The value of logK for the reaction A ⇌ B at 298K is _______ ....

We can use the relationship between the equilibrium constant ($K$) and the standard Gibbs free energy change ($\Delta G °$) to calculate the value of $logK$ at $298K$

Given,

$∆ {H}^{o} = – 54 . 07kJ{mol}^{– 1} ∆ {S}^{o} = 10{JK}^{– 1}{mol}^{– 1}$

We know,

$∆ G ° = ∆ H ° - T ∆ S °$

$= - 54 . 07 - \dfrac{298 ( 10 )}{1000}$

$= - 57 . 05kJ / mole$

$∆ G ° = - 2 . 303RT{logK}_{\text{eq}}$

$- 57 . 05 \times 1000 = - 2 . 303 \times 8 . 314 \times 298{logK}_{\text{eq}}$

$- 57 . 05 \times 1000 = - 5705\log_{eq}$

$10 = {logK}_{\text{eq}}$