JEE Main 2015 — Chemistry Physical Chemistry

2015mcqmedium

The following reaction is performed at $298K$ .

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

The standard free energy of the formation of $NO(g)$ is $86.6\mathrm{kJ}{\mathrm{mol}}^{-1}$ at $298K$ . What is the standard free energy of the formation of $N{O}_{2}(g)$ at $298K$ ? $({K}_{P}=1.6\times {10}^{12})$

Official previous-year question

Held on 4 Apr 2015 · Verified 6 Jul 2026.

Options

A
$0.5[2\times 86,600-R(298)\mathrm{ln}⁡(1.6\times {10}^{12})]$
B
$R(298)\mathrm{ln}(1.6\times {10}^{12})-86,600$
C
$86,600+R(298)\mathrm{ln}(1.6\times {10}^{12})$
D
$86,600-\frac{\mathrm{ln}(1.6\times {10}^{12})}{R(298)}$

Did you get this right?

Solution

$\Delta {\text{G}}_{\text{reac}}^{\circ } = - {\text{2.303 RT log K}}_{\text{P}}$

$= - {\text{RT ln K}}_{\text{P}}$

$=-\text{R}(298)\text{ln }(\text{1.6}\times 1{0}^{12})$

$\Delta {\text{G}}_{\text{reac}}^{\circ }=2\Delta {\text{G}}_{\text{f}({\text{NO}}_{2})}^{\circ }-2\Delta {\text{G}}_{\text{f}(\text{NO})}^{\circ }$

$\Rightarrow -\text{R}(298)\text{ln }(\text{1.6}\times {10}^{12})=2\Delta {\text{G}}_{\text{f}({\text{NO}}_{2})}^{\circ }-2\times \text{86.6}\times 1{0}^{3}$

$\Rightarrow 2\Delta {\text{G}}_{\text{f}({\text{NO}}_{2})}^{\circ }=-\text{R}(298)\text{ln}(\text{1.6}\times 1{0}^{12})+2\times \text{86,600}$

$\Delta {\text{G}}_{ \text{f} ( {\text{NO}}_{2} ) }^{\circ } = \text{86,600} - \frac{ \text{R} ( 2 9 8 ) }{ 2 } \text{ln} ( \text{1.6} \times 1 {0}^{ 1 2 } )$

$= \text{0.5} [ 2 \times \text{86,600} - \text{R} ( 2 9 8 ) \text{ ln } ( \text{1.6} \times 1 {0}^{ 1 2 } ) ]$

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Options

$0.5[2\times 86,600-R(298)\mathrm{ln}⁡(1.6\times {10}^{12})]$

$R(298)\mathrm{ln}(1.6\times {10}^{12})-86,600$

$86,600+R(298)\mathrm{ln}(1.6\times {10}^{12})$

$86,600-\frac{\mathrm{ln}(1.6\times {10}^{12})}{R(298)}$

Solution

$\Delta {\text{G}}_{\text{reac}}^{\circ } = - {\text{2.303 RT log K}}_{\text{P}}$

$= - {\text{RT ln K}}_{\text{P}}$

$=-\text{R}(298)\text{ln }(\text{1.6}\times 1{0}^{12})$

$\Delta {\text{G}}_{\text{reac}}^{\circ }=2\Delta {\text{G}}_{\text{f}({\text{NO}}_{2})}^{\circ }-2\Delta {\text{G}}_{\text{f}(\text{NO})}^{\circ }$

$\Rightarrow -\text{R}(298)\text{ln }(\text{1.6}\times {10}^{12})=2\Delta {\text{G}}_{\text{f}({\text{NO}}_{2})}^{\circ }-2\times \text{86.6}\times 1{0}^{3}$

$\Rightarrow 2\Delta {\text{G}}_{\text{f}({\text{NO}}_{2})}^{\circ }=-\text{R}(298)\text{ln}(\text{1.6}\times 1{0}^{12})+2\times \text{86,600}$

$\Delta {\text{G}}_{ \text{f} ( {\text{NO}}_{2} ) }^{\circ } = \text{86,600} - \frac{ \text{R} ( 2 9 8 ) }{ 2 } \text{ln} ( \text{1.6} \times 1 {0}^{ 1 2 } )$

$= \text{0.5} [ 2 \times \text{86,600} - \text{R} ( 2 9 8 ) \text{ ln } ( \text{1.6} \times 1 {0}^{ 1 2 } ) ]$

JEE Main 2015 — Chemistry Physical Chemistry

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JEE Main 2015 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

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Organic Chemistry

Inorganic Chemistry

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