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Question: <p>Here is the question:</p>
<blockquote>
<p>For the reaction:
<span class="math-container">$$\ce{2 A + 2 B → C + D}$$</span></p>
<p>The following data was obtained from three experiments:
<span class="math-container">\begin{array}{c|ccc}\hline
\bf{Experiment} &\textbf{[A] (mol/L)} &\textbf{[B] (mol/L)} &\textbf{Rate of Formation of C(mol min/L )} \\ \hline
1 & 0.60 &0.15 &6.3 \times10^{-3} \\
2 & 0.20 & 0.60 & 2.8 \times 10^{-3}\\
3 & 0.20 &0.15& 7.0 \times10^{-4} \\ \hline
\end{array}</span></p>
<p>a) What is the rate equation for the reaction?</p>
<p>b) What is the numerical value of the rate constant <span class="math-container">$k$</span>?</p>
<p>c) Propose a reaction mechanism for this reaction, identify any intermediates, and the rate-determining step.</p>
</blockquote>
<p>What I have done:</p>
<p><a href="https://i.sstatic.net/3DXzM.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3DXzM.png" alt="enter image description here" /></a></p>
<p>However, I don't know if this is correct and how to proceed forward :/</p>
<p>Edit:
I fixed my rate law however When determining the reaction mechanism, I am having a hard time cause I am not sure of the possible mechanism that indicates a slow step, how would I do this part?</p>
Answer: <p>As I said in the comments, your image is quite low resolution. But, if I read it correctly, you have got the exponents in the rate law wrong.</p>
<p>Solving this type of problem is simple, you just need to check if the rate law makes intuitive sense. The data given in the question contains various instantaneous rates (or initial rates maybe) with concentrations at that point.</p>
<p>First, look at the data:
<span class="math-container">\begin{array}{c|ccc}\hline
\bf{Experiment} &\textbf{[A] (mol/L)} &\textbf{[B] (mol/L)} &\textbf{Rate of Formation of C(mol min/L )} \\ \hline
1 & 0.60 &0.15 &6.3 \times10^{-3} \\
2 & 0.20 & 0.60 & 2.8 \times 10^{-3}\\
3 & 0.20 &0.15& 7.0 \times10^{-4} \\ \hline
\end{array}</span></p>
<p>From experiment 1 & 3., the rate increases by 9 times when you increase <span class="math-container">$\text{[A]}$</span> by 3 times. Clearly, the exponent of <span class="math-container">$\text{[A]}$</span> in the rate equation is <span class="math-container">$2$</span>.</p>
<p>Now from experiments 2 & 3, the rate increases by <span class="math-container">$4$</span> times when you increase <span class="math-container">$\text{[B]}$</span> by 4 times. The exponent of <span class="math-container">$\text{[B]}$</span> in rate eqn. is 1.</p>
<p>So your rate law should be <span class="math-container">$$r=k\mathrm{[A]^2[B]}$$</span></p>
<p>I think in your calculation you switched the numerator and denominator somewhere so you got negative exponents.</p>
<p>Then you can use the rate equation with the data given to find the rate constant.</p>
<p>Proposing the reaction mechanism is a bit difficult, remember that most elementary reaction steps are bimolecular. Trimolecular steps are rare. Any step where more than 3 species collide is almost impossible, so make sure your mechanism doesn't have anything like that.</p>
<p><strong>Edit</strong>: For writing a plausible mechanism, in general keep in mind that the rate law will have terms for all of the reactant species that are present in and before the slow (rate limiting) step.</p>
<p>An example mechanism for the reaction could be something like the following:</p>
<p><span class="math-container">$$\ce{A +B-> X_1}$$</span>
<span class="math-container">$$\ce{X_1 +A \overset{slow}{->} C + X_2}$$</span>
<span class="math-container">$$\ce{X_2 + B -> D}$$</span></p>
<p><span class="math-container">$\ce{X_1}$</span> and <span class="math-container">$\ce{X_2}$</span> are intermediates.</p>
<p>To get the rate law, count the number of reagent molecules (not the intermediates) for the slow step and all steps before that. So in the slow step you have one molecule of A, in the previous step you have one molecule of A and one of B. So, your rate law should be <span class="math-container">$\mathrm{r=k[A]^2[B]}$</span> which is the same as you got from experiment, so this is a possible mehcanism (this is not a proof that the mechanism is right!)</p>
<p>Note that the above way of getting the rate law from mechanism is simplistic, so there might be some unusual reactions where it fails.</p>
|
https://chemistry.stackexchange.com/questions/152863/finding-the-rate-of-a-reaction-given-experimental-data-for-reaction-kinetics
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Question: <p>It is given that when we use $50{-}\pu{60\%}\ \ce{KOH}$ in a Cannizaro reaction of formaldehyde, the reaction follows third order kinetics, but under special cases of Cannizaro reactions it was given than when formaldehyde reacts with $\pu{100\%}\ \ce{NaOH/KOH}$ the reaction follows fourth order kinetics.</p>
<p>What is the reason for this change in the order of reaction with an increase in the concentration of base?</p>
Answer:
|
https://chemistry.stackexchange.com/questions/88145/special-case-of-cannizaro-reaction-of-formaldehyde-with-100-naoh-koh-follows-4t
|
Question: <p>I am modelling a methanation process in MATLAB, where I have a rate equation for methanation</p>
<p><span class="math-container">$$\ce{CO2 + 4 H2 → CH4 + 2 H2O}$$</span></p>
<p>in which I need partial pressure of <span class="math-container">$\ce{CO2},$</span> <span class="math-container">$\ce{H2},$</span> <span class="math-container">$\ce{CH4},$</span> <span class="math-container">$\ce{H2O}.$</span></p>
<p>I am struggling to understand the link between the mass flow rate of <span class="math-container">$\ce{H2}$</span>, <span class="math-container">$\ce{CO2}$</span> and partial pressure of all other species. The required amount of catalyst and the volume for the fixed bed reactors in accordance with the chosen stimulation parameters.</p>
<p>If suppose my operational pressure is <span class="math-container">$\pu{7 atm}$</span>, <span class="math-container">$\pu{16 mol s^-1}$</span> of <span class="math-container">$\ce{H2},$</span> <span class="math-container">$\pu{4 mol s^-1}$</span> of <span class="math-container">$\ce{CO2},$</span> what will be the partial pressures of all those species and required amount of catalyst and reactor volume.</p>
<p>The paper in which I referred for rate reaction is <a href="https://doi.org/10.1016/j.apcatb.2017.11.066" rel="nofollow noreferrer">this</a>.</p>
Answer: <p>So you need three things:</p>
<ol>
<li>Partial pressures of species.</li>
<li>Amount of catalyst.</li>
<li>Reactor volume.</li>
</ol>
<p>We will design a PBR in order to calculate the mass of catalyst <span class="math-container">$ W $</span> needed to achieve a certain conversion <span class="math-container">$ X $</span>.</p>
<p><strong>Partial Pressures</strong> You have a heterogeneous reaction, since it takes place with a catalyst, but all the chemical species of the rxn are in the gas phase. It seems that you have no limiting reactant, as the stoichiometric ratio <span class="math-container">$ \ce{H2}/\ce{CO2} = 4/1 $</span> is the same as the ratio of the entering molar flow rates. I will choose <span class="math-container">$\ce{CO2}$</span> as the basis component, so the conversion refers to <span class="math-container">$ \ce{CO2} $</span>. Say the reaction is <span class="math-container">$ A + 4B \rightarrow C + 2D $</span>. The molar flow rate of any species is
<span class="math-container">$$ F_j = F_{A0} \bigg(\dfrac{F_{j0}}{F_{A0}} + \nu_j X\bigg) $$</span>
The molar flow rates of all components, and the total molar flow rate, are
<span class="math-container">\begin{align}
F_A &= F_{A0} (1 - X) \\
F_B &= 4F_{A0} (1 - X) \\
F_C &= F_{A0}X \\
F_C &= 2F_{A0}X \\
F &= \sum_j F_j = F_{A0} (5 - 2X)
\end{align}</span></p>
<p>Thus, with <span class="math-container">$ P_j = y_j P = (F_j/F) P$</span>, we have the partial pressures as a function of the conversion
<span class="math-container">\begin{align}
P_A &= \left(\dfrac{1 - X}{5 - 2X}\right)P \tag{1} \\
P_B &= \left(\dfrac{4 - 4X}{5 - 2X}\right)P \tag{2} \\
P_C &= \left(\dfrac{X}{5 - 2X}\right)P \tag{3} \\
P_C &= \left(\dfrac{2X}{5 - 2X}\right)P \tag{4}
\end{align}</span>
Now, these equations are strictly true at the entrance of the reactor, i.e., where the conversion is <span class="math-container">$ X =0 $</span>. Once we are in the reactor, the gas phase will suffer pressure drop.</p>
<p><strong>Pressure drop</strong> I will assume that the gas flow is, at every coordinate, turbulent. The Ergun equation simplifies to
<span class="math-container">$$ \dfrac{dP}{dz} = -\dfrac{7}{4}\dfrac{(1 - \phi)G^2}{\phi^3 \rho D_P} \tag{5} $$</span>
where <span class="math-container">$ G = \dot{m}/S $</span> is the mass flux (constant along the reactor), <span class="math-container">$ D_P $</span> is the catalyst particle diameter, and <span class="math-container">$ \phi $</span> the void fraction. We need to relate the density as a function of things we know, by using the ideal gas law
<span class="math-container">$$ \dfrac{dP}{dz} = -\dfrac{7}{4}\dfrac{(1 - \phi)G^2}{\phi^3D_P}
\dfrac{MRT}{P} \tag{6} $$</span>
Now we need to relate the <span class="math-container">$ z $</span> coordinate to the <span class="math-container">$ W $</span> coordinate
<span class="math-container">\begin{align}
W &= (1 - \phi) S z \rho_c \\
dW &= (1 - \phi) S dz \rho_c \\
dz &= \dfrac{dW}{(1 - \phi) S \rho_c} \tag{7}
\end{align}</span>
where <span class="math-container">$ \rho_c $</span> is the mass density of the catalyst. Combining Eqs. (6) and (7)
<span class="math-container">\begin{align}
(1 - \phi) S \rho_c \dfrac{dP}{dW} = -\dfrac{7}{4}
\bigg(\dfrac{(1 - \phi)G^2}{\phi^3D_P}\bigg)\dfrac{MRT}{P} \\
\dfrac{dP}{dW} = -\dfrac{7}{4} \left(\dfrac{G^2}{S \rho_c \phi^3D_P}\right)
\dfrac{MRT}{P} \tag{8} \\
\end{align}</span>
If you are obsessive like me, the molar mass of the gas will change along the reactor, so you should also take into account
<span class="math-container">$$ M = \sum_j y_j M_j \tag{9} $$</span></p>
<p><strong>Amount of catalyst</strong> This is obtained by a mole balance along the <span class="math-container">$ W $</span> coordinate for species <span class="math-container">$ A $</span>
<span class="math-container">\begin{align}
F_A\vert_W - F_A\vert_{W+\Delta W} + {r_A}' \Delta W = 0 \\
\dfrac{F_A\vert_W - F_A\vert_{W+\Delta W}}{\Delta W} + {r_A}' = 0 \\
-\dfrac{dF_A}{dW} + {r_A}' = 0 \tag{10}
\end{align}</span>
with <span class="math-container">$ F_A = F_{A0}(1 - X) $</span> or <span class="math-container">$ dF_A = -F_{A0}dX $</span>, Eq. (10) yields
<span class="math-container">\begin{align}
F_{A0}\dfrac{dX}{dW} + {r_A}' = 0 \\
\dfrac{dX}{dW} = \dfrac{-{r_A}'}{F_{A0}} \tag{11}
\end{align}</span>
where <span class="math-container">$ -{r_A}' $</span> is the rate of consumption of species <span class="math-container">$A$</span> in <span class="math-container">$ \text{mol A}/\text{kg cat} \cdot \text{s} $</span>.</p>
<p><strong>Energy Balance</strong> As a starting point, before studying an adiabatic reactor, assume that the reactor is isothermic. This translates the energy balance in the amount of heat <span class="math-container">$ Q(W) $</span> that you must remove at every mass coordinate in order for the temperature to remain constant. Hence, we disregard the energy balance for the first analysis, and <span class="math-container">$ T = T_0 $</span>.</p>
<p>Thus, to know the mass of catalyst in the reactor, you need to numerically solve the following ODE's
<span class="math-container">\begin{align}
\dfrac{dX}{dW} &= \dfrac{-{r_A}'}{F_{A0}} \hspace{1 cm} X(W = 0) = 0 \tag{12} \\
\dfrac{dP}{dW} &= -\dfrac{7}{4} \left(\dfrac{G^2}{S \rho_c \phi^3D_P}\right)\dfrac{MRT}{P} \hspace{1 cm} P(W = 0) = P_0 \tag{13} \\
\end{align}</span></p>
<p><strong>Other data we need to know</strong></p>
<ol>
<li>To evaluate the rate of reaction you need a kinetic law. To my surprise, there is one in the paper, and has the type of a homogenous kinetic rate law of the type
<span class="math-container">$$ -{r_A}' = f(P_A, P_B, P_C, P_D) = k'\bigg(P_A P_B^4 - \dfrac{P_C P_D^2}{K}\bigg) $$</span>
at constant temperature, where <span class="math-container">$ k' = k(T=T_0) $</span> and <span class="math-container">$ K = K(T=T_0) $</span> are fixed. In the partial pressures you plug Eqs. (1-4), this is why we had to do all that math.</li>
<li>You stop the integration depending on your objective. If equilibrium allows you, you should choose the highest conversion possible, <span class="math-container">$ X = 0.95X_{eq}(T) $</span> is a starting point.</li>
<li>Other things like the catalyst mass density <span class="math-container">$ \rho_c $</span>, the void fraction of the PBR <span class="math-container">$ \phi $</span>, the area <span class="math-container">$ S $</span>, and the particle diameter of the catalyst <span class="math-container">$ D_P $</span>.</li>
</ol>
<p><strong>Reactor Volume</strong> Once you solve the ODE this is just one single calculation, you just use the Eq. (7) but not differentiated, since the volume is <span class="math-container">$ V = Sz $</span>
<span class="math-container">$$ V = \dfrac{W}{(1 - \phi)\rho_c} $$</span></p>
<p>This is an interesting problem, once I have more time, I will do some code to solve these balances. I think there is some other guy that asked for something similar.</p>
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https://chemistry.stackexchange.com/questions/136134/kinetics-of-sabatier-reaction
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Question: <p>From the Arrhenius equation in kinetics of reactions,</p>
<p>$$k = A\exp{\left(-\frac{E_\mathrm{a}}{RT}\right)}$$</p>
<p>Which tells us about <strong>the temperature dependence on rate constant of a reaction.</strong></p>
<p><strong>Activation energy is dependent on the temperature</strong> (as it is the energy difference between the average energy of the reactants and threshold energy, and since the average energy of reactants depends on the temperature, activation energy also should depend on temperature) Is my assumption that activation energy depends on temperature correct?</p>
<p>Moreover, is there any case where activation energy is zero or negative? (like in the case of spontaneous reaction or $E_\mathrm{a} = 0$) I feel that whenever activation energy is zero, all the molecules undergoing collisions should form products successfully, but this case won't occur.</p>
<p>But when we substitute $E_\mathrm{a} = 0$, in the Arrhenius equation, we get the result as $k = A$, i.e. rate constant will be equal to pre-exponential factor.</p>
Answer: <p>The rate constant and the activation energy tell us how fast the reaction proceeds (moles/sec), not whether it is spontaneous. The higher the temperature, the higher the rate constant and, for specified concentrations of reactants, the faster the reaction. The rate of a reaction always increases with increasing temperature (higher rate of collisions), so the activation energy is always positive. As the temperature increases, the rate constant approaches the pre-exponential factor (E/T approaches zero). For some reactions, the activation energy is very low, so even at lower temperatures, the rate constant approaches the pre-exponential factor. The activation energy has a weak dependence on temperature, and, in practice, this temperature dependence is usually neglected.</p>
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https://chemistry.stackexchange.com/questions/43576/chemical-kinetics-of-a-reaction-rate-constant-and-activation-energy
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Question: <p><a href="https://i.sstatic.net/mrjg4.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/mrjg4.jpg" alt="So this is the question, hope someone can help."></a></p>
<p>For a hypothetical elementary reaction(image).
Initially, only 2 moles of A are present. The total number of moles of A, B, and C at the end of 75% of the reaction is:</p>
Answer: <p><span class="math-container">$75$</span>% of <span class="math-container">$2$</span> moles of A is <span class="math-container">$1.5$</span> mole. <span class="math-container">$1,5$</span> mol A produce twice as much of B or C. This makes <span class="math-container">$3$</span> mol of B and/or C. It remains <span class="math-container">$2$</span> - <span class="math-container">$1.5 = 0.5$</span> mol A. The total number of moles is <span class="math-container">$3 + 0.5 = 3.5$</span> mol. Where is the difficulty ?</p>
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https://chemistry.stackexchange.com/questions/125824/chemical-kineticshypothetical-elementary-reaction
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Question: <p>As an exercise in chemistry, I decided to observe the kinetics of the decay of thiosulfate ions in an acidic medium. As far as I am aware, it is a very well known exercise among chemists.</p>
<p>I found a paper in which the author uses a sodium thiosulfate solution with a concentration of 0.25 mol/L and sulfuric acid with a concentration of 0.25 mol/L. I would like to know if it is okay to use hydrochloric acid with a concentration of 0.50 mol/L, instead of sulfuric acid? In other words, can I expect the same turbidity time of the aqueous solution as the author?</p>
Answer:
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https://chemistry.stackexchange.com/questions/188374/kinetics-of-the-thiosulfate-and-hydrochloric-acid-reaction
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Question: <p>I get the concept that one can measure the rate of reaction by measuring the rate at which an equilibrium state reaches another equilibrium state after the system is perturbed (usually by heating). However, I do have an issue with it. Consider the reaction $$\ce{A <=> B}$$ </p>
<p>My lecture handout suggests that the rate of change of $[\ce{A}]$ is given by:</p>
<p>$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\text{forward}[\ce{A}]+k_\text{reverse}[B]$$</p>
<p>However, surely it's just $\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\text{forward}[\ce{A}]$. The equation in the lecture handout would make sense at equilibrium because that is when $\frac{\mathrm d[\ce{A}]}{\mathrm dt}=0$ but I think it seems to insinuate that this is always valid (without explicitly saying so). Am I right or is that equation always valid? If it is: why? </p>
Answer: <p>You have two simultaneous reactions, </p>
<p>$$\begin{array}{rcl}
\ce{A} &\xrightarrow{k_1}& \ce{B}\\[6pt]
\ce{B} &\xrightarrow{k_{-1}}& \ce{A}
\end{array}$$</p>
<p>and $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt}$ equals $-k_1 [\ce{A}]$ for the first reaction alone and $k_{-1} [\ce{B}]$ for the second alone. To get the total $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt}$ for the reactions running simultaneously, you have to add the contributions from each reaction:</p>
<p>$$\dfrac{\mathrm d[\ce{A}]}{\mathrm dt} = -k_1 [\ce{A}] + k_{-1} [\ce{B}]$$</p>
<p>This will always be true; at equilibrium you'll have $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt} = 0$ and</p>
<p>$$K = \frac{k_1}{k_{-1}} = \frac{[\ce{B}]}{[\ce{A}]}$$</p>
<p>If you did it your way, this wouldn't be the case.</p>
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https://chemistry.stackexchange.com/questions/24416/how-to-make-sense-of-the-relaxation-method-for-measuring-reaction-kinetics
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Question: <p>I am confused as to how an order of a reactant (n or m) in the rate law can be negative. This means that increasing the concentration of a reactant would actually decrease its rate of disappearance. How does this make sense? Doesn't a higher concentration always allow for more collisions? Also, when the order is 0, this means that concentration has no effect on the rate of disappearance. How does this make sense chemically/at the atomic level? </p>
Answer: <p>In principle, <strong><a href="http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Reaction_Rates/Zero-Order_Reactions" rel="nofollow">there are no zeroth-order reaction</a></strong>. Experimentally, however, zeroth-order rate law occurs because of the reaction condition. For example, in catalytic reactions such as hydrogenation of an alkene using platinum as a catalyst, the hydrogenation occurs on the surface of platinum. There is a finite surface area of platinum, so at a some point, it doesn't matter at all how much more hydrogen is put in. This is because no reaction can occur if the hydrogen cannot bind to the platinum.</p>
<p>An example of inverse first order reaction can be found <a href="https://books.google.com/books?id=VaAgihZusxwC&pg=PA369&lpg=PA369&dq=inverse+first+order+kinetics+example&source=bl&ots=bEqroqm1FM&sig=v-18e3t-uIV1ITHtZsC94Gs9pIU&hl=en&sa=X&ei=E1P2VMuyKce1oQT9vYLIBA&ved=0CB0Q6AEwADgK#v=onepage&q=inverse%20first%20order%20kinetics%20example&f=false" rel="nofollow">here</a>.</p>
<p>Basically, you have a two-step reaction:
$$\ce{A + B <=> C}$$
$$\ce{C <=> D\ (slow)}$$
Increasing concentration of $\ce{C}$ would decrease the rate of reaction.</p>
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https://chemistry.stackexchange.com/questions/26887/kinetics-order-of-reaction
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Question: <p>I am layman of chemistry. Latterly, I am asked to do an mathematical optimization task for a bio-chemical process, which contains a conditioning step to neutralize the sulfuric acid using ammonia hydroxide. I am trying to use the mathematical model of the reaction (kinetic model) to optimize the reaction time/energy/efficiency (target not determined yet). However, I could not find the kinetics data. Is there a way I could find the kinetic model of the reaction? </p>
<p>Another question is that: Should I treated this reaction as an element reaction and use third-order reaction rate formula? </p>
<p>I am confused by a former post (<a href="https://chemistry.stackexchange.com/questions/16758/finding-rate-of-acid-base-reaction">Finding rate of acid - base reaction</a>). In that post, the accepted answer says "Typical acid - base reactions that occur in solution are diffusion controlled reactions. Such reactions do not follow second order kinetics because at virtually every encounter of the acid and base, rapid proton transfer occurs and the reaction is complete. " Should I use the diffusion control model instead of the reaction rate model?</p>
<p>Any suggestion is welcomed. </p>
Answer:
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https://chemistry.stackexchange.com/questions/34340/is-there-a-way-to-find-the-kinetics-data-for-the-strong-acid-weak-base-reaction
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Question: <p>I am having trouble with the following question:</p>
<blockquote>
<p>Consider the following reversible reaction in which the reaction is first order in both
directions: $$\ce{[A] <=> [B]}$$ $k_\mathrm a$ is the rate constant for the forward reaction and $k_\mathrm b$ is the rate constant for the reverse reaction.</p>
<p>If only reagent $\ce{A}$ is present at $t = 0$, such that $[\ce{A}](t = 0)$ = $[\ce{A}]_0$ (and likewise for $[\ce{B}]_0$), then at all subsequent times $[\ce{A}]+[\ce{B}] = [\ce{A}]_0$. Write down a differential equation for $\frac{\mathrm d[\ce{A}]}{\mathrm dt}$ where $[\ce{A}] \equiv [\ce{A}](t)$ is the concentration of the reagent $\ce{A}$, and hence show that:</p>
<p>$$[\ce{A}]=[\ce{A}]_0\frac{k_\mathrm b+k_\mathrm a \exp[-(k_\mathrm a+k_\mathrm b)t]}{k_\mathrm a+k_\mathrm b}$$</p>
</blockquote>
<p>Here's my attempt:</p>
<p>$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\mathrm a[\ce{A}]+k_\mathrm b[\ce{A}]\tag{1}$$</p>
<p>Then using the identity in the question: </p>
<p>$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\mathrm a[\ce{A}]+k_\mathrm b([\ce{A}]_0-[\ce{A}])\tag{2}$$</p>
<p>Rearranging...</p>
<p>$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}+(k_\mathrm a+k_\mathrm b)[\ce{A}]=k_\mathrm b[\ce{A}]_0\tag{3}$$</p>
<p>I believe that this is a first order linear differential equation with the integrating factor: $\mathrm e^{\int (k_\mathrm a+k_\mathrm b)\,\mathrm dt}=\mathrm e^{(k_\mathrm a+k_\mathrm b)t}$ Thus:</p>
<p>$$\frac{\mathrm d}{\mathrm dt}\left([\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\right)=k_\mathrm b[\ce{A}]_0\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\tag{4}$$</p>
<p>$$[\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}=\int k_\mathrm b[\ce{A}]_0\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\,\mathrm dt\tag{5}$$</p>
<p>This gives me:</p>
<p>$$[\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}=k_\mathrm b[\ce{A}]_0\frac{\mathrm e^{(k_\mathrm a+k_\mathrm b)t}}{k_\mathrm a+k_\mathrm b}+c\tag{6}$$</p>
<p>But this doesn't give me the final result if I put in the boundary conditions. Where have I gone wrong?</p>
Answer: <p>Third equation looks right to me. That's a <a href="http://pruffle.mit.edu/3.016-2005/Lecture_20_web/node3.html" rel="nofollow noreferrer">non-homogeneous (some would say heterogeneous) first-order differential equation</a>.</p>
<p>I think the heterogeneity is what is giving you problems. Those equations have a general solution which is a <a href="http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-NonhomgenLinEqns_Stu.pdf" rel="nofollow noreferrer">sum of a "complementary" solution and a "particular" solution</a>. <strong>Don't try to use any boundary conditions until after you have combined particular and complementary solutions.</strong></p>
<p><span class="math-container">$$\frac{d[\ce{A}]}{dt}+(k_\mathrm a+k_\mathrm b)[\ce{A}]=k_\mathrm b[\ce{A}]_0 \tag{3}$$</span></p>
<p>The "complementary solution" is the solution to the corresponding homogeneous equation <span class="math-container">$\frac{d[\ce{A}]}{dt}+(k_\mathrm a+k_\mathrm b)[\ce{A}]=0$</span>, which is <span class="math-container">$[\ce{A}]_\mathrm c=K e^{-(k_\mathrm a+k_\mathrm b)t}$</span>, where <span class="math-container">$K$</span> is an unknown constant of integration.</p>
<p>The "particular solution" is a polynomial in <span class="math-container">$t$</span> of the same degree as the hetereogeneous term, which in this case is a polynomial of degree 0, <span class="math-container">$[\ce{A}]_\mathrm p=C$</span>, where <span class="math-container">$C$</span> is an undetermined constant. We solve for this constant by substitution into the original equation:</p>
<p><span class="math-container">$$0 + (k_\mathrm a+k_\mathrm b)C=k_\mathrm b[\ce{A}]_0 \tag{4}$$</span></p>
<p>This provides the value of <span class="math-container">$C$</span> as <span class="math-container">$C=\frac{k_\mathrm b[\ce{A}]_\mathrm 0}{k_\mathrm b+k_\mathrm a}$</span>.</p>
<p>The general solution is thus the sum of the particular and complementary solutions:</p>
<p><span class="math-container">$$[\ce{A}]=[\ce{A}]_\mathrm c + [\ce{A}]_\mathrm p=K e^{-(k_\mathrm a+k_\mathrm b)t} + \frac{k_\mathrm b[\ce{A}]_0}{k_\mathrm b+k_\mathrm a} \tag{5}$$</span></p>
<p>Now we can apply boundary conditions. At time 0, <span class="math-container">$[\ce{A}]=[\ce{A}]_0$</span>, which gives an algebraic equation we can solve for <span class="math-container">$K$</span></p>
<p><span class="math-container">$$[\ce{A}]_0=K + \frac{k_\mathrm b[\ce{A}]_0}{k_\mathrm b+k_\mathrm a} \tag{6}$$</span></p>
<p>If I did my algebra right, then the solution is <span class="math-container">$K=[\ce{A}]_\mathrm 0 \left (1-\frac{k_b}{k_\mathrm b+k_\mathrm a} \right ) $</span>. Substituting this value back into the general solution from my equation 5 gives:</p>
<p><span class="math-container">$$[\ce{A}]=[\ce{A}]_0 \left (1-\frac{k_\mathrm b}{k_\mathrm b+k_\mathrm a} \right )e^{-(k_a+k_b)t} + \frac{k_\mathrm b}{k_\mathrm b+k_\mathrm a}[\ce{A}]_0 \tag{7}$$</span></p>
<p>This solution satisfies the boundary condition and also the original equation. It can be simplified a bit by algebraic manipulation to:</p>
<p><span class="math-container">$$[\ce{A}]=[\ce{A}]_0\frac{k_\mathrm a e^{-(k_\mathrm a+k_\mathrm b)t}+k_\mathrm b}{k_\mathrm a+k_\mathrm b} \tag{8}$$</span></p>
<p>From this expression it is easy to check that (i) the condition at <span class="math-container">$t=0$</span> is satisfied, that (ii) the expected exponential dependence in time is obtained, and (iii) that as <span class="math-container">$t \rightarrow \infty $</span>, the reaction should go to equilibrium. The last point about equilibrium can be checked by applying the two chemical definitions for the equilibrium constant (the ratio of product <span class="math-container">$B$</span> to reactant <span class="math-container">$A$</span> vs. the ratio of rate constants <span class="math-container">$k_a/k_b$</span>), i.e. <span class="math-container">$\frac{\ce{B}_{\infty}}{\ce{A}_{\infty}}=\frac{\ce{A}_0-\ce{A}_{\infty}}{\ce{A}_{\infty}}=\frac{k_\mathrm a}{k_\mathrm b}$</span> and using equation 8 to check consistency between our obtained expression for <span class="math-container">$\ce{A}_{\infty}$</span> and the chemically intuitive definition we just made.</p>
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https://chemistry.stackexchange.com/questions/31047/chemical-kinetics-of-a-reversible-reaction
|
Question: <p>I'm trying to understand and incorporate 9-species 19-reaction H2/O2 combustion mechanism into my numerical solver. The reaction mechanism in question is <a href="https://onlinelibrary.wiley.com/doi/full/10.1002/kin.20036" rel="nofollow noreferrer">O'Conaire</a>. It seems that the paper is in some standard chemical-kinetics-paper format and thus some basic things are omitted. I have no background in chemistry and after few hours of googling there are still several unclear things that I hope you could help me with.</p>
<p>It is probably illegal to reproduce significant part of the aforementioned paper, so I'll have to only refer some thing here.</p>
<p>So far I assume the following things:</p>
<ol>
<li><p>The partial reaction rate for density of <span class="math-container">$i$</span>-th gas component in <span class="math-container">$k$</span>-th reaction can be calculated via Arrhenius and acting masses laws:
<span class="math-container">$$
\frac{d}{dt}\left(\frac{\rho_i}{\mu_i}\right)_k = \omega_i = A T^n e^{\left(\frac{E_a}{RT}\right)}\left(\frac{\rho_1}{\mu_1}\right)^a\left(\frac{\rho_2}{\mu_2}\right)^b
$$</span>
where <span class="math-container">$\mu_i$</span> is <span class="math-container">$i$</span>-th component molar mass, <span class="math-container">$A$</span> is pre-exponent constant, <span class="math-container">$T$</span> is absolute temperature, <span class="math-container">$n$</span> is temperature exponent and <span class="math-container">$E_a$</span> is activation energy; <span class="math-container">$\rho_1, \rho_2$</span> are densitites of relevant reactants. Division of densities by molar masses result in molar concentrations:
<span class="math-container">$$
\left[\rho_i\right] = \frac{kg}{m^3}, \quad \left[\mu_i\right] = \frac{kg}{mol}, \quad \left[\frac{\rho_i}{\mu_i}\right] = \frac{mol}{m^3}.
