JEE Main 2019 — Chemistry Physical Chemistry

2019mcqhard

$5.1 g {\mathrm{NH}}_{4}\mathrm{SH}$ is introduced in $3.0 L$ evacuated flask at ${327}^{o}C$ . $30%$ of the solid ${\mathrm{NH}}_{4}\mathrm{SH}$ is decomposed to ${\mathrm{NH}}_{3}$ and ${H}_{2}S$ as gases. The ${K}_{P}$ of the reaction at ${327}^{o}C$ is

$(R=0.082 L \mathrm{atm} {\mathrm{mol}}^{-1}{K}^{-1},\mathrm{Molar} \mathrm{mass} \mathrm{of} S=32 g {\mathrm{mol}}^{-1}, \mathrm{Molar} \mathrm{mass} \mathrm{of} N=14g {\mathrm{mol}}^{-1})$

Official previous-year question

Held on 10 Jan 2019 · Verified 6 Jul 2026.

Options

A
$0.242 {\mathrm{atm}}^{2}$
B
$0.242\times {10}^{-4} {\mathrm{atm}}^{2}$
C
$1\times {10}^{-4} {\mathrm{atm}}^{2}$
D
$4.9\times {10}^{-3} {\mathrm{atm}}^{2}$

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Solution

${\mathrm{NH}}_{4}\mathrm{SH}(s)\rightleftharpoons {\mathrm{NH}}_{3}(g)+{H}_{2}S(g)$

Number of moles $=\frac{5.1}{51}=0.1 \mathrm{mol}$

	${\mathrm{NH}}_{4}\mathrm{SH}$	${\mathrm{NH}}_{3}$	${H}_{2}S$
Initial concentration	$0.1\mathrm{mol}$	$0$	$0$
Equilibrium concentration	$0.1(1-\alpha )$	$0.1\alpha$	$0.1\alpha$

$(\alpha =30%=0.3)=$ Degree of dissociation

So, number of moles at equilibrium,

$[{\mathrm{NH}}_{3}]=[{H}_{2}S]=0.1\times 0.3=0.03$

$[{\mathrm{NH}}_{4}\mathrm{SH}]$ is not considered as it is a solid.

${K}_{c}=[{\mathrm{NH}}_{3}][{H}_{2}S]$

${K}_{c}=[\frac{0.03}{3}][\frac{0.03}{3}]={10}^{-4}$

${K}_{p}={10}^{-4}{(0.082\times 600)}^{2}$

${K}_{p}=0.242 {\mathrm{atm}}^{2}$

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Solution

${\mathrm{NH}}_{4}\mathrm{SH}(s)\rightleftharpoons {\mathrm{NH}}_{3}(g)+{H}_{2}S(g)$

Number of moles $=\frac{5.1}{51}=0.1 \mathrm{mol}$

	${\mathrm{NH}}_{4}\mathrm{SH}$	${\mathrm{NH}}_{3}$	${H}_{2}S$
Initial concentration	$0.1\mathrm{mol}$	$0$	$0$
Equilibrium concentration	$0.1(1-\alpha )$	$0.1\alpha$	$0.1\alpha$

$(\alpha =30%=0.3)=$ Degree of dissociation

So, number of moles at equilibrium,

$[{\mathrm{NH}}_{3}]=[{H}_{2}S]=0.1\times 0.3=0.03$

$[{\mathrm{NH}}_{4}\mathrm{SH}]$ is not considered as it is a solid.

${K}_{c}=[{\mathrm{NH}}_{3}][{H}_{2}S]$

${K}_{c}=[\frac{0.03}{3}][\frac{0.03}{3}]={10}^{-4}$

${K}_{p}={10}^{-4}{(0.082\times 600)}^{2}$

${K}_{p}=0.242 {\mathrm{atm}}^{2}$

JEE Main 2019 — Chemistry Physical Chemistry

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JEE Main 2019 — 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

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