\mu_a$</span>. This system must me "much bigger" than my reference so that it isn't disturbed by their interaction.</p>
<p>I make the separation a fixed membrane permeable only to <span class="math-container">$a$</span>, and put both in contact with a bath of <span class="math-container">$T = T^{ref}$</span>.</p>
<p>I wait for the systems to reach equilibrium, and measure <span class="math-container">$P^{ref}, V^{ref}, T^{ref}$</span> and <span class="math-container">$n^{ref}_a$</span>, which let's me calculate how <span class="math-container">$\mu^{ref}_a$</span> changed.</p>
<p>Thermodynamics say that at equilibrium <span class="math-container">$\mu^{ref}_a$</span> must have become equal to the unknown <span class="math-container">$\mu_a$</span>. So I just measured the difference between the unknown <span class="math-container">$\mu_a$</span> and the initial reference <span class="math-container">$\mu^{ref}_a$</span>.</p>
<p>This device then let's me directly measure chemical potential (relative to a reference).</p>
<p><strong>Is this correct? Can chemical potential be directly measured like this?</strong></p>
<p>I believe this is completely analogous to how thermodynamic pressure is measured, (relative to a reference like atmospheric pressure), by connecting two systems through a movable wall and letting it reach equilibrium.</p>
<p>I was thinking maybe ph meters are an example of this. Osmotic pressure tubes too.</p>
Answer:
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https://chemistry.stackexchange.com/questions/167457/can-chemical-potential-be-directly-measured-like-this
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Question: <p>I study physical chemistry on my own and I have to decide in which order I learn the following topics
These are my topics:</p>
<p>Statistical Thermodynamics (especially entropy, partition function, Boltzmann factor..)/
Maxwells Relations/
Thermodynamic Potentials/
Chemical potential/
Colligative properties/
Kinetics/
Electrochemistry/</p>
<p>What is your opinion, your preferred order</p>
Answer: <p>Take any common PC book, and follow the order given therein.</p>
<p>It usually starts with ideal gas equation, enthalpy, entropy, Gibbs, chemical potential, i.e. plain thermodynamics. (that lecture was called <strong>PC1</strong> at my alma mater)</p>
<p>On top of that, you can add kinetics and statistics (<strong>PC4a,b</strong>) and electrochemistry(<strong>PC5</strong>) at will, those do not depend on each other much. The basics of kinetics and electrochemistry we learned earlier however, I think in the introductory lecture for inorganic chemistry.</p>
<p>I´m not saying one couldn´t start with any of the latter subjects instead of plain thermodynamics, but have never heard of that being done and no idea how it would work. Would be a whole new approach to (self)teaching physical chemistry. ;)</p>
<p>Inbetween, you should however allow yourself a good dose of quantum mechanics (<strong>PC2&3</strong>), because that greatly improves understanding of the concepts and esp. their limitations. Some books and schools also start with QM (<strong>PC[zero]</strong>), I guess for that exact reason.</p>
<p>Again, I´m not saying this couldn´t be rearranged, but the above is the way most books and lectures do it. I guess that with statistics etc., you would have to stop the subject all the time to introduce another concept from basic thermodynamics, and also all the rest of chemistry remains very vague and descriptive if you don´t have a good understanding of gas laws, phase transitions, energy and temperature.</p>
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https://chemistry.stackexchange.com/questions/127738/in-which-order-should-i-learn-thermodynamic
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Question: <p>I haven’t quite reached the point where I can read a full-fledged text on chemical kinetics and thermodynamics yet, so bear with me, please. </p>
<p>I’m wondering why a value like $K_\text{eq} = \frac{[\ce{NO}]^2[\ce{O2}]}{[\ce{NO2}]^2}$ wouldn't have units of M?</p>
Answer: <p>I goofed up the first time I tried to answer this question, erroneously applying dimensional analysis to your equilibrium expression.</p>
<p>It turns out that Silberberg<sup>[1]</sup> gives a good explanation of why $K_\text{eq}$ is dimensionless, which is often glossed over as the terms of the equilibrium expression are generally taught as concentrations. In actual fact, the terms are ratios of the concentration or activity of each species with a reference concentration (1 $\mathrm{mol\cdot{L^{-1}}}$ for solutions.) For example, a concentration of 2 $\mathrm{mol\cdot{L^{-1}}}$ divided by a reference of 1 $\mathrm{mol\cdot{L^{-1}}}$ yields a ratio of 2, with no units. As each term has no units, so too does $K_\text{eq}$.</p>
<p>[1] Silberberg, M.E.; Chemistry – The Molecular Nature of Matter and Change 3e; 2003, p. 719</p>
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https://chemistry.stackexchange.com/questions/1137/why-are-equilibrium-constants-unitless
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Question: <p>Another statement of second law of thermodynamics can be formulated in terms of system properties and not properties + surroundings (isolated system). For a closed system at constant temperature and pressure:</p>
<p><span class="math-container">$$dG_{T,p,n_i} \leq 0$$</span></p>
<p><strong>Are <span class="math-container">$n_i$</span> constant during a process?</strong>
We know that during a chemical reaction Gibbs free energy take its minimum value at equilibrium. But during the reaction the number of molecules <span class="math-container">$n_i$</span> change. So would it better the inequality to take the form of:</p>
<p><span class="math-container">$$dG_{T,p} \leq 0$$</span></p>
<p>?</p>
<p><strong>Interpretation of the differential</strong>
In mathematics the differential of a function <span class="math-container">$f(x)$</span> can be interpreted as the change in <span class="math-container">$f(x)$</span> for an infinitesimal change in the argument <span class="math-container">$x$</span>, that is:</p>
<p><span class="math-container">$$df(x)= f(x+dx)-f(x)$$</span></p>
<p>I have trouble in the interpretation of <span class="math-container">$dG \leq 0$</span>. When I was introduced in thermodynamics I always thought of the spontaneity inequalities in the following manner.</p>
<p>Suppose that the system is at equilibrium state (state <span class="math-container">$1$</span>) at time <span class="math-container">$t$</span>. Something happens (e.g. a chemical reaction takes place) so the system will reach a new equilibrium state (state <span class="math-container">$2$</span>). In that case <span class="math-container">$G$</span> can be thought only as function of <span class="math-container">$ξ$</span> (the extent of reaction). Because the Gibbs energy must take its minimum value at equilibrium I interpreted the <span class="math-container">$dG \leq 0$</span> as:</p>
<p><span class="math-container">$$dG(t)= G(t+dt)- G(t) \leq 0$$</span></p>
<p>In other words Gibbs free energy decreases until it takes its minimum value. And because we can plot <span class="math-container">$G$</span> as function of <span class="math-container">$ξ$</span> the reaction will proceed in the direction where <span class="math-container">$G$</span> is minimized as shown in the following figure (the left figure):</p>
<p><a href="https://i.sstatic.net/asjGY.png" rel="noreferrer"><img src="https://i.sstatic.net/asjGY.png" alt="enter image description here" /></a></p>
<p>Although the above interpretation seems intuitive I have some problems.</p>
<p><strong><span class="math-container">$1$</span>st Problem</strong>
With the above interpretation is not clear if the <strong>equality</strong> should be interpreted as that the function have a constant value over an inteval or that it has reached a minimum value. For example if we were monitoring <span class="math-container">$G$</span> over time and we noticed that <span class="math-container">$G$</span> has a constant value would this imply that the system have reached equilibrium? Look at the following figure:</p>
<p> <img src="https://i.sstatic.net/LKYEF.png" width="350" height="200"></p>
<p>In the mid region of the function (where it is constant) the differential is zero. So how <span class="math-container">$dG \leq 0$</span> should be interpreted? In general we can't find a function <span class="math-container">$φ$</span> so that <span class="math-container">$G=φ(t)$</span> and this should be expected as Thermodynamics doesn't deal with how the state of the system varies with time.</p>
<p><strong><span class="math-container">$2$</span>nd Problem</strong> In the first scheme I have drawn an alternative path (right figure) that the system could take in order to minimize <span class="math-container">$G$</span>. In that case there is an abrupt change in the extent of the reaction but still the systems keeps decreasing its <span class="math-container">$G$</span>. To be honest I haven't seen any reaction reaching equilibrium in that way but I can't find a reason why it should not behave this way. It should be noted that the line representing the path depicts the change in <span class="math-container">$ξ$</span> in an abrupt manner and it should not be confused with a continuous change (think it like a <span class="math-container">$5$</span> step process).</p>
<p>In summary I want to make clear how to interpret the differential expression of the second law. Is a smooth incease of entropy (of surroundings + system) over time assumed?</p>
Answer:
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https://chemistry.stackexchange.com/questions/152616/do-the-differential-expressions-of-second-law-of-thermodynamics-imply-a-smooth-i
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Question: <p>One of the most fundamental equations in chemical thermodynamics states: <span class="math-container">$$ \Delta_rH_m^⦵ = \Delta_rG_m^⦵ + T\Delta_rS_m^⦵ $$</span>