$$</span>
For example, in the "<span class="math-container">$O+H_2 = H + OH$</span>" reaction (number 2 in the <a href="https://onlinelibrary.wiley.com/doi/full/10.1002/kin.20036" rel="nofollow noreferrer">paper</a>), the rate for <span class="math-container">$H_2$</span> would be
<span class="math-container">$$
\frac{d}{dt}\left(\frac{\rho_{H_2}}{\mu_{H_2}}\right) = -5.08 \times 10^4 T^{2.67} e^{\left(\frac{6.292}{RT}\right)}\left(\frac{\rho_O}{\mu_O}\right)\left(\frac{\rho_{H_2}}{\mu_{H_2}}\right).
$$</span>
with proper units. The minus sign is due to that <span class="math-container">$H_2$</span> is consumed during this reaction.</p>
</li>
<li><p>All of the reactions given in the paper are considered elementary and thus exponents for molar concentrations all equal unity. Except maybe reactions like "<span class="math-container">$O + O + M = O_2 + M$</span>" (see the questions 2, 4, 5 below).</p>
</li>
<li><p>For a particular gas component, final rate is the sum of all reaction rates that the component participate in, with "-" sign if it is consumed and with "+" sign if it is produced.</p>
</li>
</ol>
<p>So, the questions are the following:</p>
<ol>
<li>Are my assumptions correct?</li>
<li>What if there is there are two of the same molecule/atom in the left side of the reaction, what would the molar concetration exponent be? Is it just <span class="math-container">$1+1 = 2$</span>?</li>
<li>What is the physical sense of negative <span class="math-container">$E_a$</span>? Is it that participating radicals are so chemically active that the reaction would be very fast even at very low temperatures?</li>
<li>For reactions involving third bodies (denoted "M"), so called "efficiency factors" are provided. What are these?</li>
<li>How to properly take into account third body concentrations?</li>
<li>In an <a href="https://onlinelibrary.wiley.com/doi/abs/10.1002/(SICI)1097-4601(1999)31:2%3C113::AID-KIN5%3E3.0.CO;2-0" rel="nofollow noreferrer">older paper</a>, reactions with "M" are called pressure dependendent. Does that mean that all other reactions are considered independent of pressure?</li>
<li>What are the Troe parameters (given for reacxtion 9 and 15)</li>
<li>What do <span class="math-container">$k_O$</span> and <span class="math-container">$k_{\infty}$</span> in the older paper mean?</li>
<li>What would be the good textbook or article to read to understand these things better?</li>
</ol>
Answer:
|
https://chemistry.stackexchange.com/questions/166777/understanding-uni-and-termolecular-reactions-in-combustion-kinetics
|
Question: <blockquote>
<p>A reaction $\ce{A + 2 B -> C}$ is carried out in constant volume at $\pu{227 ^\circ C}$. the volume is $\pu{2 L}$. There is no $\ce{C}$ at the beginning. $[\ce{A}]_0 = \pu{0.035 mol L-1}$. $\ce{B}$ concentration changes over time.</p>
<p>\begin{array}{l|cccccccc}
t~(\pu{min})& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
[\ce{B}]~(\pu{mol L-1}) & 0.09 & 0.066 & 0.05 & 0.039 & 0.033 & 0.028 & 0.024& 0.023
\end{array}</p>
<p><strong>a</strong>) At $t = 5$, what is the conversion rate?<br>
<strong>b</strong>) Calculate the reaction rate of $\ce{C}$ by using graph method according to $\ce{C}$ at time $0$ and $6$.<br>
<strong>c</strong>) Calculate the total pressure at time $3$.<br>
<strong>d</strong>) If this reaction should have done at constant pressure what would be the volume change rate?</p>
</blockquote>
<p>I answered the <strong>a</strong>, <strong>c</strong> and <strong>d</strong> questions, but I am not sure about it. I can not solve question <strong>b</strong>, graph method, these were my exam questions and I want to know correct answers. Here are my answers:</p>
<hr>
<p>Part <strong>a</strong>.</p>
<p>$$t = \pu{5 min}$$
$$\ce{\underset{0.035}{A (g)} + 2 \underset{0.09}{B (g)} -> C (g)}$$</p>
<p>$\ce{A}$ is a limiting reactant. At constant volume</p>
<p>\begin{align}
[\ce{A}] &= [\ce{A}]_0 (1 - x) \\
[\ce{B}] &= [\ce{B}]_0 - 2x[\ce{A}]_0
\end{align}</p>
<p>$$\pu{0.028 mol L-1} = \pu{0.09 mol L-1} - 2 \cdot \pu{0.035 mol L-1} \cdot x \to x = 0.886$$</p>
<hr>
<p>Part <strong>b</strong>.</p>
<p>$$-\ln{(1 - x)} = kt$$
$$t = 6$$
$$0.024 = 0.09 - 2 \cdot 0.035 \cdot x \to x = 0.943$$
$$-\ln{(1 - 0.943)} = 6k \qquad \color{red}{?}$$</p>
<hr>
<p>Part <strong>c</strong>.</p>
<p>$$p_T = C_T RT$$
$$0.039 = 0.09 - 2 \cdot 0.035 \cdot x \to x = 0.728$$
\begin{align}
[\ce{A}] &= 0.035 (1 - 0.728) = 9.52 \cdot 10^{-3} \\
[\ce{B}] &= 0.039 \\
[\ce{C}] &= [\ce{A}]_0 \cdot x = 0.035 \cdot 0.728
\end{align}
$$C_T = 0.074$$
$$p_T = \pu{0.076 mol L-1} \cdot \pu{0.082 L atm mol-1 K-1} \cdot \pu{(273 + 227) K} = \pu{3.034 atm}$$</p>
<hr>
<p>Part <strong>d</strong>.</p>
<p>$$\epsilon = \frac{1 - 3}{3} = -\frac{2}{3}$$
$$V = V_0(1 + \epsilon x) \qquad \color{red}{?}$$</p>
Answer:
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https://chemistry.stackexchange.com/questions/86826/kinetics-of-a-2-b-c-reaction
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Question: <p>Consider two chemicals, $\ce{A}$ and $\ce{B}$ that react with each other to make $\ce{C}$ with a reaction rate $k$. The reaction can be expressed as $$\ce{A + B->C}$$ The equation expressing the rate of the reactions can be expressed as $$\frac{d[\ce{A}]}{dt}=\frac{d[\ce{B}]}{dt}=-\frac{d[\ce{C}]}{dt}=-k[\ce{A}][\ce{B}]$$</p>
<p>I can separate this equation to make a system of differential equations.$$\frac{d[\ce{A}]}{dt}=-k[\ce{A}][\ce{B}]$$ $$\frac{d[\ce{B}]}{dt}=-k[\ce{A}][\ce{B}]$$</p>
<p>With these two equations, I note that they are similar and will only work with one of these equations for the time being. Therefore, we can write one of these equations as $$\frac{d\ln([\ce{A}])}{dt}=-k[\ce{B}]$$ and by taking another derivative $$\frac{d^2 \ln([\ce{A}])}{dt}=-k\frac{[d\ce{B}]}{dt}=-k\frac{[d\ce{A}]}{dt}$$</p>
<p>I solved this equation using <a href="http://www.wolframalpha.com/input/?i=d%5E2%20ln(y(x))%2Fdx%5E2%3D-k*dy%2Fdx" rel="noreferrer">Wolfram Alpha</a> (QED) $$[\ce{A}](t)=\frac{c_1 \exp[c_1(t+c_2)]}{k \exp[c_1(t+c_2)]-1}$$ Therefore the rate of reaction can is $$[\ce{A}]'(t)=\frac{c_1^2 \exp[c_1(t+c_2)]}{k \exp[c_1(t+c_2)]-1}-\frac{k c_1^2 \exp^2[c_1(t+c_2)]}{(k \exp[c_1(t+c_2)]-1)^2}$$</p>
<p>I observed that the rate of change can be written as $$[\ce{A}]'(t)= c_1 [\ce{A}](t)-k [\ce{A}](t)^2$$ so that $c_1$ may be solved, given the initial conditions of $[\ce{A}](0)$ and $[\ce{A}]'(0)$ such that $$c_1=\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)}$$</p>
<p>Substituting the definition of $c_1$ into the equation of $[A](t)$ and $[A]'(t)$ an equation for $c_2$ can be found.</p>
<p>$$c_2=\frac{1}{c_1} \ln(1-\frac{c_1 k}{[\ce{A}](0)}) $$$$
c_2= \frac{[\ce{A}](0)}{[\ce{A}]'(0)+k[\ce{A}](0)^2} \ln(1-\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2 }{[\ce{A}](0)^2}k) $$</p>
<p>Using the equations for $c_1$ and $c_2$ an explicit equation for $[\ce{A}](t)$ can be found.</p>
<p>$$ [\ce{A}](t)=\frac{\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)} \exp[\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)}(t+\frac{[\ce{A}](0)}{[\ce{A}]'(0)+k[\ce{A}](0)^2} \ln(1-\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2 }{[\ce{A}](0)^2}k))]}{k \exp[\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)}(t+\frac{[\ce{A}](0)}{[\ce{A}]'(0)+k[\ce{A}](0)^2} \ln(1-\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2 }{[\ce{A}](0)^2}k))]-1} $$</p>
<p>Side note: since $[\ce{A}]'(0)=[\ce{B}]'(0)= -k[\ce{A}](0)[\ce{B}](0)$ then $c_1$ can be rewritten as $$c_1=\frac{-k[\ce{A}](0)[\ce{B}](0)+k[\ce{A}](0)^2}{[\ce{A}](0)}=k([\ce{A}](0)-[\ce{B}](0))$$</p>
<p>this simplifies $c_2$ to </p>
<p>$$c_2=\frac{k^{-1}}{[\ce{A}](0)-[\ce{B}](0)} \ln(\frac{[k^2 \ce{B}](0) }{[\ce{A}](0)})$$</p>
<p>which simplifes the equation for $[\ce{A}](t)$ to </p>
<p>$$[\ce{A}](t)=\frac{k([\ce{A}](0)-[\ce{B}](0)) \exp[k([\ce{A}](0)-[\ce{B}](0))(t+\frac{k^{-1}}{[\ce{A}](0)-[\ce{B}](0)} \ln(\frac{[k^2 \ce{B}](0) }{[\ce{A}](0)}))]}{k \exp[k([\ce{A}](0)-[\ce{B}](0))(t+\frac{k^{-1}}{[\ce{A}](0)-[\ce{B}](0)} \ln(\frac{[k^2 \ce{B}](0) }{[\ce{A}](0)}))]-1}$$</p>
<p>with a similar equation for $[\ce{B}](t)$</p>
<p><strong>My quesion is: Is this a valid mathematical model for a bimolecular reaction? If not, what is commonly used?</strong></p>
Answer: <p>Assuming that the <em>bimolecular</em> chemical reaction $\ce{A + B ->[\kappa] C}$ has <a href="https://en.wikipedia.org/wiki/Law_of_mass_action" rel="nofollow noreferrer">mass action</a> kinetics, we have the following pair of coupled ODEs</p>
<p>$$\begin{array}{rl} \dot a &= - \kappa \, a \, b\\ \dot b &= - \kappa \, a \, b\end{array}$$</p>
<p>where $\kappa > 0$ is the rate constant, $a := [\ce{A}]$ and $b := [\ce{B}]$. Since $\dot a = \dot b$, we have $\frac{\mathrm d}{\mathrm d t} \left( a - b \right) = 0$ and, thus, integrating, we obtain</p>
<p>$$a (t) - b (t) = a_0 - b_0$$</p>
<p>where $a_0 > 0$ and $b_0 > 0$ are the initial concentrations. Since $b (t) = a (t) - (a_0 - b_0)$, the 1st ODE can be decoupled from the 2nd, as follows</p>
<p>$$\dot a = - \kappa \, a \, \left( a - (a_0 - b_0) \right)$$</p>
<p>which can be rewritten in the form</p>
<p>$$\frac{\mathrm d a}{a \, \left( a - (a_0 - b_0) \right)} = - \kappa \, \mathrm d t$$</p>
<p>Assuming that $a_0 \neq b_0$, we have the following partial fraction expansion</p>
<p>$$\left( \frac{1}{a - (a_0 - b_0)} - \frac{1}{a} \right) \mathrm d a = - \kappa \, (a_0 - b_0) \, \mathrm d t$$</p>
<p>Integrating, we obtain</p>
<p>$$\ln \left( \frac{a (t) - (a_0 - b_0)}{a_0 - (a_0 - b_0)} \right) - \ln \left( \frac{a (t)}{a_0} \right) = - \kappa \, (a_0 - b_0) \, t$$</p>
<p>which can be rewritten as follows</p>
<p>$$\ln \left( \frac{a (t) - (a_0 - b_0)}{a (t)} \right) = \ln \left( \frac{b_0}{a_0} \right) - \kappa \, (a_0 - b_0) \, t$$</p>
<p>Exponentiating both sides, we obtain</p>
<p>$$\frac{a (t) - (a_0 - b_0)}{a (t)} = \frac{b (t)}{a (t)} = \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0 - b_0) \, t)$$</p>
<p>and, eventually, we obtain</p>
<p>$$\boxed{\begin{array}{rl} &\\ a (t) &= \dfrac{a_0 - b_0}{1 - \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0 - b_0) \, t)}\\\\ b (t) &= \dfrac{(a_0 - b_0) \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0 - b_0) \, t)}{1 - \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0 - b_0) \, t)}\\ & \end{array}}$$</p>
<p>Taking the limit,</p>
<p>$$\lim_{t \to \infty} a (t) = \begin{cases} a_0 - b_0 & \text{if } a_0 > b_0\\\\ 0 & \text{if } a_0 < b_0\end{cases}$$</p>
<p>$$\\$$</p>
<p>$$\lim_{t \to \infty} b (t) = \begin{cases} 0 & \text{if } a_0 > b_0\\\\ b_0 - a_0 & \text{if } a_0 < b_0\end{cases}$$</p>
<hr>
<h3>What if $a_0 = b_0$?</h3>
<p>Previously, we assumed that $a_0 \neq b_0$. If $a_0 = b_0$, then</p>
<p>$$\frac{\mathrm d a}{a \, \left( a - (a_0 - b_0) \right)} = - \kappa \, \mathrm d t$$</p>
<p>becomes</p>
<p>$$-\frac{\mathrm d a}{a^2} = \kappa \, \mathrm d t$$</p>
<p>Integrating, we obtain</p>
<p>$$\frac{1}{a (t)} - \frac{1}{a_0} = \kappa \, t$$</p>
<p>and, eventually, we obtain</p>
<p>$$\boxed{ a (t) = \frac{a_0}{1 + a_0 \, \kappa \, t} = b (t)} $$</p>
<p>In this case, both reactants are eventually exhausted</p>
<p>$$\lim_{t \to \infty} a (t) = \lim_{t \to \infty} b (t) = 0$$</p>
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https://chemistry.stackexchange.com/questions/83631/analytical-solution-for-kinetics-of-bimolecular-reaction
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Question: <p>The way kinetics is taught at the undergraduate level (Arrhenius and collision theory) chemical equilibrium is determined governed immensely by activation energy of the reaction. According to thermodynamics, however equilibrium is a function of free energy change. In a way, thermodynamics and kinetics seem to contradict each other. What insight am I missing. Is the energy in energy vs reaction coordinates supposed to be Gibbs free energy instead of enthalpy? Would this imply activation energy changes as a reaction proceeds? Is the way kinetics is taught wrong?</p>
Answer: <p>A great deal of chemistry is determined by the interaction of kinetics and thermodynamics. The world would be a dull place if we didn't have both.</p>
<p>While thermodynamics determines what directions a reaction can go, the kinetics often determines whether the reaction can happen. Take a simple example: diamond is not the stable from of the element carbon at room temperature; graphite is. If the world were purely determined by thermodynamics all carbon would be graphite and men would be unable to impress women with the expensive shiny jewellery they buy them. But this reaction doesn't happen.</p>
<p>The reason it doesn't is because to convert diamond into graphite we would have to have to radically restructure the chemical bonding in diamond from a tetrahedral network breaking at least one bond for every carbon atom. Breaking all those bonds requires a huge input of energy to kick off the process (thermodynamics says you will get that energy plus a little more back if you can make this happen). You can think of diamond with a whole bunch of broken carbon-carbon bonds as an <em>intermediate</em> structure on the path from diamond to graphite. But that intermediate structure needs a great deal of energy to create and there simply isn't enough energy at room temperature to get there. The reaction will proceed at very high temperatures (and, if the pressure is high enough, diamond will be the thermodynamically favoured form). This is how the earth (or industry) creates diamonds: at high temperatures there is enough energy to break the carbon-carbon bonds and the equilibrium can be established (and, at high pressures, will favour diamond and at lower pressures graphite). </p>
<p>The point of all this is that, in chemistry, what happens involves the <em>interplay</em> of thermodynamics and kinetics. If it takes a lot of energy to get to an intermediate, the thermodynamic products cannot be reached. So being able to achieve an equilibrium depends not just on the start and end products of a reaction but the intermediate structures or compounds that you have to pass though to interconvert them. If those structures are hard to get to because you don't have enough energy around then thermodynamics is irrelevant to what will happen. you always have to think through the <em>mechanism</em> behind the reaction and understand the energy required to create the intermediate states along the reaction path.</p>
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https://chemistry.stackexchange.com/questions/25077/kinetics-vs-thermodynamics
|
Question: <blockquote>
<p>Find <span class="math-container">$[\ce{C}]/[\ce{A}]$</span> for the following system at equilibrium:</p>
<p><a href="https://i.sstatic.net/tY2EA.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/tY2EA.png" alt="Cyclic reversible reaction ABC" /></a></p>
</blockquote>
<p>I know that at equilibrium</p>
<p><span class="math-container">$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-3}}{k_3} = K_3, \tag{1}$$</span></p>
<p>but my teacher told me there was another way to express it with rate constant and gave the hint that the numerator is the sum of three terms:</p>
<p><span class="math-container">$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-3}}{k_3} = \frac{\ldots + \ldots + \ldots}{\ldots}. \tag{2}$$</span></p>
<p>So, I use steady state approximation (SSA) to solve this problem:</p>
<p><span class="math-container">$$
\begin{align}
\frac{\mathrm d[\ce{A}]}{\mathrm dt} &= k_3[\ce{C}] + k_{-1}[\ce{B}] - (k_1 + k_{-3})[\ce{A}] = 0 \tag{3} \\
\frac{\mathrm d[\ce{B}]}{\mathrm dt} &= k_1[\ce{A}] + k_2[\ce{C}] - (k_{-1} + k_{-2})[\ce{B}] = 0 \tag{4} \\
\frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_{-2}[\ce{B}] + k_{-3}[\ce{A}] - (k_2 + k_3)[\ce{C}] = 0 \tag{5}
\end{align}
$$</span></p>
<p><span class="math-container">$$[\ce{C}] = \frac{k_{-2}[\ce{B}] + k_{-3}[\ce{A}]}{k_2 + k_3} \tag{6}$$</span></p>
<p><span class="math-container">$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}[\ce{B}]/[\ce{A}] + k_{-3}}{k_2 + k_3} \tag{7}$$</span></p>
<p>Since</p>
<p><span class="math-container">$$\frac{[\ce{B}]}{[\ce{A}]} = \frac{k_1}{k_{-1}}, \tag{8}$$</span></p>
<p><span class="math-container">$$
\begin{align}
\frac{[\ce{C}]}{[\ce{A}]} &= \frac{k_{-2}k_1/k_{-1} + k_{-3}}{k_2 + k_3} \\
&= \frac{k_{-2}k_1 + k_{-1}k_{-3}}{k_{-1}k_2 + k_{-1}k_3}. \tag{9}
\end{align}
$$</span></p>
<p>I don't get the correct answer term just as my teacher said. I get two or four terms in numerator, so my answer is likely wrong. Where am I mistaken?</p>
<p><span class="math-container">$
% \documentclass{article}
% \usepackage{chemfig}
% \begin{document}
%
% \schemestart
% A
% \arrow(A--C){<=>[$k_{-3}$][$k_{3}$]}[-60,1.25,,]
% C
% \arrow(@A--B){<=>[${k_{-1}}$][${k_{1}}$]}[-120,1.25,,]
% B
% \arrow(@B--@C){<=>[$k_{2}$][$k_{-2}$]}[,1.25,,]
% \schemestop
%
% \end{document}
$</span></p>
Answer: <p>Using the <a href="https://en.wikipedia.org/wiki/Steady_state_(chemistry)" rel="nofollow noreferrer">steady state approximation</a> (SSA), we can obtain the following:</p>
<p><span class="math-container">$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}\frac{[\ce{B}]}{[\ce{A}]} + k_{-3}}{k_2 + k_3} \tag{1}$$</span></p>
<p><span class="math-container">$$\frac{[\ce{B}]}{[\ce{A}]} = \frac{(k_1 + k_{-3})[\ce{A}] - k_3\frac{[\ce{C}]}{[\ce{A}]}}{k_{-1}} \tag{2}$$</span></p>
<p><span class="math-container">$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}\left((k_1 + k_{-3}) - k_3\frac{[\ce{C}]}{[\ce{A}]}\right) + k_{-1}k_{-3}}{k_{-1}(k_2 + k_3)}\tag{3}$$</span></p>
<p><span class="math-container">$$k_{-1}(k_2 + k_3)\frac{[\ce{C}]}{[\ce{A}]} = k_{-2}(k_1 + k_3) - k_{-2}k_3\frac{[\ce{C}]}{[\ce{A}]} + k_{-1}k_{-3} \tag{4}$$</span></p>
<p><span class="math-container">$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}(k_1 + k_{-3}) + k_{-1}k_{-3}}{k_{-1}(k_2 + k_3) + k_{-2}k_3} \tag{5}$$</span></p>
<p>If you want to show three terms in the numerator (to match the teacher's hint), you can also write it this way:</p>
<p><span class="math-container">$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}k_1 + k_{-2}k_{-3} + k_{-1}k_{-3}}{k_{-1}k_2 + k_{-1}k_3 + k_{-2}k_3} \tag{6}$$</span></p>
|
https://chemistry.stackexchange.com/questions/168399/chemical-kinetics-for-cyclic-reversible-reaction
|
Question: <p>In graph given below is a sequential first order reaction</p>
<p><span class="math-container">$$\ce{A ->[$k_1$] B ->[$k_2$] C}$$</span></p>
<p>For <span class="math-container">$k_2 \gg k_1$</span> the graph of concentration of <span class="math-container">$\ce{A}$</span>, <span class="math-container">$\ce{B}$</span> and <span class="math-container">$\ce{C}$</span> is as follows.</p>
<p><a href="https://i.sstatic.net/OuQ76.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/OuQ76.jpg" alt="enter image description here" /></a></p>
<p>Won't the graph of the concentrations of <span class="math-container">$\ce{B}$</span> and <span class="math-container">$\ce{C}$</span> be the same as if <span class="math-container">$\ce{B}$</span> is formed then only <span class="math-container">$\ce{C}$</span> will be formed simultaneously? Then why is <span class="math-container">$\ce{C}$</span>'s graph different?</p>
Answer: <p>This question is about the <strong>rate-determining step</strong> of multi-step reactions. Bear in mind that the overall rate of reaction is equal to the slowest step of the reaction, in which case, from <span class="math-container">$\ce{A}$</span> to <span class="math-container">$\ce{B}$</span> (<span class="math-container">$k_2 \gg k_1$</span>). Since it is a first-order reaction, the overall reaction rate = <span class="math-container">$k_1 [\ce{A}]$</span>. The graph of concentration vs time for a first-order reaction should be like <span class="math-container">$y = \frac1x$</span>.</p>
<p>On the other hand, <span class="math-container">$\ce{B}$</span> is the <strong>reaction intermediate</strong>. <span class="math-container">$[\ce{B}]$</span> initially increases when the reactant <span class="math-container">$\ce{A}$</span> is gradually consumed. But soon, <span class="math-container">$[\ce{B}]$</span> reduces to form <span class="math-container">$\ce{C}$</span>.</p>
<p>The graph of <span class="math-container">$[\ce{C}]$</span> must be symmetrical about the horizontal line <span class="math-container">$y = \frac12[\ce{A}]$</span> because as stated earlier, the overall reaction rate <span class="math-container">$= k_1 [\ce{A}]$</span>.</p>
|
https://chemistry.stackexchange.com/questions/174588/sequential-first-order-reaction-graph-in-chemical-kinetics
|
Question: <p>I would like to know according to which we can state that there is a unique equilibrium of a closed chemical reaction network (at constant circumstances) without supposing mass-action kinetics if time is increasing beyond all limits. Can we derive it from a non-equilibrium thermodynamic law or is it a postulate? Thank you for your answer! </p>
Answer: <p>The equilibrium is determined by the thermodynamics, not by the kinetics. The rate constants for the reaction kinetics must be consistent with the equilibrium determined by the thermodynamics. For example, for a single reversible reaction, the reverse reaction rate constant must be equal to the forward rate constant divided by the equilibrium constant.</p>
|
https://chemistry.stackexchange.com/questions/54519/what-guarantees-that-there-is-a-thermodynamic-equilibrium-in-a-closed-chemical-r
|
Question: <p>What does it mean when the order of reaction isn't an integer, e.g., a reaction order of 2.43. Not with something acting as a catalyst but rather as a inhibitor. Why does the inhibitor produce an unusual reaction order? Specifically fluoride ions slowing down the reaction between calcium carbonate and acid by forming calcium fluoride.</p>
Answer:
|
https://chemistry.stackexchange.com/questions/42145/how-does-a-reaction-exhibit-kinetics-with-a-non-integer-reaction-order
|
Question: <p>Can one be understood on the basis of the other or are they not interrelated at all?</p>
<p>The first thing my kinetics textbook demonstrated was how thermodynamics ignores time taken for a process whereas kinetics considers the time and rate of the process and hence kinetics is a more effective way to view change.</p>
<p>Also, I saw examples of highly thermodynamically feasible reactions which were not kinetically favourable in the sense that they were so slow that the entire process could be neglected as not happening.</p>
<p>Can one understand the thermodynamics of a reaction by studying it kinetics?</p>
<p>Or can one figure out the kinetics of a reaction by thermodynamical consideration?</p>
<p>Also, would it be a wise observation to say that thermodynamics is a largely theoretical subject while kinetics is a more experimental field?</p>
Answer: <p>No. Both I think are equally theoritical or experimental. You must study both seperately. Thermodynamics doesn't speak about rates. While kinetics doesn't tell you stability.
