Question

Two reactions, (I) A $\to$ Products and (II) B $\to$ Products, follow first order kinetics. The rate of reaction (I) is doubled when the temperature is raised from 300 K to 310 K. The half life for this reaction at 310 K is 30 min. At the same temperature B decomposes twice as fast as A. If the energy of activation for reaction (II) is twice that of reaction (I), the rate constant ($inmi{n}^{-1}$) of reaction (II) at 300 K is $x\times {10}^{-2}$. Find $x$ (nearest integer).

Answer 1

For reaction (I)
$A\to$ Products
$T_1=300K,T_2=310K$
$(K_1{)}_A$ is at 300 K and $(K_2{)}_A$ is at 310 K
Given :
${\left[\frac{k_2\left(310K\right)}{k_1\left(300K\right)}\right]}_A=2$
Using Arrhenius equation for reaction (I):
$log{\left[\frac{k_2\left(310K\right)}{k_1\left(300K\right)}\right]}_A=\frac{E_a\left(A\right)}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$
$log2=\frac{E_a\left(A\right)}{2.303R}\left(\frac{R_2-T_1}{T_1{T}_2}\right)$
(For reaction II)
$B\to Products$
Since at 310 K, B decomposes twice as fast as A.
$\because \left[k_2\right(310K){]}_A=0.0231mi{n}^{-1}$
$\because \left[k_2\right(310K){]}_B=2\times 0.0231mi{n}^{-1}$
= 0.0462 $mi{n}^{-1}$
We have to calculate
$\left[k_1\right(300K){]}_B$ = ?
Using Arrhenius equation for reaction (II)
$log{\left[\frac{k_2\left(310K\right)}{k_1\left(300K\right)}\right]}_B=\frac{E_{a\left(B\right)}}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$
Given: $(E_a{)}_B=\frac{1}{2}(E_a{)}_A$
Substittute the value of equation (v) in equation (iv),
$log{\left[\frac{k_2\left(310K\right)}{k_1\left(300K\right)}\right]}_B=\frac{1}{2}\times \frac{E_{a\left(B\right)}}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$
Comparing equations (iii) and (vi), we get
${\left[\frac{k_2\left(310K\right)}{k_1\left(300K\right)}\right]}_B=\frac{1}{2}\times log2$
$\therefore \frac{0.0462}{\left[k_1\right(300K){]}_B}=(2{)}^{\frac{1}{2}}=\sqrt{2}=1.414$
$\therefore \left[k_1\right(300K){]}_B=\frac{0.0462}{1.414}=0.03267mi{n}^{-1}$

Hence answer will be 3.