JEE PYQ: Chemical Kinetics Question 12
Question 12 - 2020 (02 Sep Shift 2)
The results given in the below table were obtained during kinetic studies of the following reaction:
$2\text{A} + \text{B} \to \text{C} + \text{D}$
| Experiment | $[\text{A}]/\text{mol L}^{-1}$ | $[\text{B}]/\text{mol L}^{-1}$ | Initial rate / $\text{mol L}^{-1}\text{min}^{-1}$ |
|---|---|---|---|
| I | 0.1 | 0.1 | $6.00 \times 10^{-3}$ |
| II | 0.1 | 0.2 | $2.40 \times 10^{-2}$ |
| III | 0.2 | 0.1 | $1.20 \times 10^{-2}$ |
| IV | X | 0.2 | $7.20 \times 10^{-2}$ |
| V | 0.3 | Y | $2.88 \times 10^{-1}$ |
X and Y in the given table are respectively:
(1) 0.4, 0.4
(2) 0.4, 0.3
(3) 0.3, 0.4
(4) 0.3, 0.3
Show Answer
Answer: (3)
Solution
Rate $= k[\text{A}]^a[\text{B}]^b$. From Exp I and II: $b = 2$. From Exp I and III: $a = 1$. From Exp II and IV: $\frac{24.0 \times 10^{-3}}{72.0 \times 10^{-3}} = \frac{k[0.1]^a[0.2]^b}{k[X]^a[0.2]^b}$. $\frac{1}{3} = \frac{0.1}{X}$, so $X = 0.3$. From Exp I and V: $\frac{6.0 \times 10^{-3}}{288 \times 10^{-3}} = \frac{(0.1)^1}{(0.3)^1} \cdot \frac{(0.1)^2}{Y^2}$. $\frac{1}{48} = \frac{1}{3} \cdot \frac{0.01}{Y^2}$. $Y = 0.4$.
BETA
