Question

Activation energy $(E_a)$ and rate constant $(k_1$ and $k_2) $ of a chemical reaction at two different temperatures $(T_1$ and $T_2)$ are related by -

• A. $ln \frac{k_2}{k_1} = - \frac{E_a}{R} \bigg( \frac{1}{T_2} - \frac{1}{T_1} \bigg) $
• B. $ln \frac{k_2}{k_1} = - \frac{E_a}{R} \bigg( \frac{1}{T_2} + \frac{1}{T_1} \bigg) $
• C. $ln \frac{k_2}{k_1} = \frac{E_a}{R} \bigg( \frac{1}{T_1} - \frac{1}{T_2} \bigg) $
• D. $ln \frac{k_2}{k_1} = - \frac{E_a}{R} \bigg( \frac{1}{T_1} - \frac{1}{T_2} \bigg) $

Answer 1

Answer: (A), (C)

Ace. to Arrheneis equation
$\ell n \frac{K_2}{K_1} = \frac{E_a}{R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$
option (1) and (3) are same