JEE Advanced 2024 Paper 2 · Chemistry · Chemical Kinetics · medium

A substance A decomposes by two parallel first-order paths, Axrightarrowk1B and Axrightarrowk2C, with k1=3k2. The half-life of A is 10 min, and…

Numerical answer

A substance

\mathrm{A}

decomposes by two parallel first-order paths,

\mathrm{A\xrightarrow{k_1}B}

and

\mathrm{A\xrightarrow{k_2}C}

, with

k_1=3k_2

. The half-life of

\mathrm{A}

10

min, and

[\mathrm{A}]_0=1.00\ \mathrm{M}

with no product initially. The time (in minutes) at which

[\mathrm{C}]

becomes

0.1875\ \mathrm{M}

Show answer and solution

Answer: 20

For parallel first-order reactions the overall rate constant is

k=k_1+k_2=4k_2

, and A decays as

[\mathrm{A}]=[\mathrm{A}]_0e^{-kt}

with

k=\ln2/10\ \mathrm{min^{-1}}

. The products form in the ratio of rate constants, so

[\mathrm{C}]=\dfrac{k_2}{k_1+k_2}\left([\mathrm{A}]_0-[\mathrm{A}]\right)=\dfrac14\left(1-e^{-kt}\right)

. Setting

0.1875=\tfrac14(1-e^{-kt})

gives

1-e^{-kt}=0.75

e^{-kt}=0.25=2^{-2}

, so

kt=2\ln2

. Therefore

t=2\times10=20

min (two half-lives of A). Check: after two half-lives, 0.75 M of A has reacted and one quarter of it, 0.1875 M, is C.

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