Question

How much faster would a reaction proceed at ${25}^{\circ }C$ than at $0^{\circ }C$ if the activation energy is 65 kJ?

• A. 11 times
• B. 9 times
• C. 6 times
• D. 2 times

Answer 1

The correct option is A

11 times

$2.303log\frac{K_2}{K_1}=\frac{Ea}{R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

$\therefore 2.303log\frac{K_2}{K_1}=\frac{65\times {10}^3}{8.314}\left[\frac{25}{298\times 273}\right]$

$\therefore \frac{K_2}{K_1}=11.05$