JEE PYQ: Chemical Kinetics Question 2
Question 2 - 2021 (16 Mar Shift 2)
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).
Show Answer
Answer: 108
Solution
Given $t_{1/2}$ of A = 54 min, $t_{1/2}$ of B = 18 min. For first order kinetics: $[A]_t = \frac{A_0}{(2)^n}$ where $n$ = number of half-lives. We need $[A]_t = 16 \times [B]_t$. $\frac{x}{(2)^{t/54}} = \frac{x}{(2)^{t/18}} \times 16$. $(2)^{t/18} = (2)^{t/54} \times (2)^4$. $\frac{t}{18} = \frac{t}{54} + 4$. $t\left(\frac{1}{18} - \frac{1}{54}\right) = 4$. $t = \frac{4 \times 18 \times 54}{36} = 108$ min.
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