Q4 / 150 · MHT CET 2026 (20 April Shift 1) · Chemistry
In a reaction,
$2\text{N}_2\text{O}_{5(g)} \rightarrow 4\,\text{NO}_{2(g)} + \text{O}_{2(g)}$
$\text{N}_2\text{O}_5$ disappears at a rate of $0.06$ mol dm$^{-3}$ s$^{-1}$
Calculate rate of formation of $\text{O}_{2(g)}$ ?
$2\text{N}_2\text{O}_{5(g)} \rightarrow 4\,\text{NO}_{2(g)} + \text{O}_{2(g)}$
$\text{N}_2\text{O}_5$ disappears at a rate of $0.06$ mol dm$^{-3}$ s$^{-1}$
Calculate rate of formation of $\text{O}_{2(g)}$ ?
- A$0.02$ mol dm$^{-3}$ s$^{-1}$
- Correct answerB$0.03$ mol dm$^{-3}$ s$^{-1}$
- C$0.04$ mol dm$^{-3}$ s$^{-1}$
- D$0.06$ mol dm$^{-3}$ s$^{-1}$
Solution
For the given reaction $2\text{N}_2\text{O}_{5(g)} \rightarrow 4\text{NO}_{2(g)} + \text{O}_{2(g)}$
The rate of reaction is given by:
$\text{Rate} = -\dfrac{1}{2}\dfrac{d[\text{N}_2\text{O}_5]}{dt} = \dfrac{1}{4}\dfrac{d[\text{NO}_2]}{dt} = \dfrac{d[\text{O}_2]}{dt}$
Given the rate of disappearance of $\text{N}_2\text{O}_5$ is $-\dfrac{d[\text{N}_2\text{O}_5]}{dt} = 0.06 \text{ mol dm}^{-3} \text{ s}^{-1}$
The rate of formation of $\text{O}_2$ is:
$\dfrac{d[\text{O}_2]}{dt} = \dfrac{1}{2} \left(-\dfrac{d[\text{N}_2\text{O}_5]}{dt}\right)$
$\dfrac{d[\text{O}_2]}{dt} = \dfrac{1}{2} \times 0.06 = 0.03 \text{ mol dm}^{-3} \text{ s}^{-1}$
The rate of reaction is given by:
$\text{Rate} = -\dfrac{1}{2}\dfrac{d[\text{N}_2\text{O}_5]}{dt} = \dfrac{1}{4}\dfrac{d[\text{NO}_2]}{dt} = \dfrac{d[\text{O}_2]}{dt}$
Given the rate of disappearance of $\text{N}_2\text{O}_5$ is $-\dfrac{d[\text{N}_2\text{O}_5]}{dt} = 0.06 \text{ mol dm}^{-3} \text{ s}^{-1}$
The rate of formation of $\text{O}_2$ is:
$\dfrac{d[\text{O}_2]}{dt} = \dfrac{1}{2} \left(-\dfrac{d[\text{N}_2\text{O}_5]}{dt}\right)$
$\dfrac{d[\text{O}_2]}{dt} = \dfrac{1}{2} \times 0.06 = 0.03 \text{ mol dm}^{-3} \text{ s}^{-1}$
Hence, the correct answer is option: 'B'.