The plot of $$\log_{10}^{K} VS \frac{1}{T}$$ gives a straight Line. The intercept and slope respectively are
(where K is equilibrium constant).

The relationship between the equilibrium constant $$K$$ and temperature $$T$$ is given by the van't Hoff equation:

$$\Delta G^{\circ} = -RT \ln K$$

and

$$\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$$

Combining these equations:

$$-RT \ln K = \Delta H^{\circ} - T \Delta S^{\circ}$$

Dividing both sides by $$-RT$$:

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

Since $$\ln K = 2.303 \log_{10} K$$, substitute:

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

Solving for $$\log_{10} K$$:

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

Simplify:

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

This equation is in the form $$y = mx + c$$, where:

• $$y = \log_{10} K$$
• $$x = \frac{1}{T}$$
• Slope $$m = -\frac{\Delta H^{\circ}}{2.303 R}$$
• Intercept $$c = \frac{\Delta S^{\circ}}{2.303 R}$$

Comparing with the options:

• Option A: $$\frac{2.303R}{\Delta H^{\circ}}, \frac{2.303R}{\Delta S^{\circ}}$$ → Incorrect
• Option B: $$\frac{\Delta S^{\circ}}{2.303R}, -\frac{\Delta H^{\circ}}{2.303R}$$ → Matches intercept and slope
• Option C: $$-\frac{\Delta S^{\circ} R}{2.303}, \frac{\Delta H^{\circ} R}{2.303}$$ → Incorrect signs and extra $$R$$
• Option D: $$-\frac{\Delta H^{\circ}}{2.303R}, \frac{\Delta S^{\circ}}{2.303R}$$ → Intercept and slope swapped

Thus, the correct answer is option B.