COMEDK · Chemistry · 19. Chemical Kinetics
How much faster would a reaction proceed at \(25^{\circ} \mathrm{C}\) than at \(0^{\circ} \mathrm{C}\) if the activation energy is \(65 \mathrm{~kJ}\) ?
- A 4 times
- B 6 times
- C 12 times
- D 11 times
Answer & Solution
Correct Answer
(D) 11 times
Step-by-step Solution
Detailed explanation
We known, \(\begin{aligned} 2.303 \log \frac{k_2}{k_1} & =\frac{E_a}{R}\left[\frac{T_2-T_1}{T_1 T_2}\right] \\ .2 .303 \log \frac{k_2}{k_1} & =\frac{65 \times 10^3}{8.314}\left[\frac{25}{298 \times 273}\right] \\ \frac{k_2}{k} & =11.5 . \end{aligned}\) The reaction would proceed…
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