I have a question regarding a reversible elementary reaction:

If we have an equilibrium where $$\ce{A<=>[k_1][k_{-1}]B}$$ , then $$-d[A]/dt= k_1[A]-k_{-1}[B]$$.

If we take $$[B]=[B]_0 + ([A]_0 - [A])$$, the equation above becomes $$d[A]/dt= -k_1[A]+k_{-1}([B]_0+([A]_0 -[A])$$

So far, I've taken the integrating factor to be $$u=\mathrm{e}^{(k_1+k_{-1})t}$$

and emerged with the following equation:

$$[A]=\frac{k_{-1}}{k_1+k_{-1}}([B]_0+[A]_0)(1-\mathrm{e}^{-t(k_1+k_{-1})}) + C \mathrm{e}^{-t(k_1+k_{-1})}$$

I have tried to put $$[A](t=0) = [A]_0$$ and the solution appears to be $$C=[A]_0$$

Are there further approximations required?

My lecturer has produced the final equation as


When we take $$t->\infty$$ our equations match, however.

  • $\begingroup$ Could you please check the braces in your equations. I corrected the formatting, so as to make it a little more pleasing to the eye and easier to read but some of your braces don't match and I'm not 100% sure where exactly they have to be matched. I could make an educated guess, but since you derived the equations it should be easier for you to correct it. $\endgroup$
    – Philipp
    Sep 27, 2014 at 20:22
  • $\begingroup$ I haven't really done the derivation myself but your equation looks reasonable. It contains only the rate constants and initial concentrations, i.e. things that have to be known to describe the kinetics of the reaction anyway or are easy to measure, so that should be ok. $\endgroup$
    – Philipp
    Sep 27, 2014 at 20:28
  • $\begingroup$ All right, thank you. I'm trying to uncover my lecturer's reasoning in this one. $\endgroup$
    – Edward
    Sep 27, 2014 at 20:44
  • $\begingroup$ I've looked at it in more detail. You seem to have used a wrong boundary condition: I think $[B]=[B]_0 - ([A]_0 - [A])$ is wrong, it must be $[B]=[B]_0 + [A]_0 - [A]$ instead. $\endgroup$
    – Philipp
    Sep 27, 2014 at 21:38
  • $\begingroup$ That was a typo, sorry. $\endgroup$
    – Edward
    Sep 27, 2014 at 22:05

1 Answer 1


I went through the derivation myself and I can reproduce the equation you got. But after using $C = [\ce{A}]_0$ and some rearrangements I get the following formula

\begin{equation} \ce{[A]} = \frac{k_{-1}}{k_{1} + k_{-1}} \left( \ce{[A]_{0}} + \ce{[B]_{0}} \right) + \frac{\mathrm{e}^{-(k_{1} + k_{-1})t} }{k_{1} + k_{-1}} \left( \ce{[A]_{0}} k_{1} - \ce{[B]_{0}} k_{-1} \right) \end{equation}

and not the one your professor got. But your professor's formula might be incorrect because for $t=0$ his formula yields the incorrect result $\ce{[A]}_{t = 0} = \ce{[A]_{0}} + \ce{[B]_{0}}$ instead of the correct result $\ce{[A]}_{t = 0} = \ce{[A]_{0}}$, so it doesn't satisfy the given boundary conditions.


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