At equilibrium, if Kp = 1, then

Question

At equilibrium, if Kp = 1, then (A) Δ G° > 1 (B) Δ G° < 1 (C) Δ G° = 0 (D) Δ G° = 1

Tardigrade Admin · Accepted Answer

At equilibrium, the relation between free energy of the reaction and equilibrium constant is as follows. Δ G°=-R T ln Kp If Kp=1 at equilibrium Δ G'=-R T ln 1 or Δ G°=0

Q. At equilibrium, if $K_{p} = 1$ , then

$Δ G^{\circ} > 1$

$Δ G^{\circ} < 1$

$Δ G^{\circ} = 0$

$Δ G^{\circ} = 1$

At equilibrium, the relation between free energy of the reaction and equilibrium constant is as follows.

$Δ G^{\circ} = - RT$ ln $K_{p}$

If $K_{p} = 1$ at equilibrium

$Δ G^{'} = - RT$ ln $1$

or $Δ G^{\circ} = 0$

Q. At equilibrium, if Kp​=1, then

ΔG∘>1

ΔG∘<1

ΔG∘=0

ΔG∘=1