Question

The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be : (R = 8.314 JK–1mol–1 and log 2 = 0.301)

• A. $\text{53.6 kJ mo}\text{l}^{\text{-1}}$
• B. $\text{48.6 kJ mo}\text{l}^{\text{-1}}$
• C. $\text{58.5 kJ mo}\text{l}^{\text{-1}}$
• D. $\text{60.5 kJ mo}\text{l}^{\text{-1}}$

Answer 1

Answer: (A)

$\text{53.6 kJ mo}\text{l}^{\text{-1}}$

Answer 2

$\log\frac{K_{2}}{K_{1}} = \frac{- E_{a}}{2.303R}\left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right)$

$\frac{K_{2}}{K_{1}} = 2;T_{2} = 310K T_{1} = 300K \Rightarrow \log{}2 = \frac{- E_{a}}{2.303 \times 8.134}\left( \frac{1}{310} - \frac{1}{300} \right) \Rightarrow E_{a} = \text{53598.6J/mol = 53.6 KJ/mol}$Ans is (1)