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My first language is not English, so there might be some parts where it's hard to read. Sorry in advance.


I have some questions regarding the derivation of the equilibrium constant and the rate-determining step.

For a hypothetical reaction $$\ce{aA + bB <=> cC + dD}$$ the equilibrium constant ${K}$ would be $${K=\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}}$$

However, my textbook states that equilibrium is when the forward reaction rate ${R_{f}=k_{f}\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}$ and the reverse reaction rate ${R_{r}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}$ becomes the same, so that $${R_{f}=R_{r}}$$ $${k_{f}\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}$$ $${\frac{k_{f}}{k_{r}}=\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}=K}$$ which is a bit unsound because there is no guarantee that $$\ce{aA + bB -> cC + dD}$$ and $$\ce{cC + dD -> aA + bB}$$ are both elementary reactions, so ${R_{f}}$ and ${R_{r}}$ might not equal ${k_{f}\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}$ and ${k_{r}\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}$.

So I started searching, and found this:

Kinetic Derivation of the Equilibrium Constant

The usual method of finding the rate equation of a multistep reaction is that it equals the rate equation of the rate-determining step with the caveat that if it contains any reactant which is a product in a previous step, one must substitute the rate factor (e.g., ${\left[\ce{A}\right]^{a}}$, ${\left[\ce{B}\right]^{b}}$) of this reactant with the rate factors of the reactants of this prior step. The idea of this proof is to write this method in a form more amenable to algebraic manipulation, and then find use it to determine the rate equation for the forward reaction, ${R_{f}}$, and then for the reverse reaction, ${R_{r}}$. Once this has done, we can equate ${R_{f}}$ and ${R_{r}}$ and hence derive the equilibrium constant expression.

Consider a general multi-step reaction, but instead of written the usual way, written so that all products appear on the left-hand side of the equation with negative coefficients:

$${x_{1}\ce{X}_{1}+x_{2}\ce{X}_{2}+x_{3}\ce{X}_{3}+⋯+x_{n}\ce{X}_{n}\ce{<=>}0}$$

Imagine that there are ${s}$ steps making up this multi-step reaction, and each elementary step is written the same way as above. Then, for an elementary step ${j}$, define ${x_{ij}}$ so that it is the coefficient of ${X_{i}}$ in the ${j}$th step. Since, as stated above, the coefficients of the elementary steps must add up to the overall coefficients,

$${\displaystyle\sum_{j=1}^{s}x_{ij}=x_{i}}\tag{1}$$

Now multiply together the terms of each elementary step, to the power of their corresponding coefficients, up to but excluding the rate-determining step (step ${r}$).

$${\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}}$$

The overall forward rate is proportional to this quantity multiplied by ${A}$, which is the reactants of the rate-determining step with their corresponding coefficients as exponents.

$${R_{f}=k_{f}A\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}}$$

Now assume that we do the same as above, except in reverse. That is, we use the negative of each coefficient, and we start from elementary step ${s}$ (that is, the last step) and go until (but exclude) the rate-determining step:

$${\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}-x_{ij}}}$$

The overall reverse rate is proportional to this quantity divided by ${B}$, which is the concentrations of the products of the rate-determining step with their corresponding coefficients as exponents.

$${R_{r}=k_{r}\frac{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}-x_{ij}}}{B}}$$

Since ${R_{f}=R_{r}}$ in an equilibrium,

$${k_{f}A\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}=k_{r}\frac{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}-x_{ij}}}{B}}$$

Rearranging this to get ${K}$,

$${\frac{k_{f}}{k_{r}}=\frac{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}-x_{ij}}}{AB\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}}}$$

Because there is a negative sign in the exponent in the numerator, we can move the numerator to the denominator if we switch the negative sign to a positive sign.

$${\frac{k_{f}}{k_{r}}=\frac{1}{\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}\right)\left(AB\right)\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}x_{ij}}\right)}}$$

Recalling that ${A}$ is the product of the concentrations of the reactants of the rate-determining step with their coefficients as exponents, and that ${B}$ is the same for the products of the rate-determining step, we can write ${AB}$ as

$${AB=\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{x_{ir}}}$$

Thus,

$${\frac{k_{f}}{k_{r}}=\frac{1}{\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}\right)\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{x_{ir}}\right)\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}x_{ij}}\right)}}$$

But then this can be written simply as,

$${\frac{k_{f}}{k_{r}}=\frac{1}{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{s}x_{ij}}}}$$

Recalling Equation 1, this is simply

$${\frac{k_{f}}{k_{r}}=\frac{1}{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{x_{i}}}}$$

$${\frac{k_{f}}{k_{r}}=\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{-x_{i}}=K}$$

I did understand this, but new questions popped up.

First of all, this derivation of the equilibrium constant assumes that the rate-determining step for the forward reaction and reverse reaction are always identical (the ${r}$th step). Is this always true?

