The correct relationship between the free energy change in a reaction and the corresponding equilibrium constant K is

Question

The correct relationship between the free energy change in a reaction and the corresponding equilibrium constant K is (A) - textΔG texto text=RTlnK (B) Δ G=RTlnK (C) -Δ G=RTlnK (D)  textΔG texto text= textRTlnK

Tardigrade Admin · Accepted Answer

textΔG text= textΔG texto text+ textRTlnk At equilibrium Δ G=0 So - textΔG texto text= textRTlnK

Q. The correct relationship between the free energy change in a reaction and the corresponding equilibrium constant K is

$- ΔG^{o} =RTlnK$

$Δ G = RTl n K$

$- Δ G = RTl n K$

$ΔG^{o} = RTlnK$

$ΔG = ΔG^{o} + RTlnk$
At equilibrium
$Δ G = 0$
So $- ΔG^{o} = RTlnK$

Q. The correct relationship between the free energy change in a reaction and the corresponding equilibrium constant K is

−ΔGo=RTlnK

ΔG=RTlnK

−ΔG=RTlnK

ΔGo=RTlnK

ΔG=ΔGo+RTlnk At equilibrium ΔG=0 So −ΔGo=RTlnK

$- ΔG^{o} =RTlnK$

$Δ G = RTl n K$

$- Δ G = RTl n K$

$ΔG^{o} = RTlnK$

$ΔG = ΔG^{o} + RTlnk$
At equilibrium
$Δ G = 0$
So $- ΔG^{o} = RTlnK$