If a reaction produces more than one product, thermodynamics will tell you which is more stable, while that doesn't necessarily mean it is the major product. It may happen that rate of formation of other product is more under certain conditionds. Read about thermodynamic vs kinetic product and you will get it.</p>
|
https://chemistry.stackexchange.com/questions/44155/what-is-the-relation-between-chemical-thermodynamics-and-chemical-kinetics
|
Question: <p>According to the referenced paper, VR has an AChE inhibition reaction rate constant almost 4 times that of VX. Interestingly, its reaction rate constant for aging of the enzyme is less than that of VX, despite the fact that, as acknowledged by the authors of the paper, aging tends to occur faster when the leaving group during aging is a branched alkoxy group. Finally, VR-inhibited AChE is also more prone to spontaneous reactivation compared to AChE inhibited by VX. What is the reason for these differences in reaction kinetics for the interactions of these two agents with AChE?</p>
<p><strong>Reference:</strong></p>
<p>Worek, F., Thiermann, H., Szinicz, L., & Eyer, P. (2004). Kinetic analysis of interactions between human acetylcholinesterase, structurally different organophosphorus compounds and oximes. Biochemical Pharmacology, 68(11), 2237–2248. doi:10.1016/j.bcp.2004.07.038</p>
Answer:
|
https://chemistry.stackexchange.com/questions/147164/differences-in-ache-inhibition-kinetics-between-vx-and-vr
|
Question: <p>When we perform a organic reaction without any addtional catalyst except that the reaction is carried out under microwave condition we can accelerate this reaction in comparison with one under heating only. So, microwaves can be considered a form of catalysis according to the concept of reaction kinetics?</p>
Answer: <p>Here is an example of an entire online GoogleBook dedicated to the topic <a href="https://www.google.com/books/edition/Microwave_Assisted_Organic_Synthesis/CrfzIiN5qlgC?hl=en&gbpv=1&dq=organic+microwave+assisted+synthesis&pg=PR5&printsec=frontcover" rel="nofollow noreferrer">Microwave Assisted Organic Synthesis</a>, apparently based on some 2,000 papers.</p>
<p>On page 3 of the cited reference, notes to quote:</p>
<blockquote>
<p>Interaction between microwave radiation and solutions of polar molecule polar molecules may be adequately described using classical models, which may be derived from Maxwell's equations.</p>
</blockquote>
<p>And further:</p>
<blockquote>
<p>Although this analysis has concentrated on the oscillating components of the electromagnetic field of the microwave radiation, it is also possible to get interaction between the oscillating magnetic field and the magnetic dipoles in the sample. The high-heating rates for Fe304 may be attributed to such interactions.</p>
</blockquote>
<p>With respect to benefits, per page 23 to quote:</p>
<blockquote>
<p>Reaction rates are generally very high and yields in many cases can be greatly improved, as competing side reactions can be minimized...</p>
</blockquote>
<p>I would also add that the use of microwave-based dielectric heating is, at times, more than just fast and efficient and heating, as it can, in the presence of activated carbon, for example, introduce surface-based radicals into the reaction mix. See, for example, <a href="https://www.researchgate.net/publication/239705822_Generation_of_hydroxyl_radical_in_aqueous_solution_by_microwave_energy_using_activated_carbon_as_catalyst_and_its_potential_in_removal_of_persistent_organic_substances" rel="nofollow noreferrer">Generation of hydroxyl radical in aqueous solution by microwave energy using activated carbon as catalyst and its potential in removal of persistent organic substances</a>.</p>
<p>As such, I described the general use of this new heating technology as microwave-assisted synthesis, and not per se, as catalytic, albeit depending on the reaction mix, it can be (per last cited reference).</p>
|
https://chemistry.stackexchange.com/questions/142850/organic-catalysis-microwaves
|
Question: <p>The rate of reaction is given by the following equation, where [Ker-S-S-Ker] is just the molecule being broken down and [RSH] is an abbreviation for thioglycolic acid</p>
<p>The thioglycolate is in large excess and so essentially remains unchanged.</p>
<p><a href="https://i.sstatic.net/eQb9Q.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/eQb9Q.png" alt="enter image description here" /></a></p>
<p>Can somebody explain why the answer is C?</p>
<p>I thought the answer was E (because the rate of change, i.e. the derivative, is equal to the concentration of the reactant itself and so, I thought this signifies a decaying curve?)</p>
<p>Many thanks.</p>
Answer: <p>Indeed, the only thing you'll need is solving the first order differential equation (but before that, let me define <span class="math-container">$R_0 := \ce{[RSH]_0}$</span> and <span class="math-container">$C :=\ce{[Ker-S-S-Ker]}$</span>.</p>
<p>Now, let's solve it assuming (that's what is given) that <span class="math-container">$R_0$</span> can be considered a constant.
Then, we have : <span class="math-container">$$\frac{\operatorname{d}C}{\operatorname{d}t} = -kR_0C \implies \frac{\operatorname{d}C}{\operatorname{d}t} + kR_0C = 0 \implies C(t) = \exp\left(-\frac{t}{kR_0}\right) + C_0$$</span> (integration constants can be determined using initial values, I won't be giving further details here on how to solve this...)</p>
<p>Hence, since the concentration of your molecule is an exponential function of time, it is inevitably a convex function (so only curves C and E can work).
Finally, a simple thought on the values of <span class="math-container">$k$</span> and <span class="math-container">$R_0$</span> show that curve E cannot be the answer as the decay is to "fast".</p>
<p>Hope this helps, if needed I can also provide a "variables separation" method of solving the differential equation.</p>
|
https://chemistry.stackexchange.com/questions/144756/rate-of-change-and-kinetics-order-of-reaction-question
|
Question: <p>I'm trying to understand what are the main factors that influence the reaction rate of an irreversible exothermic reaction. </p>
<p>I think these could be the main factors: </p>
<ol>
<li>chemical nature of the reagents and their concentration;</li>
<li>nature of the solution (homogeneous or heterogeneous)</li>
<li>presence of catalysts </li>
<li>temperature. </li>
</ol>
<p>(Let me know if I missed some factors or added one or two that had nothing to do with what I'm looking for.)</p>
<p>But I struggle understanding how the temperature influences an exothermic reaction. I can't use Le Chatelier's principle since the reaction is not reversible, so how do I know what is the effect of (for example) an increased temperature on such reaction? </p>
<p>I'm prone to think that an increased temperature will bring more products, but then again I'm not so sure since is an irreversible reaction.</p>
Answer: <p>I assume by kinetics you mean the rate of the reaction (i.e. how fast it happens), in which case the factors you mention are fine. </p>
<p>Regarding the temperature: the rate constant and therefore the reaction rate at a given concentration always increases with temperature as you can see in the Arrhenius equation. The fact the the reaction is exothermic means that it releases heat which, in the case where no adequate temperature control is applied (such as an ice bath), will increase the temperature of the solution and therefore the rate of the reaction releasing even more heat and so on, resulting in a runaway exotherm. This is the reason why exothermic reactions must be well temperature controlled especially when performed in large scale.</p>
<p>If your reaction was reversible, then heating it would not only accelerate the reaction (both forwards and backwards at equilibrium) but also shift the equilibrium towards the reactants.</p>
|
https://chemistry.stackexchange.com/questions/91147/factors-that-influence-the-kinetics-of-an-irreversible-exothermic-reaction
|
Question: <p>The answer is A, but how is that? I know B is an intermediate, so that after short time it will be bigger than C, but what does "thermodynamic control" and "kinetic control" mean?</p>
<p><a href="https://i.sstatic.net/7yoMM.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/7yoMM.png" alt="https://i.sstatic.net/7yoMM.png"></a></p>
Answer: <p>Initially no B and C exist. At short times as $k_1 > k_2$ production of B is favoured over C. (The initial amounts of B and C are so small that back reactions can be ignored wrt. forwards ones.) At long times equilibrium constants matter and as the equilibrium constant $K_2$ is greatest this is favoured to produce C over B.</p>
|
https://chemistry.stackexchange.com/questions/70344/how-do-thermodynamics-and-kinetics-control-this-reaction
|
Question: <p>What will be the order of the reaction for a chemical change having <span class="math-container">$\log t_{1/2}$</span> VS <span class="math-container">$\log a$</span> Where <span class="math-container">$a=$</span> is the initial concentration of reactant and <span class="math-container">$t_{1/2} =$</span> Half Life?</p>
<p><a href="https://i.sstatic.net/f2pT7.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/f2pT7.jpg" alt="enter image description here" /></a></p>
<ol>
<li>0 order</li>
<li>1st order</li>
<li>2nd order</li>
<li>None of These</li>
</ol>
<p>Actually the answer I found by searching in internet is option 1 but following my calculations, I am getting the answer as option 4.</p>
<p>My calculations are as follows:</p>
<p>For 0 order reaction:
<span class="math-container">$$t_{1/2} = \frac{a}{2k}$$</span></p>
<p>Taking <span class="math-container">$\log$</span> on both sides:</p>
<p><span class="math-container">$$\log t_{1/2} = \log{\frac{a}{2k}}$$</span></p>
<p><span class="math-container">$$\log t_{1/2} = \log a - \log 2k \tag1$$</span></p>
<p>This is a Straight line equation of type <span class="math-container">$y= mx - C $</span>.</p>
<p><strong>MY DOUBT IS</strong>
If you notice in graph you will get to know that the intercept is positive while in the equation <span class="math-container">$(1)$</span>, intercept is negative so how is it possible?</p>
<p>My Background- actually I didn't have kept Mathematics as my major subject in senior secondary high school so I have little knowledge with respect to graph so please forgive me if I am wrong at formulating (1).</p>
Answer: <p>The equation that you have derived is correct. For a zeroth-order reaction, <span class="math-container">$$\log t_{1/2}=\log a-\log2K$$</span>
By superficially observing, this seems to be giving a straight line with the equation <span class="math-container">$y=mx-c$</span> but actually there are several(not all) reactions in which the <span class="math-container">$K$</span> value is very small. In case of any reaction having <span class="math-container">$K\lt0.5$</span>, <span class="math-container">$2K$</span> becomes less than <span class="math-container">$1$</span> and hence <span class="math-container">$\log2K$</span> becomes negative. Then you would obtain an equation of the form <span class="math-container">$y=mx+c$</span> whose graph will have a positive ordinate and it will appear similar to the graph in the question. So that graph in your question is correct for a zero-order reaction having a small rate constant (less than <span class="math-container">$0.5$</span>).</p>
|
https://chemistry.stackexchange.com/questions/162616/finding-the-reaction-order-from-a-given-plot-of-chemical-kinetics
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Question: <p>It is always stressed that one should never mix Kinetics with Thermodynamic. How is it then possible to calculate the free energy of activation with the Arrhenius equation? As I understood, the Arrhenius equation and thus the activation energy of a reaction "belongs" to kinetics, whereas the free energy represents a term of Thermodynamics.</p>
<p>An example of the calculation I refer to can be found on page 7 of this link:
<a href="http://cbc.arizona.edu/classes/bioc460/summer/460web/lecture/EnzIntrod_11.pdf" rel="nofollow noreferrer">http://cbc.arizona.edu/classes/bioc460/summer/460web/lecture/EnzIntrod_11.pdf</a></p>
Answer:
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https://chemistry.stackexchange.com/questions/74634/arrhenius-equation-in-kinetics-and-thermodynamics
|
Question: <p>Why don't we take the effects of stirring and shaking on the rates of chemical reactions? While studying chemical kinetics, I've seen that the rates depend on the concentrations of various species, temperature, pressure, etc. But I've never seen an equation taking into account the mechanical stirring, shaking and geometry of the beaker, presumably because these things are hard to experimentally measure and form a physical model which incorporates these things. For example, if the iodine clock reaction changes its rates due to stirring, every time I do the reaction, I shall get different times for colour changes. <em>Then why do chemical kinetics equations make sense at all?</em></p>
Answer:
|
https://chemistry.stackexchange.com/questions/150043/effect-of-stirring-and-shaking-on-chemical-kinetics
|
Question: <blockquote>
<p>The reaction of tert-butyl bromide with azide ion in aqueous solution
is proposed to proceed by the following mechanism:</p>
<p>$\ce{(CH3)3CBr(aq) <=>(CH3)3C+(aq) + Br–(aq)}$</p>
<p>$\ce{(CH3)3C+(aq) + N3–(aq) \to (CH3)3CN3(aq)}$</p>
<p>Assuming that $\ce{(CH3)3C+ (aq)}$ achieves a steady-state concentration, but making no further assumptions about the relative magnitudes of the three rate constants,
what is the rate law for this reaction, where $k_1$ is the forward
reaction rate of the first equation, $k_{-1}$ is the reverse reaction
rate of the first equation, and $k_2$ is the forward reaction rate of
the second equation?</p>
<p>(A) $\ce{k_1[(CH3)3CBr]}$</p>
<p>(B) $\ce{k_2[(CH3)3CBr][N3-]}$</p>
<p>(C) $\ce{\displaystyle\frac{k_1k_2[(CH3)3CBr][N3-]}{k_{-1}[Br-]}}$</p>
<p>(D) $\ce{\displaystyle\frac{k_1k_2[(CH3)3CBr][N3-]}{k_{-1}[Br-] + k2[N3-]}}$</p>
</blockquote>
<p><strong>My thoughts:</strong> First of all, I read that $\ce{(CH3)3C+ (aq)}$ is a steady state concentration, which meant that $\frac{d\ce{[(CH3)3C+ (aq)}]}{dt} = 0,$ which prompted me to write out differential equations for each reactant. Thus, I got a system of differential equations:
$$\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} =-\frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}]$$
$$0=\frac{d\ce{[(CH3)3C+ (aq)}]}{dt} = -\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} - \frac{d\ce{[(CH3)3CN3 (aq)}]}{dt} = \frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}] -k_2 \frac{d\ce{[(CH3)3CN3 (aq)]}}{dt}$$
$$\frac{d\ce{[Br- (aq)}]}{dt} = -\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} = \frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}]$$
$$\frac{d\ce{[N3- (aq)}]}{dt} = -k_2\ce{[N3-(aq)]}$$</p>
<p>But I was unsure of how to use these differential equations to actually find the rate law. Could anyone provide any suggestions?</p>
Answer: <p>You really don't have to write all that differential equations. </p>
<p>Note: I'll write all my steps on a paper, since MathJax formatting is a little time-taking at the moment.</p>
<p>Consider the rate laws for the two reactions:
<img src="https://i.sstatic.net/6Dzil.jpg" alt="enter image description here"></p>
<p>Since the $\ce{[(CH3)3C+]}$ isn't changing with time, we get
<img src="https://i.sstatic.net/cJVw5.jpg" alt="enter image description here"></p>
<p>From which, you get</p>
<p><img src="https://i.sstatic.net/XEN5g.jpg" alt="enter image description here"></p>
<p>Plugging these in the Rate2 which I mentioned, you will get,
<img src="https://i.sstatic.net/kqAzG.jpg" alt="enter image description here"></p>
<p>Which is the same as what's given in option (D).</p>
<p>I hope I have satisfied your queries.</p>
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https://chemistry.stackexchange.com/questions/73243/chemical-kinetics-with-the-reaction-of-tert-butyl-bromide-with-azide-ion
|
Question: <p>What optical methods can be used to measure the concentration of reacting gases in a reactor during an exothermic and fast reaction? I need to monitor and measure the <em>local concentration of my gases</em> (in the methane production reaction) along a reactor bed in order to be able to clarify the <em>kinetics</em> of this reaction. Which optical techniques are able to make such measurement?</p>
Answer: <p>There are a number of techniques but what you would use depends on how you plan to start the reaction. What use to be called Flash-Photolysis and now called Pump-Probe Spectroscopy is the most general and can be used from the femtosecond to multi-second time range. This is a general method as it measured the absorption of all species produced. It works by using a suitable duration and wavelength flash of light, usually from a laser, to initiate the reaction and then at various time delays after this, the absorption spectrum of any species produced is measured. Absorption can be in the uv, visible or ir ranges. Sometimes Raman scattering or fluorescence can be used instead of absorption. What is used depends on equipment available and the species you need to measure.</p>
<p>If you cannot start the reaction with light then mixing methods can be used, but these are relatively slow, longer than microsecond time scale and more like milliseconds. In the Stopped-Flow method the reactant gases are mixed as rapidly as possible in the chamber so as to start the reaction then this is monitored with absorption etc. as above. Additionally, there are flow methods whereby the gaseous reactants are mixed at the end of a long flow tube and are monitored spectroscopically at various points as they flow along a tube. The gases are pumped to cause the flow and so distance along the tube is an alias for reaction time. Each of these methods can be used with gases or liquids.</p>
|
https://chemistry.stackexchange.com/questions/80848/optical-techniques-for-kinetics-studies
|
Question: <p>I know sodium metal reacts violently with water, but what about aqueous solutions (like sodium bicarbonate). Also, what would the reaction kinetics be? </p>
<p>I have watched YouTube videos were a small piece of elementary sodium reacts completely with water in a matter of seconds. Would that also be the case of elementary sodium in aqueous solutions (for example sodium bicarbonate)?</p>
Answer: <p>It helps to know what the reaction actually is. When sodium metal reacts with water, it is oxidised and the hydrogen in water is reduced according to the following equation:</p>
<p>$$\ce{2Na (s) + 2 H2O (l) -> 2 Na+ (aq) + 2 OH- (aq) + H2 ^ (g)}\tag{1}$$</p>
<p>Note that no other components play a role; the only thing that does is the $\mathrm{pH}$ value of the water (lower $\mathrm{pH}$ would mean a more violent reaction) — this is a spoiler and later shown in equation $(8)$. Since this is a redox reaction, we can use standard half-cell theory to analyse it. Most importantly, thanks to equation $(2)$, only the effective cell potential influences $\Delta G$ ($z$ is the number of electrons transferred while $F$ is Faraday’s constant) which in turn can be separated into two half-cells according to equation $(3)$ which can then be analysed by the Nernst equation $(4)$.</p>
<p>$$\Delta G_\text{cell} = - n_{\ce{e-}}FE_\text{cell}\tag{2}$$
$$E_\text{cell} = E_\text{cathode} - E_\text{anode}\tag{3}$$
$$E = E^0 + \frac{RT}{zF} \ln \frac{a_\text{Ox}}{a_\text{red}} = E^0 + \frac{0.059~\mathrm{V}}{z} \lg \frac{a_\text{Ox}}{a_\text{Red}}\tag{4}$$</p>
<p>The half-cell reactions are shown in equations $(5)$ (oxidation) and $(6)$ (reduction) respectively, along with their standard potentials $E^0$ (under standard conditions, i.e. most importantly $\mathrm{pH\ 0}$).</p>
<p>$$\ce{Na+ + e- <=> Na (s)} \qquad E^0 = -2.71~\mathrm{V}\tag{5}$$
$$\ce{2 H+ + 2 e- <=> H2 (g)} \qquad E^0 = 0.000~\mathrm{V}\tag{6}$$</p>
<p>Let’s take a look at the anodic oxidation first. The initial concentration of sodium ions could be, say, $3~\mathrm{M}$. Remember that the activity is defined as $1$ for solids. With the assumption that $a \approx c$, that would mean our oxidation potential is $-2.68~\mathrm{V}$ as shown in equation $(7)$.</p>
<p>$$E_\mathrm{anode} = -2.71~\mathrm{V} + \frac{0.059~\mathrm{V}}{1} \lg \frac{3}{1} = -2.71~\mathrm{V} + 2.82 \cdot 10^{-2}~\mathrm{V} = -2.68~\mathrm{V}\tag{7}$$</p>
<p>The corresponding calculation for the reductive side, equation $(8)$, is a little harder since we have no real starting value for the partial pressure (and thus activity) of hydrogen gas.</p>
<p>$$E_\text{cathode} = \frac{0.059~\mathrm{V}}{2} \lg \frac{a(\ce{H3O+})}{p(\ce{H2})} = -0.0295~\mathrm{V} \times \mathrm{pH} - 0.0295~\mathrm{V} \times \lg p(\ce{H2})\tag{8}$$</p>
<p>But we do know that a reaction will happen if $E_\text{cell} > 0~\mathrm{V}$ and we have a good idea of the anode’s potential. Thus, if we want no reaction we must set the $E_\text{cell} < 0$ (equation $(9)$):</p>
<p>$$\begin{align}0 &> E_\text{cathode} - E_\text{anode}\\
- 2.68~\mathrm{V} &> E_\text{cathode}\\
- 2.68~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH} - 0.0295~\mathrm{V} \times \lg p(\ce{H2})\end{align}\tag{9}$$</p>
<p>The second summand is negative if $p(\ce{H2}) > 1~\mathrm{bar}$. (The equation itself requires a dimensionless pressure which is typically defined to have a the same value as if it were in $\mathrm{bar}$). However, that basically requires performing the reaction under a hydrogen atmosphere. This setup is rather hard and not part of the question. Instead, let’s take the natural concentration of hydrogen which is $1~\mathrm{ppm}$ according to Wikipedia. That results in $0.177~\mathrm{V}$ for the second summand (equation $(10)$).</p>
<p>$$-0.0295~\mathrm{V} \times \lg 10^{-6} = (-6) \times -0.0295~\mathrm{V} = 0.177~\mathrm{V}\tag{10}$$</p>
<p>At least we didn’t lose too much. But there is still the first summand, which is also negative by sign. We just need a high enough $\mathrm{pH}$ value to compensate. The calculation is in equation $(11)$.</p>
<p>$$\begin{align}-2.68~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH} + 0.177~\mathrm{V}\\
-2.86~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH}\\
96.9 &< \mathrm{pH}\end{align}\tag{11}$$</p>
<p>So we would need a $\mathrm{pH}$ value of almost $97$ to prevent the reaction from happening due to too basic conditions. Of course, this value is unobtainable in water or practically any solvent known to us. Even <em>tert</em>-butyllithium, one of the most basic species known has a $\mathrm{p}K_\mathrm{a}$ value of only $\approx 50$.</p>
<p>You could now try to argue that you would just need to increase the hydrogen gas partial pressure appropriately. So how does the value change if the reaction is performed under hydrogen atmosphere? See equation $(12)$ for the answer.</p>
<p>$$\begin{align}-2.86~\mathrm{V} &> -0.0295~\mathrm{V} \mathrm{pH}\\
90.8 &< \mathrm{pH}\end{align}\tag{12}$$</p>
<p>As you may or may not have guessed, a logarithmic difference of $6$ for the partial pressure results in a lowering of the $\mathrm{pH}$ by $6$ units. So to arrive at achieveable $\mathrm{pH}$ values, your reaction would need to happen at a hydrogen pressure of $10^{75}~\mathrm{bar}$ — go figure.</p>
<hr>
<p>All things considered, the reaction is so violently endergonic that there is basically no way to inhibit it whatsoever. If metallic sodium sees a sufficient amount of water, it will start reacting, end of discussion.</p>
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https://chemistry.stackexchange.com/questions/57351/would-sodium-metal-react-differently-with-aqueous-solutions-than-with-pure-water
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Question: <p>I just read about the theory of auto-catalysis, and here's one thing which is sort of unclear to me: </p>
<blockquote>
<p>In this type of catalysis, one of the reaction product catalyses the reaction. For example, in the oxidation of oxalic acid by potassium permanganate, Mn(II) cation is formed which is known to accelerate the reaction. So when potassium permanganate is added to a warm solution of oxalic acid in presence of dilute sulphuric acid (acid medium catalysis), initially there is a time lag before decolourisation occurs. As more permanganate is added, the decolourisation is observed to be nearly instantaneous.</p>
</blockquote>
<p>Points I didn't understand:</p>
<ol>
<li><p>How does the formation of Mn(II) cation accelerate the reaction? Could someone please help me with understanding the mechanism? </p></li>
<li><p>What is the reason for <em>time lag</em> in decolourisation of the solution? Also, what exactly do they mean by decolourisation here? My guess is that it's referring to the bright purple coloured permanganate solution losing its colour after getting reduced to Mn(II) cation.</p></li>
<li><p>How does the time lag reduce? Are there ways to estimate or determine the same? Of course, it'd depend on the reaction kinetics; but I'm looking for a mathematical viewpoint on <em>how the catalyst influences reaction kinetics</em>, except for decreasing the activation energy.</p></li>
</ol>
<p>Thank you. </p>
Answer: <p>(1) There are several web sites giving details of this reaction and so this need not be copied out here.</p>
<p>(2,3) Rather than describe a complex reaction scheme it is easier to understand a generic autocatalytic reaction. The general autocatalytic reaction of species A with catalyst B has the form $\ce{A + B $\to$ P + 2B}$ and the rate equation is deceptively simple and is $da/dt =- kab$ if $a$ and $b$ are the concentration A and B. The important point here is that species B is both reactant and product. As more B is produced than that there initially, the reaction accelerates as it proceeds. This is the reason for the time lag: the reaction rate is initially small but as reaction proceeds the extra B produced speeds it up. Eventually A runs out and so the reaction stops. (The colour change you mention is just a measure of the concentration of permanganate. Note also that from the Beer-Lambert law describing transmission of light the absorbing species concentration is related exponentially to light transmission (and so absorption) and this probably complicates what you observe by eye).</p>
<p>If $x$ is species A concentration, the rate equation is, $\displaystyle \frac{dx}{dt}=-kxb=-kx(a_0+b_0-x)$ or $\displaystyle \int\frac{dx}{(a_0+b_0-x)x}=-kt+c$ which can be integrated using partial fractions;</p>
<p>$$\frac{1}{(a_0+b_0-x)x}=\frac{1}{(a_0+b_0)}\left[\frac{1}{x}+\frac{1}{a_0+b_0-x} \right]$$</p>
<p>which produces two log functions when integrated. The initial conditions are $x = a_0$ when $t$ = 0 then the integration constant is $\ln(a_0/b_0)$ and the rate equation,</p>
<p>$$ \ln \left(\left| \frac{xb_0}{a_0(a_0+b_0-x)} \right| \right) =-(a_0+b_0)kt$$</p>
<p>and the absolute value is added because the log cannot be negative. Rearranging produces $\displaystyle x = a_0\frac{a_0+b_0}{a_0+b_0e^{-(a_0+b_0)kt}} $. This and also the increase of $B =a_0+b_0-x$ are plotted.</p>
<p><a href="https://i.sstatic.net/R4YTo.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/R4YTo.png" alt="autocatalysis"></a></p>
<p>Figure Autocatalysis $\ce{A + B $\to$ P + 2B}$. Initially species A decreases slowly as the concentration of B is small. As the reaction proceeds, more B is produced and although A is reduced overall, the rate increases and A is consumed even more rapidly. At longer times, the concentration of A becomes so small that even though that of B is large the reaction rate is slow.</p>
