JEE Main 2022 · Chemistry · Chemical Kinetics · medium

A compound A decomposes by two parallel first-order paths: A→ B (k1=0.04 min^-1) and A→ C (k2=0.06 min^-1). Starting from 1.0 mol of A, the amount of…

Numerical answer

A compound A decomposes by two parallel first-order paths:

\mathrm{A\to B}

(

k_1=0.04\ \mathrm{min^{-1}}

) and

\mathrm{A\to C}

(

k_2=0.06\ \mathrm{min^{-1}}

). Starting from

1.0

mol of A, the amount of C (in millimoles) formed after

13.86

min (

\ln2=0.693

) is

Show answer and solution

Answer: 450

Both paths consume A with an overall first-order constant

k=k_1+k_2=0.10\ \mathrm{min^{-1}}

. At

t=13.86

min,

kt=1.386=\ln4

, so A remaining

=1/4

mol and A consumed

=0.75

mol. The products appear in the ratio of rate constants: fraction going to C is

k_2/(k_1+k_2)=0.6

. Hence

\mathrm C=0.75\times0.6=0.45

mol

=450

mmol.

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