The variation of equilibrium constant with temperature is given below: | Temperature | Equilibrium Constant |…

🧠 Van't Hoff gives $\Delta H^\circ$; then $\Delta G^\circ = -RT\ln K$ at each temperature The integrated Van't Hoff equation links the ratio of equilibrium constants at two temperatures to the standard enthalpy change.

🗺️ Calculation $\ln\frac{K_2}{K_1} = \frac{\Delta H^\circ}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

$\ln 10 = \frac{\Delta H^\circ}{8.314}\left(\frac{1}{298} - \frac{1}{373}\right)$

$2.303 = \frac{\Delta H^\circ}{8.314} \times 6.75 \times 10^{-4}$

$\Delta H^\circ = \frac{2.303 \times 8.314}{6.75 \times 10^{-4}} = 28{,}370\,\text{J mol}^{-1} \approx 28.4\,\text{kJ mol}^{-1}$

$\Delta G^\circ$ at $T_1 = 298\,\text{K}$: $= -(8.314)(298)(2.303) = -5706\,\text{J} \approx -5.71\,\text{kJ mol}^{-1}$

$\Delta G^\circ$ at $T_2 = 373\,\text{K}$: $= -(8.314)(373)(4.606) = -14{,}280\,\text{J} \approx -14.3\,\text{kJ mol}^{-1}$

⚠️ Data note: The calculation gives $28.4$, $-5.71$, $-14.3$ which matches option (c). The stored correct option is (a); verify against the original source.

$\boxed{\text{Answer: 28.4, -5.71, -14.3 kJ mol}^{-1}}$