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https://chemistry.stackexchange.com/questions/86979/auto-catalysis-mechanism
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Question: <p><strong>Prologue</strong>:</p>
<p>Recently started with <em>Chemical Kinetics</em> at class. Came across a couple of points in Levine's <em>Physical Chemistry</em> (the book was <a href="https://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry">recommended on Chem.SE</a>; so I thought I'd give it a read :D) that I'm having serious issues wrapping my head around. This query is about the kinetics of equilibrium (and non-equilibrium) processes, and the fact that this has puzzled me for quite a while... I attribute to my (seriously questionable) understanding of equilibrium processes (courtesy: High-school education). It would be prudent to keep my shortcomings in <em>Kinetics</em> (newbie) and <em>Equilibrium</em> (possibly flawed understanding) in mind while addressing my query in your answer or the comments section. Thanks!</p>
<hr />
<p>According to <em>Physical Chemistry</em> (Levine,I.N), Chapter 15, "<em>Kinetics</em>":</p>
<blockquote>
<p>Processes in systems in equilibrium are reversible and are comparatively easy to treat. This chapter and the next deal with non-equilibrium processes, which are irreversible and hard to treat. The rate of a reversible process is infinitesimal. Irreversible processes occur at nonzero rates.</p>
</blockquote>
<p>Which, if I understood correctly, can be paraphrased as:</p>
<blockquote>
<p><strong>Equilibrium</strong> processes are <strong>reversible</strong> and have <strong>infinitesimal</strong> (practically "non-existent") rates.</p>
<p><strong>Non-equilibrium</strong> process are <strong>irreversible</strong> and have <strong>non-zero</strong> ("finite" or "significant"?) rates.</p>
</blockquote>
<hr />
<p>From my "understanding" of <em>Equilibrium</em>; chemical equilibrium is an example of dynamic equilibrium. When a chemical reaction, <span class="math-container">$$\ce{A + B -> C + D}$$</span></p>
<p>has proceeded to equilibrium. The "forward" rate of reaction (rate at which <span class="math-container">$C$</span> + <span class="math-container">$D$</span> is spit out) must equal the "backward" rate of reaction (rate at which <span class="math-container">$A$</span> + <span class="math-container">$B$</span> is spit out). There is no <em>net</em> formation of products/reactants, but both the forward and backward rates <strong>are</strong> positive and finite.</p>
<hr />
<p>My question?</p>
<p><strong>Why does Levine say that equilibrium processes have "infinitesimal" (almost non-existent) reaction rates? I thought the forward and backward reaction rates in a reaction at equilibrium are finite...</strong></p>
Answer: <p>OK, I think I am correct about this. If not then I suppose somebody will come along and correct me anyway.</p>
<p>The author here is referring to the <strong>net</strong> rate being infinitesimal. To understand where the infinitesimal comes from, we need to figure out what is being meant by an "equilibrium process", or a "reversible process". On page 43, Levine describes a <em>reversible process</em> as</p>
<blockquote>
<p>one where the system is always infinitesimally close to equilibrium, and an infinitesimal change in conditions can reverse the process to restore both system and surroundings to their initial states.</p>
</blockquote>
<p>Often these are exemplified using thermodynamics processes, e.g. expansion/compression of a gas, but since we're talking about chemical kinetics, we could use an example where the temperature of the system is being increased in a reversible fashion. This means that the temperature is being increased in infinitesimal steps of $\mathrm{d}T$. The total change in temperature of the process is calculated via integration:</p>
<p>$$\Delta T = \int_{T_1}^{T_2}\mathrm{d}T = T_2 - T_1$$</p>
<p>and likewise for other properties. If something (e.g. the pressure $p$) depends on temperature, i.e. $p = p(T)$, then we can find the overall change in pressure</p>
<p>$$\Delta p = \int_{p(T_1)}^{p(T_2)} p(T)\,\mathrm{d}T$$</p>
<p>If the system is at equilibrium ($Q = K$), then an infinitesimal increase in temperature leads to an infinitesimal change in the equilibrium constant $K$, and hence an infinitesimal change in $Q$ (essentially this is due to Le Chatelier's principle, but more properly it is explained by the system moving in the direction of decreasing Gibbs free energy). This leads to an infinitesimal change in the concentrations of the reacting species, and hence the overall rate of reaction is infinitesimal.</p>
<p>If nothing is changing, the system would still be considered as being at equilibrium. The net rate in this case is, of course, zero. However, such a system is not particularly interesting, as there is no process actually occurring.</p>
<p>So, the bottom line is that for a system in equilibrium, its properties must be changing <em>at most</em> infinitesimally.</p>
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https://chemistry.stackexchange.com/questions/81039/conceptual-tussle-with-kinetics-of-equilibrium-and-non-equilibrium-processes
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Question: <p>What is the difference between a loaded solution and an unloaded solution? I was reading this article and I didn't understand these parts:</p>
<blockquote>
<ul>
<li>The reaction kinetics of $\ce{CO2}$ with <strong>loaded</strong> aqueous MEA solution...</li>
<li>$\ce{CO2}$ <strong>loading</strong> from $\mathrm{0-0.4~mole}~ \ce{CO2/ mole amine}$ for $30$ weight percent MEA solution...</li>
<li>kinetic constants are developed taking into account <strong>loading</strong>, temperature ...</li>
</ul>
</blockquote>
Answer: <p>Unloaded solutions are solutions without absorbed $\ce{CO2}$ and loaded solutions contain absorbed $\ce{CO2}$. Please see the <a href="https://spiral.imperial.ac.uk/bitstream/10044/1/26161/7/acs%252Ejced%252E5b00282.pdf" rel="nofollow noreferrer">this link</a>.</p>
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https://chemistry.stackexchange.com/questions/61780/loaded-and-unloaded-solutions
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Question: <p>This is for the hydrolysis of p-nitrophenyl phosphate by alkaline phosphatase, whereby the enzyme is in a solution of distilled water. We also add glycine buffer to the solution. It exhibits first-order kinetics, even though there is a second reactant (being water) - how is this possible?</p>
Answer: <p>It is because water is the solvent in the reaction and therefore its concentration remains practically constant throughout the course of the reaction. Hence, it is absorbed in the rate constant: </p>
<p>rate=k[p-nitrophenyl phosphate]$^1$[H$_2$O]$^x$=k'[p-nitrophenyl phosphate]$^1$</p>
<p>Hence, we say the reaction is peudo-first order. Also, in general, you could have a reaction with two reactants and it could still be first order overal since it could be 0 order in reactant A and first order in reactant B.</p>
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https://chemistry.stackexchange.com/questions/90475/how-can-a-reaction-exhibit-first-order-kinetics-when-there-is-more-than-one-reac
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Question: <p>My friends and I were doing some problems from this year's IChO Preparatory Problems (<a href="https://icho2017.sc.mahidol.ac.th/pdf/49th_IChO_PreparatoryProblem.pdf" rel="nofollow noreferrer">PDF from the 49th International Chemistry Olympiad (2017)</a>) when we stumbled upon a question which we had some confusion with. </p>
<blockquote>
<p><strong>Task 8.</strong> Decomposition of Nitrous Oxide<br>
Nitrous oxide decomposes exothermically into nitrogen and oxygen, at a temperature of approximately $\pu{565 ^\circ C}$.
$$\ce{2N2O (g) -> 2N2 (g) + O2 (g)}$$
This reaction follows the second-order kinetics when carried out entirely in the gas phase.</p>
<p><strong>8.1)</strong> If the reaction is initiated with $[\ce {N2O}]$ equal to $\pu{0.108 mol dm-3}$, what will its concentration be after $\pu{1250 s}$ have elapsed at $\pu{565 ^\circ C}$? The rate constant for the second order decomposition of $\ce{N2O}$ is $\pu{1.10\times10^-3 dm3 mol-1 s-1}$ at this temperature.</p>
</blockquote>
<p>Task 8 details the kinetics of the decomposition of nitrous oxide. We were confused with part 8.1 of the task which required us to find the concentration of the reactant after a particular duration of time has elapsed, given the temperature, initial concentration and rate constant of the reaction. We approached the question using two different methods:</p>
<p>I approached it using the method, as suggested in the solutions manual, of using the integrated rate law for 2nd order kinetics and substituting the values provided into that equation. My answer was exactly that stated in the solutions manual.</p>
<p>However, they approached it by first finding the half-life using the equation for the half-life for 2nd order kinetics and then, finding the concentration after knowing how many half-lives have passed. Their answer was slightly off. </p>
<p>After substituting random values into the equations for both methods, I realised that the "half-life method" which my friends used only gives a reasonable approximation when the time elapsed < half-life, gives the exact value when the time elapsed = half-life and goes completely off when the time elapsed > half-life. Why is that so? Since the half-life equation for 2nd order kinetics is derived from the integrated rate law (as shown on <a href="https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Kinetics/Reaction_Rates/Second-Order_Reactions" rel="nofollow noreferrer">Chemistry LibreTexts</a>), shouldn't they give the same answer? </p>
Answer: <p>Roughly speaking, <strong>half-life is not a thing at all for the second order</strong> (or <em>any</em> order other than first, for that matter).</p>
<p>What would they do after finding the number of half-lives? Raise 2 to that power? Too bad, because that only works for the exponential decay, which is the solution of the first-order kinetics and no other order. Really, what if the number of half-lives equals 2? After spending the first half-life, we now have a <em>different</em> half-life ahead of us, because it is concentration dependent, and the concentration has changed.</p>
<p>Following your link, we see the same statement, except they don't put enough emphasis on it, as to my taste:</p>
<blockquote>
<p>For this reason, the concept of half-life for a second-order reaction is far less useful. </p>
</blockquote>
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https://chemistry.stackexchange.com/questions/87363/half-life-equation-for-2nd-order-kinetics
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Question: <p>I have been recently studying chemical kinetics. I've understood what the rate law and the rate constant mean, and also that, kinetics depends completely on experimental observations, due to which we have zero order reactions as well as fractional order reactions.</p>
<p>Take for example, this: $$\text{Rate of a certain chemical reaction} = k [A]^p [B]^q [C]^r $$</p>
<p>Here, $[A]$, $[B]$ and $[C]$ are the molar concentrations of the three reactants involved, while $p$, $q$ and $r$ are the number of molecules of each respectively taking part in the reaction.</p>
<p>So, $\text{the order of the reaction} =p+q+r $.</p>
<p>Now, this $p$, $q$ and $r$ are completely determined experimentally, and cannot be determined theoretically.</p>
<p><strong>How</strong> are these values determined experimentally? How does one find out how many molecules of each reactant are taking part in the reaction? Is there any special instrument to do this? If not, then how do scientists find the order of a reaction?</p>
<p>As an attempt to shorten out the answering area, consider this reaction: $$\ce {NO2 +CO -> NO +CO2} $$ Now, experimentally it has been determined that $\text {the rate of this reaction} = k[\ce{NO2}]^2$. I want to know <strong>how</strong> have scientists determined that the rate depends on the square of the concentration of $\ce {NO2} $ and not on the other reactant.</p>
<p><em>N.B.: My question is <strong>how</strong> it can be experimentally determined, and <strong>not</strong> why it has to be experimentally determined.</em></p>
Answer: <p>Order of a reaction can be found by the following methods:</p>
<p><strong>1. Plotting a graph of concentration of reactant versus time</strong></p>
<p>For example, let us take decomposition of <span class="math-container">$\ce{N2O}$</span> to <span class="math-container">$\ce{N2}$</span> and <span class="math-container">$\ce{O2}$</span> on <span class="math-container">$\ce{Pt}$</span> surface.
<span class="math-container">$$\ce{2N2O ->[Pt] 2N2 + O2}$$</span></p>
<p>The graph will look similar to the following:</p>
<p><a href="https://i.sstatic.net/xCmOV.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/xCmOV.png" alt="Zeroth order decomposition of nitrous oxide on platinum surface" /></a></p>
<p>The slope of the graph is constant. That is, it does not vary with time.</p>
<p>This is possible when the reaction is zeroth order. Hence, we can conclude that the reaction is zero-order.</p>
<p>For a <span class="math-container">$\mathrm{1^{st}}$</span> order reaction, the slope of the graph drawn between <span class="math-container">$\ln[\mathrm{Reactant}]$</span> with time is constant, for a <span class="math-container">$\mathrm{2^{nd}}$</span> order reaction, the slope of the graph drawn between <span class="math-container">$\mathrm{\frac{1}{[Reactant]}}$</span> with time is constant.</p>
<p>In general, for an <span class="math-container">$n^{\text{th}}$</span> order reaction, the slope of the graph drawn between <span class="math-container">$\mathrm{\frac{1}{[Reactant]^{n-1}}}$</span> with time is constant (where <span class="math-container">$n \neq 1$</span>).</p>
<p><strong>2. Half life method</strong></p>
<p>This is useful only when a single reactant is present.</p>
<p>In this method, half life of a reaction is measured, varying initial concentration of the reactant.
<span class="math-container">$$t_{\frac{1}{2}} \propto \frac{1}{\mathrm{[A]^{n-1}}}$$</span> where <span class="math-container">$n$</span> is the order of the reaction and <span class="math-container">$\mathrm{[A]}$</span> is the initial concentration of the reactant.</p>
<p><strong>3. Initial rate method</strong></p>
<p>In this method, the initial rate of the reaction is measured at different initial concentrations of the reactant.</p>
<p>Let the reaction be <span class="math-container">$n^{th}$</span> order w.r.t. a reactant <span class="math-container">$\mathrm{A}$</span>, then
<span class="math-container">$$\mathrm{r_0 = k[A]^n}$$</span> at different initial concentration of <span class="math-container">$\mathrm{A}$</span>, say <span class="math-container">$a_1$</span> and <span class="math-container">$a_2$</span>, the observed initial rates be <span class="math-container">$r_1$</span> and <span class="math-container">$r_2$</span>, then
<span class="math-container">$$n = \frac{\log(r_1)-\log(r_2)}{\log(a_1)-\log(a_2)}$$</span></p>
<p>In case of multiple reactants, say for a hypothetical reaction
<span class="math-container">$$\ce{11A + 20 B -> 5 C + 9 D },$$</span> you want to find the order of it.</p>
<p>Let the rate equation be <span class="math-container">$$ r = k[\ce{A}]^n[\ce{B}]^m$$</span>
First take a reactant in excess and find the order of reaction with respect to other reactant by comparing initial rates or half lives.</p>
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https://chemistry.stackexchange.com/questions/82089/query-in-chemical-kinetics
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Question: <p>In an interview for post-grad level, I was asked to explain the "order" of any generic enzymatic reaction involving <a href="https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Book%3A_Biochemistry_Online_(Jakubowski)/08%3A_Transport_and_Kinetics/8.4%3A_Enzyme_Inhibition/Competitive_Inhibition" rel="nofollow noreferrer">competitive inhibition</a>. I assumed that <a href="https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Enzymes/Enzymatic_Kinetics/Michaelis-Menten_Kinetics" rel="nofollow noreferrer">Michaelis-Menten Kinetics</a> holds good and went for "first-order". I also noted that the Lineweaver Burk plots <a href="https://commons.wikimedia.org/wiki/File:Enzyme_Inhibition_lineweaver-burk_plots.gif" rel="nofollow noreferrer">remain linear</a> even in competitive inhibition.</p>
<p>However, they asserted (at the end of the interview) that it would be "second-order" - thoughts?</p>
Answer:
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https://chemistry.stackexchange.com/questions/165992/order-in-enzyme-kinetics
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Question: <p>Starting from the equations of <a href="http://en.wikipedia.org/wiki/Law_of_mass_action" rel="nofollow">mass action kinetics</a>, one may show that (for suitably chosen numbers $A_i$), the function $\sum_i x_i(A_i + \log x_i)$ is a <a href="http://en.wikipedia.org/wiki/Lyapunov_function" rel="nofollow">Lyapunov function</a>, i.e. it decreases over time.</p>
<p>Apart from an irrelevant factor of $RT$, this is the expression for the free energy of an ideal solution (or ideal gas, etc). So it seems that mass action kinetics carries with it an assumption that the reactions are taking place in an ideal solution.</p>
<p>I'm interested in how to think about kinetics in <em>non</em>-ideal solutions, but I haven't been able to find any information about it. Is there a standard set of kinetic equations that remain valid when the chemical potential is given not by $\mu_i^0 + RT\log x_i$ but by $\mu_i^0 + RT\log \gamma_i x_i$, for activity coefficients $\gamma_i$ that depend on the concentrations?</p>
<p>There are several alternatives to mass action kinetics that I know about, such as Michaelis Menton kinetics for enzymes - but these are actually <em>derived</em> from mass action kinetics applied to elementary steps, so they don't seem to be able to escape from the assumption of ideality.</p>
Answer: <p>According to <a href="http://iglesia.cchem.berkeley.edu/Publications/JournalofMolecularCatalysisA_163_189_2000.pdf" rel="nofollow">Catalytic reaction rates in thermodynamically non-ideal systems</a> Journal of Molecular Catalysis A: Chemical 163 (2000) 189–204 (citing to kinetics textbooks) non-ideal kinetics needs to consider the activity of the transition state in addition to the activities of the reactants:</p>
<p>For $\ce{A + B ->}$</p>
<p>rate $=\frac{k_BT}{h}K^{\ddagger}\frac{a_Aa_B}{\gamma_\ddagger}=k_0\frac{\gamma_A\gamma_B}{\gamma_\ddagger}C_AC_B $ </p>
<p>Where $K^{\ddagger}$ is the equilibrium constant for the formation of the transition state, $a$ is activity, $\gamma$ is activity coefficient, $C$ is concentration, $k_B$ is Blotzmann constant, $T$ is temperature, $h$ is Planck's constant and $k_0$ is the thermodynamically ideal rate constant. </p>
<p>Another reference to look at for more info is <a href="http://pubs.rsc.org/en/Content/ArticleLanding/1986/F2/f29868201297#!divAbstract" rel="nofollow">The status of transition-state theory in non-ideal solutions and application of Kirkwood–Buff theory to the transition state</a> J. Chem. Soc., Faraday Trans. 2, 1986,82, 1297-1303.</p>
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https://chemistry.stackexchange.com/questions/19234/what-is-the-non-ideal-equivalent-of-mass-action-kinetics
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Question: <p>when a chemical bond is formed, does chemical reaction occur? if i have done a coating on metal substrate then how does it adhere to the substrate? If there is a chemical bond then which reaction does takes place?</p>
Answer: <p>If a chemical bond between atoms is split, or newly generated, than there is occurrence of a chemical reaction. Not all bonds, however, are equal -- covalent bonds differ from hydrogen bonds, for example.</p>
<p>Your question <em>may</em> be interpreted as if you were unsure to discern between <a href="https://en.wikipedia.org/wiki/Physisorption" rel="nofollow noreferrer">physisorption</a>, the more general, more frequently mechanism of <a href="https://en.wikipedia.org/wiki/Adsorption" rel="nofollow noreferrer">adsorption</a>, and <a href="https://en.wikipedia.org/wiki/Chemisorption" rel="nofollow noreferrer">chemisorption</a> -- where actual bonds between substrate and coating are formed (yet: not necessarily the same bonds in terms of strength as C-C in an alkyl chain, for example). If so, look at <a href="https://en.wikipedia.org/wiki/Self-assembled_monolayer" rel="nofollow noreferrer">self assembled monolayers</a> -- it may shed some light on the matter.</p>
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https://chemistry.stackexchange.com/questions/70494/chemistry-chemical-bonding
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Question: <p>Was studying chemical bonding(Valence bond theory) but there came this topic called <em>wave equation</em> but have not studied quantum mechanics,so finding hard to understand these terms.Please help!!!</p>
Answer: <p>$\Psi$ is the wavefunction describing an electron's motion. $\Psi^2$ is the probability density function for the electron. $\Psi^2$ is usually interpreted as the probability of finding the electron at a given set of coordinates.</p>
<p>In the case that $\Psi$ is a complex function (in that it contains imaginary numbers and not that it is complicated, which it usually is), $\Psi^2$ is defined as the product of $\Psi$ with its complex conjugate: $\Psi \Psi^*=|\Psi|^2$ so that the probability density function contains only real numbers. </p>
<p>Consider a simple case of $\Psi=a+bi$. The complex conjugate would be $\Psi^*=a-bi$.</p>
<p>$$\Psi \Psi^* = (a+bi)(a-bi)=a^2+b^2$$</p>
<p>Often the probability density function is <em>normalized</em> so that its integral over all space is 1 (i.e. there is 100% chance of finding the electron in all of space). In the equation below, $A$ is the normalization constant.</p>
<p>$$1=A \int_o^{2\pi} \int_0^{2 \pi} \int_o^\infty \Psi \Psi^* dr d\theta \phi$$</p>
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https://chemistry.stackexchange.com/questions/9612/what-is-this-wave-function-in-chemical-bonding
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Question: <p>How does chemical bonding look like on the quantum scale per se since the electron is a wave, and a particle?</p>
Answer: <p>I believe that generally in chemistry the wave characteristics of the electron are far more prevalent/relevant that the particle characteristics. So it's usually easiest just to think of it as a wave (or I think "region of space" is easier).</p>
<p>A chemical bond looks like the electron wave or area spread out between two or more atoms. This shape of the wave has lower energy than any of the orbitals it could occupy on just one atom. Because it has lower energy, that means it needs energy from somewhere for the electrons & atoms to change back to a different configuration. So unless it gets some more energy it cannot change away from the bonded state. That explains why chemical bonds are stable and hold atoms together.</p>
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https://chemistry.stackexchange.com/questions/46959/what-does-chemical-bonding-look-like-at-the-quantum-level
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Question: <p>How exactly does bonding work?
For example, in CH4 orbital overlap structure, the internet shows 4 lobes with one hydrogen attached to each...</p>
<p><a href="https://i.sstatic.net/f9yPx06t.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/f9yPx06t.gif" alt="enter image description here" /></a></p>
<p>does the s orbital of the carbon we consider after hybridization look like a lobe?
usually the s orbital is spherical, which confuses me in this case.</p>
<p>if we talk about the p orbitals in carbon, 3 s-p bonds are formed, hydrogen(s) to carbon(p),</p>
<p><a href="https://i.sstatic.net/3KaZXe6l.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3KaZXe6l.png" alt="enter image description here" /></a></p>
<p>The p orbital usually consists of two lobes. One lobe forms a bond with hydrogen; what does the other one do?</p>
Answer: <blockquote>
<p>[OP] if we talk about the p orbitals in carbon, 3 s-p bonds are formed, hydrogen(s) to carbon(p), p orbital usually consists of 2 lobes, one lobe forms a bond with hydrogen, what does the other one do?</p>
</blockquote>
<p>During the early days of quantum chemistry, researchers were wondering the same thing, i.e. why do I get four equivalent bonds with a tetrahedral arrangement if I start with one "s" and three "p" orbitals on the carbon atom.</p>
<p>The two common models to explain this are called valence bond and molecular orbital theory. In one case you work with linear combinations of the s and p orbitals of carbon ("hybridization"). In the other case, you keep the s and p orbitals, and work with linear combinations of the 4 s orbitals of hydrogen (that match the symmetry of the orbitals of the central atom). Either way, the models rationalize four equidistant hydrogen atoms in a tetrahedral arrangement.</p>
<blockquote>
<p>[OP] For example, in CH4 orbital overlap structure, the internet shows 4 lobes with one hydrogen attached to each ... does the hybridized s orbital look like a lobe? usually the s orbital is spherical, which confuses me in this case.</p>
</blockquote>
<p>The hydrogen orbital combined with the sp3-hybridized orbital of carbon looks "like a lobe". As you already heard in the comments, the hybridization is not happening on the hydrogen atom (it only has a single orbital in play, the 1s).</p>
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https://chemistry.stackexchange.com/questions/186037/chemical-bonding
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Question: <p>I would like to ask how many valence electrons took part in creating chemical bond in $\ce{OH-}$? Is the minus related to $\ce{O}$ or to $\ce{H}$? Are there any spare valence electrons which didnt take part in creating chemical bond in $\ce{OH-}$?</p>
Answer: <p><img src="https://i.sstatic.net/BtYqt.jpg" alt="enter image description here"></p>
<p>In the photo the orange electrons are non bond electrons.<br>
And the black electrons are bond electrons.</p>
<p>There are 2 electrons took part in creating chemical bond in $\ce{OH}$.
The first one is from oxygen.<br>
The second is from hydrogen.<br></p>
<p>This kind of bond is called "Covalent bond".
For more about Covalent bond :
<a href="https://en.wikipedia.org/wiki/Covalent_bond" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Covalent_bond</a></p>
<p>The minus is related to the black and white electrons that are free (in Oxygen).</p>
<p>The white electron is come from other element.</p>
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https://chemistry.stackexchange.com/questions/7264/chemical-bond-covalent-valence-electrons-oh
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Question: <p>How do we compare the strength of permanent dipole permanent dipole (PDPD) interaction? Is it by bond polarity?
