Question

Two reaction (i) $A\to Products$, (ii) $B\to Products$, follow first order kinetics. The rate of the reaction (i) is doubled when temperature is raised from 300 K to 310 K. The half-life for this reaction at 310 K is 30 minute. At the same temperature B decomposes twice as fast as A. If the energy of activation for the reaction (ii) is half that for reaction (i). Calculate the rate constant of reaction (ii) at 300 K. If rate constant is $x\times {10}^{-3}$ then answer would be $x$ (nearest integer).

Answer 1

The given changes are;

I $A\to Products,t_{1/2}$ at $310K=30minute$

II $B\to Products$

$\therefore K_1$ at $310K=\left(0.693/30\right)$

$=0.0231minut{e}^{-1}$ .....(i)

$\because rate=K\left[\right]$; both reactions are of Ist order

Also given, $\frac{K_{1}at310}{K_{1}at300}=2$ ....(ii)

and $\frac{K_{1}at310}{K_{1}at310}=2$ ....(iii)

Also given that, $\frac{E_{a_2}}{E_{a_1}}=\frac{1}{2}$ ......(iv))

For I: $2.303log\frac{K_{1}at310}{K_{1}at300}=\frac{E_{a_1}}{R}\left[\frac{310-300}{310\times 300}\right]$ .....(v)

For II: $2.303log\frac{K_{2}at310}{K_{2}at300}=\frac{E_{a_2}}{R}\left[\frac{310-300}{310\times 300}\right]$ .....(vi)

On dividing (v) by eq. (vi)

$\frac{log\frac{K_{1}at310}{K_{1}at300}}{log\frac{K_{2}at310}{K_{2}at300}}=\frac{E_{a_1}}{E_{a_2}}=2$ ....(vii)

or $log\frac{K_{1}at310}{K_{1}at300}=2log\frac{K_{2}at310}{K_{2}at300}$

$=log{\left[\frac{K_{2}at310}{K_{2}at300}\right]}^2$ .....(viii)

By eqs. (ii) and (viii)

${\left[\frac{K_{2}at310}{K_{2}at300}\right]}^2=2$

or $K_2$ at $310=\sqrt{2}{K}_2$ at 300 ..... (ix)

By eqs. (iii) and (ix)

$2K_1$ at $310=\sqrt{2}{K}_2$ at 300

or $K_2$ at $300=\sqrt{2}{K}_1$ at 310 .....(x)

By eqs. (i) and (x)

$K_2$ at $300=\sqrt{2}\times 0.0231$

$=3.27\times {10}^{-2}mi{n}^{-1}$