JEE Main 2015 — Chemistry Physical Chemistry

2015mcqmedium

The reaction $2{N}_{2}{O}_{5}(g)\rightarrow 4N{O}_{2}(g)+{O}_{2}(g)$ follows first order kinetics. The pressure of a vessel containing only ${N}_{2}{O}_{5}$ was found to increase from 50 mm Hg to 87.5 mm Hg in 30 min. The pressure exerted by the gases after 60 min. Will be (Assume temperature remains constant) :

Official previous-year question

Held on 10 Apr 2015 · Verified 6 Jul 2026.

Options

A
106.25 mm Hg
B
125 mm Hg
C
116.25 mm Hg
D
150 mm Hg

Did you get this right?

Solution

$2{N}_{2}{O}_{5}(g) \rightarrow 4N{O}_{2}(g)+{O}_{2}(g)$

at t=0, P $_{0}$ 0 0

at t=30 min. (P $_{0}$ -x) 2x $\frac{x}{2}$

at t=60 min. (P $_{0}$ -y) 2y $\frac{y}{2}$

$\because {P}_{0}=50 mmHg$

$\because after 30min.$ $\Longrightarrow ({P}_{0}-x)+2x+\frac{x}{2}=87.5$

${P}_{0}+\frac{3}{2}x=87.5$

$\frac{3}{2}x=37.5 ; x=25mmHg$

${P}_{0}=50 ; {P}_{30min}=25mmHg$

That mean 30 min. is Half life of reaction.

${P}_{0}\overset{30 min.}{\rightarrow }\frac{{P}_{0}}{2}\overset{30 min.}{\rightarrow }\frac{{P}_{0}}{4}$

at t= 60 min

${P}_{{N}_{2}{O}_{5 }}=\frac{{P}_{0}}{4} \Longrightarrow \frac{50}{4}=12.5mmHg$

${P}_{0}-y=12.5 \Longrightarrow 50-y=12.5$

$y=37.5mmHg$

Total P after 60 min. $\Longrightarrow ({P}_{0}-y)+2y+\frac{y}{2}\Longrightarrow 50\times \frac{3}{2} \times 37.5$

$=106.25 mmHg$

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JEE Main 2015 — Chemistry Physical Chemistry

2015mcqmedium

Official previous-year question

Held on 10 Apr 2015 · Verified 6 Jul 2026.

Options

A
106.25 mm Hg
B
125 mm Hg
C
116.25 mm Hg
D
150 mm Hg

Did you get this right?

Solution

$2{N}_{2}{O}_{5}(g) \rightarrow 4N{O}_{2}(g)+{O}_{2}(g)$

at t=0, P $_{0}$ 0 0

at t=30 min. (P $_{0}$ -x) 2x $\frac{x}{2}$

at t=60 min. (P $_{0}$ -y) 2y $\frac{y}{2}$

$\because {P}_{0}=50 mmHg$

$\because after 30min.$ $\Longrightarrow ({P}_{0}-x)+2x+\frac{x}{2}=87.5$

${P}_{0}+\frac{3}{2}x=87.5$

$\frac{3}{2}x=37.5 ; x=25mmHg$

${P}_{0}=50 ; {P}_{30min}=25mmHg$

That mean 30 min. is Half life of reaction.

${P}_{0}\overset{30 min.}{\rightarrow }\frac{{P}_{0}}{2}\overset{30 min.}{\rightarrow }\frac{{P}_{0}}{4}$

at t= 60 min

${P}_{{N}_{2}{O}_{5 }}=\frac{{P}_{0}}{4} \Longrightarrow \frac{50}{4}=12.5mmHg$

${P}_{0}-y=12.5 \Longrightarrow 50-y=12.5$

$y=37.5mmHg$

Total P after 60 min. $\Longrightarrow ({P}_{0}-y)+2y+\frac{y}{2}\Longrightarrow 50\times \frac{3}{2} \times 37.5$

$=106.25 mmHg$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

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JEE Main 2015 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2015 — Chemistry Physical Chemistry

Options

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Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry