In the Arrhenius equation for a certain reaction the value of A and ${\mathrm{E}}_{\mathrm{a}}$ (activation energy) are $4\times {10}^{13}{\sec }^{-1}$ and $98.6 \mathrm{kJ} {\mathrm{mol}}^{-1}$ respectively. If the reaction is of first order, the temperature at which its half-life period is 10 minutes is

Calculation of K 

We know that $\mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}}=\frac{0.693}{10\times 60} ({\mathrm{t}}_{1/2}=10\times 60\sec )$

                        $=1.1558\times {10}^{-3}$

According to Arrhenius equation 

$\log \mathrm{k}=\log \mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{RT}}$

Substituting the various values in the above equation 

$\log 1.155\times {10}^{-3}=\log 4\times {10}^{13}-\frac{98.6}{2.303\times 8.314\times {10}^{-3}\times \mathrm{T}}$

On usual calculation, T = 311.35 K