Consider the reversible unimolecular reaction:

$$\ce{A <=>[k_1][k_2] B}$$

We know that the forward reaction is often considerably more thermodynamically favourable than the reverse reaction, and therefore the relationship $k_1 \gg k_2$ holds between the rate constants. The rate constants are in the same units, and so it is possible to write this relationship. A similar example would be a reaction that is bimolecular on both sides, $\ce{A + B <=> C + D}$, etc.

However, consider this reversible reaction:

$$\ce{A + B <=>[k_1][k_2] C}$$

Assume the reaction proceeds at rate $k_1 [\ce{A}][\ce{B}]$ in the forward direction and $k_2 [\ce{C}]$ in reverse. If the overall reaction rate has units $M s^{-1}$, then $k_1$ necessarily has units $M^{-1} s^{-1}$ and $k_2$ has units $s^{-1}$.

My question is: For this second case, we can no longer impose that $k_1 \gg k_2$, due to difference in units, but can we say anything about their relationship? Does setting $k_1$ to some value constrain the choice of $k_2$ in any way?


2 Answers 2


A chemical reaction may be a very complicated thing, involving many different paths and elementary reactions. Basically it always leads to the formation of an equilibrium, where the lowest possible energy will be the most favoured. Under certain conditions more than one state will be populated and hence there will be enough potential energy to interconvert molecules in certain states. We usually refer to that as an dynamic equilibrium.

Changes under equilibrium conditions will most likely be elementary reactions (or an infinitely complicated chain of these.)

The population of the various states seems to be static at equilibrium, since the overall change of enthalpy is zero. In a dynamic picture that only means, that forward and backward reaction cancel each other. In that sense they are the same reaction in different directions. Therefore they can be described as one trajectory on the potential energy surface. (If there is a chemical reaction involving many different elementary reactions and the equilibrium still occurs, then the mathematical description will be much more complicated as all elementary reactions will be correlated. However, the main conclusions should stay the same.)

\begin{aligned} \Delta G_r &= \Delta G_r^\circ + \mathcal{R}T\cdot \ln K = 0\\ \Delta G_r^\circ &= − \mathcal{R}T \cdot \ln K \end{aligned}

The equilibrium constant may be derived from the mass-action law and should be defined as a product of activities (for the standard state). For $\ce{A + B <=> C}$ this will lead to $$K^\circ= \frac{a(\ce{C})}{a(\ce{A})\cdot a(\ce{B})}$$

The activities are proportional to concentrations (in first approximation) and are unit less ($c^\circ$ being the standard concentration $1 \:\mathrm{mol/L}$)


For reasonable dilutions ($c\to0\:\mathrm{mol/L}$) one can assume that the activity coefficient becomes one $\gamma\approx1$ and therefore rewrite the equilibrium constant with concentrations. $$K^\circ= \frac{c(\ce{C})/c^\circ}{c(\ce{A})\cdot c(\ce{B})/(c^\circ)^2}$$

At equilibrium the forward and backward reaction are coupled and therefore have to reflect the equilibrium constant. $$K^\circ= \frac{k_f}{k_b} = \frac{c_{\text{eq}}(\ce{C})/c^\circ}{c_{\text{eq}}(\ce{A})\cdot c_{\text{eq}}(\ce{B})/(c^\circ)^2}$$

As one can see, the constants also need to have the same unit. As they are at equilibrium, they are the same reaction with different populated states.

If you are moving away from the equilibrium, thing will start to get a little bit more messy. On reaction will be faster than the other and rates will not be comparable any more.

  • $\begingroup$ I might be missing something here - are the units of $c_{eq}$, $(C)$, and $c^{\circ}$ all $\frac{mol}{L}$? $\endgroup$
    – thomij
    Commented May 28, 2014 at 14:51
  • $\begingroup$ I see your point now - $c_{eq}(C)$ is the equilibrium concentration of C, and has units of mol/L. The thermodynamic equilibrium coefficient is dimensionless. However, I am not sure that we can relate that to the rate constants, since they are defined from empirical rate laws based on concentration, and therefore do not necessarily have the same units. We are each making different assumptions here about the conditions, I will change my answer to clarify. $\endgroup$
    – thomij
    Commented May 28, 2014 at 17:51
  • $\begingroup$ @thomij If you are using empiric determined rate laws, then you are far away from equilibrium, and very far away from elementary reactions. Everything I have written would not apply any more, since a reaction that is not in equilibrium can not be described by an equilibrium constant. $\endgroup$ Commented May 29, 2014 at 10:36
  • $\begingroup$ if the rate constants and the equilibrium constants are really constant, then we are close enough to equilibrium to relate them. I understand your point that if you use activities, the equilibrium constant is dimensionless (although you might consider editing your answer to talk about using activities for the rate laws, otherwise the units situation gets worse). My point is that even if you don't use activities, it still works. $\endgroup$
    – thomij
    Commented May 29, 2014 at 12:48
  • $\begingroup$ @thomij I really do not understand what you are trying to say. Especially the units part. And I really do not want to spent any more thought on this, as it is not my field of expertise and I did not want to write an answer in the first place. There was a demand I tried to satisfy that. Now I am more confused than before. That should not happen. And obviously edison does not even care about an answer any more - so it might be appropriate to remove it. $\endgroup$ Commented May 30, 2014 at 3:16

The reaction rates can be related to each other through the equilibrium constant.

However, it is important to note that we are making two key assumptions in order to do this:

  1. The solution is nearly ideal - in layman's terms, that means the concentration of one substance does not depend on any of the others. When that is true, we can use an equilibrium constant defined in terms of concentrations, $K_{eq} = \frac{[C]}{[A][B]}$, instead of activities (effective concentrations).
  2. The rate laws are in fact given by $k_1[A][B]$ and $k_2[C]$ - this may be true for elementary reactions, but you cannot in the general case determine the reaction rates solely from a reaction equation. You must find them empirically, or have a good idea of what the underlying mechanism is (and even then you might not get the correct rate law.)

Assuming that these two conditions hold, we can then relate the two rate constants to each other through the equilibrium constant.

At equilibrium, we know that the rate of change in one direction is equal to the rate of change in the other direction. In other words,

$k_1[A][B] = k2[C]$.

This means that

$\frac{k_1}{k_2} = \frac{[C]}{[A][B]}$,

and since

$K_{eq} = \frac{[C]}{[A][B]}$,

$K_{eq} = \frac{k_1}{k_2}$.

When we say that a reaction is more thermodynamically favorable in the forward direction, we are also saying that $K_{eq}$ is large (relatively speaking) and that $k_1 >> k_2$.

You are right to be suspicious, since it doesn't seem like we should be able to compare two numbers with different units. The key here is that whether we explicitly look at the units or not, they are still there in the equilibrium constant. We are comparing the magnitudes alone, usually in an effort to reduce the complexity of an equation by assuming that some terms are approximately equal to zero.

  • $\begingroup$ The reaction constants do have the same unit $\mathrm{s}^{-1}$ as kinetics is defined by activities, which is unitless. With the relation $a=\gamma\frac{c}{c^{\ominus}}$ it is possible to state, that for reasonable concentrations $\gamma\approx1$ and hence $a\approx c$. We would introduce the error and the misconception, because most people forget that the used concentration is unitless. $\endgroup$ Commented May 27, 2014 at 6:48
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    $\begingroup$ @Martin, perhaps you could put together your comments to Andrea and thomij, and post a more detailed answer to this question? $\endgroup$
    – edison1093
    Commented May 27, 2014 at 10:53

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