Since H-X (where X is halogen) is polar, and for bond polarity (PDPD bond strength): HCl > HBr > HI, why is it that for boiling point: HCl < HBr < HI which is inversely proportional to bond strength?</p>
Answer:
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https://chemistry.stackexchange.com/questions/87917/chemical-bonding
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Question: <p>My chemistry teacher told me that chemical bonds are of two types: intramolecular and intermolecular. He said that intermolecular forces come under the category of intermolecular chemical bond. </p>
<p>I have never read such statement anywhere. Nor can I find anything on the Internet that would support this statement. </p>
<p>My understanding is that chemical bond is a force that holds atoms together in a chemical species. Since intermolecular forces do not hold atoms together, they should not be termed as chemical bond.</p>
<p>So, are intermolecular forces a type of chemical bond? </p>
Answer: <p>The IUPAC <a href="https://goldbook.iupac.org/html/C/CT07009.html" rel="nofollow noreferrer">definition</a> of "chemical bond" is: </p>
<blockquote>
<p>When forces acting between two atoms or groups of atoms lead to the formation of a stable independent molecular entity, a chemical bond is considered to exist between these atoms or groups. The principal characteristic of a bond in a molecule is the existence of a region between the nuclei of constant potential contours that allows the potential energy to improve substantially by atomic contraction at the expense of only a small increase in kinetic energy. Not only directed covalent bonds characteristic of organic compounds, but also bonds such as those existing between sodium cations and chloride anions in a crystal of sodium chloride or the bonds binding aluminium to six molecules of water in its environment, and <strong>even weak bonds that link two molecules of $\ce{O_2}$ into $\ce{O_4}$, are to be attributed to chemical bonds</strong>. </p>
</blockquote>
<p>So the answer is "yes" in some cases. </p>
<p>See also the IUPAC <a href="https://goldbook.iupac.org/html/M/M04002.html" rel="nofollow noreferrer">definition</a> of "molecule": </p>
<blockquote>
<p>... must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state</p>
</blockquote>
<p>So, for example, a water-water dimer, held together by hydrogen bonding, has a monomer-monomer potential energy surface that is deep enough to confine at least one vibrational state, and it would be appropriate to refer to the hydrogen bond as a chemical bond. </p>
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https://chemistry.stackexchange.com/questions/91369/are-intermolecular-forces-a-type-of-chemical-bond
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Question: <p>When oxygen is adsorbed on iron metallic surface, does it create $\ce{Fe-O}$ bonds like the one in iron oxide? Can it be considered as a monolayer of iron oxide on the surface?</p>
<p>If not, how is the atomic bond of adsorbed oxygen and iron? How much is it strong? and what is its difference with normal chemical bond in iron oxide molecule?</p>
Answer:
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https://chemistry.stackexchange.com/questions/16475/chemical-bond-of-adsorbed-atoms
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Question: <p>I am trying to find out what chemical bond holds aluminum oxide to the aluminum on which it forms. The only information I was able to find on the internet was in a Google book which suggested that the bond was ionic in nature.</p>
Answer:
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https://chemistry.stackexchange.com/questions/38505/chemical-bond-at-aluminum-aluminum-oxide-interface
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Question: <p>VSEPR theory correctly predicts the shapes of many symmetry-broken molecules such as $\ce{H2O}$ and $\ce{NH3}$. Take $\ce{NH3}$ for example. In VSEPR theory, the nitrogen atom is (approximately) at the center of a tetrahedron, the three $\ce{N-H}$ bonds point to three of the four vertices of the tetrahedron, and the lone pair of nitrogen points to the $4$th vertex. But quantum mechanically speaking, the electrons should all be delocalized in the entire $\ce{NH3}$ molecule. How do I unify the two pictures to understand the concept of chemical bonds and VSEPR theory in quantum mechanics? Does VSEPR correspond to some kind of trial wave function (e.g. antisymmetrized geminal power (AGP))?</p>
<p>Note: when I say how to understand chemical bonds in quantum mechanics, I mean in the chemical bond description of molecules, electron pairs are localized at the bonds while quantum mechanics again says everything can be delocalized. So it's the same discussion as VSEPR v.s. QM. If there are only two atoms and one bond, the quantum mechanical meaning of the chemical bond is clear.</p>
Answer: <p><strong>VSEPR is a simple and generalised approach based</strong> (mainly) <strong>on empirical observation.</strong> It is a great theory in predicting, in a first order approximation, the geometrical shape of molecules. That is a lot harder with other methods. It is of course based on a physical foundation, not only empiricism. You might want to explore electron-domains and the Pauli principle in this context.</p>
<p>There have been a lot of post-rationalisations that cause more harm than good, because they clearly go beyond the limitations of the theory. One of the most miss-taught ones is the involvement of d-orbitals in hybridisation schemes, but that comes from the lack of understanding of some instructors, and its confusion with valence bond theory.<br>
One of the most prominent examples where it (basically) fails is outlined in my answer here: <a href="https://chemistry.stackexchange.com/q/50906/4945">Are the lone pairs in water equivalent?</a> At this point it should be noted, that the general topology of the electron density is quite well reproduced.</p>
<p>With that in mind, a reconciliation of this empirical approach with quantum theory to understand bonding is dangerous, if not futile. It should be used for providing a reasonable guess for a molecular structure and its principle explanation. Anything beyond that might lead to wrong conclusions. It cannot, in any way, be used to generate a guess for a wave function, because it is not based on it.</p>
<p>You will have to use another method for that. Often confused with VSEPR is the valence bond theory (VBT). It cannot be stressed enough, that the two are completely independent (even at its crudest level). There has been a lot of development in that field, and what is often taught as VBT can only be classified as a very crude approximation. It is in principle an exact theory, but carried out at the theoretical limit is not as easily understood as that what is taught commonly. (Just have a look at resonance, and its misconceptions as are outlined here: <a href="https://chemistry.stackexchange.com/q/51632/4945">What is resonance, and are resonance structures real?</a>)<br>
Another way to describe bonding is molecular orbital theory (MOT), which already comes with the necessary properties of electron delocalisation. Unfortunately, this theory is more difficult to get started in, and does not provide an easy picture to follow. The good thing about this is, that it doesn't get much more complicated at its theoretical limit.<br>
The two approaches are at their respective theoretical limit (VBT decribes electron correlation via resonance structures, MOT need multiple determinants for this) equivalent.</p>
<p>Understanding bonding is not easy and it is by far not without controversy. Even at approximate VBT or MOT there can be many misconceptions, and incorrect deductions, conclusions, rationalisations. For everything in the <em>"grey"</em> areas, there are opinions, interpretations, and opinions about interpretations.<br>
In any and every case one should always be aware of the limitations of the model used. One should also be critical about the found results. One should always expect everything to be a lot more complicated than expected (point in case: <a href="https://chemistry.stackexchange.com/q/30797/4945">CO<sub>2</sub></a>).<br>
If you keep all of that in mind: <strong>VSEPR is an awesome model system.</strong></p>
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https://chemistry.stackexchange.com/questions/89244/vsepr-theory-chemical-bond-and-quantum-mechanics
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Question: <p>While superheavy elements are notoriously unstable, there's plenty of theoretical research and even some experimental techniques are valid for nuclides decaying in a matter of seconds. They can verify presence of relativistic effects, which may cause behavior defying expectations</p>
<p>How do the relativistic effects influence chemical bonding of hassium or other closely related super heavy elements?</p>
Answer:
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https://chemistry.stackexchange.com/questions/184390/how-do-the-relativistic-effects-influence-chemical-bonding-of-hassium-or-other-c
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Question: <p>Question: Why is ionic lattice energy inversely proportional to the radius of the atom?</p>
<p>Most heterogeneous covalent molecules are polar to some extent. The degree of polarity, or the dipole moment, depends on the difference in electronegativity difference between the two atoms. The larger the dipole moment, the higher the ionic character.</p>
<p>What I know: Electronegativity decreases as you go downward in a group, however, the size increases, usually, as you go downwards in a group. Thus, ionic character will increase upon going downward, but the ionic lattice energy will decrease? This seems contradictory. Is this true? And if it is, why is it so?</p>
Answer: <p>Considering your doubt, from <strong><em>Born-Lande</em></strong> equation:</p>
<blockquote>
<p>$$E = -\frac{N_\text{A}Mz^+z^- e^2 }{4 \pi \varepsilon_0 r_0}\left(1-\frac{1}{n}\right)$$</p>
</blockquote>
<hr>
<blockquote>
<p>$N_\text{A}$ = <em><strong>Avogadro constant</strong></em>;<br>
$M$ = <em><strong>Madelung</strong></em> constant, relating to the geometry of the crystal;<br>
$z^+$ = charge number of cation<br>
$z^-$ = charge number of anion<br>
$e$ = elementary charge, $1.6022\times10^{-19}\ \mathrm{C}$<br>
$ε_0$ = permittivity of free space<br>
$4πε_0 = 1.112\times10^{-10}\ \mathrm{C^2/(J\cdot m)}$<br>
$r_0$ = distance to closest ion<br>
$n$ = <em><strong>Born</strong></em> exponent, typically a number between 5 and 12, determined experimentally
by measuring the compressibility of the solid, or derived theoretically.</p>
</blockquote>
<p>And what you are assuming the radius is in fact the distance between the two atoms.<br>
As the distance <strong>increases</strong>, the attraction <strong>decreases</strong> from <strong><em>Coulomb's law</em></strong> :</p>
<blockquote>
<p>$$F=\frac 1{4\pi\varepsilon_0r_0^2}Q_1Q_2$$</p>
</blockquote>
<p>and thus there is <strong>less</strong> energy involved in breaking the lattice, so lattice energy <strong>decreases</strong>.</p>
<hr>
<p>As you move down, the size <strong>increases</strong> and the valence electron in the outermost shell experiences <strong>lesser</strong> repulsion and thus it can be easily removed, therefore the elements at the bottom are highly electro-positive or conversely very less electro-negative. So when forming bonds, the elements at the bottom easily remove the electron(s) and gain more developed charged rather than the ones on the top who just release it just partially. This charge variation varies the dipole moment $\vec \mu=q\times\vec d$, by altering the partial charge $q$. Thus downwards as $q$ <strong>increases</strong>, $|\mu|$ also does increase and make the compounds formed by the element more ionic. Also the ionic/covalent character may be varied depending upon the bonding partner, in sense of, elements or other entities.</p>
<hr>
<p>Also as charges $z^+$, $z^-$ <strong>increases</strong>, the magnitude of lattice energy does also <strong>increase</strong> as can be seen clearly from the <em><strong>Born-Lande</strong></em> equation.</p>
<hr>
<p>$$\begin{array}{r|c|l}
&&\text{Ionic Solid}&\text{Lattice Energy (kJ/mol)}\\\hline
&\text{Variation of}&\ce{NaF}&926\\
&r_0 \text{ i.e. distance}&\ce{NaCl}&786\\
&\text{between ions}&\ce{NaBr}&752\\
&&\ce{NaI}&702\\\hline
&\text{Variation of}&\ce{NaF}&926\\
&z^+,z^- \text{ i.e. charge}&\ce{MgO}&3800\\
&\text{on ions}&\ce{Al2O3}&15900\\
\end{array}$$</p>
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https://chemistry.stackexchange.com/questions/14132/chemical-bonding-lattice-energy
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Question: <p>I have IR spectra of polystyrene with characteristic peaks at 1450cm-1. 1500cm-1 and 1600cm-1. What chemical bond is associated with 1600cm-1 and 1500cm-1?</p>
Answer: <p>As shown <a href="https://www.spectroscopyonline.com/view/the-infrared-spectra-of-polymers-iii-hydrocarbon-polymers" rel="nofollow noreferrer">here</a> (Figure 2), the lines at 1600 and 1492 (~1500) cm<span class="math-container">$^{-1}$</span> correspond to aromatic ring modes.</p>
<p>From my doctoral dissertation (Classical and Quantum Dynamics of Vibrational Energy Flow in Benzene: the CH(<span class="math-container">$\nu = 2$</span>) Overtone. Minehardt, T. J. The University of Texas at Austin. May, 1999):
the specific modes correspond to E<span class="math-container">$_{1u}$</span> (CC stretch and in-plane CH wag) at 1494 cm<span class="math-container">$^{-1}$</span>; and E<span class="math-container">$_{2g}$</span> (CC stretch, CCC bend, and in-plane CH wag) at 1607 cm<span class="math-container">$^{-1}$</span>.</p>
<p>Thus, the corresponding bonds are carbon-carbon (stretch), carbon-carbon-carbon (bend), and carbon-carbon-hydrogen (in-plane wag).</p>
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https://chemistry.stackexchange.com/questions/167866/help-with-chemical-bond-associated-with-1500cm-1-and-1600cm-1
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Question: <p>We know that ammonia undergoes amine inversion. Why doesn't the dipole moment decrease in ammonia, since the direction of the dipole changes to the opposite direction every time there's inversion?</p>
<p>Shouldn't the effective dipole moment become zero?</p>
Answer: <p>Ivan Neretin is correct in stating that no particle in a (nondegenerate) state can have a permanent electric dipole moment in the laboratory frame. In fact, a permanent dipole moment would require a violation of time (T) and parity (P) symmetry. This is because any such dipole should be directed along the angular momentum vector of the molecule and time-reversal and parity operators have a different effect on the rotational angular momentum and direction of the dipole.</p>
<p>What chemists call a permanent dipole moment refers to the dipole moment in the molecular point group to which the molecule belongs. Only when a polar molecule is subjected to an electric field (e.g. an EM wave or DC field), this field will mix rotational states of opposite parity in such a way that the molecule "orients" itself in the field. You can only speak of the inversion of ammonia in the presence of an electric field (the eigenstates of the Hamiltonian are time independent in the absence of a field or other perturbing force). </p>
<p>Although degenerate states could in principle have a permanent dipole moment, it was shown by Klemperer et al. (J. Phys. Chem. 1993,97, 2413) that the degeneracy of these states is removed by higher-order terms in the Hamiltonian. </p>
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https://chemistry.stackexchange.com/questions/62262/chemical-bonding-and-dipole-moment
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Question: <p>For the elements such as rhenium, osmium, iridium how probable is for electrons in the inner 4f orbitals to gain energy and take part in a molecular orbital (such as an $f^2 d^2 s p^3$ hybrid orbital) when these elements are in their 6,7,8,9 oxidation state? </p>
Answer: <p>By transition metals, 4f becomes full and becomes quite deeply buried in the electron cloud. So the main orbitals that participate are 5d and 6s orbitals - especially in ion formation. However in actinides, 5f orbitals are diffuse and so can participate and so chemicals like diuranium have 5f participation. Even at the high oxidation states like +8 For osmium.</p>
<p>Also note that hybridization is a concept used to ease how covalent bonding works but it isn't fully accurate. By using a more accurate theory like Molecular orbital theory where atomic orbitals overlap to create molecular orbitals, it will provide more insight into how bonding works and whether f orbitals can be used.</p>
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https://chemistry.stackexchange.com/questions/78542/chemical-bonding-with-f-orbitals
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Question: <p>We can explain chemical anti-bonding just using the Pauli repulsion correct?Let's take He2.</p>
<p>2 atoms of He share 4 1s electrons and since the magnetic spin for electrons has 2 values there would be 2 electrons with the same wave function which would violate the Pauli exclusion principle.This makes He2 unstable and it disassociates in 2 He atoms</p>
<p>Am I correct?</p>
Answer: <p>The reason we speak of orbitals as being exclusively inhabited by 2 (max) electrons (with opposing electron spin quantum number) is because electrons are fermions and therefore observe the exclusion principle. The exclusion principle constrains the allowed electron configurations, disallowing occupation of lower E orbitals by more than 2 electrons. In the case of <span class="math-container">$\ce{He2}$</span> it requires occupation of orbitals that raise the total E above the energy of the atoms at greater separation. So, in a word, the answer is yes. </p>
<p>The wikipedia entry on the <a href="https://en.wikipedia.org/wiki/Pauli_exclusion_principle#Stability_of_matter" rel="nofollow noreferrer">Pauli exclusion principle</a> alludes to its role:</p>
<blockquote>
<p>It has been shown that the Pauli exclusion principle is responsible for the fact that ordinary bulk matter is stable and occupies volume. This suggestion was first made in 1931 by Paul Ehrenfest, who pointed out that the electrons of each atom cannot all fall into the lowest-energy orbital and must occupy successively larger shells. Atoms, therefore, occupy a volume and cannot be squeezed too closely together.[12] </p>
</blockquote>
<p>I should add that dispersion interactions mean that there actually is a very weak attractive interaction between He atoms before the repulsive term "kicks in". See e.g. <a href="https://en.wikipedia.org/wiki/London_dispersion_force" rel="nofollow noreferrer">this wikipedia</a> description of the London dispersion force, which has this illustration of the potential between two Ar atoms:</p>
<p><a href="https://i.sstatic.net/vL2dC.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/vL2dC.png" alt="enter image description here"></a></p>
<p>See also the table at the bottom of the wikipedia article, which explains that for the Ne dimer, dispersion contributes 100% of the total intermolecular interaction energy (I would modify that perhaps to <em>attractive</em> interaction).</p>
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https://chemistry.stackexchange.com/questions/123448/chemical-anti-bonding-and-pauli-repulsion
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Question: <p>What happens to the lobe of the p-orbital during sp² hybridization?</p>
<p>During the formation of <span class="math-container">$\ce{C2H4}$</span> molecule:</p>
<p>We know that both the carbon atoms will have a total of 3 p-orbitals and they will be involved in hybridization.</p>
<p>Let us number the carbon atoms as <span class="math-container">$\ce{C}1$</span> and <span class="math-container">$\ce{C}2$</span>.</p>
<p>Now the p-orbital of <span class="math-container">$\ce{C}1$</span> along <span class="math-container">$x$</span>-axis will form a <span class="math-container">$\unicode[Times]{x3C3}$</span> bond with the p-orbital of <span class="math-container">$\ce{C}2$</span> and a <span class="math-container">$\unicode[Times]{x3C3}$</span> bond with the s-orbital of hydrogen.</p>
<p>And similarly the p-orbital of <span class="math-container">$\ce{C}2$</span> along x-axis will form a <span class="math-container">$\unicode[Times]{x3C3}$</span> bond with hydrogen.</p>
<p>So both the lobes of the p-orbital of <span class="math-container">$\ce{C}1$</span> and <span class="math-container">$\ce{C}2$</span> along <span class="math-container">$x$</span>-axis are involved in sigma bond formation.</p>
<p>Lets say the p-orbital of <span class="math-container">$\ce{C}1$</span> and <span class="math-container">$\ce{C}2$</span> along the <span class="math-container">$y$</span>-axis forms a <span class="math-container">$\unicode[Times]{x3C0}$</span> bond.</p>
<p>So here also both the lobes of p-orbital of <span class="math-container">$\ce{C}1$</span> and <span class="math-container">$\ce{C}2$</span> are involved in hybridization.</p>
<p>Let say that the p-orbital of <span class="math-container">$\ce{C}1$</span> which is along the <span class="math-container">$z$</span>-axis forms a <span class="math-container">$\unicode[Times]{x3C3}$</span> bond with the s-orbital of hydrogen.</p>
<p>And similarly the p-orbital of <span class="math-container">$\ce{C}2$</span> along <span class="math-container">$z$</span>-axis forms a <span class="math-container">$\unicode[Times]{x3C3}$</span> bond with the s-orbital of the other hydrogen.</p>
<p>Now my confusion is that here two p-orbitals along the <span class="math-container">$x$</span> and <span class="math-container">$y$</span>-axes have four lobes and all of them are involved in bond formation.</p>
<p>But the p-orbital which is along <span class="math-container">$z$</span>-axis has only one of its lobes involved in bonding. What happens to the other lobe of this p-orbital?</p>
<p><img src="https://i.sstatic.net/9g8z6.jpg" alt="enter image description here"></p>
Answer: <p>You talk about hybridisation but you haven't actually looked at any hybrid orbitals in your description. Since the geometry around the carbons is trigonal planar, each carbon hybridises its $\ce{2s}$, $\ce{2p}_x$ and $\ce{2p}_y$ orbitals to form three $\ce{sp^2}$ orbitals which lie in the $xy$ plane at $120^\circ$ angles to each other. One $\ce{sp^2}$ orbital on each carbon points towards the other carbon and so these orbitals overlap to form a $\ce{C-C}$ $\sigma$ bond. The other two $\ce{sp^2}$ orbitals on each carbon overlap with the $\ce{1s}$ orbitals on the hydrogens forming $\ce{C-H}$ $\unicode[Times]{x3C3}$ bonds. The remaining unhybridised $\ce{2p}_z$ orbitals on each carbon overlap above and below the plane of the $\unicode[Times]{x3C3}$ bonds to form a $\ce{C-C}$ $\unicode[Times]{x3C0}$ bond.</p>
<p>While $\ce{sp^2}$ orbitals still have two lobes, one lobe is substantially larger than the other and only that takes part in bonding. Note that this is the usual case: even in molecules that solely use p-orbitals, such as chlorine, only <em>one</em> lobe of a p-orbital will form a $\unicode[Times]{x3C3}$ bond; the other of the same orbital <em>will not.</em></p>
<p><a href="http://www.chemtube3d.com/orbitalsethene.htm" rel="nofollow">This link</a> provides a nice visualisation of the different orbitals involved in ethene bonding.</p>
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https://chemistry.stackexchange.com/questions/24989/chemical-bonding-and-hybridization-of-organic-compounds
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Question: <p>A snippet from a textbook:</p>
<blockquote>
<p>Therefore, the hybridization model predicts that an <span class="math-container">$\mathrm{sp}$</span>-hybridized carbon atom is more electronegative than an <span class="math-container">$\mathrm{sp}^3$</span>-hybridized carbon atom. Evidence for this effect is that the positive end of the dipole moment in <span class="math-container">$\ce{N#C-Cl}$</span> is on the <span class="math-container">$\ce{Cl}$</span> atom. We conclude that the carbon atom in the cyanide group is more electronegative than a chlorine atom.</p>
</blockquote>
<p>The paragraph suggests that the <span class="math-container">$\ce{N#C}$</span> bond has a different hybrid orbital compared to the <span class="math-container">$\ce{C-Cl}$</span> bond, doesn't it? I am a bit confused since there is only <span class="math-container">$1~\ce{C}$</span> atom but somehow it has <span class="math-container">$2$</span> hybrid orbitals of different state (namely <span class="math-container">$\mathrm{sp}^3$</span> and <span class="math-container">$\mathrm{sp}^2$</span>). Is this due to different ligands with which the central atoms connect?</p>
<p>And point of interest is why the conclusion isn't</p>
<blockquote>
<p>the nitrogen atom in the cyanide group is more electronegative than a chlorine atom</p>
</blockquote>
<p>instead of carbon? Because I would imagine that the electron density of both <span class="math-container">$\ce{N#C}$</span> and <span class="math-container">$\ce{C-Cl}$</span> bonds would lie dominantly on <span class="math-container">$\ce{N}$</span> and <span class="math-container">$\ce{Cl}$</span>, respectively. Hence, it is more reasonable to me to compare the electronegativity of <span class="math-container">$\ce{N}$</span> and <span class="math-container">$\ce{Cl}$</span> instead that of <span class="math-container">$\ce{C}$</span> and <span class="math-container">$\ce{Cl}$</span>.</p>
Answer: <p>The usual explanation for <span class="math-container">$sp$</span> orbitals being more electronegative than <span class="math-container">$sp^3$</span> is the fact that more <span class="math-container">$s$</span> character means a shorter average distance from the nucleus, thus experiencing a greater effective nuclear charge. This partially explains the decreasing magnitude of the dipole moment of <span class="math-container">$\ce{CH3-CH2-Cl}$</span> to <span class="math-container">$\ce{CH2=CH-Cl}$</span> to <span class="math-container">$\ce{HC#C-Cl}$</span> and the increasing acidity of ethane to ethylene to acetylene.</p>
<p>I agree with you that the example they give doesn't really demonstrate that, since the orientation of the dipole (as you pointed out) relies on the electronegativity of the nitrogen, not the carbon. Specifically, the stability of the left resonance structure below:</p>
<p><span class="math-container">$[\ce{^-N=C=Cl+ <-> N#C-Cl}]$</span></p>
<p>which clearly shows the orientation of the dipole, even if this resonance structure is only a minor contributor.</p>
<p>As a counterexample, chloroacetylene has a dipole with the (-) end on the chlorine, even though the carbon is still <span class="math-container">$sp$</span> hybridized. The corresponding resonance structure with a negative charge on C is much higher in energy than in the nitrogen case and so contributes negligibly to the overall structure, so the inductive electron-withdrawing of Cl dominates over the resonance electron-donating.</p>
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https://chemistry.stackexchange.com/questions/109261/chemical-bonding-based-on-hybridisation-model
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Question: <p>In books it is commonly written that whenever a chemical bond is formed the energy of the molecules/atoms gets lowered and hence energy is released. This is generally explained by diagrams like this </p>
<p><a href="https://i.sstatic.net/uXJub.jpg" rel="noreferrer"><img src="https://i.sstatic.net/uXJub.jpg" alt="enter image description here"></a></p>
<p>But I'm really unable to understand this process, I accept that before bonding the potential energy of the molecules were more and after bonding the potential is less and hence this difference of energy must be released but how does this lowering of potential energy happens? Let's just take an example of electrostatic situation (and as far as I know its the electric potential energy only that get's lowered and hence some energy gets released), say I have an electron at the origin and a proton at <span class="math-container">$x=x_1$</span> and then somehow it is moved to <span class="math-container">$x=x_2$</span> like this </p>
<p><a href="https://i.sstatic.net/w2jM9.png" rel="noreferrer"><img src="https://i.sstatic.net/w2jM9.png" alt="enter image description here"></a> We know from equation of electrostatics that potential energy of proton at <span class="math-container">$x=x_1$</span> is <span class="math-container">$$ U_1 = \frac{1}{4\pi \epsilon_0}~\frac{-e^2}{x_1} \\
\\
\textrm{and at} ~x=x_2 \\
U_2 = \frac{1}{4\pi \epsilon_0}~\frac{-e^2}{x_2}$$</span>
Since, <span class="math-container">$x_1$</span> is greater than <span class="math-container">$x_2$</span> (from figure) therefore <span class="math-container">$U_1 \gt U_2$</span>. So, we energy at <span class="math-container">$x_2$</span> is less therefore there must be some release of energy, but how? If we let the field to do the work then no energy will be released because our proton gonna get the kinetic energy, if <em>we</em> do the work to move it from <span class="math-container">$x_1$</span> to <span class="math-container">$x_2$</span> then according the to the laws of physics I will do the negative work and hence energy will transfer from the proton to <em>me</em>. So, how does energy ever get <em>released</em>? </p>
<p>How in chemistry the energy gets released by the formation of bonds? I know that we can say without intricacies that potential energy gets lowered and hence the difference of energy is the energy that we get, but how does this happens? If we have some cation and then we put some anion in the reaction mixture then, of course, cations will get attracted towards the anion and hence will <em>gain</em> kinetic energy and according to me that's the only possible way to lower the potential energy. Is that kinetic energy that we measure as heat and say that the reaction is exothermic? </p>
<p>Any help or clarification will be much appreciated. </p>
Answer: <p>Let us assume the simplest case to form a bond from two atoms (assumed to be in their ground state) that approach each other from an infinite distance. For convenience, we put the potential energy of the system at zero here. If the atoms come closer, they start to interact via long-range forces (such as van der Waals forces) and the potential energy is slightly lowered. If the atoms come even closer, their wave functions start to overlap and the potential is reduced rather steeply as compared to the long-range forces. Moving the atoms even closer increases the potential as the electrons will start to repel each other. The potential energy as a function of the interatomic distance can be approximately described by a Morse potential (the blue curve in the figure below (taken from Wikipedia))</p>
<p><a href="https://upload.wikimedia.org/wikipedia/commons/thumb/7/7a/Morse-potential.png/1280px-Morse-potential.png" rel="nofollow noreferrer"><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/7/7a/Morse-potential.png/1280px-Morse-potential.png" alt="Morse potential in blue (from Wikipedia)"></a></p>
<p>In the figure you also see horizontal lines in the Morse potential which represent the energies of the vibrational states that are supported by the potential. Now that we know what the potential looks like we can take our two atoms at infinite distance and move them towards each other, that is, the total energy <span class="math-container">$E_\text{tot}(r)=E_\text{Morse}(r)+E_\text{kin}(r)>0$</span>. As the atoms get closer, the potential energy gets lower and the velocity and kinetic energy increase, the total energy is - of course - conserved. When the atoms come so close that they start to repel each other (when they hit the repulsive "wall" on the left), they fly away in opposite direction and end up at infinity again! This is the same result that you would find with an ideal marble track (no friction) that is similarly shaped to the Morse potential: the marble will roll down, first accelerate, then climbs up the wall on the left and rolls back to where it started. In order to from a bond (or to stop the marble in the middle) we need to remove part of the energy of the system. As Ivan Neretin pointed out, this can be done by emitting a photon. When the atoms are separated at <span class="math-container">$r_e$</span>, the release of a photon with energy <span class="math-container">$h\nu=D_0+E_\text{kin}$</span> will result in a loss of energy and the atoms will be trapped in the potential. (In our marble track, friction can play a similar role)</p>
