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 308K. If the value of A is 4 \times ${10}^{10}{s}^{-1}$. Calculate K at 318 K and $E_a$

Answer 1

(a) Calculation of activation energy $\left(E_a\right)$

For 1st order reaction: $k=\frac{2.303}{t}log\frac{[A{]}_0}{\left[A\right]}$

At 298K; $k_1=\frac{2.303}{t}log\frac{100}{90}\dots \dots \left(i\right)$

At 308 K; $k_2=\frac{2.303}{t}log\frac{100}{75}\dots \dots \left(ii\right)$

Dividing eq. (ii) by (i);

$\frac{k_2}{k_1}=\frac{log\frac{100}{75}}{log\frac{100}{90}}=\frac{0.1249}{0.0458}=2.73$

According to Arrhenius theory;

$log\frac{k_2}{k_1}=\frac{E_a}{2.303R}\times \frac{T_2-T_1}{T_1{T}_2}$

$log2.73=\frac{E_a}{2.303R}\left[\frac{308-298}{298\times 308}\right]$

$E_a=\frac{0.4361\times 2.303\times \left(8.314Jmo{l}^{-1}\right)\times 298\times 308}{10}$

$E_a=76640Jmo{l}^{-1}=76.640kJmo{l}^{-1}$

(b) Calculation of rate constant (k)

According to Arrhenius equation

$logk=logA\frac{-E_a}{2.303RT}$

$logk=log(4\times {10}^{10})-\frac{76640Jmo{l}^{-1}}{2.303\times \left(8.314Jmo{l}^{-1}{K}^{-1}\right)\times \left(318K\right)}$

$logk=10.6021-12.5870=-1.9849$

k = Antilog (-1.9849)

$=Antilog\left(\overline{2}.0151\right)=1.035\times {10}^{-2}{s}^{-1}$

$E_a=76.640kJmo{l}^{-1}$

$k=1.035\times {10}^{-2}{s}^{-1}$