JEE Main 2023 — Chemistry Physical Chemistry

2023integerhard

The value of $\mathrm{log}K$ for the reaction $A\rightleftharpoons B$ at $298K$ is $_______.$ (Nearest integer)

Given: $\Delta H^{\circ}=-54.07\mathrm{kJ}{\mathrm{mol}}^{-1}$

$\Delta S^{\circ}=10{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$

(Taken $2.303\times 8.314\times 298=5705$ )

Official previous-year question

Held on 6 Apr 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

We can use the relationship between the equilibrium constant ( $K$ ) and the standard Gibbs free energy change ( $\Delta G^{\circ}$ ) to calculate the value of $\mathrm{log}K$ at $298K$

Given,

$\Delta {H}^{o}=–54.07\mathrm{kJ}{\mathrm{mol}}^{–1} \Delta {S}^{o}=10{\mathrm{JK}}^{–1}{\mathrm{mol}}^{–1}$

We know,

$\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}$

$=-54.07-\frac{298(10)}{1000}$

$=-57.05\mathrm{kJ}/\mathrm{mole}$

$\Delta G^{\circ}=-2.303\mathrm{RT}{\mathrm{logK}}_{\text{eq }}$

$-57.05\times 1000=-2.303\times 8.314\times 298{\mathrm{logK}}_{\text{eq }}$

$-57.05\times 1000=-5705{\mathrm{log}}_{\mathrm{eq}}$

$10={\mathrm{logK}}_{\text{eq }}$

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JEE Main 2023 — Chemistry Physical Chemistry

2023integerhard

The value of $\mathrm{log}K$ for the reaction $A\rightleftharpoons B$ at $298K$ is $_______.$ (Nearest integer)

Given: $\Delta H^{\circ}=-54.07\mathrm{kJ}{\mathrm{mol}}^{-1}$

$\Delta S^{\circ}=10{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$

(Taken $2.303\times 8.314\times 298=5705$ )

Official previous-year question

Held on 6 Apr 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

We can use the relationship between the equilibrium constant ( $K$ ) and the standard Gibbs free energy change ( $\Delta G^{\circ}$ ) to calculate the value of $\mathrm{log}K$ at $298K$

Given,

$\Delta {H}^{o}=–54.07\mathrm{kJ}{\mathrm{mol}}^{–1} \Delta {S}^{o}=10{\mathrm{JK}}^{–1}{\mathrm{mol}}^{–1}$

We know,

$\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}$

$=-54.07-\frac{298(10)}{1000}$

$=-57.05\mathrm{kJ}/\mathrm{mole}$

$\Delta G^{\circ}=-2.303\mathrm{RT}{\mathrm{logK}}_{\text{eq }}$

$-57.05\times 1000=-2.303\times 8.314\times 298{\mathrm{logK}}_{\text{eq }}$

$-57.05\times 1000=-5705{\mathrm{log}}_{\mathrm{eq}}$

$10={\mathrm{logK}}_{\text{eq }}$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

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JEE Main 2023 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2023 — Chemistry Physical Chemistry

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