The correct equation of equilibrium constant in terms of Gibbs energy is

Question

The correct equation of equilibrium constant in terms of Gibbs energy is (A) K=e-Δ G° / R T (B) K=e-Δ G / R T (C) K=eΔ G° / R T (D) K=eΔ G / R T

Tardigrade Admin · Accepted Answer

Δ G=Δ G°+R T ln Q At equilibrium, when Δ G=0 and Q=KC, the equation becomes, Δ G =Δ G°+R T ln K=0 Δ G° =-R T ln K ln K =-Δ G° / R T Taking antilog on both sides, we get, K=e-Δ G° / R T

Q. The correct equation of equilibrium constant in terms of Gibbs energy is

$K = e^{- Δ G^{\circ} / RT}$

$K = e^{- Δ G / RT}$

$K = e^{Δ G^{\circ} / RT}$

$K = e^{Δ G / RT}$

$Δ G = Δ G^{\circ} + RT ln Q$
At equilibrium, when $Δ G = 0$ and
$Q = K_{C}$ , the equation becomes,
$Δ G = Δ G^{\circ} + RT ln K = 0$
$Δ G^{\circ} = - RT ln K$
$ln K = - Δ G^{\circ} / RT$
Taking antilog on both sides, we get,
$K = e^{- Δ G^{\circ} / RT}$

Q. The correct equation of equilibrium constant in terms of Gibbs energy is

K=e−ΔG∘/RT

K=e−ΔG/RT

K=eΔG∘/RT

K=eΔG/RT

ΔG=ΔG∘+RTlnQ At equilibrium, when ΔG=0 and Q=KC​, the equation becomes, ΔG=ΔG∘+RTlnK=0 ΔG∘=−RTlnK lnK=−ΔG∘/RT Taking antilog on both sides, we get, K=e−ΔG∘/RT

$K = e^{- Δ G^{\circ} / RT}$

$K = e^{- Δ G / RT}$

$K = e^{Δ G^{\circ} / RT}$

$K = e^{Δ G / RT}$

$Δ G = Δ G^{\circ} + RT ln Q$
At equilibrium, when $Δ G = 0$ and
$Q = K_{C}$ , the equation becomes,
$Δ G = Δ G^{\circ} + RT ln K = 0$
$Δ G^{\circ} = - RT ln K$
$ln K = - Δ G^{\circ} / RT$
Taking antilog on both sides, we get,
$K = e^{- Δ G^{\circ} / RT}$