If we look at this equation in context of net chemical reaction in electrolytic or galvanic cell, it is usually interpreted as follows: Enthalpy of reaction denotes total amount of energy at constant temperature and pressure which needs to be supplied (electrolytic cell) or which is released (galvanic cell) during a reaction, standard Gibbs energy of reaction denotes what amount needs to be supplied or which is released in form of electrical energy (electric potential difference between electrodes), the last term including standard entropy of reaction denotes what amount of heat is exchanged with surroundings during a process.</p>
<p>In electrolytic cell, as entropy of reaction is mostly positive, last term is usually interpreted as heat which comes from surroundings and as such it increases entropy of the system. As heat comes to the system, it helps us during electrolysis since we don't need to put in the whole enthalpy of reaction in form of electrical energy, but only Gibbs energy. In galvanic cell it is the other way around.</p>
<p>What I don't understand is the interpretation of that last term ,which includes entropy of reaction, which I found on hyperphysics page. According to hyperphysics, this term denotes heat exchanged with surroundings and they say that entropy change in the system is due that heat exchanged.</p>
<p>I would say that entropy change is not due to heat exchanged, but due to the fact that during chemical reactions entropy changes because products and reactants have different entropies since they are different compounds with different structure and aggregate state. Entropy of reaction denotes such entropic changes and not entropy changes due to heat exchanged. What are your thoughts?</p>
Answer: <blockquote>
<p>[OP] One of the most fundamental equations in chemical thermodynamics states: <span class="math-container">$$ \Delta_rH_m^⦵ = \Delta_rG_m^⦵ + T\Delta_rS_m^⦵ $$</span></p>
</blockquote>
<p>This is just the definition of the Gibbs energy. By itself, it does not give you any insight into chemical reactions.</p>
<blockquote>
<p>[OP] If we look at this equation in context of net chemical reaction in electrolytic or galvanic cell, it is usually interpreted as follows: Enthalpy of reaction denotes total amount of energy at constant temperature and pressure which needs to be supplied (electrolytic cell) or which is released (galvanic cell) during a reaction, standard Gibbs energy of reaction denotes what amount needs to be supplied or which is released in form of electrical energy (electric potential difference between electrodes), the last term including standard entropy of reaction denotes what amount of heat is exchanged with surroundings during a process.</p>
</blockquote>
<p>This is not the usual interpretation. Instead <span class="math-container">$ \Delta_rG_m^⦵ $</span> is the maximal work you can get out of the reaction (or, if the reaction moves away from equilibrium as written, the minimal work you have to put in to run it in that direction). <span class="math-container">$ \Delta_rH_m^⦵ $</span> is the sum of heat and non-PV work exchanged with the system. It could be negative or positive irrespective of whether this is an electrolytic or galvanic cell. Finally, the entropy of reaction is the same as for any reaction - roughly how much the dispersion of energy and particles changes within the system when the reaction proceeds.</p>
<blockquote>
<p>[OP] In electrolytic cell, as entropy of reaction is mostly positive, last term is usually interpreted as heat which comes from surroundings and as such it increases entropy of the system. As heat comes to the system, it helps us during electrolysis since we don't need to put in the whole enthalpy of reaction in form of electrical energy, but only Gibbs energy. In galvanic cell it is the other way around.</p>
</blockquote>
<p>You can't make any generalizations about the heat transfer when there is also work exchanged between system and surrounding. In many thermodynamic arguments, you talk about processes that happen in the absence of non-PV work, and in a reversible manner (i.e. the entropy of the universe is near-constant, which is the same as saying it is a near-equilibrium process).</p>
<blockquote>
<p>[OP] What I don't understand is the interpretation of that last term ,which includes entropy of reaction, which I found on hyperphysics page. According to hyperphysics, this term denotes heat exchanged with surroundings and they say that entropy change in the system is due that heat exchanged.</p>
</blockquote>
<p>Without a specific link, it is hard to understand what statement on the hyperphysics site you are referring to, and it what context it was made. In general, the entropy change of the system reflects changes in the system.</p>
<p><strong>EDIT</strong>: The remainder was written after OP provided the link.</p>
<blockquote>
<p>[<a href="http://%20hyperphysics.phy-astr.gsu.edu/hbase/thermo/electrol.html" rel="nofollow noreferrer">Hyperphysics</a>] Since the electrolysis process results in an increase in entropy, the environment "helps" the process by contributing the amount TΔS. The utility of the Gibbs free energy is that it tells you what amount of energy in other forms must be supplied to get the process to proceed.</p>
</blockquote>
<p>If there is no work and the reaction is run near equilibrium, the environment shows a change in entropy of <span class="math-container">$-\frac{\Delta H}{T}$</span>. TΔS is not directly related to the change of entropy in the surrounding. The reason that <span class="math-container">$\Delta H$</span> is related to the entropy change of the surrounding is that for a boring surrounding (i.e. one that only shows changes in temperature, and very little at that) the change in entropy is proportional to the heat transferred.</p>
<blockquote>
<p>I would say that entropy change is not due to heat exchanged, but due to the fact that during chemical reactions entropy changes because products and reactants have different entropies since they are different compounds with different structure and aggregate state. Entropy of reaction denotes such entropic changes and not entropy changes due to heat exchanged. What are your thoughts?</p>
</blockquote>
<p>That is correct. Entropy is a state function, so if the entropy of the system changes, it is because the system changes, no matter what happens in the surrounding (you can't break the second law, though). The hyperphysics quote does not say that the entropy change of the system is associated with heat exchange. Their (implicit) argument is the second law:</p>
<p><span class="math-container">$$T \Delta S_\mathrm{system} + T \Delta S_\mathrm{surrounding} > 0 $$</span></p>
<p>So if we have a positive change in entropy of the system like in the example, we can get away with cooling down the surrounding a bit (negative change in entropy of the surrounding). Typically, though, we don't want a super-slow reaction, so you would apply a higher voltage than necessary, and the heat transfer would go in the other direction (as you can tell by the heat typically dissipated from a fast battery charger).</p>
<p>Applying this to the first law, some of the energy input for the reaction can be in the form of heat transfered from the surrounding, it does not have to be all work in this case. (In other cases you have to put in more work, not for the first law energy balance but to satisfy the second law.) For a graphic illustration of the different cases, see <a href="https://chemistry.stackexchange.com/a/112958/72973">https://chemistry.stackexchange.com/a/112958/72973</a>.</p>
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https://chemistry.stackexchange.com/questions/151090/entropy-changes-in-electrolytic-galvanic-cell
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Question: <p>In the textbook Electrochemical Systems by Newman and Alyea, Chapter 14: The definition of some thermodynamic functions, chemical potential of component (ionic or neutral) is written as a function of absolute activity: <span class="math-container">$$\mu_i=RT\ln \lambda_i \tag{1}$$</span></p>
<p>where <span class="math-container">$\lambda_i$</span> is the absolute activity of the component <span class="math-container">$i$</span>. This equation is referenced to the Guggenheim: Thermodynamics textbook. I checked this textbook, but it didn't really resolve my questions.</p>
<p>Author claims that absolute activity allows us to define chemical potential without reference to the standard state, as it can be seen by the equation 1. This means that absolute activity is independent on the choice of the standard state.</p>
<p>What I know from thermodynamics is that activity is defined as a quantity which allows us to express chemical potential for real systems in the same mathematical form as for ideal systems. Chemical potential for real systems is logarithmic function of activity as it is a logarithmic function of pressure, mole fraction, concentration etc. for ideal systems.</p>
<p>In the most general form, activity is defined as: <span class="math-container">$$a_i = \frac {f_i}{f_i^⦵} \tag {2}$$</span></p>
<p>where <span class="math-container">$f_i$</span> is the fugacity of the component in the system and <span class="math-container">$f_i^⦵$</span> is the standard state fugacity of the component.</p>
<p>Definition of activity is fundamentally tied to the chemical potential and its definition equation can also be written as: <span class="math-container">$$ \mu_i = \mu_i^⦵ + RT\ln \frac {f_i}{f_i^⦵} = \mu_i^⦵ + RT\ln a_i \tag {3} $$</span></p>
<p>My question is:</p>
<p>Concept of absolute activity doesn't make sense to me. Activity by definition expresses how much is a component thermodynamically active RELATIVE to the standard state. This is why equation 2 is written as a ratio and why in equation 3, standard state chemical potential <span class="math-container">$\mu_i^⦵$</span> shows up.</p>
