If we have some reaction $\ce{A(aq) + B(aq) -> AB(aq)}$ and right now we have less $\ce{A}$ and $\ce{B}$ then we would have at equilibrium, is there ever a time where the amount of $\ce{A}$ and $\ce{B}$ would overshoot the equilibrium of $\ce{A}$ and $\ce{B}$ if the reaction proceeded without any new sources of energy or reactants. In other words, will the concentrations of $\ce{A}$ and $\ce{B}$ always increase towards their equilibrium concentrations but never actually reach those points or will they sometimes overshoot that concentration a large amount (not just an amount that can be ascribed to chance) but then fall back towards the equilibrium concentration.


2 Answers 2


In normal experiments vast numbers of molecules are involved, $10^{18}$ would not be untypical, and as a result the kinetic behaviour observed as equilibrium is approached is rather like that of a highly over-damped mechanical system. If, however, molecules could be observed one by one at each of many small time intervals then considering a very small sample it is possible that overshoot could be observed. As an example consider a cis-trans isomerisation $\ce{A <=>}B $ that is initiated by a very short light pulse of perhaps femtosecond duration. Consider this only as possible mechanism with which to start a reaction so that we might study it.

The molecule when excited into its excited state has a certain finite lifetime, its reciprocal is the rate constant. If we observe just a single molecule we cannot know exactly when it will react but only that if we repeatedly measure the times it remains in the excited state and generate a histogram of these times that an exponential decay is produced from which a lifetime can be measured. Now if we observe just a few molecules, say 10, it is possible, albeit with a small probability, that all of them will react within a small time, say, 10% of the lifetime, thus we have a situation that in such a small time interval after the reaction starts that all the molecules are now products. The 'equilibrium' at this time is now product 10, reactants 0. The product also has a lifetime with which it transforms back into reactants in just the same manner. And so depending on pure chance the s 'snap-shot' ratio of reactant to products can vary widely.

Of course the chance of all molecule reacting in the same short time-span is small, and if we observed the same molecules many times over the 'normal' equilibrium constant would be observed. The same is true if we observed very many molecules that reacted in the same short time interval as did our initial 10. Thus this variability is not observed in practice because by observing over longer times and/or larger numbers of molecules the average properties are obtained.


Not really. Overshooting doesn't happen, rather the concentrations gradually approaches the equilibrium concentration as an asymptote.

You'll soon see what I mean.

As an example, for simplicity, lets consider:

$$\ce{P <=>[k_1][k_2] Q}$$

I will assume here that the forward and backward reactions are first order, so we get the rate to be:




As a result the net rate would be:


Let's take the initial concentration of $\ce{P}$ to be $\ce{[P]_0}$, and the initial concentration of $\ce{Q}$ to be zero.

Suppose at any time $t$, the concentration of the reactant is $\ce{[P]}$. Thus, $\ce{[Q]} = \ce{[P]_0} - \ce{[P]}$. Now, we get the rate law to be:

$$\frac{d\ce{[P]}}{dt} = k_2\ce{[P]_0} - (k_1 + k_2)\ce{[P]}$$

Integrating this from $\ce{[P]_0}$ to $\ce{[P]}$:

$$\int_{\ce{[P]_0}}^{\ce{[P]}}\frac{d\ce{[P]}}{k_2{[P]_0}-(k_1 + k_2)\ce{[P]}}=\int_0^tdt$$

Solving the integral, we get:

$$\frac{-1}{(k_1 + k_2)}\log_e{\left(\frac{(k_1 + k_2)\ce{[P]} - k_2\ce{[P]_0}}{k_1\ce{[P]_0}}\right)}=t$$

$$\frac{(k_1 + k_2)\ce{[P]} - k_2\ce{[P]_0}}{k_1\ce{[P]_0}}=e^{-(k_1 + k_2)t}$$

Finally, on simplifying it, we get our integrated rate equation as:

$$\ce{[P]}=\ce{[P]_0}\frac{k_2 + k_1e^{-(k_1+k_2)t}}{k_1+k_2}$$

Let's plot this on a graph to understand what this is actually saying:

enter image description here (purple-$\ce{[P]}$ and orange-$\ce{[Q]}$)

You can clearly observe that the concentrations never overshoot the asymptotes (black dotted line). The asymptotes denote the equilibrium concentrations, so it's evident that the reaction tends toward the equilibrium concentration.

  • 4
    $\begingroup$ Then again, there is a number of clock reactions with oscillations which kinda may be interpreted as "overshooting". $\endgroup$ May 23, 2017 at 6:15

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