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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

At \( 27^{\circ}C \) in presence of a catalyst, activation energy of a reaction is lowered by 10 kJ \( mol^{-1} \). The logarithm ratio of \( \frac{k(catalysed)}{k(uncatalysed)} \) is ....
(Consider that the frequency factor for both the reactions is same)

  1. A 17.41
  2. B 1.741
  3. C 3.482
  4. D 0.1741
Verified Solution

Answer & Solution

Correct Answer

(B) 1.741

Step-by-step Solution

Detailed explanation

\( \frac{K_{catalyst}}{K_{uncatalyst}} = e^{\frac{\Delta E_{a}}{RT}} \) \(\ln \frac{ K _{\text {catalyst }}}{ K _{\text {uncatalyst }}}=\frac{\Delta E _{ a }}{ RT }\) \(\log \frac{ K _{\text {catalyst }}}{ K _{\text {uncatalyst }}}=\frac{\Delta E _{ a }}{2.303 RT }\)…
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