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Question: <p>I was wondering how to answer a question I got from my students about the possible recovery of gold from solution. We just finished the reaction between silver ions and copper metal and they wondered why copper wouldn't displace gold from solution in an electrochemical reaction. The redox potentials suggest that it can be done but there is no documentation of this reaction anywhere. I can't try the reaction because we have no gold compounds. If it could be done you would think that gold recovery but companies would use the technique since copper is a lot cheaper.</p>
<p>Please advise.</p>
Answer: <p>Short Answer: Yes, based on the electrochemical/reactivity series, it is possible to percipitate gold from solution using metals higher on the series like copper or zinc. However, it isn't done often because gold is highly unreactive, and when gold salts are formed, they are formed deliberately under atypical conditions like very high temperatures or acidities.</p>
<p>It would be possible to displace gold from it's ionic salt (gold (iii) chloride) using zinc or copper. This is because, as you mentioned copper is higher on the activity series than gold.</p>
<p>The likely reason to why companies don't use single-displacement reactions to retreive gold from solution is be cause gold salts rarely exist. Gold is extremely unreactive (see <a href="https://chemistry.stackexchange.com/questions/21810/why-is-gold-unreactive-when-only-one-electron-is-in-the-outer-shell#:%7E:text=Relativistic%20effects%20account%20for%20gold%27s,from%20falling%20into%20the%20nucleus.&text=This%20is%20why%20gold%20is%20relatively%20unreactive.">here</a>) Gold (III) chloride is normally synthesized by reacting gold metal with chlorine gas at 180 degrees Celsius. This temperature is higher than anywhere you'd find on Earth's surface, and chlorine gas is too reactive to exist in that state in the atmosphere. These conditions are typically found only in the lab and not out in nature. Another example of a man-made gold salt is aqua regia (a 3:1 ratio of hydrochloric acid and nitric acid) which can dissolve gold. To recover the gold, ions are displaced using ferrous sulfphate reducing agent (<a href="https://www.ishor.com/large-scale-gold-refining-using-aqua-regia#:%7E:text=The%20nitric%20acid%20from%20the,chlorides%20unchanged%20and%20in%20solution." rel="nofollow noreferrer">source</a>).</p>
<p>If you are interested in gold recovery, here is an excellent <a href="https://pubs.rsc.org/en/content/articlehtml/2020/ra/c9ra07607g" rel="nofollow noreferrer">article</a> on a variety of methods.</p>
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https://chemistry.stackexchange.com/questions/145552/electrochemistry-gold-recovery
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Question: <p>Unfortunately, I don't have batteries to waste to test this myself, but I am curious... </p>
<p>Let's say you have two identical batteries. AA Duracell.</p>
<p>Now let's say you run one of the batteries through an LED until the LED starts emitting at half the intensity it was emitting at when you first started (effectively draining the battery to half voltage)</p>
<p>Now, lets' say you touched the positive ends of both batteries for a reasonable amount of time, and then did the same for the negative ends.</p>
<p>Given that the used battery would have a much lower voltage, raw electrical current would travel from the unused battery into the used battery.</p>
<p>So this brings me to my question -- Would the batteries now effectively have the same amount of energy in them? I am not just talking voltage, but chemically too, or would the similarity in voltage on both batteries immediately disappear if either of them were used (because while, yes, either end of the battery does contain equal electrical charge to its counterpart, either end of each battery does NOT contain equal chemical potential)</p>
<p>To simplify this, would touching two batteries as described above effectively drain the unused one, and charge the used one?</p>
Answer: <p>First, some batteries are rechargeable, but some are not. However, even if we talk about rechargeable one, then
1) some degradation during recharge usually occurs
2) deep equality will require very long time to achieve</p>
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https://chemistry.stackexchange.com/questions/883/can-i-use-one-alkaline-battery-to-recharge-another
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Question: <p>I recently conducted an experiment where I electrolyzed a solution of sodium chloride. I was using a copper coin as the anode, and I observed a pale blue precipitate in the cup when I was done with the experiment. I then deduced the following:</p>
<p>The chloride ions get oxidized at the anode: </p>
<p>$$\ce{2Cl- -> Cl2 + 2e-}$$</p>
<p>The chlorine gas then reacts with the copper anode to form copper(II) chloride which goes into solution as it is completely soluble.</p>
<p>At the cathode, the hydrogen in the water gets reduced to form hydrogen gas:
$$\ce{2H2O + 2e- -> 2H2 + 2OH-}$$</p>
<p>I know the above half reactions exist from <a href="http://apcentral.collegeboard.com/apc/members/repository/ap04_chem_standard_9793.pdf" rel="nofollow noreferrer">the list of standard reduction potentials</a>.</p>
<p>Next, the copper(II) ions react with the hydroxide ions being formed to produce copper(II) hydroxide, which is both insoluble and pale blue.</p>
<p>However, <a href="http://www.ibchem.com/IB/ibnotes/full/red_htm/19.3.htm" rel="nofollow noreferrer">this source</a> mentions that the hydronium ions in the water get reduced at the cathode: </p>
<p>$$\ce{2H+ + 2e- -> H2}$$</p>
<p>Is this correct? In my understanding, the concentration of hydronium ions ($\pu{1e-7 M}$) would be much too low, and where would the hydroxide ions come from to form copper(II) hydroxide?</p>
Answer: <p>In answer to your last statement about the formation of Hydrogen. The actual amount of hydronium ions doesn't matter because it is at equilibrium ($\ce{H2O <=> OH- + H+}$) and so once you use an $\ce{H+}$ you'll simply have more $\ce{H2O}$ dissociate into $\ce{OH-}$ and $\ce{H+}$ (Le Chatlier states) so if you have battery current running through the system you should experience a healthy, visible stream of hydrogen bubbles. I built a similar thing myself back in high school. </p>
<p>The hydroxide ions will come from the water also. As mentioned above, the $\ce{H2O}$ will dissociate into $\ce{H+}$ and $\ce{OH-}$ and you are using both. The $\ce{H2O}$ will keep dissociating and try to keep the $\ce{H+}$ and $\ce{OH-}$ concentrations at $10^{-7} \, \small\text{M}$ as you keep using them. </p>
<p>Feel free to question my knowledge.</p>
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https://chemistry.stackexchange.com/questions/4803/electrolysis-water-to-hydrogen-gas
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Question: <p>I've got the following from a <a href="http://megglobal.com/commitment.html" rel="nofollow noreferrer">supposed new energy-storage technologist</a>. It's supposed to be a fancy new cheap, high-powered, highly scalable battery. I want to know if this is known chemistry, and if there's anything genuinely revolutionary about it:</p>
<blockquote>
<p>"Basic materials in MEG's E-SOURCE™ power cells are lead metal, absorptive glass mat paper, and sulfuric acid solution. Materials unique to the MEG advanced power cell are S2 glass fiber yarn; sodium sulfate; Teflon emulsion, glass-filled polypropylene; bituminous, petroleum-based wax; and metal-free nano-oxides. "</p>
</blockquote>
<p>The lead is apparently in the form of lead oxide.</p>
<p>The claim is that this is a radical new cheap energy-storage system - but someone makes that claim for yet another technology every week or two. Is there anything in the recent literature that would give credibility to this one?</p>
Answer: <p>There is nothing in the chemistry of what they describe that is different in any way to a conventional Lead-Acid battery. What they really seem to claim is improvement in materials in the battery not to change the electrochemistry but to improve other aspects of battery operation. <a href="http://megglobal.com/about.html" rel="nofollow noreferrer">For example</a>:</p>
<blockquote>
<p>MEG's E-SOURCE™ power cell technology is based on a lead-fiberglass composite material that is one-third the weight and 50 times the strength of metal alloys used in conventional batteries. To date, more than $100 million has been invested in research, development, manufacturing, and market testing of this technology in the United States and several international markets.</p>
</blockquote>
<p>This suggests they have built a better lead-acid battery rather than something more radical. This is not ridiculous. They claim the following advantages:</p>
<blockquote>
<ul>
<li>Extraordinary manufacturing and user cost savings</li>
<li>Significant weight reduction compared to conventional lead-acid batteries</li>
<li>Exceptionally long service life</li>
<li>Dramatically fast recharge speeds at all operating temperatures</li>
<li>Safe, simple, and low-cost maintenance</li>
</ul>
</blockquote>
<p>It is not impossible to imagine clever overall design and improved microstructural components and electrode materials could deliver these sorts of improvements and it is certainly far easier than inventing a whole new battery electrochemistry. For example, charging speed is often dominated by the physical material properties of the electrodes and not the underlying electrode chemistry, so clever design can mitigate the physical limits on charging speed such as by increasing the effective surface area.</p>
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https://chemistry.stackexchange.com/questions/5109/is-there-anything-in-the-literature-that-supports-revolutionary-claims-for-this
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Question: <p>I have been watching some videos on how batteries work and they seem to talk about the cathode being the most important. I also watched another video a long time ago and while my memory is blurred it talked about impurities in the cathode being the only reason the electrical charge doesn't flow constantly. While I found this weird I also wonder how generates seem to get this unlimited supply of electrons also so it's only fair for me to assume that the battery can also.</p>
Answer: <p>I don't think that's accurate. Even the universe doesn't have an "unlimited supply of electrons".</p>
<p>Impurities in the cathode likely give rise to internal resistance inside the battery (among other things), leading to deviation from an ideal cell. I would guess that as your cathode approached purity, your battery would approach the behavior of 'ideal cells', and that discharge rates would approach those of capacitors, and efficiency would approach 100%. To get an ideal cell though, the anode and electrolyte would also have to be 'pure' and some real-world effects would have to be ignored.</p>
<blockquote>
<p>Because of internal resistance, the terminal voltage of a cell that is discharging is smaller in magnitude than the open-circuit voltage and the terminal voltage of a cell that is charging exceeds the open-circuit voltage. An ideal cell has negligible internal resistance, so it would maintain a constant terminal voltage of <span class="math-container">$ \epsilon $</span> until exhausted, then dropping to zero. If such a cell maintained 1.5 volts and stored a charge of one coulomb then on complete discharge it would perform 1.5 joule of work. In actual cells, the internal resistance increases under discharge</p>
<p><a href="https://en.wikipedia.org/wiki/Battery_(electricity)#Battery_cell_performance#Principle_of_operation" rel="nofollow noreferrer">Wikipedia</a></p>
</blockquote>
<p>What do you mean by pure, exactly? Complete homogeneity and lack of any other elements? That sounds like a physics problem where you "ignore air resistence" and other real world effects for simplicity. If this is what you're after, search for things like 'Ideal Battery' and 'Ideal Electrochemical Cell'.</p>
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https://chemistry.stackexchange.com/questions/5446/in-a-battery-what-would-happen-if-you-had-a-perfectly-uniform-and-pure-cathode
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Question: <p>I am trying to make a copper sulfate electroplating solution. I came across a <a href="http://www2.bren.ucsb.edu/~dturney/port/papers/Modern%20Electroplating/02.pdf" rel="nofollow">document</a> (Schlesinger and Paunovic 2011) which mentions the amount of copper sulfate that should be added to water to create such a solution. In Table 2.1 on the bottom of page 35, it has two columns: one for conventional solutions and one for high throw solutions. </p>
<p>I read another <a href="http://electronics.macdermid.com/documents/resources/Innovative%20High%20Throw%20Copper%20Electrolytic%20Process.pdf" rel="nofollow">paper</a> (Nikolova and Watkowski) which talks about high throw copper electroplating process. All I could glean from the paper was that high throw solutions enable higher aspect ratios for copper PCBs.</p>
<p>Please reply if anyone knows what high throw solutions are.</p>
<p><strong>References:</strong></p>
<p>Nikolova, Maria, and Jim Watkowski. “INNOVATIVE HIGH THROW COPPER ELECTROPLATING PROCESS FOR METALLIZATION OF PCB.”</p>
<p>Schlesinger, M., and M. Paunovic. 2011. Modern Electroplating. The ECS Series of Texts and Monographs. Wiley. <a href="http://books.google.com/books?id=j3OSKTCuO00C" rel="nofollow">http://books.google.com/books?id=j3OSKTCuO00C</a>.</p>
Answer: <p><a href="http://www.plating.com/platingtechnical/poorthrowingpower.htm">This page</a> suggests that throwing power is the ability to coat a surface evenly regardless of current density differences, and thus to produce an even plating thickness across the whole of a surface.</p>
<p>A solution with high throwing power would thus have the ability to produce a good, close-to-uniform coating on more difficult surface shapes.</p>
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https://chemistry.stackexchange.com/questions/6072/what-are-high-throw-solutions
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Question: <p>Take the electrolysis of Lead(II) bromide:</p>
<p>We can write it as two half-reactions:</p>
<p>$$\ce{Pb^{2+}(l) + 2e^{-} \rightarrow Pb(l)}$$</p>
<p>$$\ce{Br^{-} \rightarrow Br + e^{-}}$$</p>
<p>In the electrolysis reaction, lead is formed at the cathode and bromine is liberated at the anode. But why is it so? I mean, before gaining or losing electrons to become neutral, lead is positive and bromide is negative. So, naturally they should be attracted to the cathode and the anode respectively. </p>
<p>But why do they become neutral after they reach the electrodes. Why do they gain electrons at only the electrodes to become neutral? </p>
Answer: <p>You have to think about what exactly a cathode and anode are. You are falling victim to this: </p>
<p><em>A widespread misconception is that anode polarity is always positive (+). This is often incorrectly inferred from the correct fact that in all electrochemical devices, negatively charged anions move towards the anode and positively charged cations move away from it. In fact anode polarity depends on the device type, and sometimes even in which mode it operates, as per the above electric current direction-based universal definition. Consequently, as can be seen from the following examples, the anode is positive in a device that consumes power, and the anode is negative in a device that provides power.</em> - Wikipedia</p>
<p>It's your conditioning from other chemical nomenclature--"cation" and "anion"--which are absolute about their charges. </p>
<p>So beyond misleading nomenclature, what does cause this to happen? </p>
<p>Take this picture for example: <a href="http://img.sparknotes.com/figures/0/02480ae8fc1a41b131a3fdb5a698e9a3/compare.gif" rel="nofollow">http://img.sparknotes.com/figures/0/02480ae8fc1a41b131a3fdb5a698e9a3/compare.gif</a></p>
<p>If E°cell > 0, then the process is spontaneous (galvanic cell)
If E°cell < 0, then the process is nonspontaneous (electrolytic cell)
where E°cell = E°(V) reduction + E°(V) oxidation
In your case, </p>
<p>Pb → Pb2+ + 2e- = 0.13 </p>
<p>2Br- → Br2 + 2e- = -1.06</p>
<p>E°cell = .13 - 1.06</p>
<p>E°cell = -.93</p>
<p>In your case, the process is nonspontaneous (makes sense, think about trying to reduce ionized bromine), so you're looking at an electrolytic cell, which is battery driven (it needs work put in to operate.)</p>
<p>So, again, to directly answer your question:</p>
<blockquote>
<p>In the electrolysis reaction, lead is formed at the cathode and bromine is liberated at the anode. But why is it so? I mean, before gaining or losing electrons to become neutral, lead is positive and bromide is negative. So, naturally they should be attracted to the cathode and the anode respectively.</p>
</blockquote>
<p>Lead is formed at the anode because work from a battery was put in that pushed the reaction in that direction. Before work was put in, the reaction would have proceeded in the opposite direction (think exothermic vs endothermic, but in this case spontaneous cell potential vs nonspontaneous cell potential.) </p>
<blockquote>
<p>But why do they become neutral after they reach the electrodes.</p>
</blockquote>
<p>They become neutral at the electrodes because that is where the reaction happens, and that is where the electrons are transferred. </p>
<blockquote>
<p>Why do they gain electrons at only the electrodes to become neutral?</p>
</blockquote>
<p>Again, where else could they gain electrons? They are sitting in a solution contain themselves and their respective anions, which they have already reacted with, and will no longer be transferring electrons with.</p>
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https://chemistry.stackexchange.com/questions/6773/during-electrolysis-why-are-the-products-attracted-to-the-cathode
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Question: <p>A current of 2.25 A is applied to $\ce{NiCl2}$ solution</p>
<p>A. Write the balanced half reaction that takes place at the anode
B. Write the balanced half reaction that takes place at the cathode</p>
<p>Can someone check if my answers seem logical? I'm not quite sure if I did the problem correctly</p>
<p>Anode: $\ce{O2 + 4H+ + 4e- -> 2H2O}$</p>
<p>Cathode: $\ce{2Ni^{2+} + 4e- -> 4Ni}$</p>
Answer: <p>At the <strong>A</strong>-node you <strong>O</strong>-xidize, at the <strong>C</strong>-athode you <strong>R</strong>-educe. (Vowel to vowel, consonant to consonant.)</p>
<p>Following this, you will have to generate electrons at the anode, which doesn't happen with your reaction.</p>
<p>What reaction will happen depends on the normal potential of each reaction, but I assume it will be something along the lines of
$$\ce{2Cl- -> Cl2 ^ + 2e-} $$</p>
<p><strong>Edit:</strong> I just noticed that the cathode reaction can be simplified to
$$ \ce{Ni^{2+} + 2e- -> Ni} $$</p>
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https://chemistry.stackexchange.com/questions/7726/balanced-reactions-at-the-anode-and-cathode
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Question: <p>The reduction of water at a cathode is represented by the following equation:</p>
<p>Cathode (reduction): $\ce{2 H2O_{(l)} + 2e^{−} → H2_{(g)} + 2 OH^{-}_{(aq)}}$</p>
<p>If this is done in a divided cell, what would happen to the $\ce{OH-}$ anions? The hydrogen gas would accumulate and bubble would form. Would the $\ce{OH-}$ anions just build up in the solution? Do they interact with each other naturally forming $\ce{H2}$ and $\ce{O2}$?</p>
Answer: <p>You only show one a half-reaction, reduction at the cathode. Write the other half-reaction, oxidation at the anode to give oxygen. A real world water electrolysis cell requires a salt electrolyte for conductivity. $\ce{NaCl}$ could be electrolyzed to $\ce{Cl2}$ and $\ce{Na}$, the latter reacting with water to give $\ce{H2}$ and $\ce{NaOH}$.</p>
<p><a href="http://www.ineris.fr/ippc/sites/default/interactive/brefca/image16.gif" rel="nofollow">http://www.ineris.fr/ippc/sites/default/interactive/brefca/image16.gif</a></p>
<p>If you want $\ce{H2}$ and $\ce{O2}$, make the electrolyte $\ce{NaOH}$. $\ce{Na2CO3}$ can be interesting.</p>
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https://chemistry.stackexchange.com/questions/8180/what-happens-to-the-hydroxide-anion-during-the-electroreduction-of-water
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Question: <p>I have a pot with some (bottled) water and added (sea) salt. It has a pencil (graphite) and tin foil (aluminum). The pencil is connected to the positive outlet of a solar panel and the tinfoil is connected to the negative outlet of the solar panel (effectively 'charging' the electrolytic cell).</p>
<p>To ensure I don't come to harm, I would like to know:</p>
<p>A) What chemical(s) gas(es) are produced at the pencil?
B) What chemical(s) gas(es) are produced at the tin foil?
