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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 \(65 \mathrm{~kJ}\) ?

  1. A 4 times
  2. B 6 times
  3. C 12 times
  4. D 11 times
Verified Solution

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…