JEE Main 2026 — Chemistry Physical Chemistry

2026integermedium

The values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below

$A(g) \rightleftharpoons B(g) + C(g)$

$\dfrac{1}{T}(\text{K}^{-1})$	$\log_{10} K_p$
$0.05$	$3.5$
$0.06$	$2.5$
$0.07$	$1.5$

The magnitude of $\dfrac{\Delta H°}{R}$ calculated from the above data is _____. (Nearest integer)

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

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Solution

According to the van 't Hoff equation:

$\log_{10} K_p = -\dfrac{\Delta H^\circ}{2.303 R} \left(\dfrac{1}{T}\right) + \dfrac{\Delta S^\circ}{2.303 R}$

This represents a linear equation $y = mx + c$ , where $y = \log_{10} K_p$ and $x = \dfrac{1}{T}$ . The slope $m$ of the line is given by:

$m = -\dfrac{\Delta H^\circ}{2.303 R}$

From the given data table, we can calculate the slope $m$ using any two points, for example $(0.05, 3.5)$ and $(0.06, 2.5)$ :

$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{2.5 - 3.5}{0.06 - 0.05} = \dfrac{-1.0}{0.01} = -100$

Equating the calculated slope to the theoretical slope:

$-\dfrac{\Delta H^\circ}{2.303 R} = -100$

$\dfrac{\Delta H^\circ}{R} = 100 \times 2.303 = 230.3$

The magnitude of $\dfrac{\Delta H^\circ}{R}$ is $230.3$ . Rounding to the nearest integer, we get $230$ .

Answer: $230$

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JEE Main 2026 — Chemistry Physical Chemistry

2026integermedium

The values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below

$A(g) \rightleftharpoons B(g) + C(g)$

$\dfrac{1}{T}(\text{K}^{-1})$	$\log_{10} K_p$
$0.05$	$3.5$
$0.06$	$2.5$
$0.07$	$1.5$

The magnitude of $\dfrac{\Delta H°}{R}$ calculated from the above data is _____. (Nearest integer)

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

According to the van 't Hoff equation:

$\log_{10} K_p = -\dfrac{\Delta H^\circ}{2.303 R} \left(\dfrac{1}{T}\right) + \dfrac{\Delta S^\circ}{2.303 R}$

This represents a linear equation $y = mx + c$ , where $y = \log_{10} K_p$ and $x = \dfrac{1}{T}$ . The slope $m$ of the line is given by:

$m = -\dfrac{\Delta H^\circ}{2.303 R}$

From the given data table, we can calculate the slope $m$ using any two points, for example $(0.05, 3.5)$ and $(0.06, 2.5)$ :

$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{2.5 - 3.5}{0.06 - 0.05} = \dfrac{-1.0}{0.01} = -100$

Equating the calculated slope to the theoretical slope:

$-\dfrac{\Delta H^\circ}{2.303 R} = -100$

$\dfrac{\Delta H^\circ}{R} = 100 \times 2.303 = 230.3$

The magnitude of $\dfrac{\Delta H^\circ}{R}$ is $230.3$ . Rounding to the nearest integer, we get $230$ .

Answer: $230$

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 2026 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2026 — 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