C) What will the chemical composition of the water generally be (assuming I've been charging it a while)? (Acidic, toxic, etc?).</p>
Answer: <p>Voltage determines the kind of reactions, amperage determines the amounts. The positive graphite electrode is the oxidizing anode. It will evolve oxygen and chlorine and be eroded to carbon dioxide, carbon monoxide, and chloro-organics. The negative aluminum electrode is the reducing cathode. It will reduce Na+ to elemental sodium that immediately reacts with water to make lye and hydrogen. The lye will erode the aluminum into more hydrogen and aluminum hydroxide gel, plus some black slime for about 1% alloy-reinforcing precipitated intermetallics in the aluminum foil (usually 8111 alloy). Lye plus chlorine gives hypochlorite bleach that will erode both electrodes, and do a little disproportionation into chloride plus chlorite, then that perhaps into chloride and chlorate. </p>
<p>Sodium carbonate (washing soda) is a more benign electrolyte, but it is an alkali hazard to eyes and skin. A glass is a more inert container. Do not short the electrodes, do it outdoors and stay upwind. Avoid electrolyte or product contact with your eyes and skin.</p>
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https://chemistry.stackexchange.com/questions/9206/electrolytic-cell-chemical-composition-during-after-charging
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Question: <p>With the following setup: plastic container, water (impure), pencil (anode), tin foil (aluminum, cathode) and solar panels supplying electrical charge (roughly 36v at about 400-800ma), with the intention of producing lye via electrochemistry:</p>
<p>1) Would it be possible to produce the right kind of lye to a level sufficient to produce handsoap with this current setup?</p>
<p>2) How would you test the lye quality to be sure?</p>
<p>If not possible to make the right quality lye, is there a way to modify the setup so it's able to produce suitable quality lye for soapmaking?</p>
Answer:
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https://chemistry.stackexchange.com/questions/9214/electrochemistry-lye-quality
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Question: <p>I am doing a lab for school and I do not know what this question is asking. In this lab we are making electrochemical cells from different metals. </p>
Answer: <p>You have a standard pot of liquid electrolyte into which are dipped non-touching same-spaced paired strips of different clean metals. The external circuit between the strips is a sensitive voltmeter (with extremely high electrical resistance - hard by zero current flow). You measure the potential (noting direction!) between all combinations of paired strips, (n)(n-1)/2, and then draw certain conclusions.</p>
<p>Why are amalgams saturated with respect to excess solid metal interesting substrates for this experiment (e.g., aluminum)? How do you construct a standard hydrogen electrode?</p>
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https://chemistry.stackexchange.com/questions/9855/how-to-determine-the-relative-differences-in-activity-between-metals
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Question: <p>Why more attractive metals will be oxidized? Aren't they being reduced, because they attract and receive electrons from negative polyatomic ions? So they will be anodes, where oxidization occurs. Why anodes are negative in voltaic cells and positive in electrolytic cells? Same for cathodes.</p>
Answer: <p>Metals want to lose electrons and become positively charged. When the metal loses the electron, it is called oxidation. The substance that gets the electron is reduced.</p>
<p>Oxidation could occur at the anode if it takes electrons from a metal (i.e. $Fe^{2+}$ to $Fe^{3+}$). </p>
<p>Anodes are negative in voltaic cells since there are excessive electrons available to power the circuit. Cathodes give electrons away to the solution.</p>
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https://chemistry.stackexchange.com/questions/10203/voltaic-and-electrolytic-cells
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Question: <p>And how do I find the diffusion coefficient from a Levich plot?</p>
Answer: <p>I would recommend you take a look at the book Electrochemical Methods fundamentals and applications.</p>
<p>For the Koutecý-Levich equation (totally irreversible one-step, one-electron reaction):</p>
<p>$\Large \frac{1}{i}=\frac{1}{i_K}+\frac{1}{i_{l,c}}=\frac{1}{i_K}+\frac{1}{0.62nFAD^{2/3}\omega^{1/2}\nu^{-1/6}C^*_O} $</p>
<p>The higher diffusion coefficient is, the higher is the current.</p>
<p>The diffusion coefficient can be computed from the Koutecý-Levich equation if you know all other coefficients. </p>
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https://chemistry.stackexchange.com/questions/10686/how-does-the-diffusion-coefficient-of-a-species-affect-the-limiting-current-for
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Question: <p>I know that the limiting current is proportional to the square root of the rotation speed, but what does the slope of the plot represent?</p>
Answer: <p>if you mean the following plot:
<img src="https://i.sstatic.net/zGGBU.jpg" alt="enter image description here"></p>
<p>(Electrochemical Methods. fundamentals and applications, p. 341). </p>
<p>If a bulk concentration is the same as the concentration at the surface of the electrode (the mass-transfer is very high) than the current will be independent on $\sqrt\omega$ and will be constant at the value $i_K$. </p>
<p>The slope of the bold line represent deviation form Levich line and tend towards the limit $\lim_{\sqrt\omega \to \infty} i = i_K $. The higher the speed of rotation is, the less is the electrode reaction limited by mass-transfer effects.</p>
<p>The Levich line (dashed; $\frac{i}{\sqrt\omega}$ ) is constant only when $i_K$ is much larger. </p>
<p><img src="https://i.sstatic.net/ertbg.jpg" alt="enter image description here"></p>
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https://chemistry.stackexchange.com/questions/10688/what-does-the-slope-of-a-levich-plot-represent
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Question: <p>Is there any difference between the terms specific conductivity and conductance. If yes, please explain.</p>
Answer: <p>Let's take a wire inside cable for example. That wire's <strong>conductance</strong> is just the inverse of the resistance this cable is making when electricity passes through it. This depends on things like the length, what is made of, the maximum/minimum width of certain areas of the wire, etc.</p>
<p>On the other hand, the <strong>conductivity</strong> of that wire is a direct property of the material it's made of. If the wire is made of copper for example, it'll have (at the same temperature) the same <em>conductivity</em> as any copper wire in the world (because they are all made of copper), but not the same <em>conductance</em> (because they are not physically equal)</p>
<p>I hope this helps, I'm sorry if I was wrong!</p>
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https://chemistry.stackexchange.com/questions/10949/confusion-regarding-specific-conductivity
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Question: <p>The electronic conductance of a metal depends on its density.Does this mean that if we increase the density of the metal,its electronic conductance will increase?How does it happen?</p>
Answer: <p>You cannot change the density of a metal without also changing other properties. </p>
<p>Brass is a mixture of several elements. It can have different densities depending on the mixture of the elements. I haven't researched it, but the electronic conductance would likely be different for each type of brass.</p>
<p>Do you have a source for your statement?</p>
<blockquote>
<p>The electronic conductance of a metal depends on its density.</p>
</blockquote>
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https://chemistry.stackexchange.com/questions/13606/how-does-electronic-conductance-of-a-metal-change-with-density
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Question: <p>I'm doing Grade 12 Chemistry and I'm unsure of this, taking the hydrogen electorde as having a potential of 0 volts.</p>
<p>I'll use the copper-zinc cell as an example. Just let me know if I've got the full logic right:</p>
<ol>
<li><p>Zinc has higher electronegativity, so due to the equilibrium $\ce{Zn <=> Zn^{2+} + 2e-}$, we have more charge building up in the Zinc electrode than in the Copper electrode. (Given both half-cells are of same concentration and in same temp)</p></li>
<li><p>We have a potential difference between the two electrodes. This means some charge from the Zinc electrode will move to the Copper electrode in such a way that both electrodes will momentarily have equal charges.</p></li>
<li><p>Since the charge in the Copper electrode has gone up, we have added electrons to the equilibrium reaction $\ce{Cu^{2+} + 2e- <=> Cu}$. We therefore create more copper.</p></li>
</ol>
<p>Is this why an electrode with a higher potential will "pull" the electrons from the electrode with a lower potential?</p>
Answer: <p>There are not many problems where electronegativity helps you find the answer. It is popular for some reason amongst newer chem students but it shouldn't be.</p>
<p>Here's what happens in the Daniell cell:</p>
<p>Metallic bonds in zinc are broken and zinc atoms each lose 2 electrons. Both of these processes are endothermic.</p>
<p>Zinc ions are hydrated. This is exothermic.</p>
<p>Copper ions are dehydrated. This is endothermic.</p>
<p>Copper ions gain 2 electrons each and metallic bonds are formed. Both of these processes are exothermic.</p>
<p>Each of the above processes also involves an entropy change.</p>
<p>Only when all of these factors are considered can the direction of the electrochemical reaction be determined. Or, you can look at a table of standard reduction potentials.</p>
<p>By the way, silver's electronegativity is 1.93 but the reduction potential for <span class="math-container">$\ce{Ag+}$</span> is <span class="math-container">$\pu{0.8 V}$</span> while <span class="math-container">$\ce{Cu^2+}$</span> is <span class="math-container">$\pu{0.34 V}$</span>.</p>
<p>In terms of charge, electrons flow from anode to cathode. There is no charge built up because of the salt bridge.</p>
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https://chemistry.stackexchange.com/questions/15002/in-a-galvanic-cell-why-does-an-electrode-with-a-higher-potential-pull-the-ele
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Question: <p>I found that reaction: HNO3 + 4.1633363423443E-17 H3O = NO2{+} + OH{-}
Please help me understand " 4.1633363423443E-17" term.</p>
Answer: <p>It seems to me like you were asked to balance a chemical equation and you used some computer program to do it for you instead.</p>
<p>4.1633363423443E-17 is the result of the limitations of floating point arithmetic (ie that computers only have finite precision). The real answer should be 0 (which you could have figure out by balancing it yourself). </p>
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https://chemistry.stackexchange.com/questions/15760/protonation-of-nitric-acid
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Question: <p>I was carrying out an experiment into the effect of temperature changes on the standard potential of a cell. </p>
<p>$$\ce{Zn(s) +2Fe^{3+}->Zn^{2+} +2Fe^{2+}}$$
I used equi-molar concentrations of $\ce{Fe(CN)_6^{4-}}$and $\ce{Fe(CN)_6^{3-}}$. Thus the Nernst equation</p>
<p>$$\mathrm{E^{\theta}_{cell}=E_{cell}-{\frac{RT}{nF}}*ln1}$$ reduces to $$\mathrm{E^{\theta}_{cell}=E_{cell}}$$</p>
<p>We plotted our $\mathrm{E^{\theta}_{cell}}$ values versus <strong>T</strong> and noticed that our $\mathrm{E^{\theta}_{cell}}$ values were actually <em>increasing</em> with temperature.</p>
<p>I would have expected that these values go down as the concentration of the reactant falls. </p>
<p>I asked my demonstrator and sourced information from different web resources, but can't access a definitive answer.</p>
<p>Any hints even? </p>
Answer: <p>A few thoughts:</p>
<p>For starters, $\ce{Zn^{2+}}$ is also part of the reaction quotient, so the log term in the Nernst equation will not necessarily be zero if the hexacyanoferrate concentrations are equal, depending on the $\ce{Zn^{2+}}$ concentration:</p>
<p>$$E = E°+\frac{RT}{zF}\ln\frac{[\ce{Zn^2+}][\ce{Fe^2+}]^2}{[\ce{Fe^3+}]^2}$$
Depending on whether the reaction quotient is greater than or less than 1, the potential can have either a positive or negative temperature dependance.</p>
<p>Second, the Nernst equation uses activities not concentrations so depending on the ionic strength and other factors, the activities of the ions in solution may not be what you think. This deviation is greater for multiply charged ions and may be significant for concentrations over 1 mM.</p>
<p>Finally, since the hexacyanoferrate species are both soluble, I'm assuming you're using some kind of reference electrode for the second electrode. This is also temperature dependant in the same way as the rest of the cell.</p>
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https://chemistry.stackexchange.com/questions/16929/electrochemical-potentials
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Question: <p>During the electrolysis of $\ce{H2SO4}$, $\ce {OH-}$ ions are more reactive than $\ce {SO4^{2-} }$ ions and so they dissociate. as $\ce {OH-}$ ions dissociate, they are the negative ions that carry current. So when the concentration of $\ce {H2SO4}$ is increased, the concentration of $\ce {OH-}$ ions actually reduces, but then the current still increases.</p>
Answer: <p>Near room temperature, considering solutions of water and sulphuric acid, the maximum conductivity is at about 30% sulfuric acid. For pure water or pure sulfuric acid the conductivity is orders of magnitude lower. </p>
<p>In pure water, there are only $10^{-7}$M of $\ce{H+}$ and $\ce{OH-}$ to carry current. </p>
<p>When a relatively small amount of sulfuric acid is added, it dissociates fully to $\ce{H+}$ and $\ce{HSO4-}$. For example if you increase the concentration of sulfuric acid from 0M to 0.01 M, the number of ions in the solution increase by a factor of $10^5$. </p>
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https://chemistry.stackexchange.com/questions/23495/in-the-electrolysis-of-h%e2%82%82so%e2%82%84-solution-when-the-concentration-is-increased-why-d
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Question: <p>Since in rusting (oxidation) of Iron, transfer of 4 electrons takes place is it possible to use this reaction under catalytic conditions to create a simple electric cell, even if it just gave a millivolt of potential.</p>
<p>$\ce {Fe + H2O -> FeOH3 + H2}$</p>
<p>Considering that we have complete control over the surroundings, like volume expansion, disposing hydrogen and other such factors.</p>
Answer: <p>Sure! A nickel-iron battery uses the oxidation of iron at the negative plate to produce about half the voltage. Though invented by Jungner, it's often called an Edison battery. See <a href="http://en.wikipedia.org/wiki/Nickel%E2%80%93iron_battery" rel="nofollow">http://en.wikipedia.org/wiki/Nickel%E2%80%93iron_battery</a>.</p>
<p>There are very low power semiconductor devices which could be powered by an iron battery. You could make three or four small cells from iron nails and a more electronegative element, e.g. lead, to power a digital watch; even without optimization to remove layers of hydrogen and rust, it should operate for months. </p>
<p>For some other low power ideas, see <a href="https://www.enocean.com/en/energy-harvesting/" rel="nofollow">https://www.enocean.com/en/energy-harvesting/</a>.</p>
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https://chemistry.stackexchange.com/questions/23720/can-we-make-a-rusting-battery
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Question: <p>as far as I've understood to measure the decomposition potential of an electrolyte using a galvanic cell we just put two electrodes, connect them to a battery, and vary the external potential till when we observe a current. The potential at which the current starts to grow linearly is the decomposition potential of the electrolyte. There is not appreciable current before the decompositon potential because the electrodes are supposed to be inert and do not provide the ionic charges needed to "close the circuit". Is it right?
An other question. How the role of the electrodes is taken into consideration in the experimental setup? Is it considered an offset due to an eventual nonzero cell potential? I made this question because I'd like to be sure that I'm understanding the experimental setup.</p>
<p>Tirrel</p>
Answer: <p>Ok I will answer your last question first. If you have two identical electrodes(i.e. same electrode material), the cell potential will be zero and there won't be any issues there.</p>
<p>You have to use inert electrodes so that you are not measuring the dissolution of the electrode itself.</p>
<p>The one part I am not sure you understand properly is the current growing. You want to vary the potential until the current raises exponentially when continuing to sweep. This will be a very sharp rise that will be hard to miss. </p>
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https://chemistry.stackexchange.com/questions/24188/offset-in-measuring-decomposition-potentials
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Question: <p>I've been planning an experiment on how to make galvanic cells and during my research I came across an article in which chloridric acid was used, along with a soluble salt and agar-agar, in salt bridges.
Since then, I've been wondering why chloridric acid was added to agar-agar.</p>
<p>Thanks in advence. </p>
Answer: <p>Any ionic solution can function as a salt bridge. Having it acid based helps with the fast conduction paths of H+ in water.</p>
<p>The critical question is the leak rate. Any salt bridge will leak some amount of the ions into the solution. Good ones leak less, bad ones leak more. </p>
<p>So if you cannot tolerate any acidity in your solution, don't use an acid. Similarly, if you cannot tolerate any chloride in your solution, this would be a bad choice.</p>
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https://chemistry.stackexchange.com/questions/25010/why-is-chloridric-acid-used-in-salt-bridges
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Question: <p>Streaming potential is an electrochemical phenomenon that a that relates
electric currents with the relative movement of solid and liquid phases in contact with
each other.</p>
<p>So when a liquid moves past a solid a potential is generated. The equations for the values of the potential and current generated can be found <a href="http://en.wikipedia.org/wiki/Streaming_current" rel="nofollow">in this Wikipedia article</a>. My question is, is the reverse true? If I applied an electrical potential difference (instead of applying a pressure difference), would a pressure difference be generated?</p>
Answer: <p>Your assumption is correct. <em>Electrokinetic pumps</em> use a potential difference across an osmotic membrane, i.e. through narrow channels, the inverse of flow causing a streaming potential difference. There are other aspects of the phenomena such as <em>electrophoresis</em>, <em>capillary osmosis</em> etc.; see:</p>
<p><a href="http://en.wikipedia.org/wiki/Electrokinetic_phenomena" rel="nofollow">Electrokinetic Phenomena</a> a</p>
<p><a href="http://web.mit.edu/andrew3/Public/Papers/2004/Min/2004_Sensors%20and%20Actuators%20B_On%20the%20ef%EF%AC%81ciency%20of%20electrokinetic_Min.pdf" rel="nofollow">On the Efficiency of Electrokinetic Pumping...</a></p>
<p>and</p>
<p><a href="https://www.youtube.com/watch?v=gWH-Hn4nBxM" rel="nofollow">short video demo</a></p>
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https://chemistry.stackexchange.com/questions/26284/an-electric-potential-is-generated-when-a-liquid-moves-with-respect-to-a-solid
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Question: <p><a href="http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=1548283&tag=1" rel="nofollow">This paper</a> describes the measurement of streaming potential using a potential divider configuration. However it is unclear where the potential difference is created. From the diagram it appears that it is created across the length of the flow channel, so at one side there is a low potential and the other a high potential. However this doesn't seem to agree with the description of streaming potential. From other sources it appears that the potential difference is across the outside of the flow channel near the surfaces, to the inner middle of the flow channel (at the point furthest away from the surface). Which is correct?</p>
<p>EDIT: there isn't a tag for electro-chemistry, so I have just put this under battery chemistry. However this is not necessarily for a battery.</p>
Answer:
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https://chemistry.stackexchange.com/questions/26287/how-is-streaming-potential-measured
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Question: <p>I want to convert resultant hours of salt spray corrosion test to estimated values in years for real life environments such as marine. In other words, How could I predict the result of real life tests using salt spray data?</p>
Answer: <p><em>Note that I'm not an expert in this field and wikipedia itself is not a scientifically relevant primary source</em>. The latter might however be different for sources cited there.</p>
<p><strong>In short, you seemingly better shouldn't.</strong></p>
<p>It seems that the German wikipedia article on the <a href="http://de.wikipedia.org/wiki/Salzspr%C3%BChtest" rel="nofollow">salt spray test</a> has collected a lot of critical remarks from different sources. In summary, standardized (ASTM B117 or DIN EN ISO 9227) salt spray tests may be used in quality control, but <strong>cannot</strong> be transfered to to real world. </p>
<p>According to the wikipedia article, DIN EN ISO 9227 explicitly states:</p>
<blockquote>
<p>Nur selten besteht ein direkter Zusammenhang zwischen der Beständigkeit gegen die Einwirkung von Salzsprühnebel und der Beständigkeit gegen Korrosion in anderen Medien. Die verschiedenen Faktoren, welche das Fortschreiten der Korrosion beeinflussen, können sich je nach den herrschenden Bedingungen sehr unterschiedlich auswirken. Dazu gehört z. B. auch die Bildung von Schutzschichten. Die Prüfergebnisse sollten deshalb nicht als direkter Hinweis auf die Korrosionsbeständigkeit der geprüften metallischen Werkstoffe in allen Umgebungsbedingungen betrachtet werden, in denen diese Werkstoffe verwendet werden können.</p>
</blockquote>
<p>I'd roughly translate this to: <em>Only rarely, a direct correlation between the resistance towards saline fog and the corrosive influence of other media can be found. Depending on the conditions, different factors promoting corrosion may have a distinctively different impact. This includes, e.g., the buildup of protective layers.
Consequently, testing results should not be taken as a direct hint on the general corrosion stability of metallic materials under all planned conditions.</em> </p>
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https://chemistry.stackexchange.com/questions/26452/how-to-use-salt-spray-data-in-real-environments
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Question: <p>This is the question extracted from my country high school level' exam.
