JEE Main 2022 — Chemistry Physical Chemistry

2022integereasy

At $600K,2\mathrm{mol}$ of $\mathrm{NO}$ are mixed with $1\mathrm{mol}$ of ${O}_{2}$ .

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

The reaction occurring as above comes to equilibrium under a total pressure of $1$ atm. Analysis of the system shows that $0.6\mathrm{mol}$ of oxygen are present at equilibrium. The equilibrium constant for the reaction is____.

Official previous-year question

Held on 28 Jul 2022 · Verified 6 Jul 2026.

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Solution

$2\mathrm{NO}+{O}_{2}\rightarrow 2\mathrm{NO}$

$\begin{matrix}2 & 1 & - \\ 2-2x & 1-x & 2x \\ 1.2 & 0.6 & 0.8\end{matrix}$

Total number of moles= $2.6$

According to Dalton's law,

Partial pressure $=$ Total pressure $\times$ Mole fraction

Mole fraction $=\frac{\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{one}\mathrm{component}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{moles}}$

${K}_{P}=\frac{{{P}^{2}}_{{\mathrm{NO}}_{2}}}{{{P}^{2}}_{\mathrm{NO}}.{P}_{{O}_{2}}}$

$=$ $\frac{{(\frac{0.8}{2.6}\times 1)}^{2}}{{(\frac{1.2}{2.6}\times 1)}^{2}(\frac{0.6}{2.6}\times 1)}$ = $=\frac{1.664}{0.864}=1.8824\simeq 2$

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JEE Main 2022 — Chemistry Physical Chemistry

2022integereasy

At $600K,2\mathrm{mol}$ of $\mathrm{NO}$ are mixed with $1\mathrm{mol}$ of ${O}_{2}$ .

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

Official previous-year question

Held on 28 Jul 2022 · Verified 6 Jul 2026.

Did you get this right?

Solution

$2\mathrm{NO}+{O}_{2}\rightarrow 2\mathrm{NO}$

$\begin{matrix}2 & 1 & - \\ 2-2x & 1-x & 2x \\ 1.2 & 0.6 & 0.8\end{matrix}$

Total number of moles= $2.6$

According to Dalton's law,

Partial pressure $=$ Total pressure $\times$ Mole fraction

Mole fraction $=\frac{\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{one}\mathrm{component}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{moles}}$

${K}_{P}=\frac{{{P}^{2}}_{{\mathrm{NO}}_{2}}}{{{P}^{2}}_{\mathrm{NO}}.{P}_{{O}_{2}}}$

$=$ $\frac{{(\frac{0.8}{2.6}\times 1)}^{2}}{{(\frac{1.2}{2.6}\times 1)}^{2}(\frac{0.6}{2.6}\times 1)}$ = $=\frac{1.664}{0.864}=1.8824\simeq 2$

JEE Main 2022 — Chemistry Physical Chemistry

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JEE Main 2022 — Chemistry Physical Chemistry

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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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