Consider the following reaction approaching equilibrium at 27 ° C and 1 atm pressure A + B ⇌ K t = 0 3 K r = 10 2 C + D The standard Gibb's energy change Δ r G ° at 27 ° C is - _____ kJmol - 1 (Nearest integer). (Given : R = 8 . 3 J   K - 1   mol - 1 and ln 10 = 2 . 3 )

Step-by-step solution

We know, K eq = K f K b ∴   K eq = 10 3 10 2 = 10 Using gibbs free energy equation, ∴     ΔG ° = - RTlnK eq     ⇒ ΔG = - RTln 10 ⇒ - 8 . 3 × 300 × 2 . 3 = - 5 . 7   kJ   mol - 1 ≈ 6   kJ   mole - 1 (nearest integer) Answer = 6 .