So it asks to discuss the feasibility of the electrochemical cell to carry out electrolysis of an aqueous solution of sodium chloride at standard condition.</p>
<p>Al(s)/Al(aq)//Cu(aq)/Cu</p>
<p>My question is how I should start to answer this question?</p>
Answer:
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https://chemistry.stackexchange.com/questions/27035/the-feasibility-of-the-electrochemical-cell
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Question: <p>Following <a href="https://byo.com/stories/issue/item/3113-etch-your-kettle-projects" rel="nofollow">this plan</a>, I am wondering what the byproducts of this method are? At the site with the q-tip, the solution turned a bright yellow, with the q-tip turning a dark yellow, almost red. Holding the q-tip in place for about 10 seconds results in a small amount of smoke, and bubbles around the contact point. What are these colors, smoke, and bubbles? I don't know if the kettle is steel or aluminum, it's an old Budweiser keg, if that helps.</p>
<p>In case the link above dies, the method is etching metal using a vinegar and salt solution, combined with a 9 volt battery. Connecting the negative lead to a piece of metal, and using the positive lead with the solution results in a type of reverse electroplating, thus removing metal ions and etching the metal.</p>
<p>Not sure on the tags for this question, so feel free to adjust them, and the description of the etching method. </p>
Answer: <p>The yellow colour in solution is likely from iron ions from the metal being oxidized and dissolved:
$$\ce{Fe_{(s)} + 2e- -> Fe^{2+}_{(aq)}}$$ (Probably $\ce{Fe^3+}$, too)</p>
<p>The gas bubbles/smoke is the product of the other half of the reaction, the reduction of something in your solution. Because you have chloride in solution, chlorine gas is a possibility:
$$\ce{2Cl^{-}_{(aq)} -> Cl_{2(g)} + 2e-}$$</p>
<p>Hydrogen may also be produced depending on what you're using as an electrode on the q-tip.</p>
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https://chemistry.stackexchange.com/questions/27126/reverse-electro-plating-byproducts
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Question: <p>According to the Nernst equation, the electrode potential $E$ is
$$E = E^\circ - \frac{RT}{nF}\ln\frac{[\ce{M}]}{[\ce{M^{$n$+}}]}$$</p>
<p>Question is, why? Assuming a simple galvanic cell, we know that in general if the temperature is increased, there should be more collisions of the solution with the electrode, resulting in a higher accumulation of the metal ions on the electrode. It should also make it easier for the metal atoms in the electrode to leave the electrode and go into the solution.</p>
<p>Also, if temperature reaches zero, the reaction should stop altogether as there would be no more collisions and no more potential difference should be developed on the electrodes. Yet according to the Nernst equation if temperature reaches zero, maximum standard electrode potential would be reached instead. Where am I going wrong?</p>
Answer:
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https://chemistry.stackexchange.com/questions/28036/why-does-an-increase-in-temperature-decrease-the-electrode-potential
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Question: <p>When recharging a lead-acid battery, how are the terminals of the battery connected to the voltage source. I mean, do we connect the positive terminal of the battery to the negative terminal of the source and vice versa? Or does the positive go with the positive and the negative with the negative? Also, are the electrodes reversed while the battery is being recharged?</p>
Answer: <p>The polarities are not reversed, basically current always flows from the positive to negative terminal (opposite the flow of electrons by definition). When a battery is discharged, the voltage difference across its terminals will drop. When you then connect the charger with the full voltage across the terminals of the batteries, the chemical process that lead to the discharge of the battery is reversed and the current flows "backwards" if you will, with the help of the charger. So you could think of the battery being discharged a little, and having only 10 volts instead of 12 volts across its terminals, connecting the 12 volts charger to those terminals (negative to negative and positive to positive) means that there's a voltage difference of 2V from the positive terminal of the charger to the positive terminal of the battery which causes the current to flow and the battery to charge.<br>
You should look at redox processes if you want to understand more of the chemistry behind it! If you just want to charge your battery, remember to connect the positive terminal of the charger to the positive terminal of the battery!</p>
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https://chemistry.stackexchange.com/questions/28634/how-is-a-car-battery-recharged
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Question: <p>Two questions: </p>
<p>First, how does the gate provide a positive charge by running a current through it? I really just don't see how this positive charge could be achieved, unless you had some sort of battery in the gate itself and attached a wire to donate some of the electrons flowing through to the cathode of some other battery. Is it possible to get electrons out of the gate by some other means?</p>
Answer: <p>The gate of a FET works much like the grid of a radio tube (valve). The gate is insulated from source and drain; the only current flow (if the insulator were perfect) is to charge the intrinsic gate capacitance. Even a MOSFET has <em>some</em> leakage, but it's pico- or femtoamperes. See <a href="https://electronics.stackexchange.com/questions/31594/mosfet-when-can-we-not-assume-that-the-gate-current-is-0">this discussion</a> on capacitance and leakage.</p>
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https://chemistry.stackexchange.com/questions/28820/transistors-how-does-the-gate-work
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Question: <p>The equation: </p>
<p>$$E^{。}_{cell}= \frac{RT}{nF}\ln K_{eq}$$</p>
<p>We all know cell potential is intensive, not affected by the amount, Because: $volt=\frac{joule}{coulomb}$. Both joule and coulomb will be doubled altogether.</p>
<p>But as seen from the equation, cell potential is affected by the number of electrons transferred.</p>
Answer: <p>No, there is no violation of "intensivity". The reason is that $K_{eq}$ depends on $n$, and changes in one cancel the other out.</p>
<p>For example, consider the electrolysis of water:</p>
<ol>
<li>$$\ce{2H2O_{(l)} -> 2H2_{(g)} + O2_{(g)}}$$</li>
</ol>
<p>The equilibrium constant for this reaction is $K_{1}=\frac{[\ce{H2}]^2 [\ce{O2}]}{1}$ and if you wrote out each electrochemical half reaction separately, $n$ would be 4.</p>
<p>Now consider this reaction:</p>
<ol start="2">
<li>$$\ce{4H2O_{(l)} -> 4H2_{(g)} + 2O2_{(g)}}$$ </li>
</ol>
<p>The equilibrium constant is now $K_{2}=\frac{[\ce{H2}]^4 [\ce{O2}]^2}{1}=(K_1)^2$. If you wrote out the half-reactions for this reaction, $n$ would be 8, twice as big. But the $\ln K_{eq}$ term would also be twice as big, since $\ln K_2 = \ln{(K_1)^2} = 2 \ln K_1$.</p>
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https://chemistry.stackexchange.com/questions/28954/does-the-relationship-equation-between-standard-cell-potential-and-equilibrium-c
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Question: <p>I mean by "combining" is to make a new half-reaction equation and not an overall equation for a reaction in whole.</p>
<p>For instance, I was trying to arrive at the following half-reaction:</p>
<p>$\ce{MnO2(s) + 4H+(aq) + 2e- -> Mn^{2+}(aq) + 2H2O(l)}\quad E^\circ= 1.23\,\mathrm{V}$</p>
<p>by combining</p>
<p>$\ce{MnO4^{-}(aq) + 8H^+(aq) + 5e- -> Mn^{2+}(aq) + 4H_2O(l)}\quad\quad E^\circ= 1.51 \,\mathrm{V}$</p>
<p>and</p>
<p>$\ce{MnO4^{-}(aq) + 4H+(aq) + 3e- -> MnO2(s) + 2H_2O(l)}\quad\quad E^\circ=
1.7\,\mathrm{V}$</p>
<p>using Hess's Law. My calculation yielded +0.19 V, and <em>not</em> +1.23 V as I would expect.</p>
<p><strong>Why caused this?</strong> I know that there are half-reactions that can be constructed in this way (as above, by combining two or more other half-reactions) that result in the correct potential being calculated.</p>
Answer: <p>In general, you cannot simply subtract electrode potentials like this to find the potential of another half-reaction. They have to be weighted by $n$ (the stoichiometric coefficient of $\ce{e-}$). In cases where your method works, it is only because $n$ is coincidentally the same for both half-reactions you are combining.</p>
<p>The foolproof way is always to convert electrode potentials to Gibbs free energy changes:
$$\Delta G^\circ = -nFE^\circ$$</p>
<p>So $\Delta G^\circ$ for the half-reaction $\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$ is</p>
<p>$$(-5)(96485 \text{ C mol}^{-1})(1.51 \text{ V}) = -728.4 \text{ kJ mol}^{-1}$$</p>
<p>and $\Delta G^\circ$ for the half-reaction $\ce{MnO4- + 4H+ + 3e- -> MnO2 + 2H2O}$ is</p>
<p>$$(-3)(96485 \text{ C mol}^{-1})(1.7 \text{ V}) = -492.1 \text{ kJ mol}^{-1}$$</p>
<p>Subtracting the second equation from the first, you will find that $\Delta G^\circ$ for the half-reaction of interest ($\ce{MnO2 + 4H+ + 2e- -> Mn2+ + 2H2O}$) is</p>
<p>$$(-728.4) - (-492.1) = -236.3 \text{ kJ mol}^{-1}$$</p>
<p>which gives:</p>
<p>$$E^\circ = -\frac{-236.3 \text{ kJ mol}^{-1}}{(2)(96485 \text{ C mol}^{-1})} = 1.22 \text{ V.}$$</p>
<p>I presume some rounding errors led to the slight discrepancy between my answer and the answer you want. You may leave out the Faraday constant in the calculations once you are more familiar with them; basically that means you work with the quantity $nE^\circ$ instead of $\Delta G^\circ$. The answer should be the same since they are directly proportional.</p>
<p>For more information you can refer to the redox chapters in any inorganic chemistry textbook (Shriver/Atkins, Housecroft, etc.). These calculations are dealt with under the sections discussing Latimer and Frost diagrams.</p>
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https://chemistry.stackexchange.com/questions/28995/why-can-some-redox-half-reactions-be-combined-and-some-cannot
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Question: <p>The classic voltaic cell has a <span class="math-container">$\ce{Zn}$</span> anode and <span class="math-container">$\ce{Cu}$</span> cathode. The reduction half reaction occurring at the cathode combines aqueous copper ions and electrons supplied by the anode to form solid <span class="math-container">$\ce{Cu}$</span>. So what is the purpose of the <span class="math-container">$\ce{Cu}$</span> cathode? <span class="math-container">$\ce{Cu}$</span> is produced anyways from the electrons supplied by the anode. Why can't we have instead of the <span class="math-container">$\ce{Cu}$</span> cathode, simply the wire connecting to the anode which will supply electrons to fuel the reduction half reaction?</p>
Answer: <p>You could certain have a zinc-copper cell in which copper was not the cathode material. Graphite or silver or whatever could be the electrode material instead. But once you close the circuit:</p>
<ol>
<li>Current flows</li>
<li>Cu is deposited as a metal solid on the electrode surface, whatever it is made out of.</li>
<li>Copper metal is, as it always is, conductive.</li>
<li>As a result of steps 2 and 3, there is now a copper metal cathode instead of a cathode made of another material.</li>
</ol>
<p>So even if you start without a copper cathode you wind up with one.</p>
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https://chemistry.stackexchange.com/questions/28732/why-does-the-cathode-have-to-be-made-out-cu-in-a-zn-cu-voltaic-cell
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Question: <h2>Question</h2>
<p>Suppose you have a cell set up between a copper metal/copper(II) ion electrode and a reference electrode.</p>
<p>Under standard conditions, the emf of this cell was −0.07 V. The standard electrode potential of the copper metal / copper(II) ion electrode is +0.34 V. Hence the standard electrode potential of the reference electrode is:</p>
<p>A: -0.41 V</p>
<p>B: -0.27 V</p>
<p>C: +0.27 V</p>
<p>D: +0.41 V</p>
<hr>
<h2>My attempt:</h2>
<p>I was taught this equation: $\mathscr{E} _{cell} = \mathscr{E} _{a} - \mathscr{E} _{b}$ </p>
<p>where: </p>
<p>$\mathscr{E} _{a}$ is the emf of the more positive electrode potential</p>
<p>$\mathscr{E} _{b}$ is the emf of the less positive electrode potential</p>
<p>But using this equation, none of the combinations of electrode potentials give -0.07V as the emf.</p>
<p>Thanks</p>
Answer: <p>Chapter 14 of Harris' Quantitative Chemical Analysis has a helpful section entitled, <em>an intuitive way to think about cell potentials.</em> Here's the corresponding graphic:</p>
<p><img src="https://i.sstatic.net/J7XRq.gif" alt="enter image description here"> </p>
<p>In this case, the author is describing a Galvanic cell with cadmium as the anode and silver as the cathode. The potentials of the two half-cells are known and marked on the line-graph. Electrons <strong>always</strong> flow to the more positive electrode, which in this figure is towards the right. The difference between the two half-cells is the cell potential.</p>
<p>In your case, you know the separation (-0.07 V) and one of two points (+0.34 V). If we ignore the sign for the moment, we can use the line-graph to narrow the possibilities down to <strong>C</strong> or <strong>D</strong>, since each of these differ from +0.34 V by 0.07. The question now is, which direction are the electrons flowing? When determining the cell potential using this line graph, we subtract the right-most number from the left-most number in order to obtain a positive potential. Since the cell potential in your case is <em>negative</em>, we must be doing the opposite. Therefore, the reference electrode potential is +0.41 V (<strong>D</strong>).</p>
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https://chemistry.stackexchange.com/questions/31921/what-is-the-meaning-of-signs-in-electrode-potentials
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Question: <p>I won this <a href="http://rads.stackoverflow.com/amzn/click/B007V5TEMW">"toy"</a> at a science fair...
Now I have to make a short presentation about it in school.
On the package description they describe its energy source as "fuel cell"...
It consists out of a metal plate(magnesium), a black metal plate with a black coating and a paper-cotton-like isolator which is between those plates.
Through putting salt water on the isolator, there's a voltage difference of 1.76V between the black and the magnesium plate.
After a couple of seconds on load the voltage stays constant at 1.31V.
My theory is that it's only a simple battery, because there's no continious source of fuel which should be required for naming it fuel cell (according to wikipedia).</p>
<p>Is my thought right?</p>
<p>My idea about the black coat is, that it is a rubberd graphite or "coal" cathode, rubberd because coal or graphite would fall apart quickly. It is described as air-cathode in the package description. My idea is that the black coal or graphite coat is used to provide a larger amount of air to the reaction...</p>
<p>The magnesium plate gets damaged through this reaction, the black thing not.</p>
<p>Which material is the black coat?</p>
<p>and what is the full reaction?</p>
<p><a href="http://fs1.directupload.net/images/150615/hdu8yaav.jpg">a picture of the components</a></p>
Answer: <p>Your analysis seems correct, i.e. it is only a <a href="https://en.wikipedia.org/wiki/Fuel_cell">fuel cell</a> if the <em>magnesium is being oxidized by the air</em>. You can prove if that is the case, and do so as part of your classroom demonstration, by excluding any air, and therefore oxygen, from the fuel cell.</p>
<ol>
<li><p>Use distilled water and fairly pure $\ce{NaCl}$ (rather than table salt) to make the salt solution.</p></li>
<li><p>Boil the salt water for a minute or so to remove air bubbles (some oxygen from air dissolves into water).</p></li>
<li><p>Put the cell with electrolyte in a plastic bag sealed around the wires and a narrow tube such as a plastic coffee stirrer or soda straw that you'll use later to admit air, carefully squeezing out air bubbles. Seal both bag and tube with plasticine modeling clay or soft wax.</p></li>
<li><p>Observe the open circuit voltage of the cell.</p></li>
<li><p>Then run it for a while with a load (e.g. a <a href="http://www.allelectronics.com/make-a-store/item/lp-41/1.5v-grain-of-wheat-lamp-w/long-leads/1.html">75 mA 1.5 V lamp</a>) or a 100 ohm resistor, which will draw 1.3 mA), depending on the current output of the cell. You would expect <em>some</em> voltage drop over time, either due to <a href="http://www.tpub.com/neets/book1/chapter2/1b.htm">cell polarization</a> or due to any residual oxygen being used up.</p></li>
<li><p><strong>To prove that this is a true fuel cell</strong> oxidizing the magnesium using air, bubble some air into the plastic bag through the (unplugged) tube. *If the voltage goes back up, then drops down until yet more air is added, you've shown that oxygen is being used and it's a true fuel cell.</p></li>
</ol>
<p>Please accept my compliments for your thoughtful question; do post your findings here!</p>
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https://chemistry.stackexchange.com/questions/32907/is-this-product-description-wrong-and-a-simple-battery-and-not-a-fuel-cell
|
Question: <p>Does flipping a reaction as written change the sign of the value of E or is it a completely different value? </p>
<p>For example if I have the value for the half reaction $\ce{Cl2 + 2e- -> 2Cl-}$ but need the value for the opposite reaction ($\ce{2Cl- -> Cl2 + 2e-}$), do I just put a negative sign and treat it like $\Delta H$ values? </p>
Answer: <p>Yes, you are correct - just flip the sign. As quoted from <a href="http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Voltaic_Cells/The_Cell_Potential" rel="nofollow">the venerable UC Davis ChemWiki</a>:</p>
<blockquote>
<p>To determine oxidation electrodes, the reduction equation can simply be flipped and its potential changed from positive to negative (and vice versa).</p>
</blockquote>
<p>They give the value for your (forward) reaction as +1.358 V, so for the reaction in question, you'll use <strong>-1.358 V</strong>.</p>
|
https://chemistry.stackexchange.com/questions/34268/electrode-potential-half-reactions
|
Question: <blockquote>
<p>What are the products of an alkane and a perchlorate reaction?</p>
<p>e.g., making the alkane methane and the perchlorate anhydrous for simplicity:</p>
</blockquote>
<p><strong>My effort</strong>:</p>
<p><span class="math-container">$$\ce{Mg(ClO4)2 + 4CH4 -> MgCl2 + 4CO2}$$</span></p>
<p>leaves 16H free so obviously incorrect, but shouldn't a salt and a gas be part of the products? I guess that the anhydrous form would make this explosive.</p>
Answer: <p>I think that there is some $\ce{H2O}$ as product:
$$\ce{2Mg(ClO4)2 + 4CH4 -> 2MgCl2 + 4CO2 + 8H2O}%edit$$</p>
|
https://chemistry.stackexchange.com/questions/44119/stoichiometry-of-alkane-perchlorate-reaction
|
Question: <p>Does "Mole of Iodine" or any other element that normally appears as diatomic molecules refer to a mole of single atoms (<span class="math-container">$6.02\times 10^{23}$</span> iodine atoms) or a mole of the molecules (<span class="math-container">$6.02\times 10^{23}$</span> molecules of <span class="math-container">$I_2$</span>)?</p>
<p>EDIT: If there is no specification of moles of atoms or moles of molecules, what do I do? Do I default to one of them?</p>
<p>e.g. in this question</p>
<blockquote>
<p>For the combustion of sucrose:</p>
<p><span class="math-container">$\ce{C12H22O11 + 12O2 -> 12CO2 + 11H2O}$</span></p>
<p>there are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent?</p>
</blockquote>
Answer: <p>When we ask, for example, "how many moles of Iodine (or Oxygen or Hydrogen) is contained in 100g?" It ALWAYS refers to the molecule (e.g. I2, O2, H2 etc). Therefore, the number of 6.02 x 10^23 refers to the number of molecules, not atomes of Iodine and therefore you actually have 12.04 x 10^23 atomes of Iodine contained in one mole of molecular Iodine. However, when you look at the periodic system, they always refer to atomic mass not molecular. In case of Oxygen, the atomic mass of 16 g/mol refers to 1 mol of O, not O2 as molecular form. Further, when you want to know mass of the Oxygen (gas form, O2) then: 2 (number of atoms) x 16 g/mol = 32 g/mol - this is the mass of 1 mol of O2 (Oxygen-diatomic form - gas present in the atmosphere).</p>
|
https://chemistry.stackexchange.com/questions/53609/moles-of-diatomic-molecules
|
Question: <blockquote>
<p>Which of the following samples contains the largest number of atoms?</p>
<p>a. 2.0 moles of <span class="math-container">$\ce{H3PO4}$</span><br />
b. 3.0 moles of <span class="math-container">$\ce{H2SO3}$</span><br />
c. 4.0 moles of <span class="math-container">$\ce{HNO3}$</span><br />
d. 6.0 moles of <span class="math-container">$\ce{HClO}$</span><br />
e. 8.0 moles of <span class="math-container">$\ce{HBr}$</span></p>
</blockquote>
<p>My logic would tell me that every mole has <span class="math-container">$6.022\times10^{23}$</span> atoms in it. Then you divide the amount of grams of each compound by the number of moles?</p>
<p>so for A I'd use the following:</p>
<p>(Avg # *2) / 98 grams/per mole of <span class="math-container">$\ce{H3PO4}$</span> = <span class="math-container">$1.22\times10^{22}$</span> atoms </p>
<p>and so on, but I don't get the right answer :/</p>
<p><span class="math-container">$(2\times(6.022\times10^{23})$</span> = number of atoms = <span class="math-container">$1.20\times10^{23}$</span>
<span class="math-container">$\ce{H3PO4}$</span> = 98 grams/mole</p>
<p>The answer is C btw, I just don't know to set up the equation right to get to this answer.</p>
Answer: <p>Geeze, NaCl has 1 mole of Na and 1 mole of Cl per mole of NaCl. So that is 2 moles of atoms per mole of NaCl. </p>
<p>$\ce{K2SO4}$ has 2 moles of potassium, 1 mole of S and 4 moles of O per mole of $\ce{K2SO4}$. So that is 7 moles of atoms per mole $\ce{K2SO4}$. </p>
<p>You don't need to convert to atoms using Avogadro's number then back to moles. That complicates the problem needlessly. Just use the stoichiometry and you can do the arithmetic in your head. </p>
|
https://chemistry.stackexchange.com/questions/59761/which-of-the-following-samples-contains-the-largest-number-of-atoms
|
Question: <p>10 g of Mg(NO3)2 was dissolved in 120 g of water. The result is a solution with a
density of 1.12 g / ml. Determine the percentage concentration
(in %) and the molar concentration (in mol/L) of the resulting
solution</p>
Answer: <p>As you are new to this forum I want to welcome you but nevertheless I would kindly ask you to <em>first</em> make your own research on a problem you have and <em>then</em> ask your question providing some information on what you do not understand. Your question seems to be a homework which you only copied without any signs of your own thoughts.</p>
<p>That said, here are some explanations regarding your assignment.</p>
<h3>Question 1</h3>
<p>If you put 10 grams of something into 120 grams of something else, than your resulting mass is 130 grams, right? So, 10 g of those 130 g are how many percent?</p>
<p><span class="math-container">$$
\frac{10\,\text{g}}{130\,\text{g}} \times 100\,\text{%} = 7.69 \%
$$</span></p>
<h3>Question 2</h3>
<p>Magnesium nitrate (<span class="math-container">$\ce{Mg(NO3)2}$</span>) has a molecular weight of 148.31 g/mol and you have used 10 g, so what is the amount of magnesium nitrate?</p>
<p><span class="math-container">$$
\begin{align}
M &= \frac{m}{n}\\
\Rightarrow n &= \frac{m}{M}\\
&= \frac{10\,\text{g}}{148.31\,\text{g/mol}}\\
&= 6.74 \times 10^{-2}\,\text{mol}
\end{align}
$$</span></p>
<p>As you know, water has a density of 1 gram per milliliter, thus 120 grams correspond to 120 milliliters, which, again, is equal to 0.12 liters.</p>
<p>To calculate the concentration in mol per liter, just divide the amount in mol by the volume in liter:</p>
<p><span class="math-container">$$
\begin{align}
c &= \frac{n}{V}\\
&= \frac{6.74 \times 10^{-2}\,\text{mol}}{0.12\,\text{L}}\\
&= 5.62 \times 10^{-1}\,\frac{\text{mol}}{\text{L}}
\end{align}
$$</span></p>
<h3>Check</h3>
<p>Let's check if our results are correct:</p>
<p>If you happen to have 120 milliliters of a 0.562 molar solution, then you can easily calculate the amount of magnesium nitrate by multiplication:</p>
<p><span class="math-container">$$
\begin{align}
n &= c \times V\\
\Rightarrow n &= 5.62 \times 10^{-1}\,\frac{\text{mol}}{\text{L}} \times 0.12\,\text{L}\\
&= 6.74 \times 10^{-2}\,\text{mol}
\end{align}
$$</span></p>
<p>If magnesium nitrate has a molecular mass of 148.31 g/mol, then the mass can, again, be easily calculated by multiplication:</p>
<p><span class="math-container">$$
\begin{align}
m &= M \times n\\
\Rightarrow m &= 148.31\,\text{g/mol} \times 6.74 \times 10^{-2}\,\text{mol}\\
&= 10\,\text{g}
\end{align}
$$</span></p>
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https://chemistry.stackexchange.com/questions/135405/stoichiometry-and-percentage-concentration
|
Question: <p>Why are figures removed from the calculation for values that are known in more detail? E.g. if you want to calculate the number of moles in 75.0 g H2 molecules. Why should one here use
for the relative atomic mass/molar mass of H 1,008 u or g/mol instead of 1,0079 u or g/mol?
What is the purpose of this?
