JEE PYQ: Chemical Kinetics Question 28
Question 28 - 2019 (08 Apr Shift 1)
For the reaction $2\text{A} + \text{B} \to \text{C}$, the values of initial rate at different reactant concentrations are given in the table below.
| $[\text{A}]$ (mol $\text{L}^{-1}$) | $[\text{B}]$ (mol $\text{L}^{-1}$) | Initial Rate (mol $\text{L}^{-1}\text{s}^{-1}$) |
|---|---|---|
| 0.05 | 0.05 | 0.045 |
| 0.10 | 0.05 | 0.090 |
| 0.20 | 0.10 | 0.72 |
The rate law for the reaction is:
(1) Rate $= k[\text{A}][\text{B}]^2$
(2) Rate $= k[\text{A}]^2[\text{B}]^2$
(3) Rate $= k[\text{A}][\text{B}]$
(4) Rate $= k[\text{A}]^2[\text{B}]$
Show Answer
Answer: (1)
Solution
Rate $= k[\text{A}]^x[\text{B}]^y$. Exp 1: $0.045 = k(0.05)^x(0.05)^y$. Exp 2: $0.090 = k(0.1)^x(0.05)^y$. Dividing: $\frac{0.045}{0.090} = \left(\frac{1}{2}\right)^x \Rightarrow x = 1$. Dividing Exp 1 by Exp 3: $\frac{0.045}{0.72} = \frac{0.05}{0.2} \cdot \left(\frac{0.05}{0.1}\right)^y$. $\frac{1}{16} = \frac{1}{4}\left(\frac{1}{2}\right)^y \Rightarrow y = 2$. Rate law $= k[\text{A}][\text{B}]^2$.
BETA
