In Arrhenius equation for a certain reaction, the values of A and ${\text{E}}_{\text{a}}$  (activation energy) are$\text{4}\times {10}^{13}{\sec }^{-1}$ and $98.6\text{KJ mo}{l}^{-1}$ respectively. The specific rate constant is$1.1\times {10}^{-3}{\text{s}}^{-1}$ then temperature is

According to Arrhenius equation,

$\text{k}=\text{A}{\text{.e}}^{-\text{Ea/RT}}$

Apply ln on both sides in the above equation.

$\ln k=\ln \text{A}-\frac{{\text{E}}_{\text{a}}}{\text{RT}}$

Now, convert ln into log by multiply with 2.303

$2.303{\text{ log}}_{10}^k=2.303{\text{ log}}_{10}^A-\frac{{\text{E}}_{\text{a}}}{\text{RT}}$

Substitute the given data in this equation to find temperature of a reaction.

$2.303\text{ log(1}\text{.1}\times {\text{10}}^{-3})=2.303\log (4\times {10}^{13})-\frac{98.6\times {10}^3}{8.314\times \text{T}}$

After solve the above equation we will get

$\therefore \text{T}=\frac{98.6\times {10}^3}{8.314\times 2.303\times 16.56}=310.96\text{ K}$