MHTCET 19 Apr 2026 Shift 2 Paper

Concept:Standard Gibbs energy change $\Delta G^\circ$ is connected to the equilibrium constant $K_p$ by the equation $\Delta G^\circ = -RT \ln K_p$.Explanation:For a gaseous reaction, the equilibrium constant is given as $K_p$.Given: $K_p = 3.5 \times 10^{17}$ and $R = 8.314 \, \mathrm{J\,K^{-1}\,mol^{-1}}$.At standard temperature, $T = 298 \, \mathrm{K}$.Substituting these values into the relation:$$\Delta G^\circ = -(8.314)(298) \ln(3.5 \times 10^{17})$$Use the logarithm conversion $\ln x = 2.303 \log x$:$$\ln(3.5 \times 10^{17}) = 2.303 \times (\log 3.5 + 17)$$Since $\log 3.5 \approx 0.544$:$$= 2.303 \times (0.544 + 17) = 2.303 \times 17.544 \approx 40.4$$Now calculate the Gibbs energy:$$\Delta G^\circ = -(8.314)(298)(40.4) = -1.001 \times 10^5 \, \mathrm{J\,mol^{-1}}$$Convert into $\mathrm{kJ\,mol^{-1}}$ by dividing by $1000$:$$\Delta G^\circ = -100.1 \, \mathrm{kJ\,mol^{-1}}$$Answer:The standard Gibbs energy for the gaseous reaction is $\boxed{-100.1 \, \mathrm{kJ\,mol^{-1}}}$.Hence, the correct option is Option A.