Q18 / 75 · JEE Main 2026 (28 January Shift 2) · Chemistry
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).
- Correct answerA$\frac{\Delta \mathrm{S}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}},-\frac{\Delta \mathrm{H}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}}$
- 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$-\frac{\Delta \mathrm{H}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}}, \frac{\Delta \mathrm{~S}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}}$
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 ($m$) = $-\frac{\Delta H^o}{2.303R}$.
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 ($m$) = $-\frac{\Delta H^o}{2.303R}$.
Hence, the correct answer is option: 'A'.