Question

The rate constant $ (k_{1}) $ of one of the reaction is found to be double that of the rate constant $ (k_{2}) $ of another reaction. The relationship between the corresponding activation energies of the two reactions $ E_{a1} $ and $ E_{a2} $ will be

• A. $ E_{a1} <\, E_{a2} $
• B. $ E_{a1} >\, E_{a2} $
• C. $ E_{a1} = E_{a2} $
• D. $ E_{a1} = 2E_{a2} $

Answer 1

Answer: (A)

From Arrhenius equation,
$K=Ae^{-E_{a}/RT}$, on taking log, we get
$log\,k=log\,A-\frac{E_{a}}{2.303\,RT}$
For reaction $I, log K_{1}=log\,A-\frac{E_{a_{1}}}{2.303\,RT} \dots(i)$
For reaction $II, log\,K_{2}=log\,A-\frac{E_{a_2}}{2.303\,RT} \dots\left(ii\right)$
Eq. (i) - Eq. (ii), we get
$log\,k_{1}-log\,k_{2}=log\,A-\frac{E_{a{1}}}{2.303RT}$
$-[log\,A-\frac{E_{a_{2}}}{2.303\,RT}]$
$log \frac{k_{1}}{k_{2}}=-\frac{E_{a_1}}{2.303\,RT}+\frac{E_{a_2}}{2.303\,RT}$
$log\,2=\frac{1}{2.303\,RT}\left(-E_{a_1}+E_{a_2}\right)$ or $E_{a_2}=E_{a_1}+2.303\,RT \,log\,2$
Thus, $E_{a_2} >\,E_{a_1}$