Question

Rate of two reactions whose rate constants are $k_{1}$ and $k_{2}$ are equal at $300 \,K$ such that :
$E_{a_{2}}-E_{a_{1}}=2 R T,$
So calculate $\ln \frac{A_{2}}{A_{1}}$

• A. $\ln 4$
• B. $2$
• C. $\log 2$
• D. $2-\ln 2$

Answer 1

Answer: (B)

$(\text { rate })_{1}=(\text { rate })_{2}$ and rate $\propto k$
Thus, $\frac{k_{2}}{k_{1}}=\frac{(\text { rate })_{2}}{(\text { rate })_{1}}=1$
$\ln k=\ln A-\frac{E_{a}}{R T}$
$\ln \frac{k_{2}}{k_{1}}=\ln \frac{A_{2}}{A_{1}}+\frac{E_{a_{2}}-E_{a_{1}}}{R T}$
$\ln 1=\ln \frac{A_{2}}{A_{1}}-\frac{2 R T}{R T} \ln \frac{A_{2}}{A_{1}}=2$