Question

A reactant (A) forms two products :
$E_{ a _{2}}=2 E_{ a _{1}}$
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 2 RT \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 } $