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JEE Main levelSingle correct optionDifficulty 2 of 3ck220

A catalyst lowers the activation energy of a reaction from $80.0$ to $60.0 kJ mo l^{- 1}$ without changing the pre-exponential factor. At $300 K$ ( $R = 8.314 J mo l^{- 1} K^{- 1}$ ), the rate constant increases by a factor of about

A $5.5 \times 1 0^{1}$
B $0.25$
C $9.2 \times 1 0^{6}$
D $3.0 \times 1 0^{3}$

Answer: D

k_{cat} / k = exp (Δ E_{a} / RT)

with

Δ E_{a} = 20000 J mo l^{- 1}

. The exponent is

20000/ (8.314 \times 300) = 8.019

, so the factor is

e^{8.019} = 3.0 \times 1 0^{3}

.

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