A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B(in s) : (Use ln 2 = 0.693)

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Main 2020 · 5 Sep · Shift 1 · Q2 with Solution

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Correct answer: D

If the initial concentrations of $A$ and $B$ are equal, let $[A]_0=[B]_0=C_0$

For first-order decomposition: $[A]=C_0 e^{-k_A t}, \qquad [B]=C_0 e^{-k_B t}$

We need the time when concentration of $A$ becomes four times that of $B$ : $[A]=4[B]$

Substitute: $C_0 e^{-k_A t}=4C_0 e^{-k_B t}$

Cancel $C_0$ : $e^{(k_B-k_A)t}=4$

Taking natural log: $(k_B-k_A)t=\ln 4=2\ln 2=2(0.693)=1.386$

For first-order kinetics: $k=\frac{\ln 2}{t_{1/2}}$

So, $k_A=\frac{0.693}{300}=0.00231\,\text{s}^{-1}$ $k_B=\frac{0.693}{180}=0.00385\,\text{s}^{-1}$

Thus, $k_B-k_A=0.00385-0.00231=0.00154\,\text{s}^{-1}$

$t=\frac{1.386}{0.00154}=900\,\text{s}$

$t=900\,\text{s}$ So the correct option is D.