JEE Chemistry 2026 Question with Solution

Given: The graph is between $\log_{10}K$ and $\frac{1}{T}$.

Find: The intercept and slope of the straight line.

Concept: Whenever you see $\log K$ vs $1/T$, immediately recall the Van’t Hoff equation.

From thermodynamics, the Van’t Hoff equation is:

$$\ln K = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R}$$

Converting natural logarithm to base $10$:

$$\log_{10}K = -\frac{\Delta H^\circ}{2.303R}\cdot\frac{1}{T} + \frac{\Delta S^\circ}{2.303R}$$

Comparing with the straight line form $y = mx + c$:

$$\text{slope} = -\frac{\Delta H^\circ}{2.303R}$$ $$\text{intercept} = \frac{\Delta S^\circ}{2.303R}$$

Therefore, the intercept is $\frac{\Delta S^\circ}{2.303R}$ and the slope is $-\frac{\Delta H^\circ}{2.303R}$. The correct option is C.

Given: $\log_{10}K$ is plotted against $\frac{1}{T}$.

Find: Intercept and slope.

Write the Van’t Hoff equation directly in the form of a straight line:

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

Now compare it term-by-term with $y = mx + c$, where $y = \log_{10}K$ and $x = \frac{1}{T}$.

So, the coefficient of $\frac{1}{T}$ is the slope, and the constant term is the intercept.

Hence, slope $= -\frac{\Delta H^\circ}{2.303R}$ and intercept $= \frac{\Delta S^\circ}{2.303R}$. The correct option is C.