JEE PYQ: P Block Elements Question 62
Question 62 - 2019 (12 Apr Shift 2)
$\text{NO}_2$ required for a reaction is produced by the decomposition of $\text{N}_2\text{O}_5$ in $\text{CCl}_4$ as per the equation, $2\text{N}_2\text{O}_5(g) \to 4\text{NO}_2(g) + \text{O}_2(g)$.
The initial concentration of $\text{N}_2\text{O}_5$ is 3.00 mol L$^{-1}$ and it is 2.75 mol L$^{-1}$ after 30 minutes. The rate of formation of $\text{NO}_2$ is:
(1) $4.167 \times 10^{-3}$ mol L$^{-1}$ min$^{-1}$
(2) $1.667 \times 10^{-2}$ mol L$^{-1}$ min$^{-1}$
(3) $8.333 \times 10^{-3}$ mol L$^{-1}$ min$^{-1}$
(4) $2.083 \times 10^{-3}$ mol L$^{-1}$ min$^{-1}$
Show Answer
Answer: (2)
Solution
Rate of reaction $= \frac{-1}{2}\frac{d[\text{N}_2\text{O}_5]}{dt} = \frac{1}{4}\frac{d[\text{NO}_2]}{dt}$. $\frac{-d[\text{N}_2\text{O}_5]}{dt} = \frac{(2.75 - 3)}{30} = \frac{0.25}{30}$ M min$^{-1}$. $\frac{d[\text{NO}_2]}{dt} = 2 \times \frac{-d[\text{N}_2\text{O}_5]}{dt} = 2 \times \frac{0.25}{30} = 1.67 \times 10^{-2}$ M min$^{-1}$.
BETA
