Chemistry - Chemical Kinetics Question with Solution | TestHub

Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here, we consider the kinetics of such reactions.

Let a reaction be represented in general terms by the scheme

$\mathrm{A}\underset{{\mathrm{k}}_{-1}}{\stackrel{{\mathrm{k}}_1}{\rightleftharpoons }}B$

where $k_1$ and $k_{-1}$ represent the rate constants for the forward and reverse reactions, respectively. The equilibrium constant for this reaction may be written as:

$K=\frac{[B{]}_{\infty }}{[A{]}_{\infty }}=\frac{k_1}{k_{-1}}\dots (1)$

where the subscript $\infty$ refers to a time $t$, sufficiently long to establish equilibrium at the given temperature.

The initial concentration of species A is $[A{]}_0$, and that of B is $[B{]}_0$. After a time $t$, let the concentration of species A be $[A{]}_t$ and that of B be $[B{]}_t$. The total rate of change of $[A{]}_t$ is given by:

$\frac{d[A{]}_t}{dt}=-k_1[A{]}_t+k_{-1}[B{]}_t$

If, as is usual, $[B{]}_0$ is initially zero, it follows from a mass balance that at any time $t$, $[B{]}_t=[A{]}_0-[A{]}_t$, whence:

$\frac{d[A{]}_t}{dt}=-k_1[A{]}_t+k_{-1}([A{]}_0-[A{]}_t)$

$\frac{d[A{]}_t}{dt}=-(k_1+k_{-1})\left([A{]}_t-\frac{k_{-1}}{k_1+k_{-1}}[A{]}_0\right).\dots \left(2\right)$

From equation (1), we have:

$\frac{[B{]}_{\infty }}{[A{]}_{\infty }}=\frac{[A{]}_0-[A{]}_{}}{[A{]}_{\infty }}=\frac{k_1}{k_{-1}}$

0r

$[A{]}_{\infty }=\frac{[A{]}_0{k}_{-1}}{k_1+k_{-1}}$

Introducing this result into (2) we obtain:

$\frac{d[A{]}_t}{dt}=-(k_1+k_{-1})([A{]}_t-[A{]}_{\infty })$

Integrating

$\ln ([A{]}_t-[A{]}_{\infty })=-(k_1+k_{-1})t+f$

For $t=0$, $f=\ln ([A{]}_0-[A{]}_{\infty })$. Hence:

$\ln \left(\frac{[A{]}_t-[A{]}_{\infty }}{[A{]}_0-[A{]}_{\infty }}\right)=-(k_1+k_{-1})t$