In addition to that, should one round interim results like the substance amount, the molar mass or should one just round the final result?</p>
<p>Many thanks in advance</p>
Answer: <p>In my experience, mostly cause we don't care.</p>
<p>In my lab if I'm measuring 3mL - 12mL of a liquid I use a syringe that only has lines on the syringe every 0.2mL. So if I were to be measuring, say, 9mL, I would only have about 2 <a href="https://en.wikipedia.org/wiki/Significant_figures" rel="nofollow noreferrer">sig figs</a>. Also, to be honest, I sometimes don't really care if I'm off by a mL. So in doing the math I'd round 1.0079 to 1 because it's easier and the extra precision will be lost when I measure out the chemical anyway.</p>
<p>As for when to round. You should always keep one or two extra sig figs during the calculation and round at the end.</p>
|
https://chemistry.stackexchange.com/questions/136321/why-use-1-008-g-mol-instead-of-1-0079-g-mol-calculating-the-number-of-moles-in-7
|
Question: <blockquote>
<p>Tin(IV) iodide is prepared by direct combination of the elements.
Add 2.00 g of granulated tin to a solution of 6.35 g of iodine. Write an equation for this reaction.</p>
</blockquote>
<p>Attempt:</p>
<p><strong>1.</strong> Determine the amount of tin using <strong>moles = mass/molar mass</strong>. This yields a result of 0.017 mol <span class="math-container">$\ce{Sn}$</span>.</p>
<p><strong>2.</strong> Determine the amount of iodine using <strong>moles = mass/molar mass</strong>. This yields a result of 0.5 mol <span class="math-container">$\ce{I}$</span>.</p>
<p><strong>3.</strong> Determine the ratio of tin to iodine. Using relative amounts, for every 1 mol <span class="math-container">$\ce{Sn}$</span> there are 3 mol <span class="math-container">$\ce{I}$</span>, so the ratio is 1:3. Because iodine is a diatomic molecule, I will adjust accordingly, so the ratio becomes 2 mol <span class="math-container">$\ce{Sn}$</span> to 3 mol <span class="math-container">$\ce{I2}$</span> or 2:6.</p>
<p><strong>4.</strong> Construct the formula:</p>
<p><span class="math-container">$$\ce{2Sn + 3I2 -> SnI4}$$</span></p>
<p>with excess tin and iodine.</p>
<p>Is my procedure correct?</p>
Answer: <p>The balanced chemical equation is independent of the amounts of each reactant added. In this case we can easily deduce the equation to be: $$\ce{Sn + 2I2 -> SnI4}$$</p>
<p>What the given masses tell us is that there is excess tin being added. We can see this because the $\ce{Sn:I2}$ ratio added is $1:1.5$ whereas the stoichiometric ratio is $1:2$ and so there is less iodine than is required to completely react with the added tin - or in other words there is an excess of tin.</p>
|
https://chemistry.stackexchange.com/questions/35309/equation-for-reaction-of-tin-and-iodine-to-produce-tiniv-iodide
|
Question: <p>What is the liquid which has the most hydrogen atoms per volume at STP ?
Is there anything better than water?</p>
<p><a href="https://chemistry.stackexchange.com/questions/117073/most-dense-hydrogen-containing-composition">This answer</a> does not answer this question because none of the compounds listed in it are liquids at STP, which contain more hydrogen per volume than water.</p>
Answer: <p><strong>Hydrazine has more hydrogen atoms per mL than water</strong></p>
<p>Some simple calculations give the moles of hydrogen per mL in some possible alternative liquids. Taking into account density and molecular mass we get the following results:</p>
<p>Water 0.056 mol/mL 2 Hydrogens -> 0.11 mol H/mL
Hydrazine 0.0313 mol/mL 4 hydrogens -> 0.125 mol H/mL</p>
<p>symmetric dimethylhydrazine 0.0138 mol/mL 8 hydrogens -> 0.111 mol H/mL
methyl hydrazine 0.0191 mol/mL 6 hydrogens -> 0.115 mol H/mL</p>
<p>So, while some other liquid at STP hydrazine derivatives beat water, it looks like hydrazine has notably more hydrogens per unit volume. This is probably the record.</p>
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https://chemistry.stackexchange.com/questions/138783/what-is-the-liquid-which-has-the-most-hydrogen-atoms-per-volume
|
Question: <p>I calculated this:
molar mass (M) of Mg(NO3)2 is 148,32 (24,3 + 2<em>14,01 + 6</em>16)
so when there is 7 grams Mg(NO3)2 there is 0,0472 mol (7/148,32)</p>
<p>But how do i find out how much of that is oxygen? Do i just divide by 8 and multiply by 6? (because 8 atoms of with 6 are oxygen)</p>
Answer: <p>No. Each mole of the compound contains 6 moles of oxygen atoms, and has a mass of about 6 * 16 grams. You don’t have a mole, so use proportional reasoning to determine the actual mass in the sample.</p>
<p>To help to understand why it is simply a factor 6 and does not depend on the total number of atoms in the compound, perhaps imagine a chemical reaction that atomizes the compound:</p>
<p><span class="math-container">$$\ce{Mg(NO3)2 ->[atomizer] Mg + 2N + 6O}$$</span></p>
<p>Irrespective of the number of other atoms in the compound, there are six moles of oxygen atoms for every mole of magnesium nitrate.</p>
<blockquote>
<p>Do i just divide by 8 and multiply by 6? (because 8 atoms of with 6 are oxygen)</p>
</blockquote>
<p>No, that would not work because the other atoms are heavier or lighter than oxygen. You already incorporated the information about the other atoms in the compound when you divided by the molar mass of the compound.</p>
|
https://chemistry.stackexchange.com/questions/141359/how-much-grams-oxygen-is-in-7-gram-magnesiumnitrate-mgno32
|
Question: <p>The problem is:</p>
<blockquote>
<p>A quantity of <span class="math-container">$\pu{35.2 g}$</span> if a certain hydrocarbon gas, occupies <span class="math-container">$\pu{13.2 L}$</span>, measured at <span class="math-container">$\pu{1 atm}$</span> and at <span class="math-container">$\pu{323 K}$</span>. Knowing that 85.5% of is C, find the molecular formula for the hydrocarbon.</p>
</blockquote>
<p><strong>Attempt.</strong> So I tried with the first step doing <span class="math-container">$0.85\cdot \pu{35.2 g} = \pu{29.92 g}$</span> of C and I'm pretty sure that's the wrong step but I'm not fully sure, so anyways, I continue. After finding the mass of C on the hydrocarbure, I subtracted <span class="math-container">$35.2 - 29.92 = \pu{5.28 g}$</span> of <span class="math-container">$\ce{H}$</span> to get the mass of <span class="math-container">$\ce{C}$</span> on the hydrocarbon. Now basically:</p>
<p><span class="math-container">$$\pu{29.92 g} \text{ of C}\cdot \frac{\pu{1 mol}~\ce{C}}{\pu{12 g}~\ce{C}} = 2.5$$</span></p>
<p><span class="math-container">$$\pu{5.28 g} \text{ of H}\cdot \frac{\pu{1 mol}~\ce{H}}{\pu{1 g}~\ce{H}} = 5.28$$</span></p>
<p>and this is why I think I'm wrong, I can't get a relationship of natural numbers for the empirical formula. May anyone give some input on this?</p>
<p><strong>Continuation.</strong> So bravely approximating to the empirical formula by diving by the smaller one, i get <span class="math-container">$\ce{(C1H2)_n}$</span> so the empirical molar mass call it, <span class="math-container">$M_\text{empirical} = 12 + 2 = \pu{14g/mol}$</span>. Getting the <span class="math-container">$n$</span>:
<span class="math-container">$$n = \frac{M_\text{real}}{M_\text{empirical}} = \frac{M_\text{real}}{\pu{14g/mol}}$$</span>
and to get the <span class="math-container">$M_\text{real}$</span>, we go to
<span class="math-container">$$PV = nRT \implies PV= \frac{m}{M_\text{real}}RT \implies M_\text{real} = \frac{mRT}{PV}$$</span>
hence
<span class="math-container">$$n=\frac{\frac{mRT}{PV}}{\pu{14g/mol}} = \frac{\frac{\pu{35.2 g}\cdot \pu{0.082 L atm//K mol}\cdot \pu{323 K}}{\pu{1 atm}\cdot \pu{13.2 L}}}{\pu{14g/mol}} = 5$$</span>
then I get <span class="math-container">$\ce{(C1H2)_5}$</span> hence <span class="math-container">$F_\text{molecular}=\ce{C5H10}$</span>. Yes, right?</p>
Answer:
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https://chemistry.stackexchange.com/questions/143757/find-the-molecular-formula-of-a-gas
|
Question: <p>How does the balanced reaction look for these compounds? I found that phosphoric acid instead of hydrogen phosphate creates a much easier problem, but the teacher said hydrogen phosphate and I can't seem to balance it.</p>
Answer: <p>Most likely that your teacher meant to say phosphoric acid ($\ce{H3PO4}$) rather than the hydrogen phosphate anion ($\ce{HPO4^{2-}}$) so lets go with that.</p>
<p>$$\ce{ 3CaCO3 + 2H3PO4 -> Ca3(PO4)2 + 3CO2 + 3H2O}$$ </p>
<p>In reality the carbon dioxide and water would exist in an equilibrium state with carbonic acid, $\ce{H2CO3}$:</p>
<p>$$\ce{ H2CO3 (aq) <=> H2O (l) + CO2(aq)}$$</p>
|
https://chemistry.stackexchange.com/questions/5018/calcium-carbonate-and-hydrogen-phosphate-producing-calcium-phosphate-carbon-dio
|
Question: <p>My book gives me the following problem:
"A mixture of $125.0 g$ $\ce{N2}$ and $32.0 g$ $\ce{H2}$ reacts to $36.5 g$ $\ce{NH3}$. Calculate the efficiency."</p>
<p>My method to do this is - one I learned from my teacher, to put every given molecule's amount of substance (moles) in a table with 3 steps: start (S), gone (G)/formed (F) and rest (R).</p>
<p>$125g\:N_2 \leftrightarrow n=4.462$ moles</p>
<p>$32g\:\left(3\right)H_2\leftrightarrow n=16.13$ moles</p>
<p>\begin{matrix} &N_2&+&3H_2&\longrightarrow &2NH_3\\S&4.462&&16.13&&\\G&4.462&&13.386&F&8.924\\R&0&&2.74&&8.924\end{matrix}</p>
<p>($\rightarrow$ abundance of $H_2$)</p>
<p>Now, what is here the efficiency? Is it $\frac{36.5\:g}{8.924\:moles\cdot M_{NH_3}}\approx 24\%$?</p>
Answer: <p>Yes, that is completely correct. Let me summarize the approach.</p>
<p>If you look at the chemical reaction:
$$\ce{N2 + 3H2 \rightarrow 2 NH3}$$</p>
<p>You can see that you can, at best, make 2 mole of $\ce{NH3}$ per mole of $\ce{N2}$ (for the case here where $\ce{H2}$ is abundant). Given that you have $4.46\; moles$ of $\ce{N2}$ you could at best make $8.92\; moles$ of $\ce{NH3}$ which would be $151.9\; g$. Since you have made only $36.5\;g$ that means that your efficiency is $36.5/151.9=24\%$</p>
|
https://chemistry.stackexchange.com/questions/6640/stoichiometry-efficiency
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Question: <p>If $15.0\ \mathrm{mol}$ of nitrogen are reacted with $30.0\ \mathrm{mol}$ of hydrogen, how much ammonia will be produced?</p>
<p>$$\ce{N2 + 3H2 -> 2NH3}$$</p>
<p>What I’ve tried is the following: </p>
<p>$$\dfrac{n(\ce{N2})}{n(\ce{H2})}=\dfrac{1}{3}=\dfrac{15.0\ \mathrm{mol}}{x}$$</p>
<p>$$\dfrac{n(\ce{N2})}{n(\ce{H2})}=\dfrac{1}{3}=\dfrac{x}{30\ \mathrm{mol}}$$</p>
<p>I’m not sure how to proceed from here. Any help would be appreciated.</p>
Answer: <p>In the reaction:</p>
<p>$$\ce{N_2 + 3H_2 \to 2NH_3}$$</p>
<p>If 15 moles of $\ce{N_2}$ is used along with 30 moles of $\ce{H_2}$, then $\ce{H_2}$ will be the limiting reagent (Reactant which is consumed completely). This is because 15 moles of $\ce{N_2}$ will need 45 moles of $\ce{H_2}$ ($15 \times 3$, by the balanced reaction) for reacting completely, but as there is only 30 moles available hence it will react in a fixed proportion, hence you will have to assume the reacting moles of $\ce{N_2}$ as $x$. By this the equation goes,</p>
<p>$$\frac{\ce{N_2}}{\ce{H_2}}=\frac{1}{3}=\frac{x}{30}$$
From this
$$x=\frac{30}{3}=10$$</p>
<p>Hence from 15 moles of $\ce{N_2}$ only 10 moles will react, hence the amount of $\ce{NH_3}$ formed will be 20 moles ($10 \times 2$, by the balanced reaction).</p>
|
https://chemistry.stackexchange.com/questions/6716/amount-of-substance-produced-in-chemical-reaction
|
Question: <p>$$C_x H_y O_z$$ </p>
<p>1.What are the units of x,y,z.I've read that they're termed as 'stoichiometric coefficients'.</p>
<p>2.What is the unit of mass proportion of carbon(c) in the fuel?</p>
<p>$$c=\frac{M_{carbon}}{M_{fuel}} .x$$</p>
Answer: <p>The stoichiometric coefficients is something else than you suggest.</p>
<p><a href="https://www.boundless.com/chemistry/definition/stoichiometric-coefficient/" rel="nofollow noreferrer">stoichiometric coefficients</a>:</p>
<p>The number of molecules of a given component that participate in the reaction as written.</p>
<p>The indexes <span class="math-container">$x, y, z$</span> represent quantities of atoms in a compound. So they do not have any units.</p>
<p>@Q2 Do you know what are <span class="math-container">$c, M$</span> and <span class="math-container">$x$</span> in the equation? If you do, you can answer your question. If you do not, than you should find it out.</p>
|
https://chemistry.stackexchange.com/questions/7727/units-in-stoichiometry
|
Question: <blockquote>
<p>There are two gases in a container: krypton and carbon dioxide. If the mass of the gases is 35 grams, and total pressure of the container is 0.708 atm, and the pressure of krypton is 0.250 atm. What is the mass of the krypton?</p>
</blockquote>
<p>I've found the mole fraction of krypton is 0.3531. First, I think if I multiply mole fraction of krypton and mass of the gases, I'd get the mass of krypton. But my teacher said if the answer is false. So what is the correct answer?</p>
Answer: <p>Let $m_{\ce{CO2}}$ be the mass of carbon dioxide and $m_{\ce{Kr}}$ be the mass of krypton in the vessel. These two are related by:</p>
<p>$$m_{\ce{CO2}}+m_{\ce{Kr}}=35\ \text{g}\ \ \ \ \ (1)$$</p>
<p>If we can get a second equation in both variables, you can solved the system for each variable.</p>
<p>You were able to determine the mole fractions $\chi$:</p>
<p>$$\chi_{\ce{Kr}}=\frac{n_{\ce{Kr}}}{n_T}=\frac{P_{\ce{Kr}}}{P_T}=\frac{0.250\ \text{atm}}{0.708 \ \text{atm}}= 0.353\ \ \ \ \ (2)$$</p>
<p>$$\chi_{\ce{CO2}}=1-\chi_{\ce{Kr}}=0.647 \ \ \ \ \ (3)$$</p>
<p>Let $n_{\ce{CO2}}$ be the number of moles of carbon dioxide and $n_{\ce{Kr}}$ be the the number of moles of krypton in the vessel (its a little easier to work in moles). We can generate two relationships between the numbers of moles using 1) the masses and 2) the mole fractions.</p>
<blockquote>
<p>From the masses, since the mass of a sample of substance is equal to the number of moles times the molar mass (or formula weight).</p>
</blockquote>
<p>$$m_x=n_x\cdot FW_x\ \ \ \ \ (4)$$</p>
<p>$$m_\ce{Kr}=n_\ce{Kr}\cdot FW_\ce{Kr}=n_\ce{Kr}\cdot (83.798\ \text{g/mol})\ \ \ \ \ (5)$$
$$m_{\ce{CO2}}=n_{\ce{CO2}}\cdot FW_{\ce{CO2}}=n_{\ce{CO2}}\cdot (44.01\ \text{g/mol})\ \ \ \ \ (6)$$
$$n_{\ce{CO2}}\cdot (44.01\ \text{g/mol})+n_\ce{Kr}\cdot (83.798\ \text{g/mol})=35\ \text{g}\ \ \ \ \ (7)$$</p>
<blockquote>
<p>From mole fractions</p>
</blockquote>
<p>$$\dfrac{\chi_{\ce{CO2}}}{\chi_{\ce{Kr}}}=\dfrac{\dfrac{n_{\ce{CO2}}}{n_T}}{\dfrac{n_{\ce{Kr}}}{n_T}}=\dfrac{n_{\ce{CO2}}}{n_{\ce{Kr}}}=\dfrac{0.647}{0.353}=1.832\ \ \ \ \ (8)$$
$$n_{\ce{CO2}}=1.832\cdot n_{\ce{Kr}} \ \ \ \ \ (9)$$</p>
<blockquote>
<p>Substitute equation (9) into equation (7) and solve for $n_{\ce{Kr}}$ and convert to $m_{\ce{Kr}}$. </p>
</blockquote>
|
https://chemistry.stackexchange.com/questions/7899/calculating-mass-of-krypton-from-mole-fraction
|
Question: <p>I'm doing revision questions and want to double check something. The opening information reads: </p>
<blockquote>
<p>Two chlorides of iron were prepared. One was prepared by reacting iron with dry chlorine
gas. 4.50 g of iron reacted with chlorine gas to produce 13.01 g of the chloride.
The other chloride was prepared by reacting iron with dry hydrogen chloride; 2.80 g of iron reacted with hydrogen chloride to produce 6.35 g of the chloride.</p>
</blockquote>
<p>The questions are then:</p>
<blockquote>
<p>(i) Calculate the mass of iron that reacted with 2.80 g of iron in each chloride.<br>
(ii) Find the ratio of the different masses of chlorine that combined with 2.80 g of iron.<br>
(iii) Which of the chemical combination laws is demonstrated using this data?</p>
</blockquote>
<p>Is that first question right? I would've thought they're asking to calculate the mass of chlorine, which you'd simply get by subtracting the mass of reactant iron from the mass of the end product.</p>
Answer: <blockquote>
<p>Two chlorides of iron were prepared.</p>
</blockquote>
<p>I do <strong>not</strong> read that as: $\ce{FeCl_{n}}$ was prepared by two different methods, but as:</p>
<ol>
<li>Reacting iron with dry chlorine gas yields $\ce{FeCl_{n}}$</li>
<li>Reacting iron with hydrogen chloride yields $\ce{FeCl_{m}}$</li>
</ol>
<p>Apparently, iron is oxidized in both cases. </p>
<ul>
<li>What are the oxidants (what is reduced)?</li>
<li>Which of them is the stronger oxidant?</li>
<li>What is the stoichiometry (what are <strong>n</strong> and <strong>m</strong>)?</li>
</ul>
<p>Could it be helpful to divide the amount of chloride produced by the amount of iron used? Would different numbers indicate which of the products has a higher chloride content? </p>
<blockquote>
<p>(i) Calculate the mass of iron that reacted with 2.80 g of iron in each chloride.</p>
</blockquote>
<p>Is it possible that they meant the same as indicated above, i.e.:</p>
<p>Calculate the amount of iron chloride that would have been produced from 2.80 g (instead of 4.50 g) iron in the first experiment and compare with the outcome of the second experiment.</p>
|
https://chemistry.stackexchange.com/questions/9958/laws-of-chemical-combination
|
Question: <p>I need to write balanced equations describing the following reactions:</p>
<ul>
<li>one mole of $\ce{Al2Me6}$ with two moles of water</li>
<li>excess of $\ce{Al2Me6}$ with silicon dioxide</li>
<li>excess of $\ce{Al2Me6}$ with tin(IV) chloride</li>
</ul>
<p>My answer so far:</p>
<ul>
<li>a) $\ce{2H2O + Al2Me6 -> Al2Me4(H2O)2}$?</li>
<li>b) $\ce{SiO2 + Al2Me6 -> SiMe4 + Al2Me2O2}$?</li>
<li>c) $\ce{SnCl4 + Al2Me6 -> SnMe4 + Al2Me2Cl4}$?</li>
</ul>
Answer: <p>For a): Many organometallic compounds are easily hydrolyzed to the respective hydrocarbons and metal (hydr)oxides. In this case, the products are $\ce{CH4}$ and more likely $\ce{Al2O3}$, because the formation of $\ce{Al(OH)3}$ requires more moles of water. </p>
<p>Reactions b) and c) should be transmetallations where the methyl groups are transferred from aluminium to the other metal. The excess of $\ce{Al2Me6}$ helps to shift the equilibrium of the reaction to the product side. $\ce{SiMe4}$ and $\ce{SnMe4}$ would be the correct products which you have identified, but aluminium is also converted into its respective oxide/chloride.</p>
|
https://chemistry.stackexchange.com/questions/10135/reactions-involving-trimethylaluminium-al2me6
|
Question: <p>here is a question that i don't know how it is solved :</p>
<p>A sample that is 75 % chloride by mass is dissolved in water and treated with an excess of AgNO3. If the mass of the AgCl precipitate that forms is 2.013 g, what was the mass of the original sample?</p>
<p>I hope you give me the the proper way of solving it </p>
Answer: <p>It is first of all important to recognise that all the chloride in the original sample will precipitate upon treatment with excess AgNO3. This will give you a starting point, which is a measurable amount of AgCl.</p>
<p>The molecular mass of AgCl = (107.87+35.45)=143.32g</p>
<p>Therefore, the proportion of chloride in AgCl = (35.45/143.32) = 0.2473</p>
<p>Therefore in 2.013g of AgCl, there is (0.2473 x 2.013) = 0.4978g of chloride.</p>
<p>If 0.4978g of chloride constitutes 75% of the mass of the original sample, then the amount of original sample is (0.4978 / 0.75) = 0.6637g</p>
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https://chemistry.stackexchange.com/questions/13751/whats-the-mass-of-the-original-sample-g
|
Question: <p>the question says :</p>
<p>Two elements, A and B, combine to form two binary compounds. In the first
compound, 3.5 g of A combines with 8.00 g of B. In the second compound,
5.0 g of A combines with 17.1 g of B. If the formula of the first compound is
AB2, then the formula of the second compound would be ?</p>
<p>I can't even make the first step :(</p>
Answer: <p>For a molecule AB2, 3.5g of A represents one molar fraction, and 8.00g B represents 2 molar fractions (or 4.0+4.0). Therefore, a direct ratio can be given as 3.5:4.0, or 1:1.14. This means a molecule A<sub>n</sub>B<sub>m</sub> will give a mass ratio for A:B of n:1.14xm</p>
<p>For a molecule AB, for every 1g of A, you will have 1.14g of B.</p>
<p>For a molecule AB<sub>2</sub>, for every 1g of A, you will have 2.28g of B.</p>
<p>For a molecule A<sub>2</sub>B<sub>3</sub>, for every 1g of A, you will have (1.14x3/2) 1.71g of B.</p>
<p>etc etc</p>
<p>So for 5.0g of A, in order to have 17.1g of B, you will have 1.14 x (m/n) = 17.1/5</p>
<p>=> m/n = 17.1 / 5 / 1.14 = 3</p>
<p>This means m/n=3, or AB3 (alternatively it could also be A2B6)</p>
|
https://chemistry.stackexchange.com/questions/13752/use-the-given-information-to-write-the-formula-of-this-compound
|
Question: <p>A sample of copper(II) sulfate pentahydrate (CuSO4·5H2O) contains 0.360 g of water.