<p>Given that, I don't understand equation 1 and where is it derived from as we can see a difference comparing equations 1 and 3.</p>
Answer: <p>Using the definition of the absolute activity provided by <a href="https://goldbook.iupac.org/terms/view/A00019" rel="nofollow noreferrer">IUPAC</a></p>
<p><span class="math-container">$\mu_i=RT\ln \lambda_i \tag{1}$</span></p>
<p><span class="math-container">$\mu_i^\circ=RT\ln \lambda_i^\circ \tag{2}$</span></p>
<p>Therefore,</p>
<p><span class="math-container">$\mu_i = \mu_i^\circ + RT\ln a_i \tag {3}$</span></p>
<p>becomes</p>
<p><span class="math-container">$RT\ln \lambda_i = RT\ln \lambda_i^\circ + RT\ln a_i \tag {4}$</span></p>
<p>or</p>
<p><span class="math-container">$$ a_i = \frac{\lambda_i}{\lambda_i^\circ} \tag {5}$$</span></p>
<p>This is equivalent to your equation (2), except that <span class="math-container">$\lambda_i^\circ$</span>, the value of the absolute activity in the reference standard state (at which <span class="math-container">$a_i=1$</span>), need not be equivalent to the value of the fugacity in that standard state. It would appear necessary mathematically, however, that these not differ by more than a multiplicative constant, provided the same standard state is being referenced, that is <span class="math-container">$\lambda_i = \textrm{constant} \times f_i$</span>.</p>
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https://chemistry.stackexchange.com/questions/172573/concept-of-absolute-thermodynamic-activity
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Question: <p>This may be a stupid question but I would like to know what prevents chemical reactions from happening among common objects in everyday life? The opposite would be, what requirements must be met for chemical reactions?</p>
<p>For example, as I am typing this what prevents no reaction between my fingers and the keyboard, or everyday objects such as chairs, tables, etc. In other words, why do we not constantly see chemical processes happening all around us constantly? Why don't my feet chemically react with the ground, my pencil reacts with the air, my fridge with the floor, etc...</p>
<p>There must be electron transfer such as conductivity, and charging but why no chemical reactions? Even as I am in AP Chemistry and understand Thermodynamics, Hess' Law, and so on, I am too embarrassed to ask my teacher with such a silly question. </p>
Answer: <p>As I see it, two factors: <em>thermodynamics</em> and <em>kinetics</em>. </p>
<p><strong>Thermodynamics</strong>: Many of the reactions that "don't happen" in everyday life are not favored thermodynamically, we call them <em>non-spontaneous</em>. This means that the products would be higher in energy than the reactants. Such reactions can certainly occur if you supply external energy (charging your cell phone battery, for instance, is a non-spontaneous process). But, without applied energy, non-spontaneous processes generally will not occur. Reactions that we observe in nature are generally <em>spontaneous</em>, favored by thermodynamics (with lower-energy products). Some thermodynamically spontaneous processes are <em>not</em> observed to occur, even when we might naively expected them to. Example: Sugars decomposing to $CO_2$ and water is a very energetically favorable reaction, in fact the energy released from this process fuels nearly all known life. However, we do not observe our table sugar decomposing in its container because without outside help like enzymes and applied heat, the process is terribly slow <em>kinetically</em>.</p>
<p><strong>Kinetics</strong>: For us to observe a reaction happening, it must proceed at rate that is fast relative to our human scale of time. Perhaps your fingers fusing to your keyboard is thermodynamically favored/spontaneous, but if it takes hundreds of years to occur then for our human purposes we may regard it as "no reaction". Indeed, the decomposition of most types of the biomolecules constituting our bodies is favored thermodynamically, but the kinetics are so slow (440 years for table sugar)<a href="https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2664835/" rel="nofollow noreferrer">°</a>. There are a few ways to increase the rate at which a reaction occurs. To occur, a reaction requires a certain amount of <em>activation energy</em>. Think of this as hill you have to climb before you can sled down to the products (or up, potentially). The reaction does not occur if the starting material molecules do not possess enough energy to climb the activation energy barrier. You may help them overcome the barrier by supplying heat; thus, heating your table sugar <em>will</em> allow it to decompose into $CO_2$ and water before your eyes (the water will be released as steam of course). The other main way to increase the kinetics of the reaction is to <em>lower the hill</em>, instead of adding the energy to climb it. That is, use a catalyst like an enzyme to lower the activation energy barrier of the reaction. Without such external factors like applied heat and use of a catalyst, many reactions that are thermodynamically favored will be disfavored by kinetics, and not proceed at an observables rate. </p>
<p>So the reason the universe isn't a useless, constantly rearranging chemical soup boils down to thermodynamics and kinetics. Even if a process is thermodynamically spontaneous, it may be kinetically slow. I should point out that in a lot of places, such as the centers of stars, there is indeed a copious abundance of uncontrollable chemical and nuclear reactions. And planets like Jupiter pretty much are a useless chemical soup (no offense to Jupiterians). But in our favorite places on Earth, at room temperature and atmospheric pressure, the carbon in our bodies isn't going to undergo nuclear fission and become lead (thankfully).</p>
<p>By the way, this is in not a silly question. In fact, if a chemistry student doesn't wonder this, one may question if they are actually learning the subject. I applaud you for your curiosity.</p>
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https://chemistry.stackexchange.com/questions/71387/what-prevents-everyday-objects-from-reacting
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Question: <p>In <em>Molecular Thermodynamics of fluid phase equilibria</em> by Prausnitz et al. [1] the authors recommend to use the following equation which gives the fugacity of component <span class="math-container">$i$</span> in terms of independent variables <span class="math-container">$V$</span> (volume) and <span class="math-container">$T$</span> (temperature):</p>
<p><span class="math-container">$$RT\ln\phi_i = \int_v^\infty\left(\left(\frac{\partial p}{\partial n_i}\right)_{T, V, n_i} - \frac{RT}{V}\right)\mathrm{d}V - RT\ln z$$</span></p>
<p>I know that I should use chemical potential definition to prove this equation, but there are many different ways to calculate it like using Helmholtz free energy or Gibbs free energy. </p>
<p>I would appreciate someone give me an idea or any references to help me proving this equation. </p>
<h3>References</h3>
<ol>
<li>Prausnitz, J. M.; Lichtenthaler, R. N.; Azevedo, E. G. de. Molecular Thermodynamics of Fluid-Phase Equilibria, 3rd edition; <em>Prentice Hall: Upper Saddle River, N.J</em>, <strong>1998</strong>. ISBN 978-0-13-977745-5.</li>
</ol>
Answer: <p>I haven't been able to work out all the details of the mathematics yet, but the derivation of this equation must start out from the following equation: </p>
<p><span class="math-container">$$nRT\mathrm{d}\ln{\phi} = (V - V^{ig})\mathrm{d}P,$$</span></p>
<p>where <span class="math-container">$\phi$</span> is the fugacity coefficient of the <em>mixture</em>, <span class="math-container">$n$</span> is the total moles of the mixture, <span class="math-container">$V$</span> is the volume of the mixture <span class="math-container">$znRT/P$</span> and <span class="math-container">$V^{ig}$</span> is the volume of the mixture under ideal gas conditions: <span class="math-container">$nRT/P$</span>. Once the equation is integrated to get <span class="math-container">$RTn\ln{\phi}$</span>, we get the fugacity of the <span class="math-container">$i$</span>th component of the mixture by evaluating the following partial derivative:</p>
<p><span class="math-container">$$RT\ln{\phi_i} = RT\frac{\partial (n\ln{\phi})}{\partial n_i}$$</span> </p>
<p>Hope that this helps a little.</p>
<p>ADDENEUM</p>
<p>If we write, <span class="math-container">$$d(nV)=\left(\frac{\partial (nV)}{\partial n_i}\right)_{p,n_k}dn_i+\left(\frac{\partial (nV)}{\partial p}\right)_{ni,n_k}dp$$</span></p>
<p>then it follows that: <span class="math-container">$$\left(\frac{\partial (nV)}{\partial n_i}\right)_{p,n_k}=-\left(\frac{\partial (nV)}{\partial p}\right)_{ni,n_k,V}\left(\frac{\partial (p)}{\partial n_i}\right)_{n_k,V}$$</span></p>
<p>The rest is easy. The subscript <span class="math-container">$n_i$</span> in the OP equation is a typo. It should be <span class="math-container">$n_k$</span>, representing all other species being held constant.</p>
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https://chemistry.stackexchange.com/questions/112989/proving-a-relation-for-fugacity-of-component-i-in-term-of-independent-variables
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Question: <p>From this question on Physics SE <a href="https://physics.stackexchange.com/questions/7284/">Thermodynamics of evaporation</a>:</p>
<blockquote>
<p>Now imagine the experiment is repeated but instead of vacuum conditions, the water is pressurized with nitrogen at 1 atm. According to the phase diagram of water, liquid is the stable form of water in these conditions. Yet it is commonly observed that the water molecules with the highest kinetic energy will escape and form a gaseous phase. The partial pressure of gaseous water will be equal to the saturation pressure at this temperature.</p>
</blockquote>
<p>The accepted answer shows a kinetic perspective. I want to understand things from classical thermodynamics.
