I get the concept that one can measure the rate of reaction by measuring the rate at which an equilibrium state reaches another equilibrium state after the system is perturbed (usually by heating). However, I do have an issue with it. Consider the reaction $$\ce{A <=> B}$$

My lecture handout suggests that the rate of change of $[\ce{A}]$ is given by:

$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\text{forward}[\ce{A}]+k_\text{reverse}[B]$$

However, surely it's just $\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\text{forward}[\ce{A}]$. The equation in the lecture handout would make sense at equilibrium because that is when $\frac{\mathrm d[\ce{A}]}{\mathrm dt}=0$ but I think it seems to insinuate that this is always valid (without explicitly saying so). Am I right or is that equation always valid? If it is: why?


1 Answer 1


You have two simultaneous reactions,

$$\begin{array}{rcl} \ce{A} &\xrightarrow{k_1}& \ce{B}\\[6pt] \ce{B} &\xrightarrow{k_{-1}}& \ce{A} \end{array}$$

and $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt}$ equals $-k_1 [\ce{A}]$ for the first reaction alone and $k_{-1} [\ce{B}]$ for the second alone. To get the total $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt}$ for the reactions running simultaneously, you have to add the contributions from each reaction:

$$\dfrac{\mathrm d[\ce{A}]}{\mathrm dt} = -k_1 [\ce{A}] + k_{-1} [\ce{B}]$$

This will always be true; at equilibrium you'll have $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt} = 0$ and

$$K = \frac{k_1}{k_{-1}} = \frac{[\ce{B}]}{[\ce{A}]}$$

If you did it your way, this wouldn't be the case.

  • $\begingroup$ Perfect. Thank you very much. Seems obvious now. $\endgroup$
    – RobChem
    Jan 31, 2015 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.