JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Chemical Kinetics

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A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolar non reactive mixture of A and B, the time taken for the concentration of

A

to become 16 times that of

B

is ______ min.
(Round off to the nearest integer).

Answer: 108

Solution

Both decays are first order, so

[A] = [A]_{0} e^{- k_{A} t}

and

[B] = [B]_{0} e^{- k_{B} t}

, and with equal starting concentrations

\frac{[ A ]}{[ B ]} = e^{(k_{B} - k_{A}) t}

. The rate constants are

k_{A} = \frac{ln 2}{54}

and

k_{B} = \frac{ln 2}{18}

, so

k_{B} - k_{A} = ln 2 (\frac{1}{18} - \frac{1}{54}) = \frac{ln 2}{27}

. Setting the ratio equal to 16,

\frac{ln 2}{27} t = ln 16 = 4 ln 2

so the time taken is 108 min.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer

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