As even an inert gas is added to the container, there would be some increase of configurational entropy in the gaseous phase compared to the initial conditions. Won't this factor will decrease the chemical potential of the gaseous phase and thus increase the vapour pressure a bit ?.</p>
Answer: <p>If the nitrogen is truly inert, its presence will have <em>no <strong>direct</strong> effect on the chemical potential of the water in the gas phase</em>. Essentially, the gaseous water is "unaware" of the nitrogen. Consistent with this, in a mixture of ideal gases, the chemical potential of each gas is determined only by the temperature, and its partial pressure. It is indepenent of the partial pressures of the other gases.</p>
<p>However, even an ideal gas can increase the vapor pressure. It doesn't do so by reducing the chemical potential of the gaseous water. Rather, by applying pressure to the liquid water, it <em>increases the chemical potential of the liquid phase</em>, thus increasing the equilibrium vapor pressure. This is a small effect.</p>
<p>Of course, gases aren't inert. As you increase the pressure of the nitrogen, you have two opposing effects on the liquid water: (1) The effect described above, which increases the liquid's chemical potential. (2) With increased pressure, the concentration of nitrogen dissolved in the water will increase, which will lower the water's chemical potential. And of course one also has to account for the non-ideal interactions between the nitrogen and the water vapor.</p>
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https://chemistry.stackexchange.com/questions/152504/will-addition-of-inert-gas-change-the-vapour-pressure-of-a-liquid
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Question: <p>Let $\sum_A \nu_A A \rightarrow \sum_B \nu_B B$ be a general reaction whose progress during time interval $dt$ is measured by $d\zeta$, so the amount of reactants consumed and products generated in mole would be $d N_i=\nu_i d\zeta$, wherein, $\nu_i$ would take negative values for the reactants being consumed.</p>
<p>From on the other hand we already know from the second law of thermodynamics that the reaction will occur in the direction in which $\sum_i \mu_i dN_i \le 0$, wherein, $\mu_i=\frac{\partial G}{\partial N_i}\bigr|_{p,T,N_j\;(j\ne i)}$ are the chemical potentials, so that we would have:
$$\left(\sum_i \mu_i \nu_i\right)\; d\zeta \le 0$$
The equality implies reversibility and should denote the equilibrium (<em>am I right?</em>). At the equilibrium it is clear that the progression of the reaction should vanish and we should have $d\zeta=0$, so that the term $\left(\sum_i \mu_i \nu_i\right)$ should be free to gain any positive, zero or negative finite value. However, this is not what Guggenheim has written in his Thermodynamics book. He has first assumed in a given direction the reaction progresses, so that in that direction $d\zeta>0$, then has canceled out this positive quantity from the inequality and achieved: $\left(\sum_i \mu_i \nu_i\right) \le 0$, and eventually concluded that at the equilibrium the equality holds and we should have $\left(\sum_i \mu_i \nu_i\right) = 0$.</p>
<p>But how is it justified to be true when we already know at the equilibrium $d\zeta>0$ doesn\t hold and we should instead write: $d\zeta=0$?</p>
<p>Thanks for bearing with me, I'm new to chemistry.</p>
Answer: <p>Chemical equilibrium is a stable equilibrium, i.e. the system returns to equilibrium to whatever perturbations. This also leads to the condition of equilibrium, as it should contain this information.</p>
<p><strong>The short answer to your question:</strong></p>
<p>Due to the stability, the system stays/returns to equilibrium at any kind of small perturbation. It also means, $\left(\sum_i \mu_i \nu_i\right)\; d\zeta = 0$ not only in the equilibrium point, but also for any small $d\zeta$. Therefore the $\left(\sum_i \mu_i \nu_i\right)$ itself must be zero.</p>
<p><strong>Somewhat longer answer:</strong></p>
<p>The reaction goes from one direction to the other ($\sum_A \nu_A A \rightarrow \sum_B \nu_B B$) spontaneously only till it reaches the equilibrium condition. In this direction $\sum_i \mu_i dN_i \le 0$. </p>
<p>We followed the reaction from a given concentration toward equilibrium. If we follow the same $\sum_A \nu_A A \rightarrow \sum_B \nu_B B$ transformation BEYOND the equilibrium point, $ \sum_i \mu_i dN_i $ must change sign at the equilibrium point, and it must be positive should be valid (since it is not a spontaneous reaction).</p>
<p>Therefore the condition of equilibrium is not $d\zeta = 0 $ , but sign change: $ \sum_i \mu_i dN_i =0$.</p>
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https://chemistry.stackexchange.com/questions/14012/how-can-we-justify-setting-the-affinity-sum-i-mu-i-nu-i-equal-to-zero-at-ch
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Question: <p>I <em>think</em> I've found a mistake in my thermodynamics textbook, <em>Chemical Thermodynamics for Process Simulation</em>, but as thermodynamics is hard and I'm a relative novice, I wanted to check here.</p>
<p>The textbook, in talking about partial molar Gibbs excess functions of species in solutions, writes that we can express the partial molar Gibbs free energy of a species $\bar{g_i}$ in solution, as the sum of the partial molar Gibbs free energy in an <em>ideal</em> solution, $\bar{g_i}^\mathrm{id}$, and an excess function, $\bar{g_{i}}^\mathrm{ex}$:</p>
<p>$$\bar{g}_i = \bar{g}_i^\mathrm{id}+\bar{g}_i^\mathrm{ex}.$$</p>
<p>It doesn't specify whether we are thinking of evaluating $\bar{g}_i^\mathrm{id}$ <em>at the system pressure</em> or <em>at a reference pressure</em>, but immediately after it 'expands' the above expression into the form:</p>
<p>$$\bar{g}_i(T,P) = \Big(\bar{g}_i^\mathrm{pure}(T,P_0) + \mathcal{R}T\ln x_i\Big) + \left(\mathcal{R}T\ln \frac{f_i}{x_if_i^0}\right)$$</p>
<p>Clearly we're comparing $\bar{g}_i$ to the ideal value, $\bar{g}_i^\mathrm{id}$, <em>at standard pressure</em> $P_0$. However, they then go on to define the activity $a_i=f_i/f_i^0$ and the activity coefficient $\gamma_i = a_i/x_i$ and substitute these in to get</p>
<p>$$\bar{g}_i(T,P) = \bar{g}_i^\mathrm{pure}(T,P_0) + \mathcal{R}T\ln x_i + \mathcal{R}T\ln \gamma_i.$$</p>
<p>Now, I'm all fine up to here. The first part, $\bar{g}_i^\mathrm{pure}(T,P_0) + \mathcal{R}T\ln x_i $, is the ideal solution Gibbs free energy at standard pressure, and the final term, $\mathcal{R}T\ln \gamma_i,$ is a correction in two ways: It corrects for the non-ideality of the solution, and it corrects for the fact that we may be at a different pressure $P$. </p>
<p>But the book then says: </p>
<blockquote>
<p>For an ideal mixture the excess part is equal to zero, since the activity coefficient is $\gamma_i=1$.</p>
</blockquote>
<p>I think this is ridiculous: we can only make this conclusion if we have an ideal solution at $P=P_0$. </p>
<p>Is this an error in the textbook? Also, I can see two possible routes to an error: Either $\bar{g}_i^\mathrm{id}$ is actually to be evaluated at system pressure, in which case their last statement is correct but the formula they give is wrong, or else their definition of excess functions is the standard one (actual value minus ideal value at ref. pressure) but they made a mistake in concluding $\gamma_i=1$ for all ideal solutions, independent of the pressure.</p>
Answer: <p>The confusion is perhaps not uncommon. However, whichever book it is has jumped from discussing the free energy of pure materials, and their variation with various things (T, P, V, ...), to discussing mixtures. In this case, all variation of the 'pure' substances is present in the ideal terms - that is because those variation exist without regard for any mixing terms. What you are focused on here are the impacts of making a solution (or binary alloy, ...).</p>
<p>So, to make the mixture, you take a certain amount of substance A, and a certain amount of substance B, and mix them together. Before mixing, the total free energy of the system is $x_A G_{A} + x_B G_{B}$ where $G_{A,B}$ are the Gibbs free energies of those substances at whatever the conditions you specify before mixing. So, what does mixing do? Well, at the very least you get the increase in entropy, $RT(x_{A}\ln{x_{A}} +x_{B}\ln{x_{B}}$. If there is any interaction between A and B, you now have non-ideal terms which can be represented in any number of different ways. However, the point is that they are "all the rest that happens" during the mixing, and those things may be functions of concentration, temperature, pressure, ...</p>
<p>But, to come back to your question - no, this is not an error in the textbook (whichever it is). But, the did not make it clear that they were leaving behind the explicit dependencies of free energy on the various parameters - they figured you had that part down.</p>
<p>I would also note that, when one calculates a phase diagram (ultimately one hopes the book might get to that, and miscibility gaps and whatnot), you are free to choose the reference points or phases of A and B yourself - that choice makes no difference in the phase boundaries and tie lines. And, this comes back to just what $G_{A,B}$ means.</p>
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https://chemistry.stackexchange.com/questions/36875/how-are-partial-molar-gibbs-excess-functions-correctly-defined
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Question: <p>I'm having difficulties understanding this problem.</p>
<p>We have in the problem that $K_\mathrm a$ for $\ce{HCN} = 6.2\cdot10^{-10}$ and $K_\mathrm b$ for $\ce{NH3} = 1.8\cdot10^{-5}$.</p>
<ol>
<li>Write chemical equations to represent the dissociation of $\ce{NH4CN}$ in water, and the acid-base equilibrium equation.</li>
<li>Will the salt react as an acid or a base?</li>
</ol>
<p>I know this is part of the solution:</p>
<p>$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$
$$\ce{CN- + H2O <=> HCN + OH-}$$</p>
<p>I understand Le Chatelier's principle and I know that the reaction will shift to the side of less energy.</p>
Answer: <p>Fun! weak acid versus weak base!</p>
<p>Ammonium cyanide, <span class="math-container">$\ce{NH4CN}$</span>, is a solid where the atoms are grouped into the same ions that are generated in solution: <span class="math-container">$\ce{NH4+}$</span> and <span class="math-container">$\ce{CN-}$</span>. Although it can be sublimed with very mild warming, it is fairly unstable. In particular it reacts with atmospheric water vapor, evolving ammonia and HCN. Dangerous stuff!</p>
<p>The first part of the answer to 1. is the decomposition of <span class="math-container">$\ce{NH4CN}$</span> into ions in aqueous solution, which should be obvious; one subtlety is that you may want to add (s) and (aq) next to each species, especially if your professor makes a point of doing so. So that's the dissociation. The data in the problem don't allow us to compute the equilibrium constant of the dissociation but it's safe to say that it is extremely high, that is, dissociation is complete: there is <strong>no</strong> <span class="math-container">$\ce{NH4CN}$</span> as such in solution.</p>
<p>The acid-base equilibrium equations can be the ones you wrote. Again, you may want to add more information, like this:</p>
<p><span class="math-container">$$\ce{NH4+ (aq) + H2O <=> NH3 (aq) + H3O+ (aq)}\quad K=5.5\times10^{-10}$$</span></p>
<p>Here the equilibrium constant <span class="math-container">$K$</span> is the acid dissociation constant of the acid <span class="math-container">$\ce{NH4+}$</span>, the protonated form (aka conjugate acid) of ammonia. I got its value by dividing <span class="math-container">$K_\mathrm w$</span>, the dissociation constant of water (<span class="math-container">$1.0\times10^{-14}$</span>), by the <span class="math-container">$K_\mathrm b$</span> of ammonia, given in the problem.</p>
<p>I encourage you to add similar information for the equilibrium involving cyanide ion. Hint: if you keep the reaction as you wrote it, the appropriate <span class="math-container">$K$</span> is not the <span class="math-container">$K_\mathrm a$</span> of <span class="math-container">$\ce{HCN}$</span> but the <span class="math-container">$K_\mathrm b$</span> of the deprotonated form (conjugate base), <span class="math-container">$\ce{CN-}$</span>.
