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JEE Main levelSingle correct optionDifficulty 2 of 3bm118

An uncatalysed reaction has $E_{a} = 75 kJ mo l^{- 1}$ and the enzyme-catalysed reaction has $E_{a} = 50 kJ mo l^{- 1}$ . Assuming the same pre-exponential factor and $T = 300 K$ with $R = 8.314 J mo l^{- 1} K^{- 1}$ , the ratio $k_{cat} / k_{uncat}$ is closest to

A $1.0$
B $3.0$
C $5.1 \times 1 0^{8}$
D $2.3 \times 1 0^{4}$

Answer: D

With equal A,

k_{c a t} / k_{u n c a t} = exp (Δ E_{a} / RT)

with

Δ E_{a} = 25000 J mo l^{- 1}

. Then

25000/ (8.314 \times 300) = 10.02

and

e^{10.02} \approx 2.3 \times 1 0^{4}

. Forgetting to convert kJ to J gives about 1.0, and squaring the result gives

5.1 \times 1 0^{8}

.

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