Question

How much faster would a reaction proceed at 25°C than at 0°C if the activation energy is 65 kJ?

• A. 2 times
• B. 16 times
• C. 11 times
• D. 6 times

Answer 1

Answer: (C)

$2.303log\frac{k_2}{k_1}=\frac{E_a}{R}\left[\frac{T_2-T_2}{T_1{T}_2}\right];$
$\therefore 2.303log\frac{k_2}{k_1}=\frac{65\times 10^3}{8.314}\left[\frac{298-273}{298\times 273}\right]$
$\therefore \frac{k_2}{k_1}=11.04\simeq 11$