JEE Main 2019 — Chemistry Physical Chemistry

2019mcqhard

What are the values of $\frac{{K}_{p}}{{K}_{c}}$ for the following reactions at $300 K$ respectively?

(At $300 K, \mathrm{RT}=24.62 {\mathrm{dm}}^{2}\mathrm{atm} {\mathrm{mol}}^{-1}$ )

${N}_{2}(g)+{O}_{2}(g)\rightleftharpoons 2\mathrm{NO}(g)$

${N}_{2}{O}_{4}(g)\rightleftharpoons 2\mathrm{NO}(g)$

${N}_{2}(g)+3{H}_{2}(g)\rightleftharpoons 2{\mathrm{NH}}_{3}(g)$

Official previous-year question

Held on 10 Jan 2019 · Verified 6 Jul 2026.

Options

A
$\begin{matrix}24.63 {\mathrm{dm}}^{3}\mathrm{atm} {\mathrm{mol}}^{-1}, \\ 606.0{\mathrm{dm}}^{6}{\mathrm{atm}}^{2}{\mathrm{mol}}^{-2}, \\ 1.65\times {10}^{-3}{\mathrm{dm}}^{-6}{\mathrm{atm}}^{-2}{\mathrm{mol}}^{2}\end{matrix}$
B
$\begin{matrix}1,24.62{\mathrm{dm}}^{6}{\mathrm{atm}}^{3}{\mathrm{mol}}^{-1} \\ 606.0{\mathrm{dm}}^{6}{\mathrm{atm}}^{2}{\mathrm{mol}}^{-2}\end{matrix}$
C
$\begin{matrix}1,24.62{\mathrm{dm}}^{3}\mathrm{atm} {\mathrm{mol}}^{-1} \\ 1.65\times {10}^{-3}{\mathrm{dm}}^{-6}{\mathrm{atm}}^{-2}{\mathrm{mol}}^{2}\end{matrix}$
D
$\begin{matrix}1, 4.1\times {10}^{-2}{\mathrm{dm}}^{-3}{\mathrm{atm}}^{-1}\mathrm{mol}, \\ 606 {\mathrm{dm}}^{6}{\mathrm{atm}}^{2}{\mathrm{mol}}^{-2}\end{matrix}$

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Solution

$\because {K}_{P}={K}_{C}{(\mathrm{RT})}^{\Delta ng}$

$\frac{{K}_{P}}{{K}_{C}}={(\mathrm{RT})}^{\Delta ng}$

(i) ${N}_{2}+{O}_{2}\rightleftharpoons 2\mathrm{NO} \Delta ng=0\therefore \frac{{K}_{P}}{{K}_{C}}=1$

(ii) ${N}_{2}{O}_{4}\rightleftharpoons 2{\mathrm{Na}}_{2} \Delta ng=1 \therefore \frac{{K}_{P}}{{K}_{C}}={(\mathrm{RT})}^{1}={(24.62)}^{1}$

(iii) ${N}_{2}+3{H}_{2}\rightleftharpoons 2 {\mathrm{NH}}_{3} \Delta ng=-2$ $\therefore \frac{{K}_{P}}{{K}_{C}}={(\mathrm{RT})}^{-2}={(24.62)}^{-2}=1.659\times {10}^{-3}$ $\simeq 1.65\times { 10}^{ -3}$

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Options

$\begin{matrix}24.63 {\mathrm{dm}}^{3}\mathrm{atm} {\mathrm{mol}}^{-1}, \\ 606.0{\mathrm{dm}}^{6}{\mathrm{atm}}^{2}{\mathrm{mol}}^{-2}, \\ 1.65\times {10}^{-3}{\mathrm{dm}}^{-6}{\mathrm{atm}}^{-2}{\mathrm{mol}}^{2}\end{matrix}$

$\begin{matrix}1,24.62{\mathrm{dm}}^{6}{\mathrm{atm}}^{3}{\mathrm{mol}}^{-1} \\ 606.0{\mathrm{dm}}^{6}{\mathrm{atm}}^{2}{\mathrm{mol}}^{-2}\end{matrix}$

$\begin{matrix}1,24.62{\mathrm{dm}}^{3}\mathrm{atm} {\mathrm{mol}}^{-1} \\ 1.65\times {10}^{-3}{\mathrm{dm}}^{-6}{\mathrm{atm}}^{-2}{\mathrm{mol}}^{2}\end{matrix}$

$\begin{matrix}1, 4.1\times {10}^{-2}{\mathrm{dm}}^{-3}{\mathrm{atm}}^{-1}\mathrm{mol}, \\ 606 {\mathrm{dm}}^{6}{\mathrm{atm}}^{2}{\mathrm{mol}}^{-2}\end{matrix}$

Solution

$\because {K}_{P}={K}_{C}{(\mathrm{RT})}^{\Delta ng}$

$\frac{{K}_{P}}{{K}_{C}}={(\mathrm{RT})}^{\Delta ng}$

(i) ${N}_{2}+{O}_{2}\rightleftharpoons 2\mathrm{NO} \Delta ng=0\therefore \frac{{K}_{P}}{{K}_{C}}=1$

(ii) ${N}_{2}{O}_{4}\rightleftharpoons 2{\mathrm{Na}}_{2} \Delta ng=1 \therefore \frac{{K}_{P}}{{K}_{C}}={(\mathrm{RT})}^{1}={(24.62)}^{1}$

JEE Main 2019 — Chemistry Physical Chemistry

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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