The value of log 10 K for a reaction A → B is (Given ∆ H r 298 K o = - 54.07 k J m o l - 1 , ∆ S r 298 K o = 10 J K - 1 m o l - 1 and R = 8.314 J K - 1 m o l - 1 , 2.303 × 8.314 × 298 = 5705 )

Step-by-step solution

For the equilibrium, A → B ΔG o = ΔH o − TΔS o ΔG o = − 2.303 RT log 10 K ( K is equilibrium constant) − 2.303 RT log 10 K = ΔH o − TΔS o 2 .303 RT log 10 K = TΔS o − ΔH o log 10 ⁡ K = T ∆ S o - ∆ H o 2.303 R T = 298 × 10 + 54.07 × 1000 2.303 × 8.314 × 298 = 10 Hence (B) is correct option.