The correct relation between equilibrium constant (K), standard free energy (Δ G° ) and temperature (T) is

Question

The correct relation between equilibrium constant (K), standard free energy (Δ G° ) and temperature (T) is (A)  Δ G° =RT ln K  (B)  K=e-Δ G° /2303RT  (C)  Δ G° =-RT log 10K  (D)  K=10-Δ G° /2.303 RT

Tardigrade Admin · Accepted Answer

The relationship between equilibrium constant (K), standard free energy (Δ G° ) and temperature (T) is given as Δ G° =- 2.303 text RT log K -(Δ G° /2.303RT)= log K K=10-Δ G° /2.303 RT

Q. The correct relation between equilibrium constant (K), standard free energy $(Δ G^{\circ})$ and temperature (T) is

$Δ G^{\circ} = RT ln K$

$K = e^{- Δ G^{\circ} /2303 RT}$

$Δ G^{\circ} = - RT lo g_{10} K$

$K = 10^{- Δ G^{\circ} /2.303 RT}$

The relationship between equilibrium constant (K), standard free energy $(Δ G^{\circ})$ and temperature $(T)$ is given as $Δ G^{\circ} = - 2.303 RT lo g K$ $- \frac{Δ G ^{\circ}}{2.303 RT} = lo g K$ $K = 10^{- Δ G^{\circ} /2.303 RT}$

Q. The correct relation between equilibrium constant (K), standard free energy (ΔG∘) and temperature (T) is

ΔG∘=RTlnK

K=e−ΔG∘/2303RT

ΔG∘=−RTlog10​K

K=10−ΔG∘/2.303RT