Which of the following statements is correct for a reversible process in a state of equilibrium?

Question

Which of the following statements is correct for a reversible process in a state of equilibrium? (A) Δ G=2.303RTlogK (B) Δ G° =-2.303RTlogK (C) Δ G° =2.303RTlogK (D) Δ G=-2.303RTlogK

Tardigrade Admin · Accepted Answer

Δ G=Δ G° +2.303RTlogQ At equilibrium Δ G=0 and Q=Ke q=K 0=Δ G° +2.303RTlogK So Δ G° =-2.30RTlogK

Q. Which of the following statements is correct for a reversible process in a state of equilibrium?

$Δ G = 2.303 RTl o g K$

$Δ G^{\circ} = - 2.303 RTl o g K$

$Δ G^{\circ} = 2.303 RTl o g K$

$Δ G = - 2.303 RTl o g K$

$Δ G = Δ G^{\circ} + 2.303 RTl o g Q$

At equilibrium $Δ G = 0$ and $Q = K_{e q} = K$

$0 = Δ G^{\circ} + 2.303 RTl o g K$

So $Δ G^{\circ} = - 2.30 RTl o g K$

Q. Which of the following statements is correct for a reversible process in a state of equilibrium?

ΔG=2.303RTlogK

ΔG∘=−2.303RTlogK

ΔG∘=2.303RTlogK

ΔG=−2.303RTlogK

ΔG=ΔG∘+2.303RTlogQ At equilibrium ΔG=0 and Q=Keq​=K 0=ΔG∘+2.303RTlogK So ΔG∘=−2.30RTlogK

$Δ G = 2.303 RTl o g K$

$Δ G^{\circ} = - 2.303 RTl o g K$

$Δ G^{\circ} = 2.303 RTl o g K$

$Δ G = - 2.303 RTl o g K$

$Δ G = Δ G^{\circ} + 2.303 RTl o g Q$

At equilibrium $Δ G = 0$ and $Q = K_{e q} = K$

$0 = Δ G^{\circ} + 2.303 RTl o g K$

So $Δ G^{\circ} = - 2.30 RTl o g K$