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.
Question:

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 is 4 × 1010 s−1. Calculate at 318 K and Ea.

Solution:

For a first order reaction,

$t=\frac{2.303}{k} \log \frac{a}{a-x}$

At $298 \mathrm{~K}, t=\frac{2.303}{k} \log \frac{100}{90}$

$=\frac{0.1054}{k}$

At $308 \mathrm{~K}, t^{\prime}=\frac{2.303}{k^{\prime}} \log \frac{100}{75}$

$=\frac{2.2877}{k^{\prime}}$

According to the question,

$t=t^{\prime}$

$\Rightarrow \frac{0.1054}{k}=\frac{0.2877}{k^{\prime}}$

$\Rightarrow \frac{k^{\prime}}{k}=2.7296$

From Arrhenius equation, we obtain

$\log \frac{k^{\prime}}{k}=\frac{E_{a}}{2.303 \mathrm{R}}\left(\frac{T^{\prime}-T}{T T^{\prime}}\right)$

$\log (2.7296)=\frac{E_{a}}{2.303 \times 8.314}\left(\frac{308-298}{298 \times 308}\right)$

$E_{a}=\frac{2.303 \times 8.314 \times 298 \times 308 \times \log (2.7296)}{308-298}$

$=76640.096 \mathrm{~J} \mathrm{~mol}^{-1}$

$=76.64 \mathrm{~kJ} \mathrm{~mol}^{-1}$

To calculate at 318 K,

It is given that, $A=4 \times 10^{10} \mathrm{~s}^{-1}, T=318 \mathrm{~K}$

Again, from Arrhenius equation, we obtain

$\log k=\log A-\frac{E_{a}}{2.303 \mathrm{R} T}$

$=\log \left(4 \times 10^{10}\right)-\frac{76.64 \times 10^{3}}{2.303 \times 8.314 \times 318}$

$=(0.6021+10)-12.5876$

$=-1.9855$

Therefore, $k=\operatorname{Antilog}(-1.9855)$

$=1.034 \times 10^{-2} \mathrm{~s}^{-1}$