Question

The activation energy for a reaction is $9.0kcal/mol$. The increase in the rate constant when its temperature is increased from $298K$ to $308K$ is:

• A. $63$%
• B. $50$%
• C. $100$%
• D. $10$%

Answer 1

The correct option is A $63$%

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

$\log \frac{K_2}{K_1}=\frac{9.0\times {10}^3}{2.303\times 2}\left[\frac{308-298}{308\times 298}\right]$

$\frac{K_2}{K_1}=1.63;K_2=1.63K_1;\frac{1.63K_1-K_1}{K_1}\times 100=63.0\%$