An SN2 reaction of an alkyl halide RX with an anionic nucleophile in methanol has rate = k[RX][Nu-] and an initial rate of 3.0x10^-5 mol L^-1 s^-1 when [RX] = 0.10 M and [Nu-] = 0.20 M. The experiment is repeated with [RX] = 0.30 M and [Nu-] = 0.50 M in DMF, where the rate constant k is 20 times larger than in methanol. By what factor is the new initial rate larger than the original rate? Give your answer as an integer.

JEE Main levelNumerical answerDifficulty 2 of 3o3131

An SN2 reaction of an alkyl halide $RX$ with an anionic nucleophile in methanol has rate $= k [RX] [Nu X^{-}]$ and an initial rate of $3.0 \times 1 0^{- 5} mol L^{- 1} s^{- 1}$ when $[RX] = 0.10 M$ and $[Nu X^{-}] = 0.20 M$ . The experiment is repeated with $[RX] = 0.30 M$ and $[Nu X^{-}] = 0.50 M$ in DMF, where the rate constant $k$ is 20 times larger than in methanol. By what factor is the new initial rate larger than the original rate? Give your answer as an integer.

Numerical answer type, tolerance 0.5.

The rate is proportional to

k [RX] [Nu X^{-}]

. The factor is

20 \times (0.30/0.10) \times (0.50/0.20) = 20 \times 3 \times 2.5 = 150

. (The original value of

k

3.0 \times 1 0^{- 5} /0.020 = 1.5 \times 1 0^{- 3} L mo l^{- 1} s^{- 1}

and the new rate is

4.5 \times 1 0^{- 3} mol L^{- 1} s^{- 1}

.) The answer is 150.