Q27 / 150 · MHT CET 2026 (16 April Shift 1) · Chemistry
The rate constant for the reaction $2\text{N}_2\text{O}_5 \rightarrow 4\text{NO}_2 + \text{O}_2$ is $3.0 \times 10^{-5}\ \text{s}^{-1}$. If the rate is $2.4 \times 10^{-5}\ \text{mole L}^{-1}\ \text{s}^{-1}$ then the concentration of $\text{N}_2\text{O}_5\,(\text{mol L}^{-1})$ is
- A$1.4$
- B$1.2$
- C$0.04$
- Correct answerD$0.8$
Solution
From the unit of the rate constant ($\text{s}^{-1}$), we can determine that the given reaction is a first-order reaction.
For a first-order reaction, the rate law is expressed as:
$\text{Rate} = k[\text{N}_2\text{O}_5]$
Given values are:
$\text{Rate} = 2.4 \times 10^{-5} \text{ mole L}^{-1} \text{s}^{-1}$
$k = 3.0 \times 10^{-5} \text{ s}^{-1}$
Substituting these values into the rate equation:
$2.4 \times 10^{-5} = 3.0 \times 10^{-5} \times [\text{N}_2\text{O}_5]$
Solving for the concentration of $\text{N}_2\text{O}_5$:
$[\text{N}_2\text{O}_5] = \dfrac{2.4 \times 10^{-5}}{3.0 \times 10^{-5}}$
$[\text{N}_2\text{O}_5] = \dfrac{2.4}{3.0}$
$[\text{N}_2\text{O}_5] = 0.8 \text{ mol L}^{-1}$
Answer: $0.8$
For a first-order reaction, the rate law is expressed as:
$\text{Rate} = k[\text{N}_2\text{O}_5]$
Given values are:
$\text{Rate} = 2.4 \times 10^{-5} \text{ mole L}^{-1} \text{s}^{-1}$
$k = 3.0 \times 10^{-5} \text{ s}^{-1}$
Substituting these values into the rate equation:
$2.4 \times 10^{-5} = 3.0 \times 10^{-5} \times [\text{N}_2\text{O}_5]$
Solving for the concentration of $\text{N}_2\text{O}_5$:
$[\text{N}_2\text{O}_5] = \dfrac{2.4 \times 10^{-5}}{3.0 \times 10^{-5}}$
$[\text{N}_2\text{O}_5] = \dfrac{2.4}{3.0}$
$[\text{N}_2\text{O}_5] = 0.8 \text{ mol L}^{-1}$
Answer: $0.8$
Hence, the correct answer is option: 'D'.