<p>This process is called radiative association and it is believed that it played an important role in the chemistry of the early Universe. However, not all molecules can emit a photon. There are some rules - selection rules - which have to be fulfilled. The molecule needs to have a dipole moment for instance. Therefore <span class="math-container">$\text{H}_2$</span> molecules are unlikely to be formed via radiative association. </p>
<p>Another, more likely, method to remove the energy is to have a third atom (or molecule) colliding with the other two atoms. This atom can then take away the excess energy in the from of kinetic energy in a so-called three-body collision. </p>
<p>When you have more complex molecules that collide to form bonds, the excess energy can be distributed over the different degrees of freedom that the molecule possesses. For instance, the molecule's vibrational and/or rotational energy is increased which results in an increase in temperature.</p>
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https://chemistry.stackexchange.com/questions/126598/how-energy-is-released-when-chemical-bond-is-formed
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Question: <p>Ionic bonds seem to be intermolecular but are classified as chemical bonds.</p>
<blockquote>
<p>"Ionic bonding is a type of chemical bond that involves the electrostatic attraction between oppositely charged ions." - <em>Wikipedia, definition of an ionic bond</em></p>
<p>"The physical force of attraction which holds atoms and molecules in a matter is called a physical bond. van der Waals' forces and coulombic forces are physical forces." - <em>Chemistry Stack Exchange, definition of an intermolecular force</em></p>
<p>"A chemical bond is an attraction between atoms that allows the formation of chemical substances that contain two or more atoms. The bond is caused by the electrostatic force of attraction between opposite charges, either between electrons and nuclei or as the result of a dipole attraction." - <em>Wikipedia, definition of a chemical bond</em></p>
</blockquote>
<p>Before an ionic bond is formed, oxidation and reduction must occur. For example, let's use sodium chloride. A chlorine atom will steal the outer electron from a sodium atom. The chlorine is gaining an electron, i.e. being reduced. The sodium is losing an electron, i.e. being oxidized. I agree that this is a purely chemical process.</p>
<p>Now that the atoms are oppositely-charged ions, they can attract and form an ionic bond:</p>
<p><span class="math-container">$$\ce{Na+ + Cl- -> NaCl}$$</span></p>
<p>My question is this: How is this bond defined as a chemical bond?</p>
<p>Let's take water; <span class="math-container">$\ce{H2O}$</span>. The oxygen forms two polar-covalent bonds with two hydrogen atoms. Each hydrogen shares two electrons with the oxygen. The electrons are being shared, so this bond an intermolecular force. The atoms stay together because they both depend on the electrons they share.</p>
<p>Now let's take more than one water molecule. These molecules will form hydrogen bonds. The hydrogens on the water molecules do not often have an electron around them because the oxygen has a greater attraction with them. This leaves the lonely proton of the hydrogen to make that end of the water positive while the oxygen is negative with its electrons. The negative end of the water (oxygen) will become attracted to the positive end of another water molecule (hydrogen), forming our anticipated intermolecular force. Does this sound familiar?</p>
<p>I hope it does. According to my knowledge, an ionic bond is that, but instead of a dipole-dipole interaction, it's a monopole-monopole interaction. If you look at the bond itself and not the reaction forming it, it acts just like an intermolecular force.</p>
Answer: <p><strong>Short version:</strong></p>
<p>We don't call bonds "physical", there are chemical bonds and other types of interactions between particles. The chemical bonds are classified this way because they make up molecules, salts, polymers and such, which are the materials chemists are interested in studying, and not particularly because of their sub-atomic/electrostatic characteristcs. </p>
<hr>
<p>Your question relates to two frequent sources of confusion: </p>
<ol>
<li><p>How do we define a chemical bond? What makes a bond between two atoms ionic or covalent?</p></li>
<li><p>Where does chemistry (hence chemical interactions) ends, and consequentially where does physics (and hence physical interactions) begins?</p></li>
</ol>
<p>Both questions have been pretty much answered here on Chemistry SE and also on Physics SE. I'll give some links to related questions and also a shrot summary.</p>
<p><strong>1. How do we define a chemical bond? What makes a bond between two atoms ionic or covalent?</strong></p>
<p>Your question didn't go much into covalent bonds, but anyone trying to understand anything past the surfaces of chemical bonds should know that the definition of a bond as covalent <strong>or</strong> ionic is simply an easier way for us to classify things, because when the bonds are categorized in groups we can say that one group behaves in a way, and the other behaves in a different way. The truth is, all bonds have some ionic and some covalent character to them. </p>
<p>Good answers here on Chemistry SE about that can be found here: </p>
<ul>
<li><a href="https://chemistry.stackexchange.com/questions/32306/can-100-covalent-bonds-exist">Can 100% covalent bonds exist?</a></li>
<li><a href="https://chemistry.stackexchange.com/questions/19568/what-happens-if-the-electronegativity-difference-is-exactly-2-1/19573">What happens if the electronegativity difference is exactly 2.1?</a></li>
</ul>
<p><strong>2. Where does chemistry (hence chemical interactions) ends, and consequentially where does physics (and hence physical interactions) begins?</strong></p>
<p>For a Physics pont of view, see: <a href="https://physics.stackexchange.com/questions/27837/what-distinguishes-between-physics-and-chemistry">What distinguishes between physics and chemistry?</a></p>
<p>For a Chemistry point of view, see: <a href="https://chemistry.stackexchange.com/questions/15607/differences-between-chemical-physics-and-physical-chemistry">Differences between chemical physics and physical chemistry?</a></p>
<p>Basically, it's clear that both sciences have a lot of overlapping, and perhaps the main difference is the approach taken and how the results will be used. While chemists usually focuses on understanding the structure and interactions between substances and materials, in order to be able to improve them or develop new ones, physicists are mostly interested in understanding the exact laws of nature and representing them with mathematical equations that can be used to explain and predict phenomena. This definitions do not seem so different, and that's not supposed to be wrong: We do have the Physical Chemistry and Chemical Physics fields.</p>
<p>Ok, but <strong>Why is an ionic bond a chemical and not a physical bond?</strong></p>
<p>Two important things were mentioned by @Mithoron in the comments. The term "Physical bond" is not frequently used by chemists, (or anyone I suppose), pretty much because <em>all interactions in whole universe are physical</em>. This single quote by @Mithoron is related to the two points I highlighted above. Both "types" bonds "follow" the same laws, and there is no hard line between them.</p>
<p>It's also worth noting that the definition you gave for Physical Bond was taken from an answer without a single upvote, which is an indicator that it isn't what the Chemistry SE community thinks (despite the fact that Google liked it very much and uses it in it's <a href="https://www.google.com.br/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=physical%20bond" rel="noreferrer">short-answer for the query</a>). Another indicator that "Physical Bond" isn't really a type of bond is that there aren't definitions of it out there (Yahoo answers doesn't count), not even a Wikipedia article.</p>
<p>So why do we call them <strong>chemical bonds</strong>? </p>
<p>That's pretty much because the bonds we call <em>chemical</em> are those which make up molecules, salts and other materials which are mainly studied by chemistry. The definition of chemical bond varies a lot (as explained <a href="https://chemistry.stackexchange.com/questions/230/what-is-the-difference-between-physical-and-chemical-bonds">here</a>), but they are the bonds which create what chemists study. </p>
<p>Other types of bonds which aren't called <em>chemical</em> are the intermolecular interactions, which are obviously of interest to chemistry, but also the <a href="http://en.wikipedia.org/wiki/Strong_interaction" rel="noreferrer">Strong Force</a>, which <em>is the force that binds protons and neutrons (nucleons) together to form the nucleus of an atom.</em> (I hope it's okay to call Strong Force a <em>bond</em>, I'm not sure to be honest).</p>
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https://chemistry.stackexchange.com/questions/32533/why-is-an-ionic-bond-a-chemical-and-not-a-physical-bond
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Question: <p>Chemical reactions are classed as endothermic or exothermic.</p>
<p>Defn: exothermic</p>
<blockquote>
<p>(of a reaction or process) accompanied by the release of heat.</p>
</blockquote>
<p>The heat is released when the chemical bonds in the product are formed.</p>
<p>Defn: heat</p>
<blockquote>
<p>heat is energy in transfer to or from a thermodynamic system, ... The
mechanisms include conduction, ...; or radiation between separated
bodies; or ....</p>
</blockquote>
<p>So, I have three questions:</p>
<p>1 - if the heat is released via conduction, is this equivalent to an increase in the kinetic energy of the product(s) compared to the kinetic energy of the reactant(s)? Also, how does the formation of a chemical bond produce kinetic energy?</p>
<p>2 - if the heat is released via radiation, how is radiation created in the formation of a chemical bond?</p>
<p>3 - are conduction and radiation the only forms of heat transfer when a bond is created?</p>
<p>Edit - to clarify the question - the issue is addresed in the OPENSTAX AP chemistry text - which covers one example of 2 H atoms coming together to form H2 in detail as follows:</p>
<blockquote>
<p>When the atoms are far apart ..... by convention the sum of their
energies is 0. As the atoms move together their orbitals begin to
overlap. Each electron begins to feel the attraction of the nucleus in
the other atom. In addition, the electrons begin to repel each other,
as do the nuclei. While the atoms are still separated, the
attractions are slightly stronger than the repulsions, and the energy
of the system decreases. (A bond begins to form.) ... At some specific
distance between the atoms, which varies with the atoms involved, the
energy reaches its lowest (most stable) value. This optimum distance
between the two bonded nuclei is the bond distance between the two
atoms.</p>
</blockquote>
<p>The text doesn't specify what energy is decreasing, and why. But I can answer these questions I think. The energy that is decreasing is the potential energies of the attractive fields of the atoms, and that energy is being transformed to kinetic energy as the atoms accelerate toward each other.</p>
<p>So we have the nuclei accelerating towards each other. What happens next?</p>
<p>According to the text ... continuing where we left off</p>
<blockquote>
<p>The bond is stable because at this point the attractive and repulsive
forces combine to create the lowest possible energy configuration.</p>
</blockquote>
<p>So there is no mention of the release of energy, and the atoms just arrive at the stable combined state.</p>
<p>Here is what I would expect, given that the nuclei are accelerating towards each other - at some point the repulsive forces become stronger than the attractive forces, and the atoms begin to decelerate. If there is no damping force on the motion the atoms will oscillate about the point where the attractive and repulsive forces are equal. So, there must be some damping force on the atoms, that causes them to come rest at the bond distance of H2.</p>
<p>Now comes the mystery, which is, how does the formation of a chemical bond release energy? </p>
<p>What happens to the kinetic energy of the two nuclei accelerating towards each other? If the two hydrogen atoms are coming together head on there will be a collision that would appear to cancel out the motion, i.e. kinetic energy, of both atoms. If there is a damping mechanism that generates the heat, what is it?</p>
<p>Since the energy is released as heat, I think it must be in the form of kinetic or electromagnetic energy, but what is the specific mechanism?</p>
<p>Or, am I trying to get too much out of the kinetic model?</p>
Answer: <p>It is easier first to think of what happens when a bond is broken in an exothermic reaction. The energy is initially released into vibrations, rotations and translational motion of the two fragments. This energy is then lost as these fragments collide with other gas molecules or with solvent molecules and so eventually ends up as heat. </p>
<p>In exothermic bond formation the extra energy vibrationally and rotationally excites the new molecule and this is then lost via collisions with surrounding gas or solvent molecules and again ends up as heat. </p>
<p>(If the new molecule were isolated in space (so no collisions) it would remain 'hot' until the energy is radiated away from vibrational levels and until it (radiatively) comes into equilibrium with the surroundings. This is a slow process, by many orders of magnitude, compared to collisions in solution or in a gas at normal atmospheric pressure )</p>
<p>In a very few types of reaction an electronically excited state is produced in the product and then photons radiate (as visible light) some of the energy away. This is called chemiluminescence. </p>
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https://chemistry.stackexchange.com/questions/111936/how-is-energy-released-when-a-chemical-bond-is-formed
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Question: <p>In general is there a convention as to how one interprets squiggly lines in chemical structure notation?</p>
<p>e.g. see the below sketch. I suppose it has something to do with cis / trans isomerism. But e.g. where do you draw the squiggly i.e. which side of the double bond? Is it used in other contexts as well? And does it mean we don't know which isomer is formed or does it mean we are talking about a cis + trans mixutre?</p>
<p><a href="https://i.sstatic.net/B317Q.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/B317Q.png" alt="enter image description here" /></a></p>
<p>Larger context here (so the explanation about variable chain length may not make sense?)</p>
<p><a href="https://i.sstatic.net/9f1Hq.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/9f1Hq.png" alt="enter image description here" /></a></p>
Answer:
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https://chemistry.stackexchange.com/questions/145043/squiggly-line-in-chemical-bond-notation
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Question: <p>I would like to ask if the 3D spacing arrangement of atoms in chemical bond configurations are always unique, please?</p>
<p>For example, glycine consists of the set of atoms HHNCHHCOOH. Does there exist a different stable 3D arrangement of this collection of atoms, other than:</p>
<p><a href="https://i.sstatic.net/VrFbZ.jpg" rel="noreferrer"><img src="https://i.sstatic.net/VrFbZ.jpg" alt="enter image description here" /></a></p>
<p>What causes this arrangement to be unique? If we throw in and react the ingredients in a different order, such as HHHHHNCCOO, will they all still end up as the same unique shape of regular glycine?</p>
<p>Thank you.</p>
Answer: <p>From a given number of atoms, you can usually construct multiple molecules (isomers), so we shall just concentrate on the case where the bond connectivity is the same.</p>
<p>The terms you are looking for are <em>conformation</em> and <em>configuration</em>. The latter is most usually related to double bonds and E/Z isomerism. If we take but-2-ene, for example, it can exist in 2 forms, E and Z. Both are stable and have the same atom connectivity. However, they cannot interconvert between each other - the activation energy is too high.</p>
<p>Two conformations of the same molecule, however, can interconvert between each other (at least at room temperature) and they are both stable (i.e., potential energy minima). A classic case of a molecule is butane, whose conformations can be seen in the picture below.</p>
<p><a href="https://i.sstatic.net/FuVZq.png" rel="noreferrer"><img src="https://i.sstatic.net/FuVZq.png" alt="enter image description here" /></a></p>
<p>The anti- and gauche- conformations are both 'butane' and both are stable (even though the anti- is more stable), but have a different 3D arrangement of atoms.</p>
<p>As a sidenote, these can be distinguished at a low enough temperature, at which they cannot interconvert.</p>
<p>Another famous molecule with various conformations is cyclohaxane, with the two stable ones being the chair and the much less common twist-boat.</p>
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https://chemistry.stackexchange.com/questions/163336/is-the-3d-spacing-arrangement-of-atoms-in-chemical-bond-always-unique
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Question: <p>I'm working on a chemistry question where I'm supposed to use average bond energies to estimate enthalpy changes for a reaction. It gives me these bond energies to work with:</p>
<p>$$\begin{align}
\ce{C-H}&=&413\\
\ce{C-C}&=&348\\
\ce{C-Br}&=&276\\
\ce{H-Br}&=&366\\
\ce{CkC}&=&614\\
\end{align}$$
I know how to do the question, but I don't what $\ce{CkC}$ is. I can't find it online or in my textbook. What does the "k" mean?</p>
Answer: <p>$\ce{C=C}$ <strong>average bond energy</strong> is somewhere around $\pu{614 kJ/mol}$ so "k" probably indicates a double bond</p>
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https://chemistry.stackexchange.com/questions/93953/what-does-ckc-mean-in-chemical-bonding
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Question: <p>I can understand that in Silicon Crystal a central Silicon atom shares its 4 valance electrons with 4 surrounding silicon atoms. And the chain continues. But then What happens to the outermost such Silicon atoms (in the drawing). They each will have incomplete so called octate.</p>
Answer: <p>In a Silicon crystal, outermost Silicon atoms are probably oxidized yielding a monoatomic layer of Oxygen atoms, chemically bound by covalences like in <span class="math-container">$\ce{SiO_2}$</span>, by analogy with Aluminium. Silicon and Aluminium are neighbors in the periodic table : they often have similar properties. And it is well known that an extremely thin layer of Aluminium Oxide <span class="math-container">$\ce{Al_2O_3}$</span> does cover Aluminium foils and pieces. It must be the same for Silicon. </p>
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https://chemistry.stackexchange.com/questions/133493/chemical-bonding-and-crystal-structure-of-silicon
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Question: <p>I recently came across a question in which an option is 'Breaking a chemical bond is the first step in any chemical reaction', which lead me to think of reactions which do not involve bond breaking. The only reaction I can think of is dimer formation, but a Google search reveals no more. Surely this cannot be the only example?</p>
Answer: <p>The reaction of a Lewis acid with a Lewis base results only in bond formation. Your question gets at the essence of chemistry, though. A chemical reaction is the rearrangement of electrons between atoms. So generally, most reactions will involve breaking bonds. Even for monomers (e.g. alkenes) to polymerize, $\pi$-bonds within the monomers have to be broken to form $\sigma$-bonds in the polymer. You seldom have one without the other.</p>
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https://chemistry.stackexchange.com/questions/99967/which-chemical-reactions-dont-involve-bond-breaking
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Question: <p>As we know, the first excited states of Hydrogen atoms is $2p$.
The wave functions in cartesian basis we can represent as
$$
p_x = \frac{1}{\sqrt{2}} \left( Y_{1, +1} + Y_{1, -1}\right) =\sqrt{\frac{3}{16\pi}} \frac{x}{r} \\
p_y = \frac{i}{\sqrt{2}} \left( Y_{1, +1} - Y_{1, -1}\right) = \sqrt{\frac{3}{16\pi}} \frac{y}{r} \\
p_z = Y_{1,0} = \sqrt{\frac{3}{4\pi}} \frac{z}{r}
$$
which is visulize in forms of dumbells:</p>
<p><a href="https://i.sstatic.net/rc6TG.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/rc6TG.jpg" alt="enter image description here"></a></p>
<p>My question as follows:</p>
<ol>
<li><p><strong>An H-atom in an excited state has the same shape?</strong></p></li>
<li><p><strong>How does an atom determines the axis along which to orient?</strong> Assuming that the directions equiprobable, the shape should be spherical. <strong>Or I'm wrong?</strong></p></li>
<li><p>Type of chemical bond ($\sigma$, $\pi$, ...) is determined by the form of the type of atomic orbital ($s$, $p$, $d$, ..., or by their hybrides), but <strong>how it is possible if an atom is spherical yet?</strong>
<a href="https://i.sstatic.net/MPdpN.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MPdpN.jpg" alt="enter image description here"></a></p></li>
</ol>
Answer:
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https://chemistry.stackexchange.com/questions/68768/spatial-form-of-an-atom-and-type-of-chemical-bond
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Question: <p>We know that if Cl and Na get too close, they produce ionic bonding.</p>
<p>Cl has 17 proton and 17 electrons and is considered stable.</p>
<p>Na has 11 protons and 11 electrons and is considered stable.</p>
<p>I understand that in Na, we have 1 valence electron and in Cl, we have 7 valence electrons. Now, if they get too close, 1 valence electron is transferred from Na to Cl. I don't understand why. Na was already stable. I get that part that Cl, since it has more protons, it would attract the electron from Na more than Na would do for itself due to less protons(11), but when ionic bond happens, Cl would end up having 17 proton and 18 electrons. Wouldn't Cl itself become unstable as there're more electrons than protons ? It seems like that before bonding, they both were stable, but after bonding, Cl ended up not stable.</p>
<p>I'm trying to understand it, but I get a point that explanation with quantum would be a waste of time as I have no knowledge in quantums. So would appreciate the explanation in terms of attraction forces and why after bonding, Cl doesn't become unstable. Note that octet rule and staff like this is something I don't really need to know as they're rules and not explanation in terms of attractions. I checked this <a href="https://chemistry.stackexchange.com/questions/16922/why-do-atoms-want-to-have-a-full-outer-shell">why do atoms want to have a full outer shell</a> but doesn't help.</p>
Answer: <p>Here is an easier explanation not relying on "stability" or quantum mechanics.</p>
<p>You start out with some sodium ions and some chloride ions in aqueous solution. Each sodium ion is surrounded by water, and each chloride ion is surrounded by water. Then, you let the water evaporate, the sodium and chloride ions come together and form a crystal, where each sodium ion is surrounded by six chloride ions, and each chloride ion is surrounded by six sodium ions.</p>
<blockquote>
<p>We know that if Cl and Na get too close, they produce ionic bonding.</p>
</blockquote>
<p>While ionic bonding is often explained as a transfer of electrons, it would be hard to find chlorine and sodium as elements. It is much more common that you already have sodium ions (as in sodium acetate or sodium hydroxide), and already have chloride ions (as in magnesium chloride) or make them from a molecular compound (as when hydrogen chloride gas comes in contact with water).</p>
<p>If you start with ions, you just have to figure out how they form an ionic solid under the right conditions (e.g. taking away water as a solvent).</p>
<blockquote>
<p>Na has 11 protons and 11 electrons and is considered stable.</p>
</blockquote>
<p>Elemental sodium is highly reactive. When it comes in contact with air or water, it will form ionic compounds or solvated ions. As you can read in the comments, "stable" becomes meaningful by saying stable with respect to some condition or in combination with which other substances.</p>
<blockquote>
<p>I'm trying to understand it, but I get a point that explanation with quantum would be a waste of time as I have no knowledge in quantums.</p>
</blockquote>
<p>If you want to have an explanation rather than a collection of rules and facts, you have to engage with quantum mechanics. All other explanations are incomplete. There is a perfect explanation of the strength of ionic bonds using <a href="https://en.wikipedia.org/wiki/History_of_Maxwell%27s_equations" rel="nofollow noreferrer">pre-1900 physics</a>, but to explain why the ions form in the first place, you need quantum mechanics.</p>
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https://chemistry.stackexchange.com/questions/173378/why-chemical-bond-between-na-and-cl-happens
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Question: <p>How can the formation of a covalent bond be described from a quantum perspective, and what implications does it have for traditional chemical bonding theory?</p>
<p>Specifically, in the context of many-electron systems where electron correlation becomes significant, how do modern quantum chemical methodologies, such as Hartree-Fock theory and Density Functional Theory, contribute to our refined understanding of covalent interactions? Furthermore, how does this quantum perspective challenge, extend, or complement classical notions of bonding, as represented by Lewis structures, valence bond theory, and molecular orbital theory?</p>
Answer: <p><strong>Quantum Chemistry and Covalent Bonding:</strong></p>
<p>Quantum chemistry, a nexus between quantum mechanics and molecular science, furnishes an in-depth framework for comprehending and predicting the nuances of chemical bonding. By incorporating quantum principles, this domain offers a granular look into electron behaviors in atoms and molecules, surpassing the sometimes oversimplified interpretations of classical models.</p>
<p><strong>Atomic and Molecular Orbitals:</strong></p>
<p>Atoms are characterized by wave functions or atomic orbitals, which are mathematical representations of the probability density of an electron, indicating the likelihood of finding an electron at a given location. As two atoms approach each other, their atomic orbitals begin to overlap. This overlap can result in the formation of molecular orbitals — either bonding (when the overlap is constructive) or antibonding (when the overlap is destructive). It's crucial to note that this phenomenon isn't driven by "intent" but by the intrinsic properties and behaviors of electrons in the atomic orbitals.</p>
<p><strong>Wave Function and Superposition:</strong></p>
<p>At its core, quantum mechanics presents the notion of the wave function, which encapsulates the probabilistic nature of electron positions. Electrons in a bond don't reside in a fixed location; they exist in a state of superposition. This means they have a probability of being in multiple locations simultaneously. The Schrödinger equation provides a time-dependent wave function, representing the state of the system, encapsulating this electron behavior.</p>
<p><strong>Electron Correlation and Many-Electron Systems:</strong></p>
<p>A particularly challenging aspect in quantum chemistry is addressing electron correlation, especially in systems with multiple electrons. Electrons, being fermions, adhere to the Pauli exclusion principle and also repel each other due to their like charges. Capturing the behavior of one electron influenced by all others is a computational challenge. This is where methods like Hartree-Fock theory, which offers an approximation by treating electrons as moving independently in an average field created by others, and Density Functional Theory (DFT), which considers the electron density rather than the wave function, come into play.</p>
<p><strong>Quantitative Approaches in Theoretical Chemistry:</strong></p>
<p>Delving deeper into bonding characterizations, there are quantitative methodologies in theoretical chemistry. Atoms in Molecules (AIM) theory provides a method to study chemical systems using the electron density, allowing for the analysis of bond critical points and bond paths. Electron Localization Function (ELF) helps in visualizing areas where electrons are likely localized.Then, Natural Bond Orbitals (NBO) analysis provides insight into bonding and antibonding interactions, giving a clearer definition and quantification to the concept of covalent bonds.</p>
<p><strong>Implications for Traditional Chemical Bonding Theory:</strong></p>
<p>Quantum chemistry refines our understanding from traditional bonding theories. Lewis structures and valence bond theory provide intuitive visual models, but they don't capture the nuanced electron behaviors that quantum mechanics does. Molecular orbital theory, rooted in quantum principles, offers a more accurate depiction, especially for molecules exhibiting resonance or delocalized electrons.</p>
<p>In essence, while classical theories provide a foundational understanding, quantum chemistry and its myriad of methodologies grant us a more intricate, quantitative, and comprehensive insight into the fascinating world of molecular interactions and electron behaviors.</p>
<hr />
<p>This answer now addresses the criticisms and integrates the suggested improvements.</p>
<p>I hope this provides a comprehensive dive into the topic!</p>
<p>For further information I recommend to read Ira.N.Levine "Quantum Chemistry".</p>
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https://chemistry.stackexchange.com/questions/175972/quantum-chemistry-and-covalent-bonding
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Question: <p>In a hypothetical (?) Og<sub>2</sub><sup>+235</sup> we would have a simple sigma bonding orbital occupied by one electron (leading to a bond order of 1/2). But how to take into account the giant repulsion of the two nuclei? Generally how are nucleis taken into account for energy calculations from molecular orbitals in MO theory? So far I've always read about the electrons but very few about the nuclei in the system.</p>
<p>Same in He<sub>2</sub><sup>3+</sup>, half bond order but very large repulsion. My intuition says that He<sub>2</sub><sup>1+</sup> should be more stable, even if both have the <strong>same bond order</strong>.</p>
<p>Where in the framework of MO theory is exactly that taken into account? How does nuclei repulsion change the MO energy scheme?</p>
Answer: <p>I would not bet on it. We apparently can't even get <span class="math-container">$\ce{He2^{3+}}$</span>, let alone higher single-electron diatomic ions. From <a href="https://en.wikipedia.org/wiki/Helium_compounds" rel="nofollow noreferrer">Wikipedia</a>:</p>
<blockquote>
<p><span class="math-container">$\ce{He2^+}$</span> was predicted to exist by Linus Pauling in 1933. It was discovered when doing mass spectroscopy on ionised helium. The dihelium cation is formed by an ionised helium atom combining with a helium atom: <span class="math-container">$\ce{He^+ + He -> He2^+}$</span>.[<a href="https://doi.org/10.1016/j.ijms.2004.07.012" rel="nofollow noreferrer">1</a>]</p>
<p>The diionised dihelium <span class="math-container">$\ce{He2^{2+}} (1Σ_g^+)$</span> is in a singlet state. It breaks up <span class="math-container">$\ce{He2^{2+}->2 He^+}$</span> releasing 200 kcal/mol of energy. It has a barrier to decomposition of 35 kcal/mol and a bond length of 0.70 Å.[1]</p>
</blockquote>
<p>Thus even with only two positive charges the helium dimer is prone to breaking up due to electrostatic repulsion. Although the dication above is metastable, we should expect the stability to only become worse with an additional positive charge and only one bonding electron instead of two.</p>
<p>A simple classical electrostatic calculation is instructive. Suppose you have a negative charge and two positive charges, the latter equidistant from the negative charge on opposite sides. The net force is attractive if all charges are one unit, but this net attraction is lost if we give two or more (and certainly 118!) units to both positive charges. The true quantum mechanical calculation is of course much more complicated, but can give only less favorable results because as the positive nuclear charges approach the electron cloud cannot stay fully between them. Thereby <span class="math-container">$\ce{H2^+}$</span> is predicted to be the only single-electron homonuclear diatomic ion that remains bound.</p>
