JEE Main 2019 — Chemistry Physical Chemistry

2019mcqmedium

$N{O}_{2}$ required for a reaction is produced by the decomposition of ${N}_{2}{O}_{5}$ in $CC{l}_{4}$ as per the equation,

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

The initial concentration of ${N}_{2}{O}_{5}$ is $3.00 mol {L}^{-1}$ and it is $2.75 mol {L}^{-1}$ after 30 minutes. The rate of formation of $N{O}_{2}$ is:

Official previous-year question

Held on 12 Apr 2019 · Verified 6 Jul 2026.

Options

A
$1.667\times {10}^{-2} mol {L}^{-1} {min}^{-1}$
B
$8.333\times {10}^{-3} mol {L}^{-1}{min}^{-1}$
C
$4.167\times {10}^{-3} mol {L}^{-1}{min}^{-1}$
D
$2.083\times {10}^{-3} mol {L}^{-1} {min}^{-1}$

Did you get this right?

Solution

Rate of reaction in terms of rate of disappearance of ${N}_{2}{O}_{5}(s)$

$-\frac{d[{N}_{2}{O}_{5}]}{dt}=\frac{1}{2}\frac{d[N{O}_{2}]}{dt}=\frac{2d[{O}_{2}]}{dt}=rate of Reaction$

$-\frac{(-3+1.75)}{30}=rate of reaction$

$-(-\frac{0.25}{30})=8.3\times {10}^{-3}mol {L}^{-1}m{in}^{-1}$

$=\frac{25}{3000}=8.3\times {10}^{-3}=\frac{1}{2}\times$ rate of formation of ${O}_{2}$

$8.3\times {10}^{-3}=\frac{1}{2}\frac{d[N{O}_{2}]}{dt}$

$1.66\times {10}^{-2}m/min =\frac{d[N{O}_{2}]}{dt}$

Your note

JEE Main Physical Chemistry in other years

Physical Chemistry 2026 Physical Chemistry 2025 Physical Chemistry 2024 Physical Chemistry 2023 Physical Chemistry 2022 Physical Chemistry 2021 Physical Chemistry 2020 Physical Chemistry 2018 Physical Chemistry 2017 Physical Chemistry 2016 Physical Chemistry 2015 Physical Chemistry 2014

Keep practicing Physical Chemistry

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

Browse all Chemistry topics

Options

$1.667\times {10}^{-2} mol {L}^{-1} {min}^{-1}$

$8.333\times {10}^{-3} mol {L}^{-1}{min}^{-1}$

$4.167\times {10}^{-3} mol {L}^{-1}{min}^{-1}$

$2.083\times {10}^{-3} mol {L}^{-1} {min}^{-1}$

Solution

Rate of reaction in terms of rate of disappearance of ${N}_{2}{O}_{5}(s)$

$-\frac{d[{N}_{2}{O}_{5}]}{dt}=\frac{1}{2}\frac{d[N{O}_{2}]}{dt}=\frac{2d[{O}_{2}]}{dt}=rate of Reaction$

$-\frac{(-3+1.75)}{30}=rate of reaction$

$-(-\frac{0.25}{30})=8.3\times {10}^{-3}mol {L}^{-1}m{in}^{-1}$

$=\frac{25}{3000}=8.3\times {10}^{-3}=\frac{1}{2}\times$ rate of formation of ${O}_{2}$

$8.3\times {10}^{-3}=\frac{1}{2}\frac{d[N{O}_{2}]}{dt}$

$1.66\times {10}^{-2}m/min =\frac{d[N{O}_{2}]}{dt}$

JEE Main 2019 — Chemistry Physical Chemistry

Options

Did you get this right?

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

JEE Main 2019 — Chemistry Physical Chemistry

Options

Did you get this right?

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