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Suppose an initial state $A$ can transition to either state $C$ or state $D$. Suppose further that both of these two processes are rate-limited by a transition to the same intermediate state $B$, as follows:

enter image description here

Note that $\Delta G_{AB}> \Delta G_{BC}>\Delta G_{BD}$.

According to transition state theory, rates are determined by the rate-limiting step. So the rate of each process will be the same,

\begin{align} r_{A\to C}&=r_{A\to B} \\ r_{A\to D}&=r_{A\to B} \end{align}

from which it follows that the number of transitions made during some large $\Delta t$ will also be the same,

\begin{align} N_{A\to C}&=r_{A\to B}\Delta t \\ N_{A\to D}&=r_{A\to B}\Delta t \end{align}

However, if we were to consider the two pathways collectively then we would predict more transitions to be made to state $D$ than $B$ since $\Delta G_{BD}<\Delta G_{BC}$.

How do we resolve this apparent discrepancy?

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In transition state theory (TST, goldbook) one of the necessary assumptions is that reactants and products are in equilibrium. In principle this gives us \begin{align}\ce{ A &<=> [AB]^{$\ddagger$} -> B\\ B &<=> [AB]^{$\ddagger$} -> A\\ A &<=> B, }\end{align} and in addition to this, also \begin{align}\ce{ B &<=> [BC]^{$\ddagger$} -> C & B &<=> [BD]^{$\ddagger$} -> D\\ C &<=> [BC]^{$\ddagger$} -> B & D &<=> [BD]^{$\ddagger$} -> B\\ B &<=> C & B &<=> D, }\end{align} as well as \begin{align}\ce{ C <=> [BC]^{$\ddagger$} -> & B <=> [BD]^{$\ddagger$} -> D\\ D <=> [BD]^{$\ddagger$} -> & B <=> [BC]^{$\ddagger$} -> C\\ C <=> & B <=> D, }\end{align} and $$\ce{ C <=> A <=> D\\ [AB]^{$\ddagger$} <=> [BC]^{$\ddagger$} <=> [BD]^{$\ddagger$} }.$$

You can use TST to estimate rate constants. $$k = \frac{\mathcal{k}_\mathrm{B} T}{\mathrm{h}} \exp\left\{ −\frac{\Delta^\ddagger{}G^\circ}{\mathcal{R} T}\right\}$$

Since all components have to be in equilibrium, you can predict a ratio between them via Boltzmann statistics, hence: $$\frac{N([BC]^{\ddagger})}{N([BD]^{\ddagger})} = \exp\left\{ −\frac{\Delta^\ddagger{}G^\circ([BC]^{\ddagger})-\Delta^\ddagger{}G^\circ([BD]^{\ddagger})}{\mathcal{R} T}\right\}=\frac{N(C)}{N(D)} $$

The preceding equilibrium does not need to be considered to determine the ratio of the products. However, it is important to note, that there are many assumptions necessary for TST to work. For example, at higher temperatures, anharmonic corrections have to be considered. More crucial is the assumption, that the transition can be described as a translatory movement, i.e. it is treated with classical mechanics.

For a system with a preceding equilibrium, it is well possible, that the TST approximation breaks down completely.

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  • $\begingroup$ Suggested edit:"one of the necessary assumptions is that reactants and products are in equilibrium" products $\to$ activation complex $\endgroup$ – Greg Aug 5 '14 at 2:36
  • $\begingroup$ The use of the phrase "activation complex" or "activated complex" is not longer recommended. It should be used transition state as it marks the translatory movement which is connected to the transition from one state into another. Saying that the reactants and the TS are in equilibrium is only half the truth, since the reverse connection is as likely in TST. It is another requirement, that all products that arrive the TS from the reactants side will react to the product and vice versa. Hence the reactants are in equilibrium with the products, making the TS a reaction coordinate only. $\endgroup$ – Martin - マーチン Aug 5 '14 at 3:28
  • $\begingroup$ Too bad then the the said Golden Book by IUPAC (that you linked) using always this term... It is even a separate entry: goldbook.iupac.org/A00092.html $\endgroup$ – Greg Aug 5 '14 at 3:36
  • $\begingroup$ If you do this calculation, and telling "everything is in equilibrium" especially the starting material and product then you fail to speak about the timeline when there is a net conversion from A to C and D (i.e. what kinetics is about). In equilibrium $$ \frac {d[A]}{dt} = 0$$ as equilibrium is a static state per definition. Any $$ \frac {d[A]}{dt} \neq 0$$ is nonequilibrium situation. $\endgroup$ – Greg Aug 5 '14 at 3:42
  • $\begingroup$ @Greg 1) You are absolutely correct about the phrasing. I was confusing it with "activated-complex theory", which is deprecated. I still do not like this phrase, as it suggests, that this state can be observed. There is also a much better definition: goldbook.iupac.org/T06468.html 2) I understand what you are saying. And it seems, that you have found one of the many problems that there are in TST. And there are plenty, and the one you pointed out is one of the lesser flaws. There are many examples, where TST fails, you can find some at the wikipedia page. $\endgroup$ – Martin - マーチン Aug 5 '14 at 4:05
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On remark: just because activation energy high/low it doesn't necessarily mean that reaction is slow or fast. You cannot tell this without the pre-exponential factor.

