A reaction was given $\ce{A<=>B}$ The rate constants for forward and backward reactions were given as $k_\mathrm{f}$ and $k_\mathrm{b}$. The initial $\ce{[A]}$ was given as $a_{0}$ and the $\ce{[B]}$ at a time t (before equilibrium is attained) was to be calculated.

I tried solving it as follows -

For A I wrote -

$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = k_\mathrm{b}\ce{[B]}-k_\mathrm{f}\ce{[A]}$$

and similarly for B -

$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = k_\mathrm{f}\ce{[A]}-k_\mathrm{b}\ce{[B]}$$

Combining these two I got a 2nd order differential equation and wasn't able to proceed further. Is there something wrong with my approach or my equations?


1 Answer 1


Your method should work as you know the boundary conditions for your scheme, but it is not the easiest method. Instead, it is easier in this case, as you are given the initial amount of A, to let $x$ be the amount of B produced at time $t$, then $a_0 - a = b = x$ where to save writing brackets, $a,\;b$ are the amounts of A and B at time $t$. The the rate equation is

$$\frac{\mathrm{d}x}{\mathrm{d}t} =k_\mathrm{A}a -k_\mathrm{B}b$$

and substituting gives

$$ \frac{\mathrm{d}x}{\mathrm{d}t}= k_\mathrm{A}a_0-(k_\mathrm{A}+k_\mathrm{B})x$$

This can now be integrated with limits $(0,x),\; (0,t)$ to give

$$\ln \left( \frac{k_\mathrm{A}a_0-(k_\mathrm{A}+k_\mathrm{B})x}{k_\mathrm{A}a_0} \right)= -(k_\mathrm{A}+k_\mathrm{B})t$$

At equilibrium, the rate of reaction is zero, then $k_\mathrm{A}a_e=k_\mathrm{B}b_e=k_\mathrm{B}x_e$ where subscript $e$ means that it is the equilibrium value. Also, we know that $\displaystyle K_\mathrm{e}=\frac{k_\mathrm{A}}{k_\mathrm{B}}=\frac{x_e}{a_0-x_e}$, where $a_0-x_e$ is the amount of A remaining at equilibrium.

Substituting and rearranging gives the amount of $x$, which is B as a function of time:

$$x=x_e -x_ee^{-(k_\mathrm{A}+k_\mathrm{B})t}$$

Both this and the previous equation show that the equilibrium is approached with a rate constant that is the sum of the two rate constants.


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