Question

Derive integrated rate equation for second order reaction

Answer 1

Answer:

The integrated rate equation for the second-order reaction is $\[\frac{1}{{{{[R]}_t}}} - \frac{1}{{{{[R]}_0}}} = kt\]$

Explanation:

Let us assume a second-order reaction:

$R\rightarrow P$

The concentration of reactant R at time t = 0 is $[R]_0$ and at t = t is $[R]_t$.

The differential rate law of this reaction in terms of the reactant can be written as:

$\frac{{ - d[R]}}{{dt}} = k{[R]^2} \hfill$

Now rearranging the differential equation to integrate it. Therefore,

$\frac{{ - d[R]}}{{{{[R]}^2}}} = kdt \hfill$

$\frac{{d[R]}}{{{{[R]}^2}}} = - kdt \hfill$

Integrating both sides, taking limit from $[R]_0$ to $[R]_t$ on the left side and from 0 to t on the right side.

$\int\limits_{{{[R]}_0}}^{{{[R]}_t}} {\frac{{d[R]}}{{{{[R]}^2}}}} = - k\int\limits_0^t {dt} \hfill$

Using the power rule of integration,

$\left[ { - \frac{1}{R}} \right]_{[R]0}^{{{[R]}_t}} = - k[t]_0^t \hfill$

$\frac{1}{{{{[R]}_t}}} - \frac{1}{{{{[R]}_0}}} = kt \hfill$

Where,

$[R]_0$  is the initial amount of reactant

$[R]_t$ is the amount of reactant after time t

k is the rate constant of the reaction

t is time

Therefore, integrating the rate equation for the second-order reaction is $\frac{1}{{{{[R]}_t}}} - \frac{1}{{{{[R]}_0}}} = kt \hfill$