(But you can instead consider the alternate reaction <span class="math-container">$\ce{HCN + H2O <=> CN- + H3O+}$</span>, and then the equilibrium constant is the <span class="math-container">$K_\mathrm a$</span> of <span class="math-container">$\ce{HCN}$</span>.)</p>
<p>How about part 2? The only correct answer to</p>
<blockquote>
<p>Will the salt react as an acid or a base?</p>
</blockquote>
<p>is "It can do either, depending on what it's reacting with". But <strong>maybe</strong> what they mean is whether the solution of the salt is acidic or basic. If so, one way to find out is to combine the two equilibrium equations and their <span class="math-container">$K$</span>s. I don't want to give problem away completely, but think of it this way:</p>
<ul>
<li><p>What is the pH of a (say) 1 M solution of <span class="math-container">$\ce{HCN}$</span>, based on the <span class="math-container">$K_\mathrm a$</span> of <span class="math-container">$\ce{HCN}$</span>?</p>
<p>(I'm assuming the calculation of the pH of a solution of a pure weak acid or a pure weak base has been covered in the course before getting to the question you posted today.)</p>
</li>
<li><p>What is the pH of a 1 M solution of <span class="math-container">$\ce{NH3}$</span>, based on the <span class="math-container">$K_\mathrm b$</span>?</p>
</li>
<li><p>Is <span class="math-container">$\ce{HCN}$</span> a stronger acid than <span class="math-container">$\ce{NH3}$</span> is a base, or is <span class="math-container">$\ce{NH3}$</span> a stronger base than <span class="math-container">$\ce{HCN}$</span> is an acid? Based on this, which would you expect to "win out" if you mix the two in equal proportions?</p>
</li>
</ul>
<p>(The point here is that what you get by dissolving <span class="math-container">$\ce{NH4CN}$</span> is indistinguishable from what you'd get by mixing equal amounts of equimolar solutions of <span class="math-container">$\ce{HCN}$</span> and ammonia. In other words, if you mix 1 L of a 2 M solution of <span class="math-container">$\ce{HCN}$</span> with 1 L of a 2 M solution of ammonia, the result is very nearly 2 L of a 1 M solution of <span class="math-container">$\ce{NH4CN}$</span>. Footnote: "very nearly" because volume is not always preserved exactly, unlike mass.)</p>
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https://chemistry.stackexchange.com/questions/14857/acid-base-equilibrium-of-nh4cn
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Question: <p>Suppose we had a closed system proceeding according to, </p>
<p>$$\ce A +2\ce B \rightleftharpoons 2 \ce C$$</p>
<p>And that three experiments produced the following concentration data. </p>
<p>Experiment 1:</p>
<p>\begin{array}{|c|c|} \hline
{\rm \small Species} & {\rm \small concentration} \\\hline
\ce A & 0.500\rm~M \\
\ce B & 0.700\rm~M \\
\ce C & 0.900\rm~M
\\\hline\end{array}</p>
<p>Experiment 2:</p>
<p>\begin{array}{|c|c|} \hline
{\rm \small Species} & {\rm \small concentration} \\\hline
\ce A & 0.300\rm~M \\
\ce B & 0.420\rm~M \\
\ce C & 0.424\rm~M \\\hline
\end{array}</p>
<p>Experiment 3:</p>
<p>\begin{array}{|c|c|} \hline
{\rm \small Species} & {\rm \small concentration} \\\hline
\ce A & 1.40\rm~M \\
\ce B & {x}\rm~M \\
\ce C & 1.49\rm~M \\\hline
\end{array}</p>
<p>How would you find the concentration of $\ce B$ from experiment 3 at equilibrium?</p>
<p>I assumed that each experiment was in equilibrium so that I could do the following, </p>
<p>$$K_c = \frac{\ce{[C]}^2}{[\ce{A}][\ce{B}]^2} = \frac{[0.900]^2}{[0.500][0.700]^2}$$</p>
<p>$$\therefore \frac{\ce{[C]}^2}{\ce{[A][B]}^2} = \frac{[0.900]^2}{[0.500][0.700]^2}$$</p>
<p>$$\therefore \ce{[B]} = \sqrt\frac{[1.490]^2[0.500][0.700]^2}{[0.900]^2[0.140]}$$</p>
<p>$$\therefore \ce{[B]} = 2.19\rm~M$$</p>
<p>However, $[\ce{B}] = 2.19\rm~M$ is not the answer. Could someone please provide some insight and direction into this question?</p>
Answer: <p>You have an error somewhere in your math.</p>
<p>Assuming you were supposed to interpolate <span class="math-container">$K_C$</span> from both experiment 1 and 2, you would do the following:
Experiment1 yields <span class="math-container">$K_C=3.31$</span>. Experiement two yields <span class="math-container">$K_C=3.40$</span>. Interpolation yields <span class="math-container">$K_C=3.35$</span>.
This would give you <span class="math-container">$$B=\sqrt{\frac{C^2}{A\cdot K_C}}=\sqrt{\frac{1.49^2}{1.4\cdot 3.35}}=0.69$$</span></p>
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https://chemistry.stackexchange.com/questions/51834/acid-base-equilibrium
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Question: <p>I'm having trouble finding anything meaningful in textbooks or searches. My question is what is the correct notation for writing acid-base reactions when their strengths are varying? I know that theoretically all reactions are reversible but when is it appropriate to use equilibrium arrows for certain acid-base reactions?</p>
<p>e.g. would a weak acid - strong base reaction be regarded reversible?</p>
<p>for example would <span class="math-container">$\ce{NaOH + CH3COOH $\leftrightharpoons$ NaCH3COO + H2O}$</span> be suitable (disregarding the lack of states and my ineptitude in writing MathJax)</p>
<p>or would <span class="math-container">$\ce{NaOH + CH3COOH -> NaCH3COO + H2O}$</span> be more suitable?</p>
<p>Do we just simplify it into only considering the forward reaction for the sake of stoichiometric calculation?</p>
Answer:
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https://chemistry.stackexchange.com/questions/165665/notation-for-acid-base-reactions-of-differing-strengths-equilibrium
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Question: <p>While reading an organic chemistry textbook I saw the following Formula for calculating equilibrium constant of an acid-base reaction</p>
<p>Consider following acid base reaction in which <span class="math-container">$\ce{HA}$</span>(reactant acid) and <span class="math-container">$\ce{HB+}$</span>(product acid) are acids in forward and backward reaction respectively.</p>
<p><span class="math-container">$$\ce{HA + B <=> A- + HB+}$$</span></p>
<p><span class="math-container">$$\mathrm pK_\mathrm{eq}= \mathrm pK_\mathrm {a_\mathrm{reactant}}-\mathrm pK_\mathrm {a_\mathrm{product}}$$</span></p>
<p><span class="math-container">$$\implies K_\mathrm{eq}=\frac{K_\mathrm{a_\mathrm{reactant}}}{K_\mathrm{a_\mathrm{product}}}$$</span></p>
<p>Now I want to show above expression is equivalent to <span class="math-container">$$K_\mathrm{eq}=\frac{\ce{[A-][HB+]}}{\ce{[HA][B]}}$$</span></p>
<p>So I have to write expressions of <span class="math-container">$K_\mathrm{a}$</span> of both acids to proceed further.</p>
<p>The way <span class="math-container">$K_\mathrm {a}$</span> is defined in my textbook is like this</p>
<blockquote>
<p><span class="math-container">$$\ce {HA + H2O <=> A- + H3O+}$$</span></p>
<p><span class="math-container">$$K_\mathrm {a}= \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}$$</span></p>
</blockquote>
<p>So In our acid base reaction we should write expressions of <span class="math-container">$K_\mathrm {a}$</span> of acids in a similar way as we have no other option left.</p>
<p><span class="math-container">$$\ce{HA + H2O <=> A- + H3O+}\tag{1}$$</span></p>
<p><span class="math-container">$$\ce{HB+ + H2O <=> B + H3O+}\tag{2}$$</span></p>
<p>The above 2 equations I have written are for defining <span class="math-container">$K_\mathrm{a}$</span> of acids (<span class="math-container">$\ce{HA}$</span> and <span class="math-container">$\ce{HB+}$</span>).</p>
<p>They are not related to our main acid-base reaction in any way.</p>
<p>As people are pointing out in the comments, concentration of <span class="math-container">$\ce{H3O+}$</span> should be same in expressions of <span class="math-container">$K_\mathrm a$</span> as they are in the same solution but I have already pointed out that equation 1 and 2 are written only for the sake of defining <span class="math-container">$K_\mathrm a$</span> of acids.</p>
<p>They are altogether different from our main acid base reaction. Then the concentration of <span class="math-container">$\ce{H3O+}$</span> can be different in equations 1 and 2 as they are two different solutions for determining <span class="math-container">$K_\mathrm a$</span></p>
<p><span class="math-container">$K_\mathrm {a}$</span> of <span class="math-container">$\ce{HA}$</span> which is the reactant is <span class="math-container">$$K_\mathrm a = \frac{\ce{[A-][H3O+]}}{\ce{[HA]}}$$</span></p>
<p><span class="math-container">$K_\mathrm {a}$</span> of <span class="math-container">$\ce{HB+}$</span> which is the acid being produced is <span class="math-container">$$K_\mathrm b = \frac{\ce{[B][H3O+]}}{[\ce{HB+}]}$$</span></p>
<p>On substituting respective <span class="math-container">$K_{a}$</span> into the obtained expression we get:</p>
<p><span class="math-container">$$K_\mathrm {eq}=\frac{\ce{[A-][HB+]}}{\ce{[HA][B]}}$$</span></p>
<p>But during this process of substitution we are required to cancel <span class="math-container">$\ce{[H3O+}$</span> term in numerator and denominator</p>
<p>My question is:</p>
<p>How can we cancel the term of concentration of <span class="math-container">$\ce{H3O+}$</span> as concentration of protonated water in expression of <span class="math-container">$K_\mathrm {a}$</span> of reactant and product can be different?</p>
