The value of log K for the reaction A ⇌ B at 298 K is _ _ _ _ _ _ _ . (Nearest integer) Given: Δ H ° = - 54 . 07 kJ mol - 1 Δ S ° = 10 JK - 1 mol - 1 (Taken 2 . 303 × 8 . 314 × 298 = 5705 )

Step-by-step solution

We can use the relationship between the equilibrium constant ( K ) and the standard Gibbs free energy change ( ΔG ° ) to calculate the value of log   K at 298   K Given, ∆ H o   =   –   54 . 07   kJ   mol – 1   ∆ S o   =   10   JK – 1   mol – 1 We know, ∆ G ° = ∆ H ° - T ∆ S ° = - 54 . 07 - 298 ( 10 ) 1000 = - 57 . 05   kJ / mole ∆ G ° = - 2 . 303 RT   logK