What is the total number of atoms in the compound</p>
<p>what's the idea of this question?</p>
<p>what i know is that atoms number is found by multiplying the Avogadro number by the mol's number</p>
<p>??</p>
Answer: <p>You have to calculate the amount of substance with the mass of water.</p>
<p>$n[H_2O]=\frac{m}{M}=\frac{0.360\,g}{18\,g} \cdot mol = 0.02\, mol$</p>
<p>$1~ n[CuSO_4\cdot 5 H_2O] = 5~n[H_2O]$</p>
<p>$1~ n[CuSO_4\cdot 5 H_2O] = 0.004\, mol$</p>
<p>In 0.004 mol copper sulfate pentahydrate you have 0.36 g water.</p>
<p>1 molecule of copper sulfate pentahydrate is composed of 21 atoms and the Avogadro constant is $N_A = 6.022 \cdot 10^{23}\, mol^{-1}$.</p>
<p>Now you can simply multiply all values.</p>
<p>$\text{#}Atoms = 0.004\, mol \cdot 6.022 \cdot 10^{23}\, mol^{-1} \cdot 21 = \underline{5.06 \cdot 10^{22}}$ </p>
|
https://chemistry.stackexchange.com/questions/13769/a-sample-of-copperii-sulfate-pentahydrate-cuso4-5h2o-contains-0-360-g-of-wat
|
Question: <blockquote>
<p>A simple mixture of $\ce{NaCl}$ and $\ce{NaBr}$ weighing $0.180~ \mathrm{g}$ is treated with $\ce{AgNO3}$ solution to give $0.3715~ \mathrm{g}$ of precipitate. Calculate the content of $\ce{NaCl}$ and $\ce{NaBr}$ in the mixture.</p>
</blockquote>
<p>The answer given in book are $\ce{NaCl}$ = $0.0682~ \mathrm{g}$ and $\ce{NaBr}$ = $0.1118~ \mathrm{g}$.</p>
<p>The answers I got are $0.1519~ \mathrm{g}$ $\ce{NaCl}$ and $0.0281~ \mathrm{g}$ $\ce{NaBr}$. Please tell me where I made mistake.</p>
<p><strong>Working:</strong></p>
<p>$$\ce{NaCl + AgNO3 -> NaNO3 + AgCl}$$
$58.5~ \mathrm{g}$ $\ce{NaCl}$ give $143.5~ \mathrm{g}$ $\ce{AgCl}$</p>
<p>$0.3715~ \mathrm{g}$ $\ce{AgCl}$ give $58.5/143.5 \cdot 0.3715~ \mathrm{g}$ $\ce{NaCl}$ = $0.1519~ \mathrm{g}$ $\ce{NaCl}$</p>
<p>$\ce{NaBr}$ = $0.180 ~ \mathrm{g}$ - $0.1519~ \mathrm{g}$ = $0.0281~ \mathrm{g}$ $\ce{NaBr}$</p>
Answer: <p>x = number of moles of chloride
y = number of moles of bromide</p>
<p>Since you know the mass of the original $\ce{NaCl}$/$\ce{NaBr }$mixture and the molar masses of $\ce{NaCl}$ and $\ce{NaBr}$, you can write an equation of the form:</p>
<p>$58.44 \pu{\frac{g}{mol}}~ x + 102.89 \pu{\frac{g}{mol}}~ y = 0.180~ \pu{g}$</p>
<p>Based on this idea, you can write a similar equation for the final products $\ce{AgCl}$ and $\ce{AgBr}$. With that in hand you will then have a system of equations you can solve (for x and y) that can be converted back into the masses of those salts.</p>
|
https://chemistry.stackexchange.com/questions/18756/given-the-mass-of-precipitate-how-can-i-calculate-the-relative-amount-of-the-or
|
Question: <blockquote>
<p><strong>Question</strong><br>
If a metallic oxide has $40\,\%$ oxygen, find the equivalent weight of
the metal.</p>
</blockquote>
<p>This amounts to finding the atomic weight of the metal and the charge on the cation (in effect identifying the metal itself). </p>
<p>I didn't know how to do this, so I looked up the solution. It goes as follows. </p>
<blockquote>
<p><strong>Solution</strong><br>
<em>Assume the oxide has the formula $\ce{MO}$</em>. Let the atomic weight of $\ce{M}$ be $x$
Then
$$\frac{16}{x + 16} = \frac{40}{100}$$
Simplifying and solving for $x$ yields
$$x = 24$$
Thus the metal is magnesium and the equivalent weight is $12$. </p>
</blockquote>
<p>My problem is, how can we assume that the metallic oxide has the formula $\ce{MO}$? For all we know, the metallic oxide may be $\ce{M2O}$ or $\ce{M2O3}$. In fact, if we assume the latter formula we get</p>
<p>$$\frac{48}{2x + 48} = \frac{40}{100}$$</p>
<p>giving</p>
<p>$$x = 72$$</p>
<p>which I believe is Hafnium. </p>
Answer: <p>Your question is a good one, and no, you can't assume that the metal is $\ce{MO}$ necessarily. However, you can find the equivalent weight of the metal without <em>any</em> assumptions about its formula. </p>
<p>The equivalent weight of a metal combining with oxygen is the mass that reacts with 8g of oxygen. We know that the compound contains 40% oxygen, so it must contain 60% metal. Suppose we had 100g of this compound; that would be 40g O and 60g M. </p>
<p>$40g\ \ce{O}:60g\ \ce{M}$ (Divide by 5)<br>
$8 g\ \ce{O}:12g\ \ce{M}$</p>
<p>The equivalent weight of the metal must be 12g. </p>
<p>If you had been asked to identify the metal you would have needed the formula (or would have needed to test the likely formulas to see if they produced reasonable results.)</p>
<p>Given an equivalent weight for M of 12g:<br>
$\ce{M2O-> 12g/mol, carbon}$ (not a metal)<br>
$\ce{MO-> 24g/mol, magnesium}$<br>
$\ce{M2O3-> 36g/mol, chlorine}$ (not a metal)<br>
$\ce{MO2->48g/mol, titanium}$</p>
<p>Both magnesium and titanium fit the information that was given, $\ce{MgO}$ and $\ce{TiO2}$ are common oxides of those metals.</p>
|
https://chemistry.stackexchange.com/questions/18939/what-does-a-metallic-oxide-has-40-oxygen-mean
|
Question: <p>If 2.30 mol of sodium reacts with 1.95 mol of water, how many moles of sodium hydroxide are produced? </p>
<p>Would it be 2.3 mol NaOH?</p>
Answer: <p>The first step is to figure out a reaction equation.</p>
<p>[I'm assuming elemental] sodium reacts with water violently because it releases a lot of energy when its only valence electron gets ripped apart. It creates \ce{NaOH} and \ce{H2}.</p>
<p>$\ce{Na + H2O -> NaOH + H2}$</p>
<p>We need to balance this.</p>
<p>$\ce{2Na + 2H2O -> 2NaOH + H2}$</p>
<p>$\ce{Na}$ and $\ce{H2O}$ are consumed at equal rates, so the $1.95$ moles of $\ce{H2O}$ are going to be used up completely (leaving $2.30 - 1.95 = 0.35$ moles of $\ce{Na}$ not reacted).</p>
<p>$\ce{NaOH}$ is formed at the same rate of consumption of the other two, so it's going to have $1.95$ moles produced.</p>
<p>--</p>
<p>Mathematically...</p>
<p>$\frac{2.30mol_\ce{Na}}{2mol_\ce{Na}} = 1.15$</p>
<p>and</p>
<p>$\frac{1.95mol_\ce{water}}{2mol_\ce{water}} = 0.975$</p>
<p>Effectively, water burns out faster in the reaction than sodium. So, we have to base our calculations off of moles of water.</p>
<p>For every 2 moles of water used, 2 moles of sodium hydroxide are formed.</p>
<p>$1.95mol_\ce{water} * \frac{2mol_\ce{NaOH}}{2mol_\ce{water}} = 1.95mol_\ce{NaOH}$</p>
|
https://chemistry.stackexchange.com/questions/19555/limiting-reactant-and-percent-yield
|
Question: <p>As part of a pre-lab exercise in Chemistry, we went through the necessary calculations to identify an unknown metal in a metal carbonate. The method was gas evolution by means of hydrochloric acid (see equation below). </p>
<p><span class="math-container">$$\ce{HCl + M2CO3 -> MCl + CO2 + H2O}$$</span></p>
<p>The unknown metal was Alkali, and we had three possibilities: Lithium, Sodium and Potassium as <span class="math-container">$\ce{M}$</span>.</p>
<p>Using data given to us, we first converted grams of released <span class="math-container">$\ce{CO2}$</span> to moles. Then here was the part I was confused on-</p>
<p>To find the molar mass of the metal carbonate, we took the mass of the sample used (the metal carbonate) and divided it by moles of <span class="math-container">$\ce{CO2}$</span> released, which was just calculated. The rest here on out was a little algebra. What is the reasoning behind dividing the mass of the sample by released <span class="math-container">$\ce{CO2}$</span>? What makes it "legal"? Yes, in the balanced equation, the ratio of <span class="math-container">$\ce{M2CO3}$</span> to <span class="math-container">$\ce{CO2}$</span> was <span class="math-container">$1:1$</span>.</p>
Answer: <blockquote>
<p>$$\ce{HCl + M2CO3 -> MCl + CO2 + H2O}$$</p>
</blockquote>
<p>Since you already figured out, that there is a one-to-one ratio of carbonate to carbon dioxide, you can simply write
$$n(\ce{CO2}) = n(\ce{M2CO3}).$$</p>
<p>With the formula for the Molar mass,
$$M(\ce{M2CO3}) = \frac{m(\ce{M2CO3})}{n(\ce{M2CO3})},$$
you can simply substitute one for the other, hence
$$M(\ce{M2CO3}) = \frac{m(\ce{M2CO3})}{n(\ce{CO2})}.$$</p>
|
https://chemistry.stackexchange.com/questions/24246/how-to-find-the-molar-mass-of-an-unknown-metal-carbonate-through-a-gas-evolution
|
Question: <p>What mass of precipitate forms when a solution containing 6.24 g of potassium sulfide is reacted with a solution containing 19.2 g of barium nitrate?</p>
<p>I have already identified the limiting reagent $\left(\text{K}_2 \text{S}\right)$ as well as the mass of the precipitate.</p>
<p>My question, however, is: why is the Barium Sulfide formed in the product a solid and not aqueous? I thought the problem translated into the molecular equation: $$\text{K}_2 \text{S}_{\text{(aq)}} + {\text{Ba}(\text{NO}_3)_2}_{\text{(aq)}} \ce {->} 2{\text{KNO}_3}_{\text{(aq)}} + \text{BaS}_{\text{(aq)}}$$</p>
<p>But according to the answer key I was given, $\text{BaS}$ is a solid precipitate, not aqueous. How can that compound be a solid if, according to solubility rules, all sulfides plus an alkali earth metal are soluble? Shouldn't it be aqueous, not solid? </p>
<p>Please help, thanks :)</p>
Answer: <p>So basically this problem is bad and incomplete (they tend to be at this level) in that it doesn't give you the concentrations/volumes of the reagents/solutions, so we end up making several assumptions. First of all, <strong>precipitates cannot be aqueous</strong>, by definition they are solids. So the way you end up solving this question is by assuming one of the products precipitates out completely, and it is the least soluble one that precipitates out (barium sulfide in this case). Unfortunately, this isn't exactly accurate.</p>
<p>In a real world scenario, given enough water you could have no precipitate at all, but most likely you would have a mix (not all of the barium sulfide would precipitate out of solution). At higher levels of chemistry (AP/IB -> post-secondary) you'll learn that none of these reactions (including the solubility ones) ever go to completion (some just go so far that you can assume they go to completion), and what you end up doing is using solubility constants to find out how much of each product is dissolved and/or precipitated. If you're interested in more info google "solubility constants" and "common ion effect".</p>
|
https://chemistry.stackexchange.com/questions/24656/limiting-reagent-stoichiometry
|
Question: <blockquote>
<p>One gram of Hydrogen reacts with exactly (almost) 8 grams of oxygen to produce $\ce{H2O}$. Another single gram of Hydrogen reacts with 16 grams of Oxygen to produce $\ce{H2O2}$. </p>
</blockquote>
<p>Can the ratio $\frac{m_O}{m_H}$ be determined from this information? If not, what else do chemists need to know to find this ratio?</p>
Answer: <p>Are you asking whether we can figure out the ratio of the mass of an oxygen atom to a hydrogen atom from this information? If so, then the answer is yes, assuming that you know that $\ce{H2O}$ and $\ce{H2O2}$ are the formulas for each compound, and that atoms exist and are indivisible in chemical reactions.</p>
<p>$$2n\space \rm{atoms \space H} = 1 \space g \space H$$
$$1n \space \rm{atoms \space O} = 8 \space g \space O$$</p>
<p>Divide the second equation by the first:</p>
<p>$$\frac{1n \space \rm{atoms \space O}}{2n\space \rm{atoms \space H}} = \frac{8 \space g \space O}{1 \space g \space H}$$</p>
<p>Cross multiply to get a 16:1 mass ratio. You can confirm this by looking at $\ce{H2O2}$, where the ratio of atoms is 1:1 and mass is again 16:1.</p>
<p>If you didn't know the formulae, then you could assume that the ratio was some integer <em>multiple</em> of 8:1, but without more data you wouldn't know which. This is because you would not know whether water had a 1:1 atom (or mole) ratio of $\ce{H}$ to $\ce{O}$, or if it was something else.</p>
<p>Incidentally, what you are describing is the <a href="http://en.wikipedia.org/wiki/Law_of_multiple_proportions" rel="nofollow">law of multiple proportions</a>, and it is one of the key findings that led to the development of <a href="http://en.wikipedia.org/wiki/Atomic_theory" rel="nofollow">atomic theory</a>.</p>
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https://chemistry.stackexchange.com/questions/25027/atomic-mass-ratios-is-the-problem-providing-enough-info
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Question: <p>A student mixed 260ml of 1.2 M lead(II) nitrate with 300ml of 1.90M potassium iodide. What is the final concentration of $\ce{NO3^{-}}$ ?</p>
<p>the answer is 1.11M , but he used the moles of lead(II) nitrate to find out the moles of $\ce{NO3^{-}}$ ,, and I think he should had used the moles of potassium iodide since it is the limiting reactant.</p>
<p>could anyone explain ?</p>
Answer: <blockquote>
<p>[…] he should had used the moles of potassium iodide since it is the limiting reactant.</p>
</blockquote>
<ul>
<li>Did any reaction take place upon mixing solutions of $\ce{Pb(NO3)2}$ and $\ce{KI}$?</li>
<li>If so, did the reaction change the number of moles of $\ce{NO3-}$ in solution by precipitation, oxidation, reduction, etc.?</li>
<li>Would the result of the calculation be different if just 300 mL of water were added to the $\ce{Pb(NO3)2}$ solution?</li>
</ul>
<p><strong>Update I</strong></p>
<p>Now for the rocket science (solubilities at 20 °C):</p>
<p>\begin{array}{cr}
\mathrm{compound} & \mathrm{solubility\ [g\cdot L^{-1}]} \\ \hline
\ce{Pb(NO3)2} & 522\\
\ce{KI} & 1430\\
\ce{KNO3} & 316\\
\ce{PbI2} & 0.76\\
\end{array}</p>
<p>The rather low solubility of $\ce{PbI2}$ suggests that on combination of the two solutions, the following may happen:</p>
<p>$$\ce{Pb^{2+} + 2 NO3- + 2 K+ + 2 I- -> 2 K+ + 2 NO3- + PbI2 v}$$</p>
<p>So again: </p>
<ol>
<li>Does this in any way change the absolute amount of $\ce{NO3-}$ <strong>in solution</strong>?</li>
<li>What does change for the concentration of $\ce{NO3-}$ when 300 mL of another solution is added to 260ml of 1.2 M $\ce{Pb(NO3)2}$ solution?</li>
</ol>
<p><strong>Update II</strong></p>
<ul>
<li>Initial $\ce{Pb(NO3)2}$ solution</li>
</ul>
<blockquote class="spoiler">
<p> 260 mL of a 1.2M solution of $\ce{Pb(NO3)2}$ contain $1.2\ \mathrm{\frac{mol}{L}}\ \cdot\ 2\cdot\ 0.26\ \mathrm{L} = 0.624\ \mathrm{mol}\ \ce{NO3-}$ <br></p>
</blockquote>
<ul>
<li>After mixing</li>
</ul>
<blockquote class="spoiler">
<p> Addition of 300 mL of the $\ce{KI}$ solution results in the precipitation of $\ce{PbI2}$, but <strong>does not change the absolute amount</strong> of $\ce{NO3-}$ in solution. However, it <strong>changes the total volume</strong> and hence the concentration, which calculates to $\frac{0.624\ \mathrm{mol}}{(0.26 + 0.3) \mathrm{L}} = 1.11 \ M$</p>
</blockquote>
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https://chemistry.stackexchange.com/questions/26726/calculate-the-final-concentration-of-this-equation
|
Question: <p>A question says:</p>
<p>Find the empirical formula of an organic compound from the following composition:</p>
<p>34.62% C, 3.88% H, 61.50% O.</p>
<p>The answer is $\ce{C3H4O4}$.
