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,
$K=A{e}^{-E_a/RT}$, on taking log, we get
$logk=logA-\frac{E_a}{2.303RT}$
For reaction $I,log{K}_1=logA-\frac{E_{a_1}}{2.303RT}\dots (i)$
For reaction $II,log{K}_2=logA-\frac{E_{a_2}}{2.303RT}\dots \left(ii\right)$
Eq. (i) - Eq. (ii), we get
$log{k}_1-log{k}_2=logA-\frac{E_{a1}}{2.303RT}$
$-[logA-\frac{E_{a_2}}{2.303RT}]$
$log\frac{k_1}{k_2}=-\frac{E_{a_1}}{2.303RT}+\frac{E_{a_2}}{2.303RT}$
$log2=\frac{1}{2.303RT}\left(-E_{a_1}+E_{a_2}\right)$ or $E_{a_2}=E_{a_1}+2.303RTlog2$
Thus, $E_{a_2}>E_{a_1}$