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In this special reaction coordinate diagram with two reaction mechanisms, I tried to analyze it in two other ways, one with kinetics, another with equilibrium.

A. kinetics, I tried to find reaction rate of forward, and reverse. By the diagram, rate determining step of forward reaction is the first one (because activation energy is the highest and let's say it is one and only factor for RDS in this case). so Vf=k1[A][B]. RDS in reverse reaction is second step of the reaction judging by the diagram activation energy (M->A+B). So I used pre-equilibrium approximation for that. That's Vr in the picture.

B. equilibrium, I wrote all the equilibrium constant and simply added the reactions.

now here is the question. Expression of the rate of forward and reverse reaction in dynamic equilibrium is quite different from each case, analyzing in a way of kinetics and equilibrium constant, as you see in the picture below. Why is it different? What's right expression? What am I missing here?

  • $\begingroup$ I don't understand your question. What do you mean by "different"? The two expressions at the bottom left and right e.g. are equivalent. $\endgroup$
    – aventurin
    Apr 7, 2018 at 19:18

2 Answers 2


You have chosen an unusual scheme in that A appears to be an intermediate as well as a reactant. If you use the simpler scheme things are easier and so rather than try to follow yours I have examined the scheme below which is very nonetheless very similar

$$RX\underset{k_{-1}} {\stackrel{k_1}{\leftrightharpoons}}R+X$$ $$R+Y\underset{k_{-2}} {\stackrel{k_2}{\leftrightharpoons}}RY$$

where species RX looses X and is replaced by Y. The intermediate species is R. The overall reaction is $RX+Y=RY+X$ and the equilibrium constant for the two steps are

$$K_1=\frac{k_1}{k_{-1}}=\frac{\mathrm{[R]_e[X]_e}}{\mathrm{[RX]_e}}$$ $$K_2=\frac{k_2}{k_{-2}}=\frac{\mathrm{[RY]_e}}{\mathrm{[R]_e[Y]_e}}$$

and the overall equilibrium constant

$$K=K_1K_2=\frac{k_{1}k_{2}}{k_{-1}k_{-2}}=\frac{\mathrm{[RY]_e[X]_e} }{\mathrm{[RX]_e[Y]_e }}$$

and all the concentrations in square brackets with subscript $e$ are equilibrium values. [This type of equation is true if there are many equilibria one after the other, $\displaystyle K=K_1K_2K_3K_4\cdots=\frac{k_1k_2k_3k_4}{k_{-1}k_{-2}k_{-3}k_{-4}}\cdots$].

In a rate equation approach we can apply a steady state approach to the intermediate species R. The steady state assumes that the rate of change of R is zero;

$$\frac{d[R]}{dt}= k_1[RX]-k_{-1}[R][X]-k_2[R][Y]+k_{-2}[RY]=0$$

from which

$$ [R]_{ss}= \frac{k_{-2}[RY]+k_1[RX]}{k_{-1}[X]+k_2[Y]}$$

The rate $r$ can be given by

$$\begin{align} r=-\frac{d[Y]}{dt}=k_2[R][Y]-k_{-2}[RY] &= \frac{k_2(k_{-2}[RY]+k_1[RX])[Y]-k_{-2}[RY](k_{-1}[X]+k_2[Y])}{k_{-1}[X]+k_2[Y]}\\&=\frac{k_1k_2[RX][Y]-k_{-1}k_{-2}[RY][X]}{k_{-1}[X]+k_2[Y]}\\ \end{align}$$

and the numerator is zero when equilibrium concentrations are used making the rate equal to zero also. So this connects the equilibrium to the rate equations.

Initially just after the reactants are mixed the amount of product is very small and $k_1k_2[RX][Y]>>k_{-1}k_{-2}[RY][X] $ and the the rate is


which can be confirmed by experiment. Hope this helps.

  • $\begingroup$ It's beautiful. Thank you so much. It helped me a lot. I just need a little more time to organize. $\endgroup$
    – Zillai
    Apr 8, 2018 at 13:49
  • $\begingroup$ so overall k(forward) is k1 x k2 x .... anyway. But by the aspects of reaction rates, it's not really important. What's important is to find it by suggesting some cases with experiments. If k-1 is small in your last equation, erase it and finally k(forward) is similar to k1. (even though k(forward) is k1xk2 thermodynamically) Am I right? anything missing? $\endgroup$
    – Zillai
    Apr 8, 2018 at 13:57
  • $\begingroup$ I'm not sure I understand your last question, but the last equation only applies at very short times, i.e. just as reaction starts; its not just the rate constants but also the low concentration of products that are important in removing the last term in the numerator. $\endgroup$
    – porphyrin
    Apr 10, 2018 at 15:14

I think you get confused by thinking reaction rate and equilibrium constant are same. However, they are in different concepts: The rate of a reaction is dependent on reaction mechanism, which may have different steps, orders, different activation energies for different steps, etc. Unlike reaction rates, equilibrium constant is a state function, which is a function defined for a system relating several state variables or state quantities that depends only on the current equilibrium state of the system.


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