<p><strong>Cited Reference</strong></p>
<ol>
<li>Grandinetti, Felice (October 2004). "Helium chemistry: a survey of the role of the ionic species". <em>International Journal of Mass Spectrometry</em>. <strong>237</strong> (2–3): 243–267. Bibcode:2004IJMSp.237..243G. <a href="https://doi.org/10.1016/j.ijms.2004.07.012" rel="nofollow noreferrer">https://doi.org/10.1016/j.ijms.2004.07.012</a>.</li>
</ol>
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https://chemistry.stackexchange.com/questions/169595/would-a-hypothetical-og2-235-form-a-chemical-bond
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Question: <p>Here's a "simple" bond diagram of 1,1,1,2-Tetrafluoroethane which I believe is a Lewis structure:</p>
<p><a href="https://upload.wikimedia.org/wikipedia/commons/thumb/a/a4/1,1,1,2-Tetrafluoroethane.svg/100px-1,1,1,2-Tetrafluoroethane.svg.png" rel="nofollow noreferrer"><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/a/a4/1,1,1,2-Tetrafluoroethane.svg/100px-1,1,1,2-Tetrafluoroethane.svg.png" alt="1,1,1,2-Tetrafluoroethane"></a></p>
<p>OK, that's easy enough - four bonds from each carbon atom, and each fluorine atom having one bond.</p>
<p>Here are some more "complicated" bonds which I have yet to do in my high school chemistry class, so I don't understand how to interpret them.</p>
<p>Here's caffeine:</p>
<p><a href="https://upload.wikimedia.org/wikipedia/commons/thumb/5/5e/Caffeine-2D-skeletal.svg/150px-Caffeine-2D-skeletal.svg.png" rel="nofollow noreferrer"><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/5/5e/Caffeine-2D-skeletal.svg/150px-Caffeine-2D-skeletal.svg.png" alt="Caffeine"></a>
<a href="https://upload.wikimedia.org/wikipedia/commons/thumb/0/0e/Caffeine_(1)_3D_spacefill.png/150px-Caffeine_(1)_3D_spacefill.png" rel="nofollow noreferrer"><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/0/0e/Caffeine_(1)_3D_spacefill.png/150px-Caffeine_(1)_3D_spacefill.png" alt="Caffeine"></a></p>
<p>And chloroform:</p>
<p><a href="https://i.sstatic.net/u7ZZL.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/u7ZZL.png" alt="Chloroform"></a>
<a href="https://upload.wikimedia.org/wikipedia/commons/thumb/f/fe/Chloroform_3D.svg/100px-Chloroform_3D.svg.png" rel="nofollow noreferrer"><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/f/fe/Chloroform_3D.svg/100px-Chloroform_3D.svg.png" alt="Chloroform"></a></p>
<p>How would I go on interpreting these diagrams? Or is there missing information? </p>
<p><sub>Image sources: Wikimedia</sub></p>
Answer: <p>The second structure you posted is a line angle diagram. In this diagram, methyl groups (-CH3) are implied using sticks with nothing at the end.</p>
<p><img src="https://i.sstatic.net/kXRjA.png" alt="enter image description here"></p>
<p>Carbon atoms are also implied; carbons exist at every "corner" of the polygon which doesn't have an element explicitly identified. Hydrogens <em>attached to carbons</em> are also implied. So at the intersection of two lines, if there are no other attachments and no charge, you can expect there to be two hydrogens (carbon is generally tetravalent). </p>
<p><img src="https://i.sstatic.net/weIxP.jpg" alt="enter image description here"></p>
<p>The third structure you posted is a wedge-dash diagram. Wedges depict substituents angled toward you; dashes are substituents pointing away from you; the lines are coplanar. The advantage of this diagram is that it shows the 3D nature of the molecule. Your first picture of the fluoro-carbon compound implies 90 degree bond angles. However, you should realize that the carbon atoms are sp3 hybridized and thus should not have 90 degree bond angles. </p>
<p><img src="https://i.sstatic.net/JPxgg.jpg" alt="enter image description here"></p>
<p><img src="https://i.sstatic.net/Nuw8k.png" alt="enter image description here"></p>
<p>Further reading and exploration of other depictions of molecular structure:</p>
<p><a href="http://www.masterorganicchemistry.com/2010/10/15/the-many-many-ways-to-draw-butane/" rel="noreferrer">http://www.masterorganicchemistry.com/2010/10/15/the-many-many-ways-to-draw-butane/</a></p>
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https://chemistry.stackexchange.com/questions/14460/explanation-of-certain-chemical-bond-diagrams
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Question: <p>Both in fine art, decorative and in utilitarian painting, sanding or roughening a substrate to provide a so-called “mechanical bond” has often been recommended as a preliminary step before applying paint.</p>
<ol>
<li><p>Interestingly, sanding is said to both roughen a surface, giving it some “tooth” for paint to adhere to, but is also used to <em>smooth</em> out a surface. I wonder whether roughening and smoothing might be considered to be two diametrically opposed concepts being performed by the same sandpaper, pumice stone or other abrasive product. If they are not inherently opposing concepts, how might their similarities be reconciled or related?</p>
</li>
<li><p>Why would a chemical bond based on tiny molecular interactions of paint to substrate molecules make a “mechanical bond” based on the interlocking of much larger pigment-and-binder particles to substrate particles necessary or advisable? Even if a “mechanical bond” promoted greater hills and valleys for surfaces to interlock, wouldn’t a chemical bond based on much smaller molecular adhesion obviate the need for larger hill and valley interlocking? Can the interlocking of large sizes make a bond more fragile as in, say, a fallen rock zone? Can it be that the chemical bond is minor in, say, linseed oil to hide glue or acrylic paint to acrylic ground -- and that further, larger hill and valley mechanical interlocking is thus advisable? I wonder how strong chemical bonds in such respects might be or how strong chemical bonds might be without “mechanical bonds.”</p>
</li>
</ol>
Answer:
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https://chemistry.stackexchange.com/questions/187939/is-a-chemical-molecular-bond-considered-more-effective-for-paint-adhesion-than
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Question: <p>I'm working with an amorphous system. With oxygen, my system has both covalent and ionic bond forming cations. I utilise the <a href="https://en.wikipedia.org/wiki/Wannier_function" rel="nofollow noreferrer">Wannier centre</a> to define a covalent bond, and my theory is that if the Wannier centre is near to the line connecting the cation and anion, it is a covalent bond.</p>
<p>However, this method fails to identify an ionic bond as there is no region of high electronic charge density which can be localized to obtain Wannier center. How can an ionic bond be defined using a computational technique?</p>
Answer:
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https://chemistry.stackexchange.com/questions/165918/how-to-define-a-chemical-bond-computationally
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Question: <p>In a paper by Peng <em>et. al.</em>,<sup>[1]</sup> I read about the valence band maximum of <span class="math-container">$\ce{TiO2}$</span> consists of <strong>non-bonding O p<sub>π</sub></strong> orbitals</p>
<blockquote>
<p>... the valence band maximum (VBM) consists of non-bonding O p<sub>π</sub> states</p>
</blockquote>
<p>What does non-bonding p<sub>π</sub> orbital mean? If there is not a chemical bond, why did they call it a π orbital?</p>
<h3>References:</h3>
<ol>
<li>Peng, H.; Li, J.; Li, S.; Xia, J. First-principles study of the electronic structures and magnetic properties of 3d transition metal-doped anatase TiO2. <em>J. Phys.: Condens. Matter</em> <strong>2008,</strong> <em>20</em> (12), 125207. <a href="https://doi.org/10.1088/0953-8984/20/12/125207" rel="nofollow noreferrer">DOI: 10.1088/0953-8984/20/12/125207</a>.</li>
</ol>
Answer: <p>When you have atoms bonded all in one plane, there will be <span class="math-container">$p$</span> orbitals oriented perpendicular to the plane which may not interact significantly with adjacent atoms. Such orbitals would then be called nonbonding.</p>
<p>We may compare water with carbon dioxide. Introductory textbooks often describe the oxygen in water as having a distorted <span class="math-container">$sp^3$</span> hybridization, but in reality only two oxygen <span class="math-container">$p$</span> orbitals mix with the hydrogen and oxygen <span class="math-container">$s$</span> orbitals to form the bonds. The third, perpendicularly oriented <span class="math-container">$p$</span> orbital is a nonbonding <span class="math-container">$p$</span> orbital. We can see this orbital (HOMO) in the middle of this diagram from <a href="https://www.flickr.com/photos/69057297@N04/44441397812/" rel="nofollow noreferrer">flickr.com</a>:</p>
<p><img src="https://i.sstatic.net/5gZRX.jpg" alt="enter image description here" /></p>
<p>In contrast, carbon dioxide does have the <span class="math-container">$p$</span> orbitals on the central carbon atom interacting strongly with out-of-plane orbitals on the oxygen atoms forming the familiar <span class="math-container">$\pi$</span> bonds.</p>
<p>Now let's look at titanium dioxide. Titanium dioxide may come in different phases depending on temperature and pressure, but the phase we generally see under ambient conditions is the <a href="https://commons.m.wikimedia.org/wiki/File:Rutile-unit-cell-3D-balls.png#mw-jump-to-license" rel="nofollow noreferrer">rutile structure</a> by WP user Ben Mills:</p>
<p><img src="https://i.sstatic.net/kCdZ6.png" alt="enter image description here" /></p>
<p>The red atoms are the oxygen atoms in titanium dioxide, and these are seen to have a distorted, but planar, triangular coordination with the gray titanium atoms (an equilateral triangular coordination does not fit with the octahedral coordination of the titanium). The anatase form described in the OP's reference has a more complex atomic arrangement but similar triangular coordination around each oxygen atom. So, there are perpendicularly oriented <span class="math-container">$p$</span> orbitals. The titanium atoms also have out-of-plane orbitals, but unlike the carbon in carbon dioxide these do not interact strongly with the out-of-plane oxygen <span class="math-container">$p$</span> orbitals. The difference in orbital energy and size is too great. So the out-of-plane oxygen <span class="math-container">$p$</span> orbitals are more like their counterparts in water, essentially nonbonding.</p>
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https://chemistry.stackexchange.com/questions/157594/what-does-it-mean-by-non-bonding-p-pi-orbital
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Question: <p>My main question is why is the boiling point of methanol so much different from that of water? </p>
<p>I understand that both compound are able to develop hydrogen bonding, and obviously water can develop one more hydrogen bond than methanol. </p>
<p>But on the other hand:</p>
<blockquote>
<p>Due to presence of an electron-releasing alkyl group, the oxygen atom of an alcohol molecule is less-electron withdrawing on the remaining hydrogen atom. Therefore the electron deficiency of the hydrogen atom is less than that of a water molecule resulting in the formation of a weaker hydrogen bond. </p>
</blockquote>
<p>What does this actually mean? </p>
<p>Both hydrogen, in methanol and water, are connected to oxygen, which has the same electronegativity and takes electrons off the hydrogen making it have a partial positive charge. </p>
<p>Why does a "less electron withdrawing oxygen atom" lead to a weaker hydrogen bond? </p>
<p>I thought that a less electron withdrawing oxygen atom would mean that the H is more negatively charged, meaning it would be stronger for hydrogen bonding, since F, O, N want more electrons as they are more electronegative? </p>
Answer: <p>A hydrogen bond is a particularly strong dipole:dipole interaction with some covalent character. If we ignore the covalent character for a moment, we can say the higher the partial positive charge on the hydrogen, the stronger the hydrogen bond (keeping the O-H bond distance constant). If you relate "electron deficient" to positive partial charge, it might help you to make sense of the textbook answer.</p>
<p>You write:</p>
<blockquote>
<p>Both hydrogen, in methanol and water, are connected to oxygen, which has the same electronegativity and takes electrons off the hydrogen making it have a partial positive charge.</p>
</blockquote>
<p>Yes, that is the primary reason compounds like water, methanol and - say - hydrogen peroxide and acetic acid are hydrogen bond donors. But the chemistry beyond the OH group also plays a role. To illustrate, just compare the acidity of the proton connected to the oxygen in these four cases. They are very different. In a similar manner, the hydrogen bonding will be different (not as different, though).</p>
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https://chemistry.stackexchange.com/questions/76727/how-does-hydrogen-bonding-affect-the-boiling-points-of-chemical-compounds
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Question: <p>This question has to do with the idea of stability and energy. The premise is that systems will tend towards lower energy states, that’s why bonding happens and electrons prefer single orbitals. And that makes sense in all considering the second law of thermodynamics, but it still doesn’t answer what causes the release of energy when bonding. Clearly, the potential energy of each electron is lower when there is a greater charge and a stronger force of attraction between things—it makes sense, as when you have a stronger force of attraction, you have less “pathways” or degrees of freedom, hence less potential energy. It also checks out with the math of Coulomb’s law, as when the radius decreases from bonding, potential energy decreases as well. So by an atom exhibiting a magnetic pull on another and bonding, potential energy of each electron decreases. But the thing is I just don’t get where that loss of potential energy is going—in this scenario, it wouldn’t make sense for it to be randomly emitted. Or is that not even the reason bonds loose energy?</p>
<p>Consider this analogy, and correct me if I’m wrong:
Consider I have 2 magnets A and B that are oppositely charged. If I take magnet A and pull it away from magnet B, now I just did work on magnet A and increased the magnetic potential energy of it as well. Now lets say I released magnet A. The potential energy would be converted into kinetic energy and it would tend towards the other magnet. Now, let’s say I catch magnet A before it hits magnet B. The <em>potential energy of Magnet A just decreased in the field because the kinetic energy transferred to my hand, right?</em></p>
<p>So instead in atoms, the “hand” that stops atoms from ramming into each other completely is other electrons and the repulsion of the nucleus. So it seems all energy is conserved. Each atom is just transferring energy between each other by both pulling on each other. <strong>So how does energy get released?</strong></p>
Answer: <p>Consider <span class="math-container">$$\ce{2 H <=> H2^{*} ->[collision][emission] H2 + energy}.$$</span></p>
<p>Imagine a hydrogen atom figuratively rolling down to the potential pit of the bond with another hydrogen atom.</p>
<p>Without passing energy elsewhere, it will bounce and appear free again, keeping its initial energy. It more or less performs one molecule vibration at the top energy level which has just enough energy to break the bond.</p>
<p>Like a skateboarder on U-ramp in lossless scenario would appear on the other side of the ramp.</p>
<p>Now imagine the atom collides with another molecular entity during the time it has increased its kinetic energy and decreased the potential one. Or, it emits a photon. The atom then passes a part of its energy away and is then trapped in the potential pit of the bond.</p>
<p>Like if the skateboarder accidentally hits something/somebody during his ride and loses energy. He will not get in such a case to the other side and is "trapped" in the ramp.</p>
<hr />
<p>In reactions with more than one product, the released potential energy is converted to kinetic energy of products as there is always something handy to push to release the energy, products flying away of each other.</p>
<p>In this way, there is not needed an inert object to pass momentum as in the prior case.</p>
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https://chemistry.stackexchange.com/questions/176232/how-does-a-reduction-in-potential-energy-in-a-chemical-bond-release-energy
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Question: <p>My question related with another one <a href="https://physics.stackexchange.com/questions/387723/shape-of-hydrogen-atom-in-excited-state-with-nonzeros-angular-momentum">here</a>.</p>
<p>For explanation valence bond direction chemists are used real wave functions, for example $p_x, p_y, p_z, \ldots$. But this functions do not describe the preffred direction of angular momentum, they are not the eigenfunctions of $\hat L_z$.
The functions, that are describe the states with some preffered directions are not real, they are complex. For hydrogen, for example, they looks like: <a href="https://i.sstatic.net/HcRKE.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/HcRKE.png" alt="https://i.sstatic.net/S2BJJ.png"></a></p>
<p>If some other atom nearby first one, then there is some prefered direction, and good funtcions for this situation should be not real ones, but complex ones. Then why chemists used real functions for bond explanation?</p>
Answer: <p>Chemists use the linear combinations of hydrogen atom wavefunctions which are real because this is simply much more convenient to think about. Notice that for hydrogen, all the orbitals for a given $n$ are degenerate, so taking these linear combinations does not affect the energy, and thus thinking about the shapes of these orbitals is potentially more useful than thinking about the shape of something which is complex.</p>
<p>You are correct that taking the real-valued wavefunctions causes them to no longer be eigenfunctions of $L_z$. This is a bit frustrating, but in practice we never use these orbitals for anything, so it does not really matter. I used to think organic chemists took the shape of orbitals a bit too seriously (and they maybe do), but the notion of the shape of an orbital is actually recovered in a more rigorous manner.</p>
<p>That is, molecular orbital theory is just a qualitative form of Hartree-Fock (HF) theory. In HF, the Fock matrix is unitarily invariant, so we are free to transform our orbitals by any unitary rotation. These orbitals can be written as functions in $\mathbb{R}^3$ as desired. Thus, there exists schemes by which molecules orbitals can be "localized" and they often take the shape of orbitals which look a whole lot like the real-valued hydrogen atom orbitals you mention.</p>
<p>Thus, the use of these real-valued orbitals as a notion of shape is justified after the fact from HF theory and beyond.</p>
<p>In practice, this never really matters if one is being rigorous because any attempt at saying exactly which orbitals are involved in a bond is thwarted by unitarily transforming those orbitals. Thus, many chemists use the shape of an orbital as a heuristic because thus far, it has proven to be a very useful conceptual tool. I think pretty much all chemists know this is not completely correct, but neither is anything in physics.</p>
<p>Basically, the shape of orbitals provide a useful model.</p>
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https://chemistry.stackexchange.com/questions/91050/using-real-wave-functions-for-chemical-bond-explanation
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Question: <p>I understand the geometry and calculation of a dihedral angle, discussed in <a href="https://math.stackexchange.com/q/47059/221321">this question</a>. It is not clear how the direction is defined for a chemical bond. On the <a href="https://en.wikipedia.org/wiki/Dihedral_angle" rel="nofollow noreferrer">wiki page for dihedral angles</a>, it states that, "the dihedral angle $\phi$ is the counterclockwise angle..." between the planes $P_{123}$ and $P_{234}$ (where $P_{123}$ is defined as the plane containing atoms 1, 2, and 3 in the diagram below, and likewise for $P_{234}$).</p>
<p>$\hspace{45 mm}$<img src="https://i.sstatic.net/DiwKp.png" alt="bonds"></p>
<p>This counterclockwise definition means that the direction of the dihedral angle is defined using the left-hand rule about the line connecting atom 2 to atom 3. The left-hand rule is unusual in mathematics and geometry. Alternatively, this example may be defining the directionality of $\phi$ simply so the result is a positive angle less than $180^\circ$. </p>
<p>I have not been able to find any chemistry standards on this. Is there a common convention?</p>
Answer: <p>The <a href="http://goldbook.iupac.org/T06406.html" rel="nofollow">IUPAC Gold Book</a> states that, </p>
<blockquote>
<p>In a <a href="http://goldbook.iupac.org/N04134.html" rel="nofollow">Newman projection</a> the torsion angle is the angle (having an absolute value between 0° and 180°) between bonds to two specified (<a href="http://goldbook.iupac.org/F02356.html" rel="nofollow">fiducial</a>) groups, one from the atom nearer (proximal) to the observer and the other from the further (distal) atom. The torsion angle between groups A and D is then considered to be positive if the bond A-B is rotated in a clockwise direction through less than 180° in order that it may eclipse the bond C-D: a negative torsion angle requires rotation in the opposite sense.</p>
</blockquote>
<p>This implies a range of -180 to 180, which is by far the most typical range used, but I certainly have seen reports ranging from 0..360, e.g., 225 degree dihedrals. </p>
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https://chemistry.stackexchange.com/questions/40181/when-calculating-a-dihedral-angle-for-a-chemical-bond-how-is-the-direction-defi
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Question: <p>With respect to chemical bonding does the $\ce{I2-}$ ion exist?</p>
Answer: <p>I found a <a href="http://pubs.acs.org/doi/pdf/10.1021/jp0671087" rel="nofollow noreferrer">paper</a><span class="math-container">$\ce{^{[1]}}$</span> regarding the formation of diiodide anion species(<span class="math-container">$\ce{I2^-}$</span>). It is assumed to be an unstable species which forms during the formation of triiodide anion (excess of iodide in solution which makes it brown). This is a part of the paper which describes the formation of the species. For more information, read the full paper. Also, you can check the abstract which @andselisk linked.</p>
<blockquote>
<p>The <span class="math-container">$\ce{I2^-}$</span> species is formed by the 248-nm laser photolysis of
iodide through the following reactions:</p>
<p><span class="math-container">$$\ce{I- + hν -> I + e_s-}$$</span></p>
<p><span class="math-container">$$\ce{I + I- ->[k2]I2^-}$$</span></p>
<p><span class="math-container">$$\ce{I2- + I2- ->[k3]I3- + I-}$$</span></p>
<p>As discussed later, we found that photodetachment of electrons from
<span class="math-container">$\ce{I-}$</span> ions produces solvated electrons (<span class="math-container">$\ce{e_s-}$</span>) in ionic
liquids as well as in molecular solvents. After photodetachment,
iodine atoms react with <span class="math-container">$\ce{I-}$</span> to form diiodide anion radicals. In
aqueous solution, the transient absorption maxima of <span class="math-container">$\ce{I2-}$</span> are
located around 400 and 720 nm. The extinction coefficients of
<span class="math-container">$\ce{I2-}$</span> in aqueous solution at 385 and 725 nm are 10000 and 2560
M-1cm-1, respectively.[...]</p>
</blockquote>
<p><span class="math-container">$\ce{^{[1]}}$</span>: J. Phys. Chem. B, <strong>2007</strong>, 111 (18), pp 4807–4811</p>
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https://chemistry.stackexchange.com/questions/87971/does-the-diiodide1%e2%88%92-anion-exist
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Question: <p>My textbook says:</p>
<blockquote>
<p>It is appropriate to remark that the trick of rationalizing apparent violations of the octet rule by invoking the participation of d-orbitals in the bonding scheme is a matter of some controversy. Some scientists feel that the d-orbitals lie so high in energy that they should not be treated as valence orbitals, but as excited orbitals which cannot confer any appreciable stability to chemical bonds. Others feel that while d-orbitals are of high energy in free atoms, their energy decreases as other atoms approach to make bonds.</p>
<p>This author has not been convinced of the absolute validity of either side of this argument, and interprets the controversy merely as more evidence that the description of chemical bonds in terms of a few atomic orbitals is a highly approximate procedure. The most convincing reason for invoking d-orbitals in chemical bonding is that it is a simple idea which works, if it is used carefully and not over-interpreted.</p>
</blockquote>
<p>Since this book is quite an old edition, I wanted to know whether there are any newer developments in this direction. Which of the two views is generally acceptable?</p>
Answer: <p>Yes it works, according to result they have with organometallic chemistry. You may not know $e_g$ and $t_{2g}$ orbitals. Have a look at this diagram of orbitals : </p>
<p>Its diagram for an octaedric complex, where M is the metal. Positions of $e_g$ and $t_{2g}$ depends of what L is. </p>
<p><img src="https://i.sstatic.net/ZblCc.png" alt="enter image description here"></p>
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https://chemistry.stackexchange.com/questions/33869/can-d-orbitals-participate-in-bonding
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Question: <p>Do the antimatter molecules have the same chemical properties with matter molecules?Should we define electropositivity for antimatter molecules since the chemical bonding would be between positrons?</p>
Answer: <p>The following is an approximation based on the most accepted theories about cosmology (that I know of).</p>
<p>We currently consider that our observable universe is composed only of normal matter, by opposition with antimatter. Also, matter and antimatter should have the exact same physical and chemical properties.</p>
<p>However, there is a catch! According to the current theories, the Big Bang should have produced exactly the same amount of matter and antimatter, and both should have annihilated together by now. The fact that our universe is composed only(?) of matter implies that there is something fundamentally different between normal matter and antimatter which left an excess of normal matter.</p>
<p>Because we think that there must be a fundamental difference between the 2, we try to isolate antimatter and mesure some of its physicochemical properties to be compared with normal matter so we can determine where the current theory is wrong.</p>
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https://chemistry.stackexchange.com/questions/116704/antimatter-molecules-properties
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Question: <p>This is more of a terminology question. How are phase changes considered a physical property, not a chemical property when hydrogen bonds break. A chemical property can only be observed during a chemical change. Aren't hydrogen bonds a type of chemical bond, and those bonds are being broken during a phase change from ice to liquid water, for example. I understand that no intramolecular bonds are being broken. Water molecules remain intact. But we teach that a chemical bond has to break during a chemical change, but phase changes are considered to be physical.</p>
Answer: <p>You must realize that the line between physical and chemical properties is very thin. Most bonds work via the same principle, which has mostly to do with electronegativity. When looked at in reactions it has probably also to do with the according entropy and enthalpy change. Hydrogenbonds are basically bonds based on electronegative attraction between two atoms (which is a simplified few of the reality). A phase change basically when a certain energy point has been reached: since heating op a substance is equal to giving it more kinetic energy (since heat is the same as the movement of individual atoms or molecules). When you reach a certain heat there is too much kinetic energy to form hydrogenbonds the same way it as it did before it changed phase. You probably can agree that molecules or atoms moving quickly is an obvious physical property. When you look at great detail to how hydrogenbonds work, it has something in common with many other way of two objects attracting each other (for example: the formula for the gravitional force between to large heavenly bodies, is very similar to the formula for the electrical force between an electron and a nucleus). </p>
<p>In simple: Many of these chemical properties seem very physical when you look at them in detail. So saying that a certain property should actually be chemical because the changing of this property seems to correspond with the changing of chemical bonds, is not a realistic claim looking at how "physical"/physic based these "chemical" bonds are.</p>
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https://chemistry.stackexchange.com/questions/85906/if-hydrogen-bonds-are-chemical-bonds-and-these-bonds-are-broken-when-water-chan
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Question: <p>Hi so I was studying chemical bonding where i encountered a problem which is stated below.</p>
<p>When we talk about <span class="math-container">${NO_3}^-$</span> we draw its structure as following
<img src="https://i.sstatic.net/Q8cda.jpg" alt="enter link description here">
But the thing which I dont understand is that in two of the O atoms there are three lone pairs(ie: <span class="math-container">$6 e^-$</span>) but as we know from O's <span class="math-container">$e^-$</span> config. (which is <span class="math-container">$1s^2 2s^2 2p^4$</span>) there are two unpaired <span class="math-container">$e^-$</span> which will take part in bonding as there is no vacant orbitals for <span class="math-container">$e^- $</span> to get excited to . Now if in this O atom only one sigma bond is present then technically we are left with <span class="math-container">$5e^-$</span> and not <span class="math-container">$6e^-$</span> which is shown in the figure.</p>
<p>Similarly in the N atom (whose <span class="math-container">$e^-$</span> is <span class="math-container">$1s^2 2s^2 2p^3$</span> ) does not have any other vacant orbital to excite its <span class="math-container">$e^-$</span> thus showing that it can make only 3 bonds then why is it making 4 bonds( 1 pi and 3 sigma) ?</p>
Answer: <p>The problem lies in the electron count. Each bond contains 2 electrons and the pair is counted twice, once for being around each atom in the bond. And don't forget the extra electron to make the ion.</p>
<p>The + and - are from the point of view of the atom nucleus: e.g., consider the nitrogen to share 2 electrons with the oxygen on the right, to give 2 to the oxygen above, and to give 1 to the oxygen on the left. The other electron in that bond is one that makes the anion negative overall.</p>
<p>Then nitrogen gains its octet of 8 electrons, but formally only shares 50% of those electrons, so it started out with 5 but only controls 4 - it now has a positive charge. The right oxygen shares, so no charge. The top oxygen got 2 from nitrogen, so is negative, and the left oxygen got 1 from nitrogen and one from the metal of which this is the anionic part. And then, draw up the other two diagrams that have the "neutral oxygen" in the other positions, add them all and average them. Then each oxygen winds up with a 2/3 negative charge (but the nitrogen is still +1), while the anion is -1 and planar.</p>
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https://chemistry.stackexchange.com/questions/118527/bonding-in-the-nitrate-anion
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Question: <p>I don't know much about quantum mechanics, but I do know that it's quite far removed from the easily observable and that anything "quantum" involves things that are discreet. So, how close is the analogy between molecular bending and vibration with that of springs? Is there maybe a better way to think, maybe a better analogy, to describe molecular bonds?
