JEE PYQ: Chemical Kinetics Question 34
Question 34 - 2019 (09 Jan Shift 1)
The following results were obtained during kinetic studies of the reaction: $2\text{A} + \text{B} \to$ Products
| Experiment | $[\text{A}]$ (in mol $\text{L}^{-1}$) | $[\text{B}]$ (in mol $\text{L}^{-1}$) | Initial Rate of reaction (in mol $\text{L}^{-1}\text{min}^{-1}$) |
|---|---|---|---|
| I | 0.10 | 0.20 | $6.93 \times 10^{-3}$ |
| II | 0.10 | 0.25 | $6.93 \times 10^{-3}$ |
| III | 0.20 | 0.30 | $1.386 \times 10^{-2}$ |
The time (in minutes) required to consume half of A is:
(1) 5
(2) 10
(3) 1
(4) 100
Show Answer
Answer: (1)
Solution
From experiments I and II, the order of reaction w.r.t. B is zero. From experiments II and III, $\alpha$ can be calculated: $\frac{1.386 \times 10^{-2}}{6.93 \times 10^{-3}} = \left(\frac{0.2}{0.1}\right)^\alpha$, so $\alpha = 1$. Rate $= K[\text{A}]^1$. $6.93 \times 10^{-3} = K(0.1)$, $K = 6.93 \times 10^{-2}$. For the reaction $2\text{A} + \text{B} \to$ Products: $t_{1/2} = \frac{0.693}{2K} = \frac{0.693}{0.693 \times 10^{-1} \times 2} = 5$ min.
BETA
