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)
- A 17.41
- B 1.741
- C 3.482
- D 0.1741
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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