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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2025 · 23 January · Morning Shift · Chemistry · Chemical Kinetics

For the thermal decomposition of N_2 O_5( g) at constant volume,...

For the thermal decomposition of N2O5(g)\mathrm{N}_2 \mathrm{O}_5(\mathrm{ g}) at constant volume, the following table can be formed, for the reaction mentioned below. 2N2O5(g)→2N2O4(g)+O2(g)x=…×10−3atm [nearest integer] Given:Rateconstantforthereactionis2 \mathrm{ N}_2 \mathrm{O}_5(\mathrm{ g}) \rightarrow 2 \mathrm{ N}_2 \mathrm{O}_4(\mathrm{ g})+\mathrm{O}_2(\mathrm{ g}) \mathrm{x}=\ldots \times 10^{-3} \mathrm{ atm} \text{ [nearest integer] } Given : Rate constant for the reaction is4.606 \times 10^{-2} s−1\mathrm{ s}^{-1}.

Correct answer

0

Step-by-step solution

2 N2O5( g)→2 N2O4( g)+O2( g)2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g}) \rightarrow 2 \mathrm{~N}_2 \mathrm{O}_4(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})k=2.303tlog⁡0.9−0.6(0.9−x)2×10−2×100=log⁡0.3(0.9−x)100=0.3(0.9−x)=0.9−x0.3=0.010.9−X=0.003=897×10−3\begin{aligned} & k=\frac{2.303}{t} \log \frac{0.9-0.6}{(0.9-x)} \\ & 2 \times 10^{-2} \times 100=\log \frac{0.3}{(0.9-x)} \\ & 100=\frac{0.3}{(0.9-x)} \\ & =\frac{0.9-x}{0.3}=0.01 \\ & 0.9-X=0.003 \\ & =897 \times 10^{-3}\end{aligned}$$
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