The correct relationship between free energy and equilibrium constant K of a reaction is

Question

The correct relationship between free energy and equilibrium constant K of a reaction is (A) Δ G° = -RTlnK (B) Δ G = RTln K (C) Δ G° = RTlnK (D) Δ G = - RTln K

Tardigrade Admin · Accepted Answer

The relation between free energy change and equilibrium constant is given by Nernst equation

E = E^{\circ} - \frac{RT}{n F} l n Q

At equilibrium,

E_{C} = 0

and

Q = K_{C}

∴ E^{\circ} = \frac{RT}{n F} l n K

......(i)
Again

Δ G^{\circ} = n F E^{\circ}

.....(ii)
Put in (i)

\frac{- Δ G ^{\circ}}{n F} = \frac{RT}{n F} l n K; Δ G^{\circ} = - RTl n K

Q. The correct relationship between free energy and equilibrium constant K of a reaction is

$Δ G^{\circ} = - RTl n K$

$Δ G = RTl n K$

$Δ G^{\circ} = RTl n K$

$Δ G = - RTl n K$

Q. The correct relationship between free energy and equilibrium constant K of a reaction is

ΔG∘=−RTlnK

ΔG=RTlnK

ΔG∘=RTlnK

ΔG=−RTlnK

The relation between free energy change and equilibrium constant is given by Nernst equation E=E∘−nFRT​lnQ At equilibrium, EC​=0 and Q=KC​ ∴E∘=nFRT​lnK ......(i) Again ΔG∘=nFE∘ .....(ii) Put in (i) nF−ΔG∘​=nFRT​lnK;ΔG∘=−RTlnK

$Δ G^{\circ} = - RTl n K$

$Δ G = RTl n K$

$Δ G^{\circ} = RTl n K$

$Δ G = - RTl n K$