<hr />
Answer: <p>The issue as I can see it here is a simple misunderstanding that can rectified:</p>
<p><span class="math-container">$K_\mathrm a$</span> is a constant even if <span class="math-container">$\ce{[A]}$</span> is not. (which is how it was defined)</p>
<p>The definition of <span class="math-container">$k_\mathrm a$</span> is as follows:</p>
<blockquote>
<p><span class="math-container">$K_\mathrm a$</span> , the acid ionization constant, is the equilibrium constant for chemical reactions involving weak acids in aqueous solution.</p>
</blockquote>
<p>Nowhere does this definition state that the concentration taken for water has to be constant, in fact, only <span class="math-container">$K_\mathrm a$</span> is constant.</p>
<p>Once we find the value of <span class="math-container">$K_\mathrm a$</span>, the value cannot change for a given temperature. This means that no matter what solution you add the acid to - in the end, the value of <span class="math-container">$K_\mathrm a$</span> remains constant. i.e:</p>
<p><span class="math-container">$$K_\mathrm a = \frac{\ce{[A-][H3O+]}}{\ce{[HA]}} \tag{constant}$$</span></p>
<p>Now, in your solution, the value of <span class="math-container">$\ce{H3O+}$</span> may not be the same as in the place where we determined the <span class="math-container">$K_\mathrm a$</span> in the beginning but that only means that are <span class="math-container">$K_\mathrm a$</span> (which is still constant) is the same ratio but with different numbers.</p>
<p>Now, as to the reason why <span class="math-container">$\ce{[H3O+]}$</span> can be cancelled. Imagine you take a glass of water and add two acids <span class="math-container">$\ce{HA}$</span> and <span class="math-container">$\ce{HB}$</span>.</p>
<p>What is the concentration of protonated water when we try to find the <span class="math-container">$K_\mathrm a$</span> of these acids in this solution, we get:</p>
<p><span class="math-container">$$K_{\mathrm a_\mathrm A} = \frac{\ce{[A-][H3O+]}}{\ce{[HA]}} \tag{1}\label{1}$$</span></p>
<p><span class="math-container">$$K_{\mathrm a_\mathrm B} = \frac{\ce{[B-][H3O+]}}{\ce{[HB]}} \tag{2}\label{2}$$</span></p>
<p>Now what is <span class="math-container">$\ce{[H3O+]}$</span> in (<span class="math-container">$\ref{1}$</span>) and (<span class="math-container">$\ref{2}$</span>)?</p>
<p>They will be the same since you are measuring the <span class="math-container">$K_\mathrm a$</span>s in the same glass of water.</p>
<p>Therefore you can cancel <span class="math-container">$\ce{[H3O+]}$</span> out since the value is same for the one glass of water that you have.</p>
|
https://chemistry.stackexchange.com/questions/150940/equilibrium-constant-expression-for-an-acid-base-reaction
|
Question: <p>I am trying to find the equilibrium value for different acid-base reactions. I have understood that when an acid and base are in an aqueous solution, many different reactions will occur. By combining these reactions, and multiplying their equilibrium constants, we can find the equilibrium value for an acid-base reaction.</p>
<p><strong>Example 1: <span class="math-container">$NH_3$</span> and <span class="math-container">$HCl$</span></strong></p>
<p>To illustrate my question I will take an example: if <span class="math-container">$NH_3$</span> and <span class="math-container">$HCl$</span> are in an aqueous solution, what reactions will occur? I believe the following reactions are those that will occur:</p>
<p><span class="math-container">$HCl + H_2O ⇌ Cl^- + H_3O^+$</span>, <span class="math-container">$K_1 = K_a(HCl) = 1.3 * 10^6$</span></p>
<p><span class="math-container">$Cl^- + H_2O ⇌ HCl + OH^-$</span>, <span class="math-container">$K_2 = K_b(Cl^-) = K_w/K_a(HCl)$</span></p>
<p><span class="math-container">$NH_3 + H_2O ⇌ NH_4^+ + OH^-$</span>, <span class="math-container">$K_3 = K_b(NH_3) = 1.8 * 10^{-5}$</span></p>
<p><span class="math-container">$NH_4^+ + H_2O ⇌ NH_3 + H_3O^+$</span>, <span class="math-container">$K_4 = K_a(NH_4^+) = K_w/K_b(NH_3)$</span></p>
<p><span class="math-container">$2H_2O ⇌ H_3O^+ + OH^-$</span>, <span class="math-container">$K_5 = K_w = 10^{-14}$</span> (the temperature is 25 degrees Celsius)</p>
<p>These are all the reactions that will occur in this solution, right? I noticed that adding all these will be of no use since the combined reaction will be the autoprotolysis of water. Instead, we only add reactions 1 and 3. That gives:</p>
<p><span class="math-container">$HCl + NH_3 + 2H_2O ⇌ Cl^- + NH_4^+ + H_3O^+ + OH^-$</span>, <span class="math-container">$K_6 = K_1 * K_3 = K_a(HCl) * K_b(NH_3)$</span></p>
<p>We can subtract the autoprotolysis of water from this reaction to get:</p>
<p><span class="math-container">$HCl + NH_3 ⇌ Cl^- + NH_4^+$</span>, <span class="math-container">$K_7 = K_6 / K_w = K_a(HCl) * K_b(NH_3) / K_w = 1.3 * 10^6 * 1.8 * 10^{-5} * 10^{14} = 2.34 * 10^{15}$</span></p>
<p>So the equilibrium constant for the reaction between <span class="math-container">$HCl$</span> and <span class="math-container">$NH_3$</span> is <span class="math-container">$2.34 * 10^{15}$</span>; in other words, it essentially goes to completion. Is this correct?</p>
<p><strong>Example 2: <span class="math-container">$HCl$</span> and $NaOH</strong></p>
<p>Another example is the neutralization reaction between <span class="math-container">$HCl$</span> and <span class="math-container">$NaOH$</span>. How should one deal with <span class="math-container">$NaOH$</span>? Here is my attempt. The reactions that occur are:</p>
<p><span class="math-container">$ HCl + H_2O ⇌ Cl^- + H_3O^+ $</span>, <span class="math-container">$K_8 = 1.3 * 10^6$</span></p>
<p><span class="math-container">$ NaOH → Na^+ + OH^- $</span>, <span class="math-container">$K_9$</span>, what should <span class="math-container">$K_9$</span> be?</p>
<p>In addition to these, <span class="math-container">$Cl^- + H_2O$</span> and the autoprotolysis of water will occur. Similarly to the first example, these reactions will be of no use to find the equilibrium constant for the neutralization reaction.</p>
<p>Adding reactions 8 and 9 gives:</p>
<p><span class="math-container">$HCl + NaOH + H_2O ⇌ Cl^- + OH^- + Na^+ + H_3O^+$</span>, <span class="math-container">$K_{10} = K_8 * K_9$</span></p>
<p>Can we here subtract the autoprotolysis of water as in the first example (since there is no "2" in front of "H_2O" on the left side of the reaction)? If we can do that then the reaction becomes:</p>
<p><span class="math-container">$HCl + NaOH ⇌ Cl^- + Na^+$</span>, <span class="math-container">$K_{11} = K_8 * K_9 / K_w$</span></p>
<p>I know that reaction 11 is unbalanced. <span class="math-container">$K_{11}$</span> is very large since <span class="math-container">$K_a(HCl)$</span> is very large and <span class="math-container">$K_9$</span> also is very large. So the reaction will essentially go to completion. However, it feels like something is wrong because reaction 11 is not balanced. What have I done wrong?</p>
<p><strong>Summary</strong></p>
<p>My goal is to understand the underlying theory of acid-base reactions and equilibrium. I have here presented in these two examples what I understand. What am I misunderstanding about how to calculate the equilibrium value for general acid-base reactions?</p>
<p>Thank you in advance!</p>
Answer: <p><strong>The rule No 0:</strong> Learn to enumerate chemical reactions. The prerequisite is the knowledge of arithmetic of small natural numbers. Total charges and total atom counts must be equal on both reaction sites. No atom is created nor destroyed during any chemical reaction.</p>
<p><strong>The rule No 1:</strong> Simplify everything where simplification brings negligible error, otherwise you often end with too complicated math. Considering strong acids as HCl or strong bases as NaOH as completely dissociated brings very negligible error ( with much bigger ignored error due activity versus concentration).</p>
<p><strong>The rule No 2:</strong> After the calculation, verify if conditions for the simplification are met. If they are not, you have to reject simplification.</p>
<p>E.g., we want to derive an equation of pH of diluted solution of weak acid. The dissociation constant for a weak acid is:</p>
<p><span class="math-container">$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$</span></p>
<p>If we assume <span class="math-container">$\ce{H+}$</span> from water autodissociation is negligible, compared to the acid dissociation (<span class="math-container">$[\ce{H+}] \gg [\ce{OH-}]$</span>), we can simplify it using <span class="math-container">$[\ce{H+}] \approx [\ce{A-}]$</span> to:</p>
<p><span class="math-container">$$K_\mathrm{a} = \frac{[\ce{H+}]^2}{[\ce{HA}]}$$</span></p>
<p>If we then also assume near all acid is not dissociated (<span class="math-container">$c \approx [\ce{HA}]$</span>), we can simplify it to:</p>
<p><span class="math-container">$$K_\mathrm{a} = \frac{[\ce{H+}]^2}{c}$$</span></p>
<p>leading to</p>
<p><span class="math-container">$$c \gg [\ce{H+}] \gg [\ce{OH-}] \implies \mathrm{pH} = \frac 12(\mathrm{p}K_\mathrm{a} - \log{c})$$</span></p>
