Question

A reactant (A) forms two products :
$E_{a_2}=2E_{a_1}$
Frequency factors for both the reactions are equal. Therefore,

• A. $k_2=k_1{e}^{-E_{a_1}/RT}$
• B. $k_2=k_1^2/A$
• C. $k_1=A{k}_2{e}^{E_{a_1}/RT}$
• D. Both (a) and (b) are correct

Answer 1

Answer: (D)

$k_2=A{e}^{-E_{a_2}/RT};k_1=A{e}^{-E_{a_1}/RT}$
$\ln k_2=\ln A-\frac{E_{a_2}}{RT}\&\ln k_1=\ln A-\frac{E_{a_1}}{RT}$
or $E_{a_2}=RT\ln A/k_2\&E_{a_1}=RT\ln \frac{A}{k_1}$
$\therefore 2RT\ln \frac{A}{k_1}=RT\ln \frac{A}{k_2}$
$\therefore 2\ln \left(\frac{A}{k_1}\right)=\ln \frac{A}{k_2}$
$\therefore {\left(\frac{A}{k_1}\right)}^2=\frac{A}{k_2}$
or $\frac{A^2}{k_1^2}=\frac{A}{k_2}$
$\therefore k_2=\frac{k_1^2}{A}$
And $k_2=A{e}^{-E_{a_2}/RT};k_1=A{e}^{-E_{a_1}/RT}$
$\frac{k_2}{k_1}=e^{E_{a_1}-E_{a_2}/RT}=e^{-E_{a_1}/RT}$
or $k_2=k_1{e}^{-E_{a_1}/RT}$