JEE Main · 2012 · 29 Apr 2012 · Offline · Chemical Kinetics

For a first order reaction,

(A) \to

products, the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of

A

is 0.01 M is :

Answer: (B)

Solution

For a first-order reaction

k = \frac{2.303}{t} lo g \frac{[ A ] _{0}}{[ A ]}

. The fall from

0.1

M to

0.025

M is a drop to one quarter, that is two half-lives, over

40

minutes, so

k = \frac{2.303}{40} lo g 4 \approx 0.0347 mi n^{- 1}

. The rate at the moment when

[A] = 0.01

M is

k [A] = 0.0347 \times 0.01 \approx 3.47 \times 1 0^{- 4} M/min

JEE Main · 2012 · 29 Apr 2012 · Offline · Single correct option