JEE Main 2026 — Chemical Equilibrium Question with Solution

According to the van 't Hoff equation:

${\log }_{10}{K}_p=-\frac{\Delta H^{\circ }}{2.303R}\left(\frac{1}{T}\right)+\frac{\Delta S^{\circ }}{2.303R}$

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

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

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

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

Equating the calculated slope to the theoretical slope:

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

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

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

Answer: $230$