The equilibrium constant for certain reaction is 100. If the value R is given to be 2 cal K-1 mol-1 then standard Gibb's free energy change will be

Question

The equilibrium constant for certain reaction is 100. If the value R is given to be 2 cal K-1 mol-1 then standard Gibb's free energy change will be (A) -2.764 kcal (B) 2.674 kcal (C) 2.764 kcal (D) none of these

Tardigrade Admin · Accepted Answer

Δ G° = - 2.303 RT log K = - 2.303 × 2 × 298 × log 100 = -2.303 × 2 × 298 × 2 = -2745.2 cal = -2.745 kcal

Q. The equilibrium constant for certain reaction is 100. If the value R is given to be 2 $c a l K^{- 1} m o l^{- 1}$ then standard Gibb's free energy change will be

-2.764 kcal

2.674 kcal

2.764 kcal

none of these

$Δ$ G $^{\circ}$ = - 2.303 RT log K = - 2.303 $\times$ 2 $\times$ 298 $\times$ log 100 = -2.303 $\times$ 2 $\times$ 298 $\times$ 2 = -2745.2 cal = -2.745 kcal

Q. The equilibrium constant for certain reaction is 100. If the value R is given to be 2 calK−1mol−1 then standard Gibb's free energy change will be