Which is the correct relationship between Δ G° and equilibrium constant Kp ?

Question

Which is the correct relationship between Δ G° and equilibrium constant Kp ? (A) Kp = -RT log Δ G° (B) Kp = [e/RT]Δ G° (C) Kp = -Δ G° /RT (D) Kp = e-Δ G°/RT

Tardigrade Admin · Accepted Answer

Correct answer is (d) Kp = e-Δ G°/RT

Q. Which is the correct relationship between $Δ G^{°}$ and equilibrium constant $K_{p}$ ?

$K_{p} = - RT l o g Δ G^{°}$

$K_{p} = [e / RT]^{Δ G^{°}}$

$K_{p} = - Δ G^{°} / RT$

$K_{p} = e^{- Δ G^{°} / RT}$

Correct answer is (d) $K_{p} = e^{- Δ G^{°} / RT}$

Q. Which is the correct relationship between ΔG° and equilibrium constant Kp​ ?

Kp​=−RTlogΔG°

Kp​=[e/RT]ΔG°

Kp​=−ΔG°/RT

Kp​=e−ΔG°/RT

Correct answer is (d) Kp​=e−ΔG°/RT

Q. Which is the correct relationship between $Δ G^{°}$ and equilibrium constant $K_{p}$ ?

$K_{p} = - RT l o g Δ G^{°}$

$K_{p} = [e / RT]^{Δ G^{°}}$

$K_{p} = - Δ G^{°} / RT$

$K_{p} = e^{- Δ G^{°} / RT}$

Correct answer is (d) $K_{p} = e^{- Δ G^{°} / RT}$