Question

The rate of the chemical reaction doubles for an increase of $10K$ in absolute temperature from $298K$ . Calculate $E_a$

• A. 52.988 kJ
• B. 42.024 kJ
• C. 52.898 kJ
• D. 51.898 kJ

Answer 1

Answer: (C)

$\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
$\frac{k_2}{k_1}=2,T_1=298K,T_2=308K$
$R=8.314J{K}^{-1}mo{l}^{-1}$
$\log 2=\frac{E_a}{2.303\times 8.314}\times \left[\frac{308-298}{308\times 298}\right]$
$0.3010=\frac{E_a}{2.303\times 8.314}\times \frac{10}{298\times 308}$
$\therefore E_a=\frac{0.3010\times 2.303\times 8.314\times 298\times 308}{10}$
$=52.898kJ$