It was found by using the mass (percentages) divided by the molar mass of each element.</p>
<p>But they didn't consider the (2moles) of hydrogen and oxygen in calculations. Why?
in this case it will be $\ce{CH3O3}$</p>
<p>On the other hand, in a similar question in the text book, they consider the moles of hydrogen and oxygen in calculations by multiplying by 2 after getting the moles.</p>
<p>Could anybody explain?</p>
Answer: <blockquote>
<p>The answer is $\ce{C3H4O4}$. It was found by using the mass (percentages) divided by the molar mass of each element. </p>
</blockquote>
<p>Exactly! That's the way to do it.</p>
<blockquote>
<p>But they didn't consider the (2moles) of hydrogen and oxygen in calculations. Why? In this case it will be $\ce{CH3O3}$</p>
</blockquote>
<p>No, they did not! As you have pointed out yourself, it is about the empirical formula of a compound in question, such as one of the following:</p>
<p><img src="https://i.sstatic.net/82IkD.png" alt="enter image description here"></p>
<p>It is completely irrelevant whether an element usually exists as a diatomic molecule in nature. If that would matter, how would you then treat sulfur-containing compounds? Note that natural sulfur typically consists of $\ce{S8}$ molecules! How would you treat elements that only appear in the form of compounds in nature but never as pure elements.</p>
<p>In summary, the "natural appearance" of elements in nature does not matter here, it is only about the relative atomic abundance of an element in the compound in question. </p>
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https://chemistry.stackexchange.com/questions/26848/why-not-to-consider-hydrogen-and-oxygen-moles-to-determine-an-empirical-formula
|
Question: <p>A student analysed a hydrocarbon X and found it to be containing 43g carbon and 7.2g hydrogen.The relative molecular mass of X was found to be 42.What is the molecular formula of X?</p>
Answer: <p>$43\ \mathrm g$ of carbon for $7.2\ \mathrm g$ hydrogen.<br>
Molar mass of carbon is $12\ \mathrm{g/mol}$ and that of hydrogen is $1\ \mathrm{g/mol}$.</p>
<p>Hence there are $\frac{43\ \mathrm g}{12\ \mathrm{g/mol}} = 3.58\ \mathrm{mol} \approx 3.6\ \mathrm{mol}$ of carbon for every $\frac{7.2\ \mathrm g}{1\ \mathrm{g/mol}}=7.2\ \mathrm {mol}$ of hydrogen.</p>
<p>Hence the empirical formula of the compound is $\ce{CH2}$.</p>
<p>Relative molecular mass of $\ce{CH2}$ is $12+2\times1 = 14$. Given relative molecular mass is $42$.</p>
<p>$\displaystyle\frac{42}{14}=3$</p>
<p>Hence the formula of the compound is $\ce{C3H6}$</p>
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https://chemistry.stackexchange.com/questions/27046/the-molecular-formula-of-an-unknown-hydrocarbon
|
Question: <p>I was once told on this site that it was incorrect form to use units that specify the chemical being referred to in dimensional analysis. For example: </p>
<p>$$150.~\mathrm{g}~~\ce{KNO3} \cdot \frac{1~\mathrm{mol}~~\ce{KNO3}}{101.103~\mathrm{g}~~\ce{KNO3}} \cdot{} \frac{1~\mathrm{mol}~~\ce{C7H4O}}{6~\mathrm{mol}~~\ce{KNO3}} \cdot \frac{104.106~\mathrm{g}~~\ce{C7H4O}}{1~\mathrm{mol}~~\ce{C7H4O}}=25.7~\mathrm{g}~~\ce{C7H4O}$$</p>
<p>This is the way I learned it, and I feel that this clears up confusion that might otherwise arise when from performing such a calculation.</p>
<p>My overall question is: is it acceptable to do dimensional analysis as I presented it? If not, please explain why it is not correct to do it this way and show me what I should be doing instead.</p>
Answer: <p>The way I would perform the dimensional analysis you give in your question would be slightly different. That is mainly due to the fact that I don't consider chemical names or formulae as units, but as designators.</p>
<p>As an aside, I think you were starting from a question like the following:</p>
<blockquote>
<p>How much $\ce{C7H4O}$ would you have to weigh in for it to fully react
with $150~\mathrm{g}$ of $\ce{KNO3}$ if the reaction is described by
the following equation? $$ \ce{C7H4O + 6KNO3 -> Fumes^ + Tar} $$</p>
</blockquote>
<p>The result is $m_{\ce{C7H4O}}$, given is $m_{\ce{KNO3}} = 150~\mathrm{g}$.</p>
<p>The amount of $\ce{C7H4O}$ used per $\ce{KNO3}$ is
$$|\nu_{\ce{C7H4O}} / \nu_{\ce{KNO3}}| = \frac{1}{6}$$</p>
<p>As such, the result is composed as follows:
$$\begin{align}
n_{\ce{C7H4O}} &= \frac{1}{6} n_{\ce{KNO3}} \\
\tag{$n=m/M$}\frac{m_{\ce{C7H4O}}}{M_{\ce{C7H4O}}} &= \frac{1}{6} \frac{m_{\ce{KNO3}}}{M_{\ce{KNO3}}} \\ \hline
m_{\ce{C7H4O}} &= \frac{1}{6} \frac{m_{\ce{KNO3}}}{M_{\ce{KNO3}}} M_{\ce{C7H4O}}
\end{align}$$</p>
<p>If we now apply dimension analysis to what we think is the result, we can very quickly see how the two molar masses cancel each others units out and only the mass of $\ce{KNO3}$ remains to give the right hand side of the equation a dimension of $\mathrm{g}$. As the only symbol left on the left hand side is a $m$ we know that it needs a dimension $\mathrm{g}$ and as such we can happily plug in the numbers into our calculator for the result.</p>
<p>Frankly, I am so used to "my" method that I had to look at your example quite closely to see what was actually going on. To me that method just seems unwieldy, as you clutter up equation space with chemical formulae.</p>
<p>That being said; find the method that works the best for you (meaning it is robust enough to also solve your chemical engineering, biology and other problems using dimensional analysis) and stick to it.</p>
|
https://chemistry.stackexchange.com/questions/29711/dimensional-analysis-in-chemistry
|
Question: <blockquote>
<p>What is the amount of nitrate ions in <span class="math-container">$20.0\:\mathrm{g}$</span> of <span class="math-container">$\ce{Fe(NO3)3}$</span>?</p>
</blockquote>
<p>The chemical formula for the nitrate ion is <span class="math-container">$\ce{NO3}$</span>, I think.</p>
<p>The molar mass of <span class="math-container">$\ce{Fe(NO3)3}$</span> is <span class="math-container">$242~\mathrm{g~mol^{-1}}$</span> of which <span class="math-container">$186\ \mathrm{g\ mol^{-1}}$</span> belong to <span class="math-container">$\ce{NO3}$</span>. That means that around <span class="math-container">$76.85\:\%$</span> of the substance is nitrate ion.</p>
<p><span class="math-container">$20.0\ \mathrm g$</span> of the substance are equivalent to <span class="math-container">$20.0\ \mathrm g/242\ \mathrm{g\ mol^{-1}} = 0.0826~\mathrm{mol}$</span> of <span class="math-container">$\ce{Fe(NO3)3}$</span>.</p>
<p>Since roughly <span class="math-container">$76.85\:\%$</span> are nitrate ions, there are about <span class="math-container">$0.06~\mathrm{mol}$</span> of nitrate ions in those <span class="math-container">$20\ \mathrm g$</span> of substance.</p>
<p>One mole is <span class="math-container">$6.02\times 10^{23}$</span>, so if I multiply <span class="math-container">$(0.06\ \mathrm{mol})(6.02\times 10^{23}\ \mathrm{mol^{-1}}) = 3.612\times10^{22}$</span>.</p>
<p><strong>My Answer:</strong> There are <span class="math-container">$3.612\times10^{22}$</span> nitrate ions in <span class="math-container">$20.0\:\mathrm{g}$</span> of <span class="math-container">$\ce{Fe(NO3)3}$</span>.</p>
<p>However, that is <strong>wrong</strong>. The options in the website are:</p>
<ul>
<li><span class="math-container">$1.49\times 10^{23}$</span></li>
<li><span class="math-container">$4.98\times 10^{21}$</span></li>
<li><span class="math-container">$60.0$</span></li>
<li><span class="math-container">$8.25\times10^{21}$</span></li>
</ul>
<p>And my answer is not even close to any of them.</p>
<p>What did I do wrong?</p>
Answer: <p>The molar mass of iron (III) nitrate is $m(\ce{Fe(NO3)3}\approx 242~\mathrm{g\, mol^{-1}}$. Pay close attention to the unit.</p>
<p>You correctly calculated the amount of substance of iron (III) nitrate to be $n(\ce{Fe(NO3)3}= 0.0826~\mathrm{mol}$</p>
<p>Now you should ask yourself the question: How many nitrate ions are in one formula unit of iron (III) nitrate?</p>
<blockquote class="spoiler">
<p> There are three $\ce{NO3^-}$ per every $\ce{Fe(NO3)3}$.</p>
</blockquote>
<p>What does that mean for the number of moles of nitrate ions?</p>
<blockquote class="spoiler">
<p> It means that $n(\ce{Fe(NO3)3} = \frac13\cdot n(\ce{NO3^-})$, so you have to multiply the number of moles by three. $n(\ce{NO3^-}) = 0.248~\mathrm{mol}$</p>
</blockquote>
<p>Now you know the number of moles of nitrate ions and you simply have to multiply with Avogadro's constant.</p>
<blockquote class="spoiler">
<p> $N(\ce{NO3^-}) = n(\ce{NO3^-}) \cdot L = 1.49\cdot10^{23}$</p>
</blockquote>
|
https://chemistry.stackexchange.com/questions/33040/what-is-the-amount-of-nitrate-ions-in-20-g-of-feno33
|
Question: <p>The mass of one molecule of a compound is $2.19\times10^{-22}\ \mathrm g$. What is the molar mass of the compound?</p>
<p>My attempt:
Using the formula <strong>moles=mass/molar mass</strong>, I found how to calculate molar mass using the formula <strong>molar mass=moles/mass</strong>.
Therefore, I was under the impression that, surely, $6.02\times10^{23}$ divided by the mass of the compound would have resulted in the molar mass. However, the answer was wrong.</p>
Answer: <p>To get the molar mass of your mystery compound, make use of the fact that there are $N_{\rm A}$ (Avogadro's number) of molecules of it in one mole of it. </p>
<p>To solve the problem, we multiply the molecular mass (given) by Avogadro's number (known) to get the mass of one mole of the mystery compound:</p>
<p>$${\rm molar \;mass} = N_{\rm A}\times 2.19\times 10^{-22}\,{\rm g} = (6.022\times 10^{23}\,{\rm mol}^{-1}) \times (2.19\times 10^{-22}\,{\rm g}) = 131.88\;{{\rm g}\over{\rm mol}}$$</p>
|
https://chemistry.stackexchange.com/questions/35257/calculate-molar-mass-of-a-compound-from-a-given-molecular-mass
|
Question: <p>I will calculate the change in enthalpy of combustion of 1.12g of hexane.</p>
<p><strong>1)</strong> Calculate the energy transferred to 200g of water using the equation: <em>J = mass of water in grams (<strong>200g</strong>) x specific heat capacity of water (<strong>4.18 J/gK</strong>) x temperature increase in Kelvin (24C = <strong>297.15K</strong>)</em>. This results in the amount of energy transferred to the water of a total of 248,417.4J or <strong>248.4kJ</strong>.</p>
<p><strong>2)</strong> Calculate number of moles of hexane burnt using <em>n=m/M</em>. Weight of hexane before experiment = 222.07g. Weight of hexane after experiment = 220.92g. Therefore <strong>1.15g</strong> of hexane was burnt during the experiment. 1.15g of hexane = <strong>0.013mol</strong> of hexane.</p>
<p><strong>3)</strong> Find enthalpy change of reaction. Combustion of 1mol of hexane produces (by my calculations) <strong>-4194 kJ/mol.</strong> Therefore, 0.013mol will produce (0.013/-4194) <strong>-3.1x10-6 kJ/mol</strong>.</p>
<p>So this is the procedure that I used to find out the answer! Was it correct?
Because, apparently, you can use the values of steps 1 and 2 to find out the enthalpy change?</p>
Answer: <p>I can't figure out exactly what you are trying to calculate, but if you want to know how much energy is created when burning 1.15 g of hexane by measuring an temperature increase of water, then this is the right way:</p>
<p>1.) 0.2 kg Water was icreased by 24 °C:
U = 0.2 kg * 4187 J/(kgK) * 24 K = 20,100 J = 20.100 kJ </p>
<p>2.) n(Hexane) = 1.15 g / (86.18 g/mol) = 0.013 mol </p>
<p>3.) 1.15 g caused the temperature increase of 24 °C:
20.100 kJ / 0.013 mol = 1,510 kJ/mol</p>
<p>So you will get 1.510 MJ out of burning 1 mol (86.18 g) hexane. </p>
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https://chemistry.stackexchange.com/questions/36923/calculating-change-in-enthalpy-of-combustion-of-1-12g-of-hexane
|
Question: <blockquote>
<p>A <span class="math-container">$\pu{4.250 g}$</span> sample of <span class="math-container">$\ce{Na2SO4.nH2O}$</span> the sample loses <span class="math-container">$\pu{2.388 g}$</span> upon heating. What is <span class="math-container">$n$</span> for this hydrate?</p>
</blockquote>
<p>Do I first determine the mass of the water than the amount of substance of the sample and than the whole number ratio?</p>
Answer: <ol>
<li>Determine what the <span class="math-container">$\pu{2.388 g}$</span> is.
<blockquote class="spoiler">
<p> It's the amount of water lost from the crystal structure, assuming the sample was heated enough for this purpose.</p>
</blockquote></li>
<li>What does 'mole' mean? Isn't it just an indicator of the number of atoms/ions/molecules? So, what would be the relation between the quantity sodium sulfate and water molecules?
<blockquote class="spoiler">
<p> Yes, 'mole' is an indicator of quantity. Take a look at the formula: <span class="math-container">$\ce{Na2SO4.nH2O}$</span>. What does that mean? It means in a unit, there are <span class="math-container">$n$</span> water molecules trapped inside the crystal framework, which consists of sodium and sulfate ions.</p>
</blockquote></li>
<li>How do you proceed into finding the amount of substance (in mole)?
<blockquote class="spoiler">
<p> You need to somehow 'remove' any unknown variables to calculate the amount of substance, so you need to remove <span class="math-container">$n$</span>. Try subtracting the amount of water from the weight of the whole sample:<br>
<span class="math-container">$$m(\ce{Na2SO4.nH2O}) - m(\ce{H2O}) = m(\ce{Na2SO4})\\ \pu{4.250 g} - \pu{2.388 g} = \pu{1.862 g}$$</span></p>
</blockquote></li>
<li><p>How many grams of anhydrous <span class="math-container">$\ce{Na2SO4}$</span> are there? What is the formula to convert that into moles? How many moles of <span class="math-container">$\ce{Na2SO4}$</span> are there?</p>
<blockquote class="spoiler">
<p> The formula is <span class="math-container">$\displaystyle \mathfrak{n}=\frac{m}{M}$</span> where <span class="math-container">$m$</span> is mass, <span class="math-container">$M$</span> is <a href="https://en.wikipedia.org/wiki/Molar_mass" rel="nofollow noreferrer">molar mass</a> and <span class="math-container">$\mathfrak{n}$</span> is the amount of substance. Thus <span class="math-container">$\displaystyle \mathfrak{n}=\frac{\text{mass of anhydrous sample}}{\text{molar mass of sodium sulfate}} =\frac{1.862}{142} = \pu{0.0131 mol}$</span>.</p>
</blockquote></li>
<li><p>You found how many moles of sodium sulfate are there. Isn't that going to be the same for water? Plug into the formula; and find <span class="math-container">$n$</span>.</p>
<blockquote class="spoiler">
<p> We know that for in any unit of Gluber salt, there are 2 sodium ions, one sulfate ion, and <span class="math-container">$n$</span> molecules of water. Thus, it's safe to assume that <span class="math-container">$\mathfrak{n}$</span> for water is <span class="math-container">$n$</span> times that of what we found for sodium sulfate.<br>
<span class="math-container">$$\begin{align}0.0131\times n &=\frac{2.388}{18} \\ &= 0.132\bar{6} \\ \Rightarrow n&=10\end{align}$$</span></p>
</blockquote></li>
</ol>
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https://chemistry.stackexchange.com/questions/37592/how-to-calculate-the-number-of-water-molecules-of-a-hydrate-based-on-mass-loss-u
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Question: <p>We have to figure out what the water hardness in mg/L or ppm is for a $\pu{20ml}$ solution of $\pu{0.400M}$ $\ce{CaCl2}$. </p>
<p>We learned that the formula for hardness is mg/L of calcium carbonate per liter. I started by calculating that there would be $\pu{0.801 g}$ of $\ce{CaCO3}$ precipitate if reacted with $\ce{Na2CO3}$. I then converted $\pu{0.801 g}$ to mg and got $\pu{801 mg}$. I finally divides this by the original $\pu{20ml}$ or $\pu{0.02 L}$ but this gives $\pu{40050 mg/L}$ but the correct answer is $\pu{400 mg/L}$. </p>
<p>I feel like I might be making a mistake with units as my answer is 100 times greater than the correct answer but I don't see the problem.</p>
Answer: <p>Unless I'm missing something, I don't see the problem with your working either. Perhaps check whether you got the original concentration of $\ce{CaCl2}$ given in the problem right (the volume of the solution does not affect your final answer). The given answer may well be wrong, it's irritating but it does happen.</p>
<p>There is one thing I want to comment on though: <strong>never round off your answer too early.</strong> Since you obtained the intermediate answer of $801 \text{ mg}$, I am assuming you used a precise molar mass for $\ce{CaCO3}$ - I will take this to be $100.09 \text{ g mol}^{-1}$. (If you used the value $100$ you would have obtained $800 \text{ mg}$.) Your calculations should be:</p>
<ol>
<li>Number of moles of $\ce{CaCO3}$ formed upon addition of excess $\ce{CO3^2-}$</li>
</ol>
<p>$$\eta_{\ce{CaCO3}} = (0.400 \text{ mol dm}^{-3})(0.02 \text{ dm}^{3}) = 0.008 \text{ mol}$$</p>
<ol start="2">
<li>Mass of said $\ce{CaCO3}$</li>
</ol>
<p>$$m_{\ce{CaCO3}} = (0.008 \text{ mol})(100.09 \text{ g mol}^{-1}) = 0.80072 \text{ g} = 800.72 \text{ mg}$$</p>
<ol start="3">
<li>Hardness of water in $\text{mg L}^{-1}$</li>
</ol>
<p>$$[\ce{CaCO3}] = \frac{800.72 \text{ mg}}{0.02 \text{ L}} = 40036 \text{ mg L}^{-1}$$</p>
<p>This is your final answer and if you want to round to the appropriate number of significant figures, you should do so only at this point. Your problem is that you rounded too early (after step 2) and this means you obtained an imprecise final answer, $40050 \text{ mg L}^{-1}$.</p>
<p>Here, the appropriate number of sf is technically 3, since the original concentration of $\ce{CaCl2}$ is given to you in 3 sf. Since you obtained the imprecise answer $40050 \text{ mg L}^{-1}$, you would have rounded this to $40100$. But if you stuck to the precise value at step 2 and obtained the answer $40036 \text{ mg L}^{-1}$, you would round this to $40000$.</p>
|
https://chemistry.stackexchange.com/questions/37944/calculate-water-hardness-from-grams-of-caco3
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Question: <p>I am a bit confused on how to account for the stoichiometry of a reaction as follows:</p>
<p>$$\ce{A}(s) + \delta \ce{B}(g) \ce{->} \ce{C}(s) + \delta \ce{D}(g)$$</p>
<p>The solid $\ce{A}$ is stationary in the reactor, and gas $\ce{B}$ flows through it with known inlet molar flow rate. Since 1 mol of $\ce{A}$ produces $\delta$ mol of $\ce{D}$, but the solids are static, how can I determine the amount of $\ce{D}$ formed (molar flow rate)? The other known piece of information is the <em>normalized extent of reaction</em>, $\alpha$, which can be defined as the amount of $\ce{D}$ formed at time $t$ divided by its total amount.</p>
<p>Please let me know if it's not clear enough. It's been difficult for me to find clear explanations about the modeling of such type of reaction in the papers and theses I've read. The closest I've found to a definition of $\alpha$ is as follows:</p>
<p>$$\alpha = \int \frac{\ce{D}(t)}{\ce{D}_\text{total}} \mathrm{d}t$$</p>
<p>(yes, written exactly like that)</p>
<p><strong>My Approach So Far</strong></p>
<p>I believe I have to somehow consider both the flow rate of $\ce{B}$ and the amount of $\ce{A}$ in the reactor. If the molar flow rate of $\ce{B}$ is "too large", then the stoichiometric coefficient $\delta$ will limit the flow rate of $\ce{D}$. I've thought of doing something like this:</p>
<p>$$\dot{n}_{\ce{D}} = \min\{\dot{n}_{\ce{B}},\delta n_{\ce{A}}\}$$</p>
<p>where $\dot{n}_i$ is the molar flow rate of species $i$ and $n_\ce{A}$ is the number of moles of solid $\ce{A}$, but as you can see one of the arguments of the $\min$ operator is a molar flow rate and the other is just moles, which is problematic. Any suggestions?</p>
<p><strong>Edit</strong></p>
<p>Clearly, I have to use $\alpha$ somewhere. But I'm still unsure what to do here. Generically, I'd write something like,</p>
<p>$$\dot{n}_i = \dot{n}_{i,0} + \dot{n}_0 \nu_i \alpha$$</p>
<p>but I'm not sure if this is correct in this case (again, how do I use the amount of solid in the reactor?).</p>
Answer:
|
https://chemistry.stackexchange.com/questions/39498/stoichiometry-in-gas-solid-flow-system
|
Question: <p>The equation for the burning of octane:</p>
<p>$$\ce{2C8H18 + 25O2 -> 16CO2 + 18H2O}$$</p>
<ol>
<li>How do i find the number of moles of carbon dioxide which is produced when one mole of octane burns ?</li>
<li>How do i find , at what volume , at STP , is occupied by the number of moles determined in the answer to my first question ?</li>
<li>If the relative molecular mass of carbon dioxide is 44 , what is the mass of carbon dioxide produced by burning 2 moles of octane?</li>
</ol>
Answer: <p>This is a homework problem, so I'll not spell the answer outright, but try and give as much as I would if you came to me in office hours.</p>
<ol>
<li><p>Your equation tells you what the ratio of moles of $\ce{C8H18}$ to moles of $\ce{CO2}$. For every two moles of octane burned, how many moles of $\ce{CO2}$ are produced? Now, what would it be for only one mole of octane?</p></li>
<li><p>What have you tried so far on this? You have the ideal gas law (PV = nRT), and you know the P, the n from the first question, and the T - so just solve for the V.</p></li>
<li><p>So you know one mole of $\ce{CO2}$ masses 44 grams, so just multiply this by the number of moles of $\ce{CO2}$ produced by 2 moles of octane (see my hint on point 1 above).</p></li>
</ol>
|
https://chemistry.stackexchange.com/questions/40921/stoichiometric-calculations-for-the-combustion-of-octane
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Question: <p>The reason given for balancing chemical equations is the law of conservation of mass. They say that we balance equations with keeping in mind that atoms of every element included in the reaction must be equal on both sides of equations. And they just add some coefficients for this purpose. </p>
<p>But I apprehend why is the amount of substance not brought under consideration while doing this?</p>
<p>For example when we write,</p>
<p>$$\ce{H2 +O2->H2O}$$</p>
<p>That is; until we don't know how much of $\ce{H2}$ and $\ce{O2}$ was included in the reaction, how can start to balance it? How can we decide that atoms are not equal on both sides? However, the molecular formulas given above are only representing the substances. </p>
Answer: <p>When you write the equation</p>
<p>$$\ce{H2 + O2 -> H2O}$$</p>
<p>you are implicitly saying that the stoichiometric coefficient of every substance is $1$. It is like writing the equation $a + b = c$. By not having coefficients in front of $a$, $b$, or $c$, you are implying that $1a + 1b = 1c$ and not $2a + 3b = 4c$ or anything like that.</p>
<p>Obviously that cannot be true for the reaction above because if you react one $\ce{H2}$ molecule and one $\ce{O2}$ molecule, you get one $\ce{H2O}$ molecule and one leftover oxygen atom that doesn't know what to do with itself. The only balanced equation that does not lead to stray atoms is</p>
<p>$$\ce{2H2 + O2 -> 2H2O}$$</p>
