JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

The plot of $\log _{10} \mathrm{~K}$ vs $\frac{1}{\mathrm{~T}}$ gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).

Official previous-year question

Held on 28 Jan 2026 · Verified 6 Jul 2026.

Options

A
$\frac{\Delta \mathrm{S}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}},-\frac{\Delta \mathrm{H}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}}$
B
$-\frac{\Delta \mathrm{S}^{\circ} \mathrm{R}}{2 \cdot 303}, \frac{\Delta \mathrm{H}^{\circ} \mathrm{R}}{2 \cdot 303}$
C
$\frac{2 \cdot 303 \mathrm{R}}{\Delta \mathrm{H}^{\mathrm{o}}}, \frac{2 \cdot 303 \mathrm{R}}{\Delta \mathrm{~S}^{\mathrm{o}}}$
D
$-\frac{\Delta \mathrm{H}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}}, \frac{\Delta \mathrm{~S}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}}$

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Solution

The relationship between the standard Gibbs free energy change and the equilibrium constant is given by $\Delta G^o = -RT \ln K$ .

We also know that $\Delta G^o = \Delta H^o - T\Delta S^o$ .

Equating the two expressions: $-RT \ln K = \Delta H^o - T\Delta S^o$ .

Dividing by $-RT$ , we get: $\ln K = -\frac{\Delta H^o}{RT} + \frac{\Delta S^o}{R}$ .

To convert natural log to base 10, use $\ln K = 2.303 \log_{10} K$ :

$2.303 \log_{10} K = -\frac{\Delta H^o}{RT} + \frac{\Delta S^o}{R}$ .

Dividing by $2.303$ : $\log_{10} K = -\frac{\Delta H^o}{2.303RT} + \frac{\Delta S^o}{2.303R}$ .

Rearranging in the form of a straight line equation $y = mx + c$ where $y = \log_{10} K$ and $x = \frac{1}{T}$ :

$\log_{10} K = \left( -\frac{\Delta H^o}{2.303R} \right) \frac{1}{T} + \frac{\Delta S^o}{2.303R}$ .

Comparing the terms:

Intercept ( $c$ ) = $\frac{\Delta S^o}{2.303R}$

Slope ( $m$ ) = $-\frac{\Delta H^o}{2.303R}$ .

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Options

$\frac{\Delta \mathrm{S}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}},-\frac{\Delta \mathrm{H}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}}$

$-\frac{\Delta \mathrm{S}^{\circ} \mathrm{R}}{2 \cdot 303}, \frac{\Delta \mathrm{H}^{\circ} \mathrm{R}}{2 \cdot 303}$

$\frac{2 \cdot 303 \mathrm{R}}{\Delta \mathrm{H}^{\mathrm{o}}}, \frac{2 \cdot 303 \mathrm{R}}{\Delta \mathrm{~S}^{\mathrm{o}}}$

$-\frac{\Delta \mathrm{H}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}}, \frac{\Delta \mathrm{~S}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}}$

Solution

The relationship between the standard Gibbs free energy change and the equilibrium constant is given by $\Delta G^o = -RT \ln K$ .

We also know that $\Delta G^o = \Delta H^o - T\Delta S^o$ .

Equating the two expressions: $-RT \ln K = \Delta H^o - T\Delta S^o$ .

Dividing by $-RT$ , we get: $\ln K = -\frac{\Delta H^o}{RT} + \frac{\Delta S^o}{R}$ .

To convert natural log to base 10, use $\ln K = 2.303 \log_{10} K$ :

$2.303 \log_{10} K = -\frac{\Delta H^o}{RT} + \frac{\Delta S^o}{R}$ .

Dividing by $2.303$ : $\log_{10} K = -\frac{\Delta H^o}{2.303RT} + \frac{\Delta S^o}{2.303R}$ .

Rearranging in the form of a straight line equation $y = mx + c$ where $y = \log_{10} K$ and $x = \frac{1}{T}$ :

$\log_{10} K = \left( -\frac{\Delta H^o}{2.303R} \right) \frac{1}{T} + \frac{\Delta S^o}{2.303R}$ .

Comparing the terms:

Intercept ( $c$ ) = $\frac{\Delta S^o}{2.303R}$

Slope ( $m$ ) = $-\frac{\Delta H^o}{2.303R}$ .

JEE Main 2026 — Chemistry Physical Chemistry

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JEE Main 2026 — Chemistry Physical Chemistry

Options

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