The time required for $10 \%$ completion of a first order reaction at $298 \,K$ is equal to that required for its $25 \%$ completion at $308 \,K$. If the value of $A$ is $4 \times 10^{10} \,s^{-1}$,calculate $k$ at $318 \,K$ and $E_a$.

(N/A) For a first order reaction,$t = \frac{2.303}{k} \log \frac{a}{a-x}$.
At $298 \,K$,$t = \frac{2.303}{k} \log \frac{100}{90} = \frac{0.1054}{k}$.
At $308 \,K$,$t' = \frac{2.303}{k'} \log \frac{100}{75} = \frac{0.2877}{k'}$.
Given $t = t'$,so $\frac{0.1054}{k} = \frac{0.2877}{k'}$,which gives $\frac{k'}{k} = 2.7296$.
Using the Arrhenius equation: $\log \frac{k'}{k} = \frac{E_a}{2.303 \,R} \left( \frac{T' - T}{T \,T'} \right)$.
$\log (2.7296) = \frac{E_a}{2.303 \times 8.314} \left( \frac{308 - 298}{298 \times 308} \right)$.
$E_a = \frac{2.303 \times 8.314 \times 298 \times 308 \times \log (2.7296)}{10} = 76640.1 \,J \,mol^{-1} = 76.64 \,kJ \,mol^{-1}$.
To calculate $k$ at $318 \,K$ using $\log k = \log A - \frac{E_a}{2.303 \,R \,T}$:
$\log k = \log (4 \times 10^{10}) - \frac{76640.1}{2.303 \times 8.314 \times 318} = 10.6021 - 12.5876 = -1.9855$.
$k = \text{antilog}(-1.9855) = 1.034 \times 10^{-2} \,s^{-1}$.