<p><strong>The rule No 3:</strong> Any respective conjugate pair acid/base (like <span class="math-container">$\ce{NH4+}$</span>/<span class="math-container">$\ce{NH3}$</span>) is in water solutions in first place in equilibrium with <span class="math-container">$\ce{H2O}$</span>/<span class="math-container">$\ce{H+(aq)}$</span>/<span class="math-container">$\ce{OH-(aq)}$</span> respectively. Any equilibrium with other acid/base system is indirect, dependent and redundant.</p>
<p><strong>The rule No 4:</strong> Any acid-base reaction in water is generally reaction of some acid passing a proton to some base, the former becoming a base and vice versa.</p>
<p><span class="math-container">$$\ce{\mathrm{acid}_1 + \mathrm{base}_2 <=> \mathrm{base}_1 + \mathrm{acid}_2}$$</span></p>
<hr />
<blockquote>
<p>How to calculate the equilibrium constant for the reaction between <span class="math-container">$\ce{H2O}$</span> and <span class="math-container">$\ce{NH3}$</span> (in my example nr 1) and between <span class="math-container">$\ce{HCl}$</span> and <span class="math-container">$\ce{NaOH}$</span> (in my example nr 2). Could you please show me how you would do that using your method (if my method is wrong?).</p>
</blockquote>
<ul>
<li>Example No 1: The equilibrium constant is experimentally determined for NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq).
You have already listed it in your question as <span class="math-container">$K_\mathrm{b} = \pu{1.8E-5}$</span>.</li>
<li>Example No 2: It is equilibrium <span class="math-container">$\ce{H2O <=> H+(aq) + OH-(aq)}$</span> with <span class="math-container">$K_\mathrm{w} = \pu{e-14}$</span> at <span class="math-container">$\pu{25 ^\circ C}$</span>, as HCl and NaOH are in water (with negligible error) fully dissociated. <span class="math-container">$\ce{Na+}$</span> and <span class="math-container">$\ce{Cl-}$</span> ions are often called spectator/bystander ions that do not take part in the reaction.</li>
</ul>
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https://chemistry.stackexchange.com/questions/171444/calculating-equilibrium-constant-for-a-general-acid-base-reaction
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Question: <p>Should the equilibrium constant <span class="math-container">$K$</span> be calculated at the end of an acid-base titration or can we calculate it in the middle of the titration for example? If so, will the calculated equilibrium constant be equal to the equilibrium constant calculated from the final state of the titration?</p>
Answer: <p>Because proton transfers are very fast, the equation <span class="math-container">$K_a=\frac{[\text{conjugate base}][\ce{H+}]}{[\text{acid}]}$</span> will always hold as long as your solution is well mixed. So if you are able to measure those three concentrations, you can determine <span class="math-container">$K_a$</span> using any mixture. As the concentrations change during the titration, the value of <span class="math-container">$K_a$</span> will remain constant. </p>
<p>Be aware that the value of <span class="math-container">$K_a$</span> will vary with temperature, so you should maintain a constant temperature during the experiments. </p>
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https://chemistry.stackexchange.com/questions/108602/can-equilibrium-be-calculated-at-any-time-of-an-acid-base-reaction-and-still-be
|
Question: <p>Let's talk about the following compounds - water and methanol. </p>
<ol>
<li><p>We know that <em>methanol is slightly more acidic than water</em>, because water's ability to donate a proton as an acid is reduced due to extensive hydrogen bonding.
The $\mathrm{p}K_\mathrm{a}$ values are approximately $15.5$ and $15.7$, for methanol and water respectively. </p></li>
<li><p>Now consider their conjugate bases, i.e. methoxide ion and hydroxide ion. It is very clear that <em>methoxide is a stronger base compared to hydroxide</em>, due to the +I (inductive) effect of the methyl group (electron releasing).</p></li>
</ol>
<p>How is this possible? Isn't it contradictory?
It is a well known fact that a strong acid has a weak conjugate base, and a weak acid has a strong conjugate base. </p>
<p>This is a consequence of $\mathrm{p}K_\mathrm{a}$ + $\mathrm{p}K_\mathrm{b}$ = $\mathrm{p}K_\mathrm{w}$, where K<sub>w</sub> is the ionic product of water, <em>a constant at a given temperature.</em></p>
Answer: <blockquote>
<ol>
<li><p>We know that methanol is slightly more acidic than water, because water's ability to donate a proton as an acid is reduced due to extensive hydrogen bonding. [...]</p>
</li>
<li><p>[...] It is very clear that methoxide is a stronger base compared to hydroxide, due to the +I (inductive) effect of the methyl group (electron releasing).</p>
</li>
</ol>
</blockquote>
<p>You are pointing out <em>competing factors</em> that, in abstract, should lead to either methanol or water being more acidic than the other (that is. lead to either hydroxide or methoxide to be more basic than the other, <em>respectively</em>). Once we move from considering properties in abstract to an actual system in which an acid-base equilibrium is present, the competing factors are superimposed, with a non-contradictory result: either methanol will be more acidic (and hydroxide more basic) or water will be more acidic (and methoxide more basic), depending on which factors dominate. (In particular, as pointed out in <a href="https://chemistry.stackexchange.com/q/42535/3683"><em>Why is methanol more acidic than water?</em></a> and <a href="https://chemistry.stackexchange.com/q/57986/3683"><em>Is methanol really more acidic than water?</em></a>, the relative acidities depend on the solvent.)</p>
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https://chemistry.stackexchange.com/questions/94188/how-to-justify-this-contradiction-in-acid-base-equilibrium-of-methanol-and-water
|
Question: <p>While deriving the relation between $K_a$ and $K_b$ in ionic equilibrium topic, say I take an example of</p>
<p>$$
\ce{NH4+ + H2O -> NH3 + H3O+} \tag{1}\label{1}
$$</p>
<p>The $K_a$ of this reaction is:</p>
<p>$$
\frac{\ce{[NH3][H3O+]}}{\ce{[NH4+][H2O]}}
$$</p>
<p>Now $\ce{NH3}$ is a conjugate base of $\ce{NH4+}$. Its $K_b$ can be written as:</p>
<p>$$
\ce{NH3 + H2O -> NH4+ + OH-} \tag{2}\label{2}
$$</p>
<p>$K_b$ of the reaction:</p>
<p>$$
\frac{\ce{[NH4+][OH-]}}{\ce{[NH3][H2O]}}
$$</p>
<p>In my textbooks while deriving relation between $K_a$ and $K_b$, it added the two equations. In the addition of two equations we get:</p>
<p>$$
\ce{NH4+ + 2H2O + NH3 -> NH3 + NH4+ + H3O+ + OH-} \tag{1 + 2}\label{1p2}
$$</p>
<p>My textbook cancelled out both the $\ce{[NH4+]}$ and $\ce{[NH3]}$ concentration terms from the LHS and RHS. $\ce{[NH4+]}$ on the LHS is from equation $\eqref{1}$ and $\ce{[NH4+]}$ on the RHS is from equation $\eqref{2}$.</p>
<p>My question is, how can we cancel them out? How can you say the $\ce{[NH4+]}$ formed from hydrolysis of $\ce{NH3}$ in reaction $\eqref{2}$ is equal to that of $\ce{[NH4+]}$ initially taken in reaction $\eqref{1}$? Even if you initially take same concentration of $\ce{[NH3]}$ for reaction $\eqref{2}$ as formed from reaction $\eqref{1}$, that concentration of $\ce{NH3}$ on hydrolysis can never give same concentration of $\ce{[NH4+]}$ as initially taken in reaction $\eqref{1}$. Then, how can you cancel these different $\ce{[NH4+]}$ concentrations when you are adding reaction $\eqref{1}$ and $\eqref{2}$?</p>
Answer: <ol>
<li><p>$K_\mathrm{a}$ and $K_\mathrm{b}$ should not have a term for the concentration of water inside.</p></li>
<li><p>When you dissolve $x$ moles of $\ce{NH3}$ in water (doesn't matter how much $x$ is), both relations</p></li>
</ol>
<p>$$K_\mathrm{a} = \frac{[\ce{NH3}][\ce{H3O+}]}{[\ce{NH4+}]} \tag{1}$$</p>
<p>as well as</p>
<p>$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} \tag{2}$$</p>
<p>must <strong>simultaneously</strong> hold true. You agree that both of these relations must independently hold no matter what concentration of $\ce{NH3}$ (or $\ce{NH4+}$, or $\ce{H+}$, or $\ce{OH-}$) there is, right? That is why they are called equilibrium <em>constants</em>, after all.</p>
<p>So, in any given <strong>single</strong> solution with <em>one</em> well-defined value of $[\ce{NH3}]$, one well-defined value of $[\ce{NH4+}]$, ... both equations must simultaneously hold. You can then cancel out the terms as necessary to obtain</p>
<p>$$K_\mathrm{a}K_\mathrm{b} = K_\mathrm{w} \tag{3}$$</p>
<hr>
<p>If you use two different solutions, with different concentrations of all the relevant species, the relation still holds true. However, the maths will be much harder because for example if you have solution 1 with pH = 1, and solution 2 with pH = 10, then $[\ce{H+}]_\mathrm{sol,1}[\ce{OH-}]_\mathrm{sol,2}$ is no longer equal to $K_\mathrm{w}$. The relation $[\ce{H+}][\ce{OH-}] = K_\mathrm{w}$ is only necessarily true if we are talking about one single solution.</p>
|
https://chemistry.stackexchange.com/questions/65255/derivation-of-relation-between-acid-and-base-equilibrium-constants
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Question: <p>Suppose I want to make a buffer solution in any of the three ways possible with dihydrogen phosphate as my weak acid. In the first way, a certain ratio of the weak acid and its conjugate base are added at once to form the buffer. Why does this not return to the equilibrium described by its Ka value instead of forming a stable buffer? With a pka of 7.2, shouldn't the mixture shift to the weak acid to reestablish equilibrium.
