NEET (UG) 2023 — Question 74

Write the expression for the equilibrium constant: $K_c = \frac{[C][D]}{[A][B]}$. Substitute the given concentrations: $[A] = 2\ \mathrm{mol\,L^{-1}}, [B] = 3\ \mathrm{mol\,L^{-1}}, [C] = 10\ \mathrm{mol\,L^{-1}}, [D] = 6\ \mathrm{mol\,L^{-1}}$. Calculate $K_c = \frac{10 \times 6}{2 \times 3} = \frac{60}{6} = 10$. Use the relation $\Delta G^\circ = -RT\ln K_c$. Since $R = 2\ \mathrm{cal\,mol^{-1}\,K^{-1}}$, $T = 300\ \mathrm{K}$, and $\ln K_c = \ln 10 \approx 2.3026$. Alternatively, $\Delta G^\circ = -2.303 RT \log_{10} K_c$. Substitute: $\Delta G^\circ = -2.303 \times 2 \times 300 \times \log_{10}10 = -2.303 \times 2 \times 300 \times 1$. Compute: $\Delta G^\circ = -1381.8\ \mathrm{cal}$; the closest option is B) -1381.50 cal.