The answer: Off course, the original statement should not applied as you do it. "According to transition state theory, rates are determined by the rate-limiting step." doesn't mean automatically that all the reaction rates are equal.

A to B conversion is slow compared to other processes, is true, but still B to C and B to D are competing processes. The original statement only means that B will not build up in bigger quantities, so during the reaction all the final products (C and D together) are approximately equal with the used A after you consider stoichimetries.

Edited for clarification (sorry, pretty new to LaTex):

You should consider all the transformations as independent equations. Black magic is not involved, "rate determining step" and such only helps making approximations so one can solve the coupled differential equations on napkin.

Example 1.

Let us assume a reaction system with $A \to B \to C$. The kinetics of the system can be describe by two independent equations

$A \to B$ with a corresponding $k1$ and $B \to C$ with a corresponding $k2$

To describe conversions we can get different equations:

$$ \frac{d[A]}{dt} = -k1 [A] $$

$$ \frac{d[B]}{dt} = k1 [A] - k2 [B]$$

$$ \frac{d[C]}{dt} = k2 [B] $$

We see that it is already pretty messy, so assumptions like $k1 << k2$ can help. The general way is that we say, this difference in the rate constants allows us to asume $$ \frac{d[B]}{dt} \approx 0 $$. So we have pipeline, and in first approximation have that much $B \to C$ conversion going as $A \to B$. I.e. the is no "clogging" in the middle. Not much magic here.

Example 2.

Now what if we have e.g. two slow reactions as second step instead of one? Noting special: $A\to B$ , with $k1$, $B\to C$ , with $k2$, $B\to D$ , with $k3$ which translates to

$$ \frac{d[A]}{dt} = -k1 [A] $$

$$ \frac{d[B]}{dt} = k1 [A] -( k2 [B] + k3 [B])$$

$$ \frac{d[C]}{dt} = k2 [B] $$

$$ \frac{d[D]}{dt} = k3 [B] $$

Now we have the same assumption that the first step is much slower than the others than

$$ \frac{d[B]}{dt} = k1 [A] -( k2 [B] + k3 [B]) \approx 0$$

I think you can try it from here and see how the rates comes out.

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Consider the following hourglass-analogy. The radius of mouth of each sphere indicates slowness in the sense, the slowest reaction has smallest radius. enter image description here

Even though there is difference between the mouths at $B-C$ and $B-D$ junctions both are happening at same rate.This is what you thought.

But, the energy from $TS_1$has been* released in form of heat as an exotheromic reaction.Also when there is a difference in rates of r/d step this effect is diminished.

*Thanks to @Martin for pointing out an error

Also, the rate of B-D and B-C are different and the products must be proportional to the rates.Otherways, the product must increase in which reaction is fast. Consider this:enter image description here

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  • $\begingroup$ It is wrong to state that the activation energy is used in bonds. It will be released after the transition state. If $E(A)<E(B)$, the reaction would not take place. After the transition state there should always be more energy in the system. $\endgroup$ – Martin - マーチン Aug 4 '14 at 4:52
  • $\begingroup$ it might be that my answer is wrong or maybe right, but please tell me why you consider that,I say, they need to gain energy again by collision/etc. $\endgroup$ – RE60K Aug 5 '14 at 8:44
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    $\begingroup$ On the path towards the transition state kinetic energy is converted to potential energy. After the transition state the potential energy is converted back to kinetic energy, which could be lost to the system or to overcome the next transition state. The energy is not used up in bonds, but released to the system eventually (in one form or another). The reaction would not proceed if this was not the case. $\endgroup$ – Martin - マーチン Aug 5 '14 at 9:02
  • $\begingroup$ @Martin Aha my bad, on second thought what must be done of a wrong answer(I know I must be knowing this beforehand).. $\endgroup$ – RE60K Aug 5 '14 at 9:09
  • $\begingroup$ Your answer is not wrong per se, only this little statement. Just change " has been has been used up in bonds and " to " is released to the system. Additional ..." After that I think it will be a correct answer. $\endgroup$ – Martin - マーチン Aug 5 '14 at 9:28

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