For example, in the reaction shown below, the 2nd step is the rate-determining step for both forward reaction and reverse reaction.

In this reaction, the 2nd step is the rate-determining step for both forward reaction and reverse reaction

However, in this reaction below, the 2nd step is the rate-determining step for the forward reaction, while the 1st step is the rate-determining step for the reverse reaction.

In this reaction, the 2nd step is the rate-determining step for the forward reaction, while the 1st step is the rate-determining step for the reverse reaction.

I made the second graph myself, and I do not know if such a reaction is thermodynamically possible. But if reactions like this exist, then the derivation process written above would be flawed.


And here's my second question. Let's bring back the reaction $$\ce{aA + bB <=> cC + dD}$$ and suppose that this reaction is a 2-step reaction: $$\ce{aA <=> cC + eE}\tag{Step 1}$$ $$\ce{bB + eE <=> dD}\tag{Step 2}$$

If the 1st step is the rate-determining step, then the forward reaction rate would be ${R_{f}=k_{f}\left[\ce{A}\right]^{a}}$ and the reverse reaction rate would be... uh... ${R_{r}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}}$? I tried to derive ${K}$ from ${R_{f}}$ and ${R_{r}}$, but I got stuck in the middle:

The equilibrium constant of Step 2 of the reverse reaction is

$${K_{r2}=\frac{\left[\ce{B}\right]^{b}\left[\ce{E}\right]^{e}}{\left[\ce{D}\right]^{d}}}$$

so

$${\left[\ce{E}\right]^{e}=\frac{K_{r2}\left[\ce{D}\right]^{d}}{\left[\ce{B}\right]^{b}}}$$

Equilibrium is when the forward reaction rate ${R_{f}=k_{f}\left[\ce{A}\right]^{a}}$ and the reverse reaction rate ${R_{r}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}}$ becomes the same, so

$${R_{f}=R_{r}}$$

$${k_{f}\left[\ce{A}\right]^{a}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}=k_{r}\left[\ce{C}\right]^{c}\frac{K_{r2}\left[\ce{D}\right]^{d}}{\left[\ce{B}\right]^{b}}}$$

Thus

$${K=\frac{k_{f}}{k_{r}}=K_{r2}\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}}\tag{???????}$$

What mistake did I make here? I have been trying to solve this for a few days but I can't seem to obtain a clear answer.

(Sorry for such a long question, and thanks in advance.)


EDIT

So about the second question... Thanks to @porphyrin, I've found my mistake. I shouldn't have assumed that the reverse reaction rate is ${R_{r}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}}$. It should be

$${R_{r}=k_{r}\left(\frac{\left[\ce{D}\right]^{d}}{\left[\ce{B}\right]^{b}\left[\ce{E}\right]^{e}}×\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}\right)=k_{r}\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{B}\right]^{b}}}$$

Knowing this and solving again, I was able to get

$${K=\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}}$$


EDIT 2

Thanks to @thewitness, I think I got the answer for the first question. So apparently, it is possible for the rate-determining step of the reverse reaction to be different from that of the forward reaction.

Though I got the explanations I needed, feel free to add an answer and give me more insight into these matters. Thanks.

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  • $\begingroup$ The equation you seek is only true for each barrier, not two as you have in your pictures. The equilibrium applies to each barrier. Secondly you have to assume that $a$ A molecules and $b$ B molecules react in a single step, which is extremely unlikely (i.e. does not happen to any extent) unless $a=b=1$, same for C and D. So you have to apply chemical knowledge as well as using the maths. If $a, b $ etc are not one there is invariably an intermediate step involved. $\endgroup$ – porphyrin Sep 29 at 10:10
  • $\begingroup$ @porphyrin Of course, the true values of ${a}$, ${b}$, ${c}$, and ${d}$ would be 1, but why would they matter herein just deriving ${K}$ in a hypothetical example? If the values of ${a}$, ${b}$, ${c}$, and ${d}$ becomes an actual problem in the derivation, let me know. Moreover, could you indicate my mistake more clearly? Which equation is only true for each barrier, and how should I revise it? $\endgroup$ – LucasShin Sep 29 at 11:06
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    $\begingroup$ Well it does not matter in a theoretical calculation because you can assume as many molecules as you like react in a single step. K will relate to starting and ending species only and will give no information about mechanism in just the same sense as $aA+bB\ cdots $ does not tell you anything about what actually happens in a reaction. In multiple equilibria the equilibrium constants generally multiply. The forward/reverse rate constants should refer only to each barrier, i.e each step not the whole reaction. $\endgroup$ – porphyrin Sep 29 at 11:45
  • $\begingroup$ @porphyrin I'm sorry I might be lacking of understanding, what do you mean that the forward/reverse rate constants ${k_{f}}$ and ${k_{r}}$ should refer only to each barrier and not the whole reaction? In general cases, doesn't the rate constant of the RDS become the rate constant of the whole reaction? (Except for some cases, of course.) $\endgroup$ – LucasShin Sep 29 at 13:46
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    $\begingroup$ Yes just each barrier, of course you could do this for the whole of a complex reaction but such a thing does not make sense. Same for RDS as you don't know what this step is for a complex reaction unless you also know all the other species present and rate constants. $\endgroup$ – porphyrin Sep 29 at 14:31
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I will first try to answer your second question.