<p>or any integer multiple of that, i.e.</p>
<p>$$\ce{$(2n)$H2 + $(n)$O2 -> $(2n)$H2O}$$</p>
<p>Humans prefer to work with small integers or half-integers, so $n$ is usually either chosen to be $1$ or $1/2$.</p>
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https://chemistry.stackexchange.com/questions/41390/validity-of-rationale-for-balancing-chemical-equations
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Question: <blockquote>
<p>Combustion analysis of a $\rm1.500~g$ sample of ascorbic acid yields $4.023\rm~g$ of $\ce{CO2}$ and $\rm0.96~g$ of $\ce{H2O}$. What is the empirical formula of ascorbic acid?</p>
</blockquote>
<p>Actually my question is that in many solutions I saw it's solved like that:</p>
<p>In $\ce{CO2}$ the moles of C is calculated as $$\frac{4.023}{44.02}=0.9139\rm~mass=1.0976~g$$</p>
<p>and the $\ce{H2O}$ is solved with getting out the H moles with same way that C was solved and the mass is
$$\rm m~H=0.10761~g$$
the mass of O will be $$\rm 1.500-(0.10761~g-1.0976~g)=0.29489~g~moles=0.1842$$
My question is why didnt we use the same procedure we did with H and C to compute the moles with O? Why did we make it in the last step?</p>
Answer: <p>You have done combustion analysis, so you have introduced a large amount of oxygen to the ascorbic acid sample. Because of this, you can't know how much of the oxygen in the $\ce{CO2}$ and $\ce{H2O}$ you produced actually came from the acid sample and how much came from the atmosphere.</p>
<p>But you do know that the mass of your ascorbic acid is the mass of all the carbon plus the mass of all the hydrogen, which you calculated, plus the mass of oxygen. The <em>difference in mass</em> between what you have calculated and what you started with is the mass of oxygen that was in the ascorbic acid.</p>
<p>If there were any other elements in your sample, such as sulphur or nitrogen, those amounts would also have to be calculated before finally working out how much mass you haven't accounted for which must have been oxygen.</p>
|
https://chemistry.stackexchange.com/questions/46408/why-werent-the-moles-of-oxygen-calculated-the-same-way-as-c
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Question: <p><strong>How many atoms of Lithium are there in 14.3 grams of Lithium Fluoride?</strong>
I can't seem to single out how many atoms of Lithium are in lithium fluoride.</p>
<p>I know that the first step is to convert 14.3 grams to moles, so 14.3 multiplied by the molar mass of LiF divided by 1 is 0.55 moles. But where do I go after this? Do I just multiply 0.55 by Avogadro's number?</p>
Answer: <p>Convert 14.3 grams of Lithium Fluoride to moles which is the same for lithium and lithium fluoride since there is one atom of lithium per molecule of lithium fluoride (carry an extra significant digit to round later to 3 significant figures). </p>
<blockquote>
<p>$\mathrm{moles(Li) = \dfrac{g(LiF)}{mw(LiF)} = \dfrac{14.3}{25.939} = 0.5513\text{ }moles(Li)}$</p>
</blockquote>
<p>Convert moles to atoms by multiplying with Avogadro's number, and round to 3 significant figures. </p>
<blockquote>
<p>$\ce{atoms(Li) = moles(Li)\text{ }(N_{a})} \mathrm{ = 0.5513(6.022E23) = 3.32E23}$</p>
</blockquote>
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https://chemistry.stackexchange.com/questions/47849/how-many-atoms-of-lithium-are-there-in-14-3-grams-of-lithium-fluoride
|
Question: <p>$\ce{2Ca3(PO4)2 +6SiO2 +10C->6CaSiO3 +P4 +10CO}$</p>
<p>What is the maximum amount of P4 that can be produced from 1.0 kg of phosphorite if the phosphorite sample is 75% $\ce{2Ca3(PO4)2}$ by mass</p>
<hr>
<p>The wording is confusing. I do not know what to do. Would you start by finding the limiting reactant? </p>
Answer: <p>First you find the number of moles of $\ce{Ca3(PO4)2}$ reacting. You have 75% of 1 kg, i.e. 750 grams, of calcium phosphate. The rest of reagents are assumed to be in excess as we have to find the maximum amount of $\ce{P4}$ obtained, hence reaction should not have other limiting reagents.</p>
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https://chemistry.stackexchange.com/questions/48157/maximum-amount-of-p4
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Question: <p>$\ce{2Ca3(PO4)2 +6SiO2 +10C->6CaSiO3 +P4 +10CO}$</p>
<p>What is the maximum amount of P4 that can be produced from 1.0 kg of phosphorite if the phosphorite sample is 75% $\ce{2Ca3(PO4)2}$ by mass</p>
<hr>
<p>I am confused. How do I find a maximum amount of a reaction</p>
Answer: <p>Phosphorite is a mineral containing calcium phosphate. This particular 1kg sample has 75% percent of it by weight i.e. 750 grams. All others reagents are assumed to be in excess as you have to find the maximum p4 obtained which would be at a point where all of our mineral is used.</p>
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https://chemistry.stackexchange.com/questions/48178/stoichiometry-assignment
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Question: <blockquote>
<p>Predict how the total pressure varies during the gas-phase reaction
$\ce{N_2}(g)+ 3 \ce{H2}(g) \rightarrow 2 \ce{NH3} (g) $ in a constant volume container. </p>
</blockquote>
<p>Reference: Atkin's Physical Chemistry, Tenth Edition, Chapter 20, Exercise 20A.1 (b) on page 870</p>
<p>My attempt at a solution: Assuming ideal gas behaviour we know that pressure is proportional to number of moles (volume and temperature held constant). Also, by observing the reaction stoichiometry the final pressure should be half the initial pressure as the no. of gas moles changes from 4 to 2</p>
<p>Initial no. of moles $n_i = n_{\ce{N2}} + n_{\ce{H2}}$ with no ammonia present.</p>
<p>at an arbitrary instant in the reaction $\alpha$ is the fraction of $\ce{N2}$ that has reacted. </p>
<p>So total no. of gas moles: $n = [n_{\ce{N2}}(1-\alpha)] + [n_{\ce{H2}}-3\alpha n_{\ce{N2}}]+ 2\alpha n_{\ce{N2}} $ ; the first two terms being the change in no. of moles of reactant and the last one being the no. of moles of ammonia produced.</p>
<p>This simplifies to, $n = n_{\ce{N2}} + n_{\ce{H2}}-2\alpha n_{\ce{N2}}$</p>
<p>What I think the question wants from me is a $P$ at an arbitrary instant given as a fraction of $P_i$ i.e $P_t= $$P_i$*[term depending on $\alpha$]. Also substituting $\alpha$ = 0 or 1 should yield $P_i$ and $\frac{1}{2}P_i$ respectively.</p>
<p>For that, I am guessing the expression for $n$ should be expressed as a fraction of $n_i$. However, I can't seem to think of the necessary algebra to obtain such an expression. </p>
<p><strong>EDIT:</strong> Inspiration struck, what if I do the following:</p>
<p>$n = n_{\ce{N2}} + n_{\ce{H2}}-2\alpha n_{\ce{N2}}$ </p>
<p>gives,</p>
<p>$n = (n_{\ce{N2}} + n_{\ce{H2}})(1-2\alpha \frac{n_{\ce{N2}}}{n_{\ce{N2}} + n_{\ce{H2}}})$</p>
<p>which is the same as, $(n_{\ce{N2}} + n_{\ce{H2}})(1-2\alpha .\chi_{\ce{N2}})$ where $\chi_{\ce{N2}}$ is the initial mole fraction of nitrogen in the reaction mixture</p>
<p>so, $n = n_i(1-2\alpha .\chi_{\ce{N2}})$
and consequently, $P_t = P_i(1-2\alpha .\chi_{\ce{N2}})$</p>
<p>Would this be a valid expression for describing the variation in pressure during the gas phase reaction described above?</p>
Answer:
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https://chemistry.stackexchange.com/questions/54547/variation-in-pressure-for-a-gas-phase-reaction
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Question: <p>Problem : An organic compound containing carbon, nitrogen and oxygen will have a weight ratio $9:1:3$. The molecular weight of the compound is $108$. What is the formula of the compound?</p>
<p>I thought of calculating the empirical formula, but percentages are given for that, and not weight. </p>
Answer: <p>It's an algebra problem.</p>
<p>108 = 9x + 1x + 3x</p>
<p>Solve for x.</p>
<p>Weight of carbon per molecule = 9x
Weight of nitrogen per molecule = 1x
Weight of oxygen per molecule = 3x</p>
<p>Then convert from weights to numbers (I assume you know how to do that :) ).</p>
<p>Note that the result is rather nonsensical, as it gives you fractional atoms.</p>
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https://chemistry.stackexchange.com/questions/57897/how-do-i-find-the-formula-of-the-compound-given-the-ratio-of-its-constituents-by
|
Question: <p>I know that the relation between Avogadro’s number and amu is a reciprocal relationship but the relation is slightly unclear.</p>
<p>Could anyone give me more clarification?</p>
Answer: <p>$\pu{1amu}$ is defined as one twelfth of the mass of one carbon-12 atom.</p>
<p>Avogadro's number is defined as the total number of entities in $\pu{12g}$ of carbon-12. </p>
<p>$N_\mathrm A=6.022\times10^{23}$</p>
<p>It is a measuring criteria, just like a dozen, which is used to put a number to a certain item. A dozen means 12 items (items like bananas, atoms, molecules, etc) and 1 mole means $6.022\times10^{23}$ items (items like bananas, atoms, molecules, etc).</p>
<p>So, $\pu{12g}\ \ce{^12C}$ contains $N_\mathrm A$ atoms</p>
<p>Therefore, 1 atom of carbon-12 weighs $\frac{12}{N_\mathrm A}\ \mathrm{g}$.</p>
<p>As per the defintion of atomic mass unit,
$$\pu{1amu}=\frac{1}{12}\frac{12}{N_\mathrm A}\mathrm{\ g}=\frac{1}{6.022\times10^{23}}\ \mathrm{g}$$</p>
<p>Hence, $\pu{1amu}=\pu{1.6\times10^{-27}kg}$</p>
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https://chemistry.stackexchange.com/questions/70135/what-is-the-relation-between-the-amu-and-avogadro-s-number
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Question: <p>My textbook had this question on balancing chemical reactions </p>
<blockquote>
<p>Ozone reacts with nitric oxide to give nitrogen dioxide and oxygen gas</p>
</blockquote>
<p>Here's how I balanced it: $$\ce{O3 + NO -> NO2 + O2}$$</p>
<p>However I realised that the equation is also balanced if I write it as: $$\ce{2O3 + 4NO -> 4NO2 + O2}$$</p>
<p>and when it is written as: $$\ce{3O3 + NO -> NO2 + 4O2}$$</p>
<p>This seems wrong as I am sure that there cannot be more than one way of balancing a reaction but I do not have a satisfactory explanation. What is going wrong here?</p>
Answer: <p>This happens when you have a <strong>sum of two</strong> (or more) independent reactions. You can balance each one, and then add them together in an <em>arbitrary</em> proportion. People usually run into this when trying to come up with an equation for combustion of gunpowder, where the oxidation of carbon and oxidation of sulfur are pretty much independent.</p>
<p>Now, in your case the reactions are:
$$\ce{2O3 -> 3O2}\tag A$$
and
$$\ce{2NO + O2 -> 2NO2}\tag B$$
Your first reaction is a combination of ${1\over2}A+{1\over2}B$, the second is $A+2B$, and the third is ${3\over2}A+{1\over2}B$. There are infinitely many possible combinations.</p>
<p>Ozone, being an extremely powerful oxidant, is known to participate in certain reactions where it spends <strong>one</strong> atom of oxygen and the rest falls off as the less reactive $\ce {O2}$, but this is not the case here. $\ce {O2}$ reacts with $\ce {NO}$ instantly, as well as $\ce {O3}$.</p>
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https://chemistry.stackexchange.com/questions/64202/more-than-one-way-of-balancing-a-chemical-equation
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Question: <p>How many liters of hydrogen, H2, are needed to react with 10 liters of nitrogen gas in the reaction forming ammonia?</p>
<p>$$\ce{3 H2(g) + N2(g) -> 2 NH3(g)}$$</p>
<p><strong>My try::</strong> Because we have 10 L of nitrogen gas, we have 10/22.4 moles = 0.446 moles of nitrogen gas, and thus need 0.446 * 3 = 1.338 moles of hydrogen gas. As such, we need 2.676 grams of hydrogen gas, or 2.676/0.09 = 29.8 L of hydrogen gas. However this is not correct. Can someone find my problem?</p>
<p>Note: This reaction occurs at STP.</p>
Answer: <p>You are on the right track. Dimensional Analysis is your friend:</p>
<p>$$\frac{10\,\text{L $\ce{N2}$}}{1} \cdot \frac{1\,\text{mol $\ce{N2}$}}{22.4 \,\text{L $\ce{N2}$}}\cdot \frac{3\,\text{mol $\ce{H2}$}}{1\,\text {mol $\ce{N2}$}}\cdot \frac{22.4\,\text{L $\ce{H2}$}}{1\,\text {mol $\ce{H2}$}}=\,?\,\, \text{L $\ce{H2}$}$$</p>
<p>If you do this, you will realize that the quantities cancel exactly and give you a nice clean number. </p>
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https://chemistry.stackexchange.com/questions/64333/how-many-liters-of-hydrogen-h2-are-needed-to-react-with-10-liters-of-nitrogen
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Question: <p>When diluting a mixture of two reactants with lets say a 100ml of H2O, how do you add this to the uncertainties of the concentrations of the reactants? Should the relative uncertainty of the added H2O be added to the relative uncertainty of the volume in both reactants? </p>
Answer: <p>When adding values together, you should add together their absolute uncertainties. For example, if you started with $500\pm5\mathrm{mL}$ and add $100\pm1\mathrm{mL}$, your new volume value is $600\pm6\mathrm{mL}$. </p>
<p>You would add the relative uncertainties if you are multiplying values. For example, if you started with $1\pm.06\text{ moles}$ and want to find the molarity in a solution with a volume of $500\pm5\mathrm{mL}$, your molarity will be $2\pm.14\text{M}$. The uncertainty was obtained by finding the sum of the relative uncertainties and multiplying it by the concentration to obtain an absolute uncertainty, as shown below. $$(\frac{.06}{1}+\frac{5}{500})(2\text{M})=\frac{7}{100}(2\text{M})=.14\text{M}$$ </p>
<p>For more details on uncertainty rules, this <a href="http://web.uvic.ca/~jalexndr/192UncertRules.pdf" rel="nofollow noreferrer">page</a> from the University of Victoria gives some of the basics.</p>
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https://chemistry.stackexchange.com/questions/71658/help-with-uncertainties
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Question: <ul>
<li>I was wondering how I'd go around calculating the equivalent weight
of CaO. From my knowledge, the EW of a compound is its (given mass
)/(valence factor, n). Since CaO has no charge and no change in<br>
the oxidation number takes place, my valence factor comes out to be
0, which I know is wrong because the compound has to have some EW, which is 28 as I know that 28grams of CaO will be formed by 8g O</li>
</ul>
Answer: <p>Yes, Dev, you know the correct process of calculating EW, but the problem you are facing is how to calculate the valence factor?, right?</p>
<p>So, in this case you have to understand that EW for any compound is reaction-specific. So, you can't calculate the EW of $\ce{CaO}$ without any reaction reference.</p>
<p>Still, usually $\ce{CaO}$ reacts through 2 electron exchange mechanism, in other words, $\ce{CaO}$, on dissociation, gives total cationic charge +2 and total anionic charge -2.</p>
<p>$$\ce{CaO = Ca^2+ + O^2-}$$</p>
<p>Hence, in this case of calculating EW, you have to consider $n=2$ and so the EW of $\ce{CaO}$ is $(40+16)/2 = 28$.</p>
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https://chemistry.stackexchange.com/questions/74410/equivalent-weight-for-cao
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Question: <p>I'm pretty new to chemistry and I've been stuck on it for hours.</p>
<p>Question: $\pu{10g}$ of the hydroxide of a metal on ignition gave $\pu{8g}$ of oxide. The equivalent weight of the metal is:<br>
a) $\pu{136g}$<br>
b) $\pu{40g}$<br>
c) $\pu{56g}$<br>
d) $\pu{28g}$ </p>
<p>I used the law of equivalence but the answer I got was $\pu{3g}$. what am I doing wrong?</p>
<p>How I went about it:<br>
10/E=8/16;<br>
E=20;<br>
E=M(equivalent mass of metal)+16+1;<br>
M=3</p>
Answer: <p>I found your numbers confusing, but I think the first equation you wrote is not correct. This is how I'd solve it:</p>
<p>Let's say we have $\mathrm{n}$ equivalents of metal in the sample, and let's call the equivalent weight of the metal $\mathrm{e}$. For the hydroxide sample, we have:</p>
<p>$n(e+17)=10$</p>
<p>where $\mathrm{e+17}$ is the molar mass of the hydroxide $\ce{MOH}$.</p>
<p>For the oxide, we have:</p>
<p>$n/2(2e+16)=8$</p>
<p>where $\mathrm{2e+16}$ is the molar mass of the oxide $\ce{M_2O}$. Note that $\mathrm{n}$ becomes $\mathrm{n/2}$ because of stoichiometry:</p>
<p>$\ce{2MOH->M2O + H2O}$</p>
<p>So you have a system of two equations with two unkowns ($\mathrm{n}$ and $\mathrm{e}$), which you can solve normally. For instance, if you divide the equations:</p>
<p>$\cfrac{n(e+17)}{n/2(2e+16)}=\cfrac{10}{8}$</p>
<p>$\cfrac{e+17}{e+8}=\cfrac{10}{8}$</p>
<p>$8e+136=10e+80$</p>
<p>$2e=56$</p>
<p>$e=28$</p>
<p>So the equivalent weight of the metal is 28g.</p>
<p>Makes sense?</p>
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https://chemistry.stackexchange.com/questions/76327/need-help-with-stoichiometry-problem
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Question: <blockquote>
<p>The equation below represents combustion of methane ($\ce{CH4}$, $\pu{16.04 g/mol}$). Balance the equation, and calculate the mass of water ($\pu{18.02 g/mol}$) formed when $\pu{40.0g}$ of methane is burned.
$$\ce{CH4(g) + O2(g) -> CO2 + H2O(g)}$$</p>
</blockquote>
<p>I have started the problem by balancing:</p>
<p>$$\ce{1CH4(g) + 2O2(g) -> 1CO2 + 2H2O}$$</p>
<p>I then continued by calculating the amount of substance of methane:</p>
<p>$$\ce{C ($\pu{12g/mol}$) + H ($\pu{4 \times 1g/mol}$) = $\pu{16 g/mol}$ \implies ($\pu{40 g}$ methane) / ($\pu{16 g/mol}$) = $\pu{2.5 mol}$}$$</p>
<p>From here I wold like to know how to convert or get to amount of substance of $\ce{H2O}$ to get the mass of $\ce{H2O}$ produced in this reaction.</p>
Answer: <p>You have already balanced the equation, so we can get the stoichiometric coefficients. From the equation we can say that 1 mole $\ce{CH4}$ gives 2 moles of $\ce{H2O}$ because their coefficients are 1 and 2, correspondigly. So 2.5 moles of $\ce{CH4}$ would give 5 moles of $\ce{H2O}$, that is $\pu{90 g}$ of $\ce{H2O}$.</p>
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https://chemistry.stackexchange.com/questions/76548/how-to-calculate-mass-produced-when-a-given-gas-is-burned
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Question: <p>A colleague said we can’t dissolve a salt (whose solvation enthalpy is exothermic) faster if we increase the temperature (the solubility equilibrium product is not reached) because Le Chatelier‘s principle would favor the reactants.</p>
<p>For example, imagine dissolving NaOH(s) in distilled water, can’t this be accelerated by elevating the temperature?</p>
<p>I read a similar question on here, see: <a href="https://chemistry.stackexchange.com/questions/91147/factors-that-influence-the-kinetics-of-an-irreversible-exothermic-reaction">Factors that influence the kinetics of an irreversible exothermic reaction</a></p>
<p>How I see it is the following:
As Le Chatelier‘s principle only applies to systems in equilibrium, and we only dissolve a salt (whose enthalpy happens to be exothermic), no equilibrium is established. The same goes for exothermic reactions in general. As long as no equilibrium is yet established, the process should be accelerated by an increase in temperature, regardless whether it’s exothermic or endothermic, as more molecules can overcome the activation energy barrier in general (for example in the Arrhenius equation).</p>
<p>It is important to note that this is about kinetics, that is, how quickly the NaOH dissolves, in a non-saturated solution, that is about 1M, not about how much NaOH a saturated solution can contain at a given temperature.</p>
<p>Is my reasoning correct?</p>
Answer: <p>Typically, rates of uncatalyzed simple reactions increase with temperature. There are well-known examples where this is not the case (such as enzyme-catalyzed reactions where the enzyme denatures at high temperatures, or reactions with an intermediate that is at rapid equilibrium with the reactant in an exothermic step).</p>
<p>In this case, if the <span class="math-container">$\ce{NaOH}$</span> is soluble at both temperatures, it is likely that the increasing rate in the forward direction will make a bigger difference that the also increasing rate in the reverse direction.</p>
<blockquote>
<p>As long as no equilibrium is yet established, the process should be accelerated by an increase in temperature, regardless if it’s exothermic or endothermic, as more molecules can overcome the activation energy barrier in general (for example in Arrhenius equation).</p>
</blockquote>
<p>In this case, the reaction goes to completion rather than attaining equilibrium (if you stay below the solubility limit). Even for reactions that attain equilibrium, however, the rate with which they approach the equilibrium constant is proportional to the <a href="https://en.wikipedia.org/wiki/Temperature_jump" rel="nofollow noreferrer">sum of the forward and reverse reaction</a> (for a one-step reaction closes to equilibrium, it is possible to show this in a quick derivation).</p>
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https://chemistry.stackexchange.com/questions/161619/reaction-kinetics-of-exothermic-reaction
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Question: <p><a href="https://i.sstatic.net/cMItP.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/cMItP.png" alt="enter image description here"></a></p>
<p>On an exam in organic chemistry I was asked why the reaction B proceeds faster than reaction A. </p>
<p>I answered that: in reaction B, I- is a better leaving group than Cl-, the reason being it is a weaker base. </p>
<p>The answer our teacher gave after correcting the exam was:
I- is a good nucleophile and at the same time a good leaving group. </p>
<p>I scored 1,5p/2p for my answer. </p>
<p>Are the two answers very different or is there some incorrect information in mine? </p>
Answer: <p>Your answer is simply missing something. Yes, I is a better leaving group, therefore an Sn reaction of an iodide proceeds faster than the one of the chloride. But your starting material in both cases is the chloride. You first need to exchange that chloride for an iodide and that only works well if the iodide is a good nucleophile.</p>
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https://chemistry.stackexchange.com/questions/100861/reaction-kinetics-organic-chemistry
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Question: <p>The addition of a catalyst is known to increase the rate constant of a reaction by providing an easier pathway for the reaction to occur, one with lesser activation energy. </p>
<p><em>My doubt is:</em></p>
<p>Is it possible that the addition of a catalyst, alters the <strong>order of a reaction</strong>?</p>
<p>I haven't come across any such reaction so far, so it'd be great if someone could provide an example as well; <em>if</em> the reaction order can be changed upon addition of catalyst</p>
<p><em>My thoughts:</em></p>
<p>Sometimes, catalysts participate in the reaction; but are regenerated in the end without change in mass and chemical composition.