<a href="https://i.sstatic.net/NlK6k.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/NlK6k.jpg" alt="enter image description here"></a></p>
Answer: <p>That's a good question, but remember, an analogy is just that: comparing <em>dissimilar</em> things on the basis of structure or function. Though there may not be a <em>physical</em> similarity, the analogy is effective in describing what happens, and can even be used to calculate outcomes, within limits.</p>
<p>In a similar way, the <a href="https://www.youtube.com/watch?v=N8L4sjFkCcg" rel="nofollow">concept of a <em>mass</em> and a <em>spring</em> are used to describe an electrical R-L-C tuned circuit</a>. Both have resonant frequencies and other properties that can be calculated from the physical properties of the components.</p>
<p>One of the difficulties in describing quantum phenomena is that <em>until the physical properties are measured</em>, they may be <em>indeterminate</em>. An electron is not in a specific place, but has a <em>probability of being detected</em> in a specific place. So though the analogy is not physically comparable, it helps understanding of the chemical bond, and can even produce useful results, such as IR spectra, and its use is valid.</p>
<p>This begs the question, though; perhaps there is another analogy that <em>you</em> might find more useful... Any suggestions from others are welcome!</p>
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https://chemistry.stackexchange.com/questions/41889/if-chemical-bonding-is-quantum-mechanic-is-saying-they-bend-and-vibrate-like-sp
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Question: <p>I am reading an introductory semiconductor physics textbook. The textbook states the following:</p>
<blockquote>
<p>Several types of atomic bonding have been identified, including ionic, covalent, van der Waals, hydrogen, and metallic. Whatever the name given to the type of bonding, in all cases it is the electrostatic force acting between charged particles that is responsible for all the forms of bonding.</p>
</blockquote>
<p>My concern is with the validity of the latter sentence; that is, that all cases of bonding are electrostatic forces acting between charged particles.</p>
<p><a href="https://en.wikipedia.org/wiki/Chemical_bond" rel="nofollow noreferrer">Wikipedia</a> says the following:</p>
<blockquote>
<p>A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds. The bond may result from the electrostatic force of attraction between oppositely charged ions as in ionic bonds or through the sharing of electrons as in covalent bonds.</p>
</blockquote>
<p>If electrostatic forces act between charged particles (ions), and atoms are, by definition, <em>neutral</em> particles, then, since atoms do bond, how can it make sense to say that all bonding is electrostatic forces acting between charged particles?</p>
<p>And my interpretation of the Wikipedia article is that the electrostatic force of attraction is responsible for bonding between oppositely charged ions, as in ionic bonding, whereas it is not responsible for bonding through the sharing of electrons, as in covalent bonding.</p>
<p>So is the textbook's claim that all types of bonding are electrostatic forces acting between charged particles false?</p>
<p>I would appreciate it if people could please take the time to clarify this.</p>
Answer: <p>You ask how bonding between two neutral atoms can be electrostatic. In the case of covalent and van der waals the electrostatic basis is not so obvious as it is for ionic compounds. A covalent bond forms when two neutral atoms establish a shared molecular orbital which is much larger than the separate atomic orbitals that the bonding electrons contributed by each atom formerly occupied. Electrostatic repulsion favors electron pairs being in larger orbitals, because this increases average distance between the electrons. In the case of Van der Waals attraction, the neutral atoms are electrostatically attracted because when their electrons are not symmetrically distributed, there is a partial positive charge at one end of a molecule and a partial negative at the other, even though overall charge is zero. Electrons in adjoining molecules will tend to synchronize their movements so that the positive end of one molecule will be near the negative end of the adjoining molecule; this is called induced charge. </p>
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https://chemistry.stackexchange.com/questions/111708/textbook-claim-in-all-cases-it-is-the-electrostatic-force-acting-between-c
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Question: <p>In general when we talk about chemical bonding, we say that it is "nature's way to stabilize the systems" and the energy of the molecule formed by atoms will be lower than that of the individual atoms; hence leading to stabilization. </p>
<p>However, while reading about the Molecular Orbital Theory in my NCERT Chemistry Textbook, I came across a statement - </p>
<blockquote>
<p>The total energy of the molecular orbitals formed remains same as that of the original atomic orbitals.</p>
</blockquote>
<p>How can we then justify through the Molecular Orbital Theory that the process of chemical bonding leads to stabilization, if no energy is released?</p>
Answer:
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https://chemistry.stackexchange.com/questions/125090/energy-levels-of-molecular-orbitals-formed-by-combination-of-atomic-orbitals
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Question: <p><a href="https://simple.wikipedia.org/wiki/Molecule" rel="nofollow noreferrer">Wikipedia</a> says so:</p>
<blockquote>
<p>Bonds can also be broken apart. Since most bonds require energy to form, they also give off energy when they are broken. But before most bonds break, the molecule has to be heated. Then the atoms start to move, and when they move too much, the bond breaks. Molecules that require less energy to break than they give off when broken are called fuels. For example, a candle will just sit there and nothing happens. But when you use a match to light it, it will burn for a long time. The match brings the energy to break the first bonds, which release enough energy to break the bonds below them, until the candle has burned down.
But what I have learned is bonds release energy when formed.</p>
</blockquote>
Answer: <p>"Molecules that require less energy to break than they give off when broken are called fuels." is the most imprecise sentence in the paragraph, because fuels don't give off energy when their bonds are broken, but when these broken bonds form new bonds (with oxygen) that are much more stable and release more energy than it took to break the original (carbon-hydrogen) bonds.</p>
<p>A candle is stable in air, even tho it does not have complete stability in an oxygen-containing atmosphere. But if you put sufficient energy into a few bonds, the excited atoms will combine with oxygen, which gives off enough energy to excite a few more bonds, which etc., etc. It's this new reaction, which is overlooked in the paragraph, which provides the increased temperature during the reaction.</p>
<p>The energy that "bonds require to form" is an activation energy which destabilizes a molecule enough so that it can reorganize with other reagents (like O2) to form more stable products. This activation energy may be supplied initially from outside sources (the match), but in a continuing reaction, the exotherm from the reaction provides the activation energy and then some. </p>
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https://chemistry.stackexchange.com/questions/118911/does-it-sometimes-take-energy-to-create-a-chemical-bond
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Question: <p>My teacher was teaching about chemical bonding and she said that pi bond needs sigma bond to exist, but she didn't know why. It is a doubt which I get in my mind, can someone explain it?</p>
Answer: <p>I think it would be more correct to say that pi bonds tend to exist in conjunction with sigma bonds, rather than that they <em>need</em> sigma bonds to exist - as pointed out in the comments. </p>
<p>This can be explained rather easily by basic molecular orbital (MO) theory, which seems like an appropriate framework for the question. Let's consider the simple case of a homoatomic bond such as C=C. When you have two C atoms coming into contact so as to form MOs, you form sigma MOs (bonding and antibonding) from the set of atomic S orbitals, and pi MOs (again, bonding and antibonding) from the Px and Py atomic orbitals. </p>
<p>Among the (molecular) orbitals, the sigma bonding MO <em>generally</em> has the lowest energy, followed by the pi bonding MOs. This means that the sigma bonding orbitals will 'fill' before the pi bonding orbitals. Because electrons preferentially occupy MOs from lower to higher energy, as long as the sigma bonding MOs are lower energy than the pi bonding MOs you will see sigma bonding orbitals populated before pi bonding MOs. Thus, to have a pi bond (occupied pi bonding MOs), you generally also have a sigma bond (occupied sigma bonding MO). </p>
<p>As pointed out in the comments to your question, this is somewhat of a generalization - but I think it's an accurate description of the phenomenon your instructor described. </p>
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https://chemistry.stackexchange.com/questions/58300/pi-bond-need-sigma-bond-to-exist
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Question: <p>My chemical bonding professor says that the $\ce{O-S-O}$ bond angles in $\ce{SO4^2-}$ are ideal (109.5°).</p>
<p>Why? Is this because all bonds are equivalent, and electron distribution is shared among all oxygens?</p>
Answer: <p>Yes. All the bonds are equivalent as in the 4 resonating structures, each with 1.5 S and O bonds. So the resonant hybrid has four 1.5 double bonds (sigma and pi) with -0.5 formal charge on each O. This is why, the ion has a perfect tetrahedral geometry and shape as it is very symmetric and equivalent.
Hope it helps.</p>
<p>The picture below on the right is what is thought to be the most accurate representation nowadays. However, the resonance structure on the left (and all of its equivalent representations) are more than likely acceptable for introductory chemistry classes. </p>
<p><strong>Note that no matter which resonance structure you think is a better representation of the sulfate ion, the bonds are all equivalent and therefore sulfate ion has the ideal tetrahedral bond angle of 109.5 degrees.</strong></p>
<p><a href="https://i.sstatic.net/kVbgm.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/kVbgm.png" alt="enter image description here"></a></p>
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https://chemistry.stackexchange.com/questions/37875/bond-angles-in-sulfate
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Question: <p>If you characterize the chemical bonds to two categories physical and chemical bonds, how do you do it? Aren't all bonds <strong><em>chemical and physical</em></strong>?</p>
<p>From the freedictionary.com, chemical bond:</p>
<blockquote>
<p>Any of several forces, especially the ionic bond, covalent bond, and metallic bond, by which atoms or ions are bound in a molecule or crystal.</p>
</blockquote>
<p>What is the physical bond then? How should the atom be bonded to molecule if these bonds are not the ones. What are these "several forces" excactly?</p>
<p>In the answer I would like to see the lists of chemical and physical bonds.</p>
Answer: <p>In short: the definition of a chemical bond is not unique and a clearly-drawn line. The simplest and most common definition is <strong>the sharing of electrons between two or more nuclei</strong>. In contrast, other interactions are often said to be <em>intermolecular</em> (which is somewhat more specific than the term “physical”.</p>
<hr>
<p>In a longer commentary, I see can have five different types of definition of the chemical bond (vs. intermolecular interactions).</p>
<ol>
<li><p>Let's start from the beginning, in this case using the words of Linus Pauling, winner of the 1954 Nobel Prize for “determining the nature of the chemical bond linking atoms into molecules”. In <a href="http://books.google.com/books?id=L-1K9HmKmUUC" rel="nofollow noreferrer"><em>The Nature of the Chemical Bond</em></a> (1960), he gives the following definition:</p>
<blockquote>
<p><strong>The Chemical Bond Defined</strong>.—We shall say that <em>there is a chemical bond between two atoms or groups of atoms in case that the forces acting between them are such as to lead to the formation of an aggregate with sufficient stability to make it convenient for the chemist to consider it as an independent molecular species.</em></p>
</blockquote>
<p><strong>A bond is what links atoms into molecules</strong>, and molecules are defined at the discretion of the chemist. You can find the same definition still in use in some high-school textbooks, but it isn't very helpful…</p></li>
<li><p>The complete opposite: <strong>consider all interactions as chemical bonds</strong>, whose strenght can vary. I actually hadn't heard that one before I researched textbooks to write this answer, but you can find it in some textbooks, like <a href="http://books.google.com/books?id=bLgkS8pE1N0C&pg=PT175" rel="nofollow noreferrer">this one</a>:</p>
<blockquote>
<p>A chemical bond is the physical phenomenon of chemical species being held together by attraction of atoms to each other through sharing, as well as exchanging, of electrons and is a phenomenon that is fully described by the laws of quantum electrodynamics. In general, strong chemical bonds are found in molecules, crystals or in solid metal and they organize the atoms in ordered structures. Weak chemical bonds are classically explained to be effects of polarity between molecules which contain strong polar bonds. Some very weak bond-like interactions also result from induced polarity London forces between the electron clouds of atoms, or molecules. Such forces allow the liquification and solidification of inert gases. At the very lowest strengths of such interactions, there is no good operational definition of what constitutes a proper definitional "bond".</p>
</blockquote>
<p>This view has some grounding, because all interatomic interactions stem from the behaviour of the system's electrons (in addition to nuclei–nuclei Coulombic forces). However, it does not allow to make a strong distinction between interactions whose energies differ by orders of magnitude. Chemists like molecules, and they like categorizing things between intra- and inter-molecular, as it's a nice model (making it easier for our mind to handle).</p></li>
<li><p>You can classify interactions by energy: decide that <strong>chemical bonds are those interactions that have an energy higher than a certain threshold</strong>, let's say 50 kJ/mol. This makes things clean, and makes sure that you can easily classify interactions. However, the choice of a threshold is problematic.</p></li>
<li><p>Finally, what I believe is the most common description is to look at <strong>the nature of the interaction</strong>, and classify it following a certain <strong>convention</strong>. The two other answers so far have focused on this part and listed the various “usual” types of bonds and intermolecular interactions, so I won't say more on that.</p></li>
<li><p>I said five types, right? Well, the fifth is mine, of course. Not only mine, but that of the <em>New Oxford American Dictionary</em> as well, which I quite like:</p>
<blockquote>
<p><strong>chemical bond</strong><br> a strong force of attraction holding atoms together in a molecule or crystal, resulting from the sharing or transfer of electrons.</p>
</blockquote>
<p>Short and powerful. What I like in that is that it gives a general prescription, allowing one to argue individual cases and not based too much on convention. What are the features of a chemical bond? Well, it has to be an attractive force between atoms, sure… but I think the most relevant criterion of all is <strong>sharing (or transfer) of electrons</strong>. That is, after all, what chemistry is about: description of electronic clouds around two or more atoms. And I think when this criterion is applied to the list of interaction types commonly classified, it works quite well (whithout being dogmatic).</p>
<p>Also, what I like in it is that a given interaction type can be considered one way or another depending on its strength. The best example for that may be the hydrogen bond, which is the archetype of the almost-chemical-bond…</p></li>
</ol>
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https://chemistry.stackexchange.com/questions/230/what-is-the-difference-between-physical-and-chemical-bonds
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Question: <p>So, if I have multiple FTIR spectra each taken on different location of the sample and averaged from ~20 spectra, how can I check if the sample is homogeneous, i.e., if there are statistically significant differences between the spectra based on location? For visuals, I tried to compute at each wavenumber confidence upper and lower limits, but I am not sure if this method is valid even for visualisation and it does not give anything exact or quantitative.</p>
Answer:
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https://chemistry.stackexchange.com/questions/139661/statistical-methods-used-for-testing-the-homogeneity-of-the-sample
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Question: <p>Surface-enhanced Raman spectroscopy or surface-enhanced Raman scattering (SERS) is a surface-sensitive technique that enhances Raman scattering by molecules adsorbed on rough metal surfaces or by nanostructures such as plasmonic-magnetic silica nanotubes.</p>
<p>Does this require a Raman microscope? Remember you are not scanning on a large object or liquid in vials but molecules in the rough metal surfaces. If you just use a normal Raman spectrometer. Will it work on the SERS sample?</p>
Answer:
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https://chemistry.stackexchange.com/questions/142749/can-you-do-sers-without-a-raman-microscope
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Question: <p>I am computing Franck-Condon Factors between the neutral ground state and cationic states to simulate the photoelectron spectra (using Gaussian 16, freq=FCHT). The target cationic state is a pi state and has 3/2 and 1/2 spin-orbit components. I have done optimization and frequency calculation for the pi state in TDDFT. Thus I have FCF for the pi state before spin-orbit splitting. How do I compute FCF two spin-orbit components ? In another word, how can one optimize the two different components of a spin-orbit state. I used the same FCFs for both the components separated by the SOC. It provided reasonable agreement with experiment, however, I wonder if there is more accurate way of doing it.</p>
<p>Thank you for your help.</p>
Answer:
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https://chemistry.stackexchange.com/questions/145328/how-to-compute-franck-condon-factors-for-spin-orbit-states
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Question: <p>I was thinking back to my chemistry major days and remembering the bunch of cool spectra we produced by various means (HNMR, IR, UV/Vis) and I remember all the cool 'self-interference' like couplets, triplets doublets, etc. </p>
<p>My question is basically motivated by the complexity of spectra, and the reductionist instinct to 'break it down' into simpler components. The FFT or DFT is known to separate out some periodic components of signals, as are SVD or tensor decompositions known to isolate independent components. </p>
<p>I am completely not an expert in any of this but I am interested to hear from any coal-face chemists if they regularly use some 'spectra decomposition' method to aid in interpretation or analysis. I remember being taught how to read various peaks in spectra and analyze them, but I can't help thinking, for some very complex molecules, surely it gets pretty hard. And what if, we don't actually know the structure beforehand? </p>
<p>So given that Fourier transform is used in NMR, my question is : is there ever a need to take a spectrum and do a computer analysis of it that isn't just about matching peaks to a database of known signals? What other nifty things can be done with a spectrum once it is obtained?</p>
Answer: <p>As Mad Scientist said, many spectroscopy methods are already based on Fourier transform of a time-dependent observable. (This is not true of the basic <a href="http://en.wikipedia.org/wiki/Spectrophotometry" rel="noreferrer">spectrophotometry</a> measurements, where you directly measure the transmittance as a function of frequency.) NMR is a method based on Fourier transform, as is <a href="http://en.wikipedia.org/wiki/Fourier_transform_infrared_spectroscopy" rel="noreferrer">FTIR</a>.</p>
<p>Regarding NMR signal processing, it's a very wide field, which includes interesting research ranging from academic interest to software development. So, it's hard to answer your question other than saying “yes, of course, further processing of NMR signals is sometime possible”. One thing you have to realize is: while you have probably learnt <a href="http://en.wikipedia.org/wiki/Proton_NMR" rel="noreferrer">proton NMR</a> during your studies, this is only one NMR technique, and the simplest of them all. <strong>There are plenty of different NMR techniques</strong>, depending on the type of excitation and sequence of RF pulses, many of them being multidimensional. Each method can use a variety of signal processing techniques.</p>
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https://chemistry.stackexchange.com/questions/3957/fourier-transform-for-spectroscopy-spectra
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Question: <p>I am in mathematics, but I am writing a paper on molecular vibrations (because you can use representation theory to handle this).</p>
<p>I want to prove the selection-rule, but I also have to give a presentation of my intermediate results. Can someone give a little (maybe trivial) explanation (in one or maximal two small sentences) of what the selection-rule does? (for example for the IR-spectrum). This is for my presentation if someone doesn't know something about the mathematical and chemical background.</p>
Answer: <p>The operator for interaction with infrared light is the permanent dipole of the molecule, $\mu$. I don't know anything about representation theory, but in group theory, you decompose the molecule by its symmetry elements to find the eigenmodes. The infrared active modes are those modes in which the permanent dipole changes over the course of a vibration. An excellent reference is Symmetry and Spectroscopy by Harris and Bertolucci which should be available in your library.</p>
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https://chemistry.stackexchange.com/questions/4488/explanation-of-the-selection-rule-in-light-of-ir-spectroscopy
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Question: <p>I was trying to figure out the different peaks in a XPS spectrum for Molybdenum, but I could not understand them. About XPS, I know that this technique is based on the binding energy of the electrons of different elements</p>
Answer: <p>you are quite right, XPS does measure the 'binding energy' of electrons. More specifically, XPS bombards a surface with X-rays and when those X-rays have sufficient energy they are able to sufficiently excite electrons to be emitted from the sample. Each peak in an XPS spectrum corresponds to the energy necessary to emit a given electron.</p>
<p>So what would you expect to see for Mo?</p>
<ul>
<li>Outer lying electrons (for Mo, we are talking about electrons in the
4p and 4s orbitals) will emit at a lower energy. You should observe
low-energy peaks that correspond to these emissions (approx. 30 - 60
eV). </li>
<li>you should see a cluster of peaks at approx. 230 eV that correspond to the d-orbital electrons.</li>
<li>Finally, you might see discrete peaks at very high energy, approx. 2600 - 2800 eV that correspond to emission from the 2s and 2p orbitals</li>
</ul>
<p>In general, outer orbitals emit at lower energy, and inner orbitals emit at higher energy. </p>
<p>Hope this helps!</p>
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https://chemistry.stackexchange.com/questions/4772/what-does-the-signals-in-a-xps-spectrum-mean
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Question: <blockquote>
<p>Using a state-of-the-art atomic force microscope, the scientists have taken the first atom-by-atom pictures, including images of the chemical bonds between atoms, clearly depicting how a molecule's structure changed during a reaction. Until now, scientists have only been able to infer this type of information from spectroscopic analysis. — <a href="http://www.sciencedaily.com/releases/2013/05/130530142007.htm" rel="nofollow">Source</a></p>
</blockquote>
<p>Are there any papers how this thing actually works?</p>
Answer: <p>There are a ton of papers on how AFM works, but I'll have a go at explaining it anyhow.</p>
<p>Atomic Force Microscopy (AFM) is a venerable technology for imaging nanostructures down to individual molecules by physically contacting atoms on surfaces. The whole process is very similar to a finger reading Braille.</p>
<p>What does it mean to physically contact an atom? It means that the AFM works by detecting some kind of repulsion between an atom and a very, very sharp probe due to their respective electron clouds repelling. The forces between surfaces on the nanoscale are a complicated mixture of attraction due to Casimir-Polder/Van der Waals interactions, dipole-dipole interactions, electrostatic repulsion, Pauli exclusion, <em>etc. etc.</em>, but the key point is that if two objects get too close together they will experience a rapidly increasing repulsive interaction, which the AFM picks up.</p>
<p>This repulsion is detected through the deflection of a cantilever (a thin, long, elastic beam of material). The probe is mounted on the end of this cantilever, and the cantilever bends when the probe is pushed up against a substrate. The deflection of this cantilever is measured by bouncing a laser beam off the end, the change in position of the reflected beam indicating the force.</p>
<p>The AFM probe has an atomically sharp tip that can localise single atoms on a surface, and the tip is moved around using exquisitely precise actuators so that the tip can be scanned back and forth over a surface, picking up the bumps and valleys of the sample that indicate the positions of atoms. I've seen ads for AFM probes that actually have flared tips - these can obtain even more 3D information about nanostructures as they can hook into nanostructures from the side and measure the texture and relief parallel to the surface.</p>
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https://chemistry.stackexchange.com/questions/5170/how-did-scientists-capture-the-first-images-of-molecules-before-and-after-reacti
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Question: <p>I know that elements (ex. Ag) have spectral/emission/absorption lines, but do chemical compounds (ex. NaCl) have spectral lines as well? Ones that when seen you can tell that that compound (NaCl) is made up of certain elements (Na and Cl).</p>
Answer: <p>Yes, but for "spectral lines" the gaseous state would generally be the implication. In solids and liquids photon interactions are generally over "bands" of energy not a very discrete wavelength as in a "spectral line."</p>
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https://chemistry.stackexchange.com/questions/5235/do-chemical-compounds-have-spectral-lines-or-only-elements
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Question: <p>Are there any two distinct types of atom or molecule that have identical emission or absorption spectra?</p>
Answer: <p>In addition to the R/S chiral molecules mentioned by @user26143. There are also molecules with a distinct atomic composition that have experimentally indistinguishable spectra, even though theoretically there will be minor differences. One example, according to <a href="http://link.aps.org/doi/10.1103/PhysRevLett.102.253001" rel="nofollow">Rabitz et al. 2009 PRL</a> (freely available for viewing) are <a href="http://en.wikipedia.org/wiki/Flavin_mononucleotide" rel="nofollow">flavin mononucleotide</a> and <a href="http://en.wikipedia.org/wiki/Riboflavin" rel="nofollow">riboflavine</a> and in fact many other flavins. Due to there very similar structure you need unconventional techniques to be able to distinguish them.</p>
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https://chemistry.stackexchange.com/questions/6994/are-there-any-molecules-with-the-same-spectrum
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Question: <p>I want to do some spectrometry experiments with glow in the dark paints. The red paints I've read product details for all seem to require a uv light to charge. Are there any red glow in the dark paints that can be charged with visible light?</p>
<p>I was also wondering why most red phosphorescent pigments don't have the ability to glow as long as green pigments.</p>
Answer: <p>The wikipedia page on <a href="http://en.m.wikipedia.org/wiki/Phosphorescence" rel="nofollow">Phosphoresence</a> has a red pigment. Calcium Sulfide. I can imagine that the lower energy (higher wavelength, lower energy) is more stable because the triplet is not as high in state energy as lower wavelength emitters. Because of the stability the time to emit is longer and the intensity lower.</p>
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https://chemistry.stackexchange.com/questions/7096/red-glow-in-the-dark-pigments
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Question: <p>Question is rather self-explanatory. Came up during a lecture without a concrete answer. </p>
<p>I understand that the differences in emission wavelengths is due to relaxation to the lowest energy level of S1, but <strong>why do fluorescent molecules necessarily overlap in their excitation and emission spectra?</strong></p>
Answer: <p><a href="https://i.sstatic.net/mnC9H.png" rel="nofollow noreferrer">Energy-Level Diagram</a>
<br>There is overlap because between the excitation and the emission spectra because there exists a Range of orbitals at different energies. You'll always notice that the energy for excitation is always higher, or the lambda value is always lower (these say the same thing since E=planck's constant * frequency).
<br>
<br>I've attached a HE-Ne laser schematic from <a href="http://books.google.com/books/about/Laser_electronics.html?id=WvFRAAAAMAAJ" rel="nofollow noreferrer">Verdeyen's Laser Electronics book</a>. For a better visual. What you need to know is let's say your gain medium, in this case He-Ne was excited at it's lowest wavelength (we'll call it Lambda1) then it's emission wavelength may be only 2nm higher (or red shifted), Lambda2.
<br>
<br>However, as you'll notice in this schematic there are many wavelengths at which this gain medium (He-Ne) can emit light. These other possible excitation orbitals can have lower energy levels that correspond to higher wavelengths and thus overlap on the spectrum. </p>
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https://chemistry.stackexchange.com/questions/7235/why-does-the-excitation-and-emission-spectrum-of-a-fluorescent-molecule-have-ove
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Question: <p>Is visible spectroscopy the only non-electronic method of all the spectroscopy method?</p>
<p>I know some of the spectroscopy method is possible being used because of electronics.</p>
Answer: <p>No, Ultraviolet spectroscopy and Infra-red spectroscopy were practiced without any use of electricity. </p>
<p>Infra-red and ultraviolet light were both known by 1801.</p>
<p>Ultraviolet absorption lines had been observed by 1843. </p>
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https://chemistry.stackexchange.com/questions/10928/is-visible-spectroscopy-the-only-non-electronic-method-of-all-the-spectroscopy-m
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