Question

A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction.

(Given : log 2 = 0.3010, log 4 = 0.6021, R = 8.314 $J{K}^{-1}mo{l}^{-1})$

Answer 1

K = $\frac{0.693}{t_{\frac{1}{2}}},$ So

$K_1=\frac{0.693}{40}$ and $K_2=\frac{0.693}{20}$

$log\frac{K_2}{K_1}=\frac{E_a}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

$log\frac{\frac{0.693}{20}}{\frac{0.693}{40}}=\frac{E_a}{2.303\times 8.314}[\frac{1}{300}-\frac{1}{320}]$

$log2=\frac{E_a}{19.24}\left[\frac{20}{320\times 300}\right]$

0.3010 = $\frac{E_a}{19.24}\times \frac{1}{4800}$

$E_a=0.3010\times 19.24\times 4800$

=27663.8 J $mo{l}^{-1}$ = 27.7 kJ/mol