Question

In the pressence of a catalyst, the activation energy of a reaction is lowered by $2kcal$ at ${27}^{o}C$. The rate of reaction will increase by: $(Antilog1.43=28)$

• A. $2$ times
• B. $14$ times
• C. $28$ times
• D. $20$ times

Answer 1

Answer: (C)

$28$ times

$\frac{k_2}{k_1}$ = $e^{\frac{E_{a\left(2\right)}-E_{a\left(1\right)}}{RT}}$

$\frac{k_2}{k_1}$ = $e^{\frac{2000}{2\times 300}}$

$k_2=28\times k_1$

The Rate of reaction will increase by 28 times