There's absolutely no mistake you made! As you assumed first step is RDS(rate determining step), You would expect $K=K_{f1}$.

Continuing from where you left, $$K=K_{r2}\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}$$

$$=\frac{\left[\ce{B}\right]^{b}\left[\ce{E}\right]^{e}}{\left[\ce{D}\right]^{d}} \times \frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}$$

$$ = \frac{\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}}{\left[\ce{A}\right]^{a}} =K_{f1}$$ as expected.

You seem to assume that the the slowest step is the RDS always, and that it only governs the rate of the reaction. This isn't always true. Only if the slowest step(RDS) is MUCH slower that all the steps this (approximation) that rate of reaction = rate of RDS is valid.(hopefully this answers your first question).

More generally(ALWAYS TRUE, no approximation), for your second question, (assuming no RDS anywhere)

$K_{f1}=\frac{\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}}{\left[\ce{A}\right]^{a}}$

$K_{f2}=\frac{\left[\ce{D}\right]^{d}}{\left[\ce{B}\right]^{b}\left[\ce{E}\right]^{e}}$

$K_{f1}K_{f2} =\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}} = K$

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  • $\begingroup$ Thank you for the detailed explanation! Actually, I've found my mistake on the second question. I'm still confused about the first question, though. Could you elaborate, please? $\endgroup$ – LucasShin Sep 30 at 8:34
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    $\begingroup$ The rate of the reaction depends only on the concentration. Energy doesn't affect it directly. You mentioned that "the 2nd step is the rate-determining step for the forward reaction, while the 1st step is the rate-determining step for the reverse reaction." which isn't necessarily true(we can't conclude this from energy profile diagram). The difference in energy is what we call the "Activation energy" and difference between activation energy of reactants and products is called "enthalpy of the reaction". This is constant for a given reaction. $\endgroup$ – user600016 Sep 30 at 9:10
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    $\begingroup$ There are two equations: Arrhenius equation and vant Hoff equations. The former relates rate constant with activation energy, temperature and a constant called Arrhenius factor or frequency factor. The vant Hoff equation relates equilibrium constant with temperature and enthalpy for the reaction. You would learn more about this in chemical kinetics chapter. $\endgroup$ – user600016 Sep 30 at 9:10
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    $\begingroup$ Your diagram is of a first order reaction which takes place in two steps. It's been long since I revised chemical kinetics so extremely sorry if I mentioned anything wrong. $\endgroup$ – user600016 Sep 30 at 9:14
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    $\begingroup$ Hmmm... Then the "Kinetic Derivation of the Equilibrium Constant" I quoted above would be extremely flawed... I guess the method you stated at the bottom of your post (${K}$=${K_{1}K_{2}K_{3}×⋯×K_{s}}$) is the most accurate for deriving the equilibrium constant. Thanks. $\endgroup$ – LucasShin Sep 30 at 13:08
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There is no reason to invoke a rate-determining step in making the connection between kinetic constants and equilibrium. In fact, you might get the wrong answer.

The easiest way to make the connection is to separate the reaction into elementary steps. The principle of micro-reversibility states that any elementary reaction will also have a elementary reverse reaction. At equilibrium, the rate of each step is equal to the rate of the reverse step (all intermediates at constant concentration).

Once this is set up, you can combine all the kinetic equations to arrive at the equilibrium constant equation. The equilibrium constant itself will be a ratio of the product of the forward rate constants divided by the product of the reverse rate constants. Note that even rate constants of steps after the "rate-determining step" will be part of that expression.

The math is shown and illustrated for a simple example in this answer. Alternatively, you could first derive the equilibrium constant expression for each elementary step (as the ratio of forward and reverse rate), and then combine these to arrive at the equilibrium constant expression of the net reaction (see bottom of answer by @thewitness).

Some remarks about the rate-determining step

In a multi-step reaction, the overall rate law will change when concentrations change. For example, if under some condition the rate is zero order in reactant X, if I lower the concentration of X (or leave it out completely), the reaction with X as reactant will become rate-determining step. However, the equilibrium constant expression and its value are independent of the rate-determining step. Therefor, deriving the equilibrium constant expression by relying on a rate-determining step is logically flawed.

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  • $\begingroup$ I have carefully read "Kinetic Derivation of the Equilibrium Constant" I quoted in the question once again after reading your answer. What I've realized is that even though the whole process is logically flawed, it is algebraically accurate because it basically shares the same method with the examples you provided in your answer. Thanks for sharing your explanation. $\endgroup$ – LucasShin Oct 2 at 4:37

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