The other two ways involve adding a strong acid or base to a weak base or weak acid respectively, but I still see the same issue. Say a strong base is added to a solution of the dihydrogen phosphate and the same moles added react and form an equal number of moles of the conjugate base. After this is complete, why doesn't equilibrium get restored. Is the weak acid/base equilibrium out-competed by the seperate strong base/weak acid to weak acid/weak base equilibrium?</p>
<p>Hope this made sense. Thanks in advance!</p>
Answer: <p>Let us consider the standard acetic acid/sodium acetate buffer, with, say, 0.1M of acetic acid and 0.1M of sodium acetate. These are not the equilibrium concentrations of acetic acid and sodium acetate; the buffer solution will equilibrate. In this case, the pKa of acetic acid being less than 7, we know that some acetic acid will dissociate.</p>
<p>Quantitatively, the equilibrium expression can be rearranged into the Henderson-Hasselbalch equation $$\text{pH} = \text{pKa} + \log\left(\frac{\ce{[A-]}}{\ce{[HA]}}\right) \approx \text{pKa} + \log\left(\frac{0.1+x}{0.1-x}\right) \approx \text{pKa} + \log\left(1+20x\right) \approx \text{pKa} + 20x \approx \text{pKa},$$ where $x$ represents the concentration of HA that has dissociated, assumed small. The third equality follows from a Taylor expansion of $0.1-x$, and the fourth from a Taylor expansion of $\log(1+20x)$. We can check whether this quantity is indeed small by solving the equilibrium expression for $x$ exactly; intuitively, it should be small because the difference in proton concentration at a pH of 5 and at a pH of 7 is small relative to the concentration of the acid.</p>
<p>Crucially, <em>the extent of dissociation is small enough to be neglected as a first approximation.</em></p>
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https://chemistry.stackexchange.com/questions/101085/why-dont-buffer-solutions-return-to-equilibrium
|
Question: <p>My former question <a href="https://chemistry.stackexchange.com/questions/188136/is-deprotonation-limiting-the-product-formation-in-this-thiohydantoin-synthesis">Is deprotonation limiting the product formation in this thiohydantoin synthesis?</a>
made me wonder to what extent the KOH deprotonates the imide (pka 8)? in ethanol as solvent. Assuming that there is 1 equivalent base to product.</p>
<p>Answer:</p>
<p>Can be calculated by difference in pKa values
Keq=10^(pKa difference)
In this case the reaction equilibrium is fully on the product side.</p>
<p><a href="https://chem-libretexts-org.translate.goog/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.08%3A_Predicting_Acid-Base_Reactions_from_pKa_Values?_x_tr_sl=en&_x_tr_tl=de&_x_tr_hl=de&_x_tr_pto=rq" rel="nofollow noreferrer">https://chem-libretexts-org.translate.goog/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.08%3A_Predicting_Acid-Base_Reactions_from_pKa_Values?_x_tr_sl=en&_x_tr_tl=de&_x_tr_hl=de&_x_tr_pto=rq</a></p>
<p><a href="https://i.sstatic.net/oqt6GA4i.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/oqt6GA4i.png" alt="enter image description here" /></a></p>
Answer:
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https://chemistry.stackexchange.com/questions/188241/can-we-estimate-acid-base-reaction-equilibrium-in-non-aqueous-media
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Question: <p>My question is in regards to acid base reaction that take place in larger chemical reaction in orgo. Like, I understand acid base reaction, how to qualitatively check which is stronger acid or base or where equilibrium lies, very well. But i feel that when I am trying to study the mechanism of series of other reaction like grignard mechanism, oxidation of alcohol, reaction with aldehydes etc… I see many intermediate acid base reaction in these mechanism that don’t make sense in terms of what I learned earlier when I was exclusively studying acid base reaction. </p>
<p>For instance, I have seen many intermediate acid base reaction in larger reaction where an oxygen atom of one molecule attack H atom attached to oxygen atom in another molecule. And then I am thinking what the point of this reaction because equilibrium is gonna be in middle and not favor products since the conjugate base create will be acid right away anyways. </p>
<p>Other times I have seen H20 act as an acid while molecule with sulfur atom act as a base. And again I am thinking how this is possible because isn’t sulfur a stronger acid then water. When I asked two of my teacher about this. One of them said she actually made a mistake. But the main professor actually said that she is correct and to explain this, she said that water would even be hydronium ion here and when there is lot of water around, this water can serve as acid. However I couldn't fully understand what she meant by that? Can anyone please explain what she meant by this or how these acid base reaction possible. I have attached picture of what I am mean below to explain it my problem more clearly.
<a href="https://i.sstatic.net/dJX2R.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/dJX2R.jpg" alt="enter image description here"></a></p>
Answer: <p>The answer to both questions lies in the fact that a reaction can be directed to a certain extent by controlling the concentration of the reagents. The first reaction is perfectly fine since <strong><em>an alkoxide ion is more basic than a hydroxide ion.</em></strong> As for the second reaction, <strong><em>a carbon attached to an electronegative element is quite acidic.</em></strong> Hence is capable of possessing a negative charge. This along with the excess amount of water we supply for this reaction and the fact that water is neutral in nature adds to the reaction progressing in the forward direction. </p>
<p>Hope this helps</p>
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https://chemistry.stackexchange.com/questions/71300/how-do-you-make-sense-of-acid-base-reactions-in-organic-chemistry
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Question: <p>From what I understand, the concentration of hydronium and hydroxide molecules is constant in pure water (and equals $10^{-7}\ \text{M}$, which is measured experimentally). </p>
<p>What I don't understand is why this remains true for an aqueous solution with an acid or base in it. Shouldn't it affect these concentrations?</p>
<p>When you write the equilibrium reactions for a weak/weak acid/base pairs (one with its conjugate), and get that $K_\mathrm a\cdot K_\mathrm b = [\ce{H+}][\ce{OH- }]$, what's the explanation for it? Why does their multiplication (either as pure water or in a acidic/basic solution) remain constant?</p>
Answer: <p>The concentrations do change in the presence of acid or base and are no longer $10^{-7}$M.</p>
<p>Instead, [H3O+][OH-] $= 10^{-14}M^2$</p>
<p>This comes from: </p>
<p>$\ce{2H2O <=> H3O+ + OH-}$</p>
<p>$K = \frac{a(\ce{H3O+})a(\ce{OH-})}{(a(\ce{H2O}))^2}$</p>
<p>Where "a(x)" is activity of species "x". </p>
<p>Then, approximating activity of $\ce{H2O}$ as 1, and the activity of the ions as the concentration of the ions there is a constant $K_w$, such that:</p>
<p>[H3O+][OH-] $= 10^{-14}M^2 = K_w$</p>
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https://chemistry.stackexchange.com/questions/41696/acid-base-dissociation-constants-relationship
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