The very fact that a catalyst can <em>participate</em> in a reaction, should mean that it can influence the reaction's order, right? </p>
Answer:
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https://chemistry.stackexchange.com/questions/86980/reaction-kinetics-and-catalyst-addition
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Question: <p><a href="https://i.sstatic.net/2srvk.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/2srvk.jpg" alt="enter image description here"></a></p>
<p>In this special reaction coordinate diagram with two reaction mechanisms, I tried to analyze it in two other ways, one with kinetics, another with equilibrium. </p>
<p>A. kinetics, I tried to find reaction rate of forward, and reverse. By the diagram, rate determining step of forward reaction is the first one (because activation energy is the highest and let's say it is one and only factor for RDS in this case). so Vf=k1[A][B].
RDS in reverse reaction is second step of the reaction judging by the diagram activation energy (M->A+B). So I used pre-equilibrium approximation for that. That's Vr in the picture.</p>
<p>B. equilibrium, I wrote all the equilibrium constant and simply added the reactions.</p>
<p>now here is the question. Expression of the rate of forward and reverse reaction in dynamic equilibrium is quite different from each case, analyzing in a way of kinetics and equilibrium constant, as you see in the picture below. Why is it different? What's right expression? What am I missing here? </p>
Answer: <p>You have chosen an unusual scheme in that A appears to be an intermediate as well as a reactant. If you use the simpler scheme things are easier and so rather than try to follow yours I have examined the scheme below which is very nonetheless very similar</p>
<p>$$RX\underset{k_{-1}} {\stackrel{k_1}{\leftrightharpoons}}R+X$$
$$R+Y\underset{k_{-2}} {\stackrel{k_2}{\leftrightharpoons}}RY$$</p>
<p>where species RX looses X and is replaced by Y. The intermediate species is R. The overall reaction is $RX+Y=RY+X$ and the equilibrium constant for the two steps are</p>
<p>$$K_1=\frac{k_1}{k_{-1}}=\frac{\mathrm{[R]_e[X]_e}}{\mathrm{[RX]_e}}$$
$$K_2=\frac{k_2}{k_{-2}}=\frac{\mathrm{[RY]_e}}{\mathrm{[R]_e[Y]_e}}$$</p>
<p>and the overall equilibrium constant</p>
<p>$$K=K_1K_2=\frac{k_{1}k_{2}}{k_{-1}k_{-2}}=\frac{\mathrm{[RY]_e[X]_e} }{\mathrm{[RX]_e[Y]_e }}$$</p>
<p>and all the concentrations in square brackets with subscript $e$ are equilibrium values. [This type of equation is true if there are many equilibria one after the other, $\displaystyle K=K_1K_2K_3K_4\cdots=\frac{k_1k_2k_3k_4}{k_{-1}k_{-2}k_{-3}k_{-4}}\cdots$].</p>
<p>In a rate equation approach we can apply a steady state approach to the intermediate species R. The steady state assumes that the rate of change of R is zero;</p>
<p>$$\frac{d[R]}{dt}= k_1[RX]-k_{-1}[R][X]-k_2[R][Y]+k_{-2}[RY]=0$$</p>
<p>from which </p>
<p>$$ [R]_{ss}= \frac{k_{-2}[RY]+k_1[RX]}{k_{-1}[X]+k_2[Y]}$$</p>
<p>The rate $r$ can be given by</p>
<p>$$\begin{align}
r=-\frac{d[Y]}{dt}=k_2[R][Y]-k_{-2}[RY] &= \frac{k_2(k_{-2}[RY]+k_1[RX])[Y]-k_{-2}[RY](k_{-1}[X]+k_2[Y])}{k_{-1}[X]+k_2[Y]}\\&=\frac{k_1k_2[RX][Y]-k_{-1}k_{-2}[RY][X]}{k_{-1}[X]+k_2[Y]}\\
\end{align}$$</p>
<p>and the numerator is zero when equilibrium concentrations are used making the rate equal to zero also. So this connects the equilibrium to the rate equations.</p>
<p>Initially just after the reactants are mixed the amount of product is very small and $k_1k_2[RX][Y]>>k_{-1}k_{-2}[RY][X] $ and the the rate is </p>
<p>$$r=\frac{k_1k_2[RX][Y]}{k_{-1}[X]+k_2[Y]}$$</p>
<p>which can be confirmed by experiment. Hope this helps.</p>
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https://chemistry.stackexchange.com/questions/94751/reaction-coordinate-kinetics-equilibrium-in-example
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Question: <p>Specifically, I am interested in the reaction of different hydroxides (in aqueous form) with <span class="math-container">$\ce{CO_2}$</span>. I would like to determine which hydroxide would conduct the reaction the fastest. My initial thought would be that spectator ions would have no effect, and since the concentration of the hydroxide has a direct relationship with the reaction rate, therefore the most soluble hydroxides would react the fastest.</p>
<p>However, after seeing <a href="https://pubs.acs.org/doi/10.1021/ed1001703" rel="nofollow noreferrer">this article</a>, I am confused as to whether spectator ions affect reaction rate or not. The abstract mentions "halide-acceleration effect," so I don't think the principle would apply to hydroxides, but I'm not entirely sure.</p>
Answer: <p>Everything present has an effect. The question is, is it distinguishable and measurable? Your proposal has many variables to consider; here are some. The amount of hydroxide, analysis by formulation or by titration, Activity of OH-, a good pH meter can help here, solubility of the hydroxide, carbonate, and bicarbonate; even the viscosity of the solutions. You will think up some more.</p>
<p>You intend to study the effect of different hydroxides in reacting with CO2. Since the differences are the counter-ions they will most likely have an effect. You will determine what the effect is and its importance. [avoid adding any halides]</p>
<p>There is much in the literature about this, check some out, a fresh look can be important.;</p>
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https://chemistry.stackexchange.com/questions/175496/can-spectator-ions-affect-reaction-kinetics
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Question: <p>This question has been bothering me for some time, and I can't seem to find a good answer online. </p>
<p>Say I have four chemical species $\ce{A}$, $\ce{B}$, $\ce{C}$, $\ce{D}$, and these four react in the following ways: </p>
<p>\begin{align}\ce{
A + A &-> B\\
A + B &-> C\\
A + C &-> D\\
A + D &-> B + C\\
}\end{align}</p>
<p>The kinetic reaction equations for these four species should be:
\begin{align}
\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &= -k_1[\ce{A}]^2-k_2[\ce{A}][\ce{B}]-k_3[\ce{A}][\ce{C}]-k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= +k_1[\ce{A}]^2 -k_2[\ce{A}][\ce{B}] +k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &= +k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &= +k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\
\end{align}</p>
<p>where $k_i$ is the Arrhenius coefficient for that reaction. </p>
<p>In my mind, the stoichiometry does not seem to add up right. It seems that when the system undergoes $\ce{A + A -> B}$, the concentration of $\ce{A}$ should decrease by twice as much as a reaction like $\ce{A + B -> C}$. Similarly, the products of $\ce{A + D}$ should be split evenly between $\ce{B}$ and $\ce{C}$. </p>
<p>Therefore, stoichiometrically, I want to write:
\begin{align}
\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &= -2k_1[\ce{A}]^2-k_2[\ce{A}][\ce{B}]-k_3[\ce{A}][\ce{C}]-k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= +k_1[\ce{A}]^2 -k_2[\ce{A}][\ce{B}] +\tfrac{1}{2}k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &= +k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +\tfrac{1}{2}k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &= +k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\
\end{align}</p>
<p>Is this wrong-headed? What conceptual issue am I missing? </p>
Answer: <p>Reaction rates should be defined in terms of the <a href="http://en.wikipedia.org/wiki/Extent_of_reaction" rel="nofollow">Extent of reaction</a> ($\xi$) that corresponds to the number of moles (or the molarity for reactions in solution) of specie $i$ divided by the the stoichiometric number, $\nu_i$:
$$\xi = \frac{[i]}{\nu_i}$$
Therefore, considering that the first reaction should be:
$$\ce{A + A -> B ~=~ 2A -> B}$$
The corresponding reaction rates are:
$$\frac{\mathrm{d}\xi}{\mathrm{d}t} = -\frac{1}{2} \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}$$
In this way, the reaction rate is always the same (positive) number, independently on the specie we are referring to.
The global rate equations are:
\begin{align}
\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &= −2k_1[\ce{A}]^2−k_2[\ce{A}][\ce{B}]−k_3[\ce{A}][\ce{C}]−k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= 2k_1[\ce{A}]^2−k_2[\ce{A}][\ce{B}]+k_4[\ce{A}][D]\\
\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &= k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &= k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\
\end{align}</p>
<p>It is conceptually wrong to assume that </p>
<blockquote>
<p>the products of $\ce{A + D}$ should be split evenly between $\ce{B}$ and $\ce{C}$.</p>
</blockquote>
<p>The reaction scheme you described is the "parallel reactions" model. The amount of $\ce{A}$ consumed by each reaction depends on the specific rate. The stoichiometry of the first step says that every time 1 molecule of $\ce{B}$ is produced 2 molecules of $\ce{A}$ are consumed. But maybe $k_1$ is very small and so all the other process will consume $\ce{A}$ much more rapidly than this one.</p>
<p>You cannot predict what will happen in a reaction system just looking at the stoichiometry of each step. If you have a large amount of $\ce{C}$, it will probably consume a lot of $\ce{A}$ in the third step, producing a lot of $\ce{D}$ that, in turn, will produce a lot of $\ce{B}$, etc. Stoichiometry is just one piece of information. But kinetics is much more that this.</p>
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https://chemistry.stackexchange.com/questions/24506/reaction-kinetics-and-stoichiometry-mass-conservation
|
Question: <p>According to the Arrhenius equation, the rate of reaction is proportional to $e^{-1/RT}$.</p>
<p>Here is a plot of the equation and its derivative with respect to temperature, both with respect to temperature. All constants are set to unity.</p>
<p><a href="https://i.sstatic.net/23Hut.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/23Hut.png" alt="enter image description here"></a></p>
<p>It seems to suggest that there is almost no change in the rate of reaction with increasing temperature except at very-low-temperature range.</p>
<p>Then, why is temperature considered an important factor in deciding the kinetics of a chemical reaction?</p>
<p>
</p>
<p>
</p>
<p>
</p>
Answer:
|
https://chemistry.stackexchange.com/questions/67798/why-is-temperature-considered-as-an-important-factor-in-deciding-chemical-reacti
|
Question: <p><img src="https://i.sstatic.net/8fKfm.png" alt="enter image description here"></p>
<p>How do I relate the half life to the overall rate of reaction? </p>
<p>I argued that from the data, doubling the partial pressure of either reactant, keeping the other constant, will half the half life. </p>
<p>So try t1/2 =
$\frac{\ce{[1]}}{\ce{[k][PA_0][PB_0]}}$ and since pressure of A is always greater than B, so t1/2 = $\frac{\ce{[1]}}{\ce{[k'][PB_0]}}$ ?</p>
<p>I then wrote:
$$\frac{-\mathrm{d}[PB]}{\mathrm{d}t} = k [P_B]^m $$, where <strong>k</strong> = k[PA] </p>
Answer: <p>Recall the meaning of the half life $t_{1/2}$: At $t_{1/2}$, a concentration (or partial pressure) is decreased to the half of its initial value.</p>
<p><strong>It is crucial to realize that this represents exactly the $\mathbf{-\frac{dp_A}{dt}}$ in your rate equation!</strong></p>
<p>With other words, $-\frac{dp_A}{dt} = k\cdot p_A\cdot p_B^2$ becomes $-\frac{54}{2\cdot100} = k\cdot54\cdot(1.0)^2$ for your first set of data. </p>
<p>Solve for $k$ and do the same for the other data sets. What is the result?</p>
|
https://chemistry.stackexchange.com/questions/32561/reaction-kinetics-relating-half-life-to-reaction-rate
|
Question: <p>We were taught that the kinetics of a reaction is determined by the reactions ‘rate determining step’ which is also the slowest step of the reaction. For E1 as well as E2 reactions the slowest steps are easily determined by the mechanism and I derived the rate laws for those. </p>
<p>In the case of E1cb reactions, which step can be taken as the slowest step? </p>
<p>Is it the one in which the acidic hydrogen is extracted by the base or is it the one where the elimination takes place (carbanion intermediate)?</p>
Answer:
|
https://chemistry.stackexchange.com/questions/112482/kinetics-for-e1cb-reactions
|
Question: <p>The rate constant for the reaction of hydrogen with iodine is <span class="math-container">$\pu{2.45E-4 M-1 s-1}$</span> at 302 °C and <span class="math-container">$\pu{0.905 M-1 s-1}$</span> at 508 °C.</p>
<p>a. calculate the activation energy and Arrhenius preexponential factor for this reaction.</p>
<p>b. What is the value of the rate constant at 400 °C ?</p>
<hr>
<p>I'm a bit confused because I get a negative value for the activation energy <span class="math-container">$E_a$</span></p>
<p>We have at 302 °C and 508 °C respectively
<span class="math-container">$$k_1 = \pu{2.45E-4 M-1 s-1}$$</span>
<span class="math-container">$$k_2 = \pu{0.905 M-1 s-1}$$</span></p>
<p>We convert in Kelvin:
<span class="math-container">$$T_1 = 302 °C = 575.15 K $$</span>
<span class="math-container">$$T_2 = 508 °C = 781.15 $$</span></p>
<p>The Arrhenius Equation is
<span class="math-container">$$\ln{k} = \ln{A} - \frac{E_a}{RT}$$</span></p>
<p>So we have
<span class="math-container">$$\ln{k_1} = \ln{A} - \frac{E_a}{RT_1}$$</span>
<span class="math-container">$$\ln{k_2} = \ln{A} - \frac{E_a}{RT_2}$$</span></p>
<p>We want to solve for <span class="math-container">$E_a$</span>. We subtract the second equation from the first:
<span class="math-container">$$\ln{k_1} - \ln{k_2} = - \frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_2})$$</span></p>
<p>Solving for <span class="math-container">$E_a$</span>, we get:
<span class="math-container">$$E_a = - R \times \frac{\ln{k_1} - \ln{k_2}}{\frac{1}{T_1} - \frac{1}{T_2}}$$</span></p>
<p>With <span class="math-container">$R = 8.314 J/mol K$</span>, <span class="math-container">$T_1 = 575.15 K$</span>, <span class="math-container">$T_2 = 781.15$</span>, <span class="math-container">$k_1 = \pu{2.45E-4 M-1 s-1}$</span>, <span class="math-container">$k_2 = \pu{0.905 M-1 s-1}$</span></p>
<p>(Link to <a href="https://www.wolframalpha.com/input?i=8.314%20*%20%28ln%282.45%20*%2010%5E%28-4%29%29%20-%20ln%280.905%29%29%2F%281%2F575.15%20-%201%2F781.15%29" rel="nofollow noreferrer">WolframAlpha calculation</a>)</p>
<p>I get <span class="math-container">$E_a = 1.48948 \times 10^5 J/mol$</span></p>
<p>Then using the Arrhenius equation with <span class="math-container">$k_1$</span> or <span class="math-container">$k_2$</span>:
<span class="math-container">$$A = k_1 \times \exp{\frac{E_a}{RT_1}}$$</span></p>
<p>(Link to <a href="https://www.wolframalpha.com/input?i=2.45%20%2010%5E%7B-4%7D%20%20e%5E%28-148948%2F%288.314%20*%20575.15%29%29" rel="nofollow noreferrer">WolframAlpha calculation</a>)</p>
<p>I get <span class="math-container">$A = \pu{8.26E9 M-1 s-1}$</span></p>
<p>But I'm confused, why the Activation Energy is negative ? Or did I miss something ? I checked all my calculations</p>
<p>EDIT: the error was due to a missing negative sign for <span class="math-container">$E_a = ...$</span>. Now it makes more sense.</p>
Answer: <p>A mistake must have happened in your calculations, because, when I do them, I obtain :</p>
<p>ln<span class="math-container">$k_1 = - 8.314$</span>;</p>
<p>ln<span class="math-container">$k_2 = - 0.100$</span>;</p>
<p>ln<span class="math-container">$k_1$</span> - ln<span class="math-container">$k_2$</span> = <span class="math-container">$-8.314 + 0.100 = -8.214$</span></p>
<p><span class="math-container">$\frac{1}{T_1} - \frac{1}{T_2}$</span> = <span class="math-container">$ \frac{1}{575} - \frac{1}{781} = 1.76·10^{-3} - 1.97·10^{-3} = - 0.21 ·10^{-3} $</span> K<span class="math-container">$^{-1}$</span></p>
<p><span class="math-container">$E/R = \frac{-8.214}{-0.21·10^{-3} \mathrm{K}^{-1}} = + 39 100$</span> K.</p>
<p><span class="math-container">$E = 39100$</span>K<span class="math-container">$ · 8,314$</span> <span class="math-container">$\frac{J}{mol·K} = + 325.2$</span> kJ/mol</p>
<p>It is a positive result.</p>
|
https://chemistry.stackexchange.com/questions/178022/reaction-kinetics-exercise-for-hydrogen-iodide-synthesis
|
Question: <p>In kinetics, zero order reactions are those reactions who are independent from the concentration of the reactant or the product. So their rate is equal to the constant <strong>k</strong>. But still we build the diagram where we find the correlation between concentration and time, where the slope is negative.
Why do we even write a diagram to find the correlation between <strong>concentration</strong> and time, since these reactions are completely independent from concentration ? </p>
Answer: <blockquote>
<p>So their rate is equal to the constant k.</p>
</blockquote>
<p>With other words, the <strong>change</strong> of concentration over time is linear:</p>
<p>\[ \frac{d[\ce{A}]}{dt} = k\]</p>
|
https://chemistry.stackexchange.com/questions/30849/order-of-reactions-in-kinetics
|
Question: <p>I was solving some problems on chemical kinetics, then I was just struck at some ques of sequential reaction kinetics. </p>
<p>Given reactions $$\ce{O_3 + Cl \rightarrow O_2 + ClO ~~~ k_1=5.2 \times 10^9~Lmol^{-1}s^{-1}}$$</p>
<p>and $$\ce{ClO + O \rightarrow O_2 + Cl ~~~ ~~~ k_2=2.6 \times 10^{10} ~Lmol^{-1}s^{-1}}$$</p>
<p>So, which value is closest to rate const of overall net reaction? $$\ce{O_3 + O \rightarrow 2O_2}$$</p>
<p>and the answer given was $=5.2 \times 10^9 ~\mathrm{Lmol^{-1}s^{-1}}$</p>
<p>Therefore I am confused, how did they find the rate const of the net reaction?</p>
Answer: <p>The rate constant of the overall reaction is determined by the rate constant of the RDS, Rate Determining Step.</p>
<p>The RDS is the one with the lowest rate constant. Hence in this case the overall rate of the reaction is $5.2 \times 10^9 Lmol^{-1}s^{-1}$</p>
|
https://chemistry.stackexchange.com/questions/27466/sequential-kinetics
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