NEET Chemistry Surface Chemistry Questions Solved

At 400K energy of activation of a reaction is decreased by 0.8 kcal in presence of catalyst. Hence rate will be

(A) Increased by 2.71 times

(B) Increased by 1.18 times

(D) Increased by 6.26 times

$(A) . K_{1} (no catalyst) = {Ae}^{- E_{a} / RT} K_{2} (catalyst) = {Ae}^{- E_{a} - 0.8 / RT} \frac{K_{2}}{K_{1}} = e^{0.8 / RT} = e^{0.8 / 0.002 \times 400} = 2.72$

Difficulty Level:

(1)
38%
(2)
33%
(3)
23%
(4)
8%