Question

The standard Gibbs free energy change, $\Delta G^0$ is related to equilibrium constant, $K_p$ as:

• A. $K_p=-RTln\Delta G^0$
• B. $K_p={\left[\frac{e}{RT}\right]}^{\Delta G^0}$
• C. $K_p=-\frac{\Delta G}{RT}$
• D. $K_p=e^{-\Delta G^0/RT}$

Answer 1

Answer: (D)

$K_p=e^{-\Delta G^0/RT}$

According to Gibb's free energy. We know that

$\Delta G=\Delta G^0+RTlnKp$

Where,

$\Delta G$ = Change in Gibb's free energy,

$\Delta G^0$ = Standard Change in Gibb's free energy ,

$R=$ Universal Gas constant ,

$T=$ temprature in kelvin ,

$Kp=$ Equlibrium constant

For equilibrium condition $\Delta G=0$

So,

$0=\Delta G^0+RTlnKp$

$lnKp=-\Delta G^0/RT$

Now

$Kp=e^{-\Delta G^0/RT}$