The standard Gibbs energy… - JEE Main 2015 Chemistry PYQ

JEE Main 2015 — Chemistry Physical Chemistry

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The standard Gibbs energy change at 300 K for the reaction $2A\rightleftharpoons B+C$ is 2494.2 J. At a given time, the composition of the reaction mixture is $[A]=\frac{1}{2}, [B]=2$ and $[C]=\frac{1}{2}$ . The reaction proceeds in the: $[R=8.314 J/K-mol , e=2.718]$ {Given antilog (-0.44)=0.36}

Official previous-year question

Held on 4 Apr 2015 · Verified 6 Jul 2026.

Options

A
Reverse direction because $Q<{K}_{C}$
B
Forward direction because $Q>{K}_{C}$
C
Reverse direction because $Q>{K}_{C}$
D
Forward direction because $Q<{K}_{C}$

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Solution

$\Delta {\text{G}}^{\circ } = - {\text{RT ln K}}_{\text{C}}$

$\text{2494.2} = - \text{8.314} \times {\text{300 ln K}}_{\text{C}}$

$\text{2494.2} = - \text{8.314} \times 300 \times {\text{2.303 log K}}_{\text{C}}$

$- \frac{ \text{2494.2} }{ \text{2.303} \times 300 \times \text{8.314} } = - \text{0.44} = {\text{log K}}_{\text{C}}$

${\text{log K}}_{\text{C}}=-\text{0.44}$

${\text{K}}_{\text{C}} = \text{0.36}$

$\text{Q}=\frac{[B][C]}{[A{]}^{2}}=\frac{2\times \frac{1}{2}}{{(\frac{1}{2})}^{2}}=4$

$\text{Q}>{\text{K}}_{\text{C}}$ , reaction will shift in reverse direction.

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Solution

$\Delta {\text{G}}^{\circ } = - {\text{RT ln K}}_{\text{C}}$

$\text{2494.2} = - \text{8.314} \times {\text{300 ln K}}_{\text{C}}$

$\text{2494.2} = - \text{8.314} \times 300 \times {\text{2.303 log K}}_{\text{C}}$

$- \frac{ \text{2494.2} }{ \text{2.303} \times 300 \times \text{8.314} } = - \text{0.44} = {\text{log K}}_{\text{C}}$

${\text{log K}}_{\text{C}}=-\text{0.44}$

${\text{K}}_{\text{C}} = \text{0.36}$

$\text{Q}=\frac{[B][C]}{[A{]}^{2}}=\frac{2\times \frac{1}{2}}{{(\frac{1}{2})}^{2}}=4$

$\text{Q}>{\text{K}}_{\text{C}}$ , reaction will shift in reverse direction.

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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

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