At in presence of a catalyst, activation energy of a reaction is lowered by . The logarithm of ratio of is... (Consider that the frequency factor for both the reactions is same) [2026]

Question

At $27$ ${}^{o}C$ in presence of a catalyst, activation energy of a reaction is lowered by $10 {kJ mol}^{- 1}$ . The logarithm of ratio of $\frac{k (catalysed)}{k (uncatalysed)}$ is...
(Consider that the frequency factor for both the reactions is same) [2026]

Smart Achievers · Accepted Answer

(3)

$\frac{K_{catalyst}}{K_{uncatalyst}} = e^{\frac{Δ E_{a}}{R T}}$

$\ln (\frac{K_{catalyst}}{K_{uncatalyst}}) = \frac{Δ E_{a}}{R T}$

$\log (\frac{K_{catalyst}}{K_{uncatalyst}}) = \frac{Δ E_{a}}{2.303 R T}$

$= \frac{10 \times 1000}{2.303 \times 8.314 \times 300}$

$\log (\frac{K_{catalyst}}{K_{uncatalyst}}) = 1.741$

Solution

Q.
At $27$ ${}^{o}C$ in presence of a catalyst, activation energy of a reaction is lowered by $10 {kJ mol}^{- 1}$ . The logarithm of ratio of $\frac{k (catalysed)}{k (uncatalysed)}$ is...
(Consider that the frequency factor for both the reactions is same) [2026]

1 3.482

2 17.41

3 1.741

4 0.1741

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