Question

In presence of a catalyst, the activation energy is lowered by 2 kcal at $127^{\circ} C$. The rate as compared to old rate

• A. $e ^{2}$ times
• B. $e ^{-2}$ times
• C. $e ^{2.5}$ times
• D. $e ^{-4.5}$ times

Answer 1

Answer: (C)

$K _{1}= Ae ^{- E _{a_{1}} / RT }, K _{2}= Ae ^{- E _{a_{2}} / RT }$

$\left.\frac{K_{1}}{K_{2}}=e^{\left(-E_{a_{1}}-E_{a_{2}}\right) / R T}=e^{-\left[\left(2 \times 10^{3}\right) /(2 \times 400)\right.}\right]$

$\frac{K_{1}}{K_{2}}=e^{-2.5}=\frac{r_{1}}{r_{2}}$

$\therefore \frac{ r _{1}}{ r _{2}}= e ^{2.5}$