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JEE Advanced levelSingle correct optionDifficulty 3 of 3bm128

For a reaction at $T = 310 K$ the uncatalysed activation energy is $100 kJ mo l^{- 1}$ and the activation energy with an enzyme is $40 kJ mo l^{- 1}$ . Assuming equal pre-exponential factors and $R = 8.314 J mo l^{- 1} K^{- 1}$ , the rate enhancement $k_{cat} / k_{uncat}$ is closest to

A $1.0$
B $1.3 \times 1 0^{10}$
C $2.5$
D $1.7 \times 1 0^{20}$

Answer: B

The enhancement is

exp (Δ E_{a} / RT)

with

Δ E_{a} = 60000 J mo l^{- 1}

. The exponent is

60000/ (8.314 \times 310) = 23.28

, so

e^{23.28} \approx 1.3 \times 1 0^{10}

. Using kJ directly gives about 1.0, the ratio of energies gives 2.5, and squaring gives

1.7 \times 1 0^{20}

.

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