JEE Main 2025 Jan 23 Shift 1, Chemistry Q24: Reactions of Various Orders

JEE Main2025Jan 23, Shift 1Chemistry

For the thermal decomposition of $N_{2} O_{5} (g)$ at constant volume, the following table can be formed, for the reaction mentioned below :

$2 N_{2} O_{5} (g) \to 2 N_{2} O_{4} (g) + O_{2} (g)$

S.No.	Time/s	Total pressure / (atm)
1.	0	0.6
2.	100	'x'

x = ________ $\times 1 0^{- 3}$ atm [nearest integer]

Given : Rate constant for the reaction is $4.606 \times 1 0^{- 2}$ s $^{- 1}$ .

Your answer

Solution

Let the partial pressure of $N_{2} O_{5}$ consumed in 100 s be $P$ . By stoichiometry, $N_{2} O_{4}$ formed $= P$ and $O_{2}$ formed $= P /2$ . Hence:

$P_{total} = (0.6 - P) + P + \frac{P}{2} = 0.6 + \frac{P}{2}$ .

Treating decomposition as first order in $N_{2} O_{5}$ with $k = 4.606 \times 1 0^{- 2}$ s $^{- 1}$ :

$k t = ln \frac{P _{0}}{P _{0} - P} = 2.303 lo g \frac{0.6}{0.6 - P}$

$(4.606 \times 1 0^{- 2}) (100) = 4.606 \Rightarrow lo g \frac{0.6}{0.6 - P} = 2 \Rightarrow \frac{0.6}{0.6 - P} = 100$ .

Solving: $0.6 - P = 0.006 \Rightarrow P = 0.594$ atm.

$P_{total} = 0.6 + \frac{0.594}{2} = 0.6 + 0.297 = 0.897$ atm $\approx 897 \times 1 0^{- 3}$ atm.

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JEE Main 2025 Jan 23 Shift 1, Chemistry Q24: Reactions of Various Orders

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