91Ó°ÊÓ

Vant Hoff's equation:

The equilibrium constant in chemical reactions is given by:

$K=e^{-\mathrm{Δ}G°/RT}\left(1\right)$

Where

$\mathrm{Δ}G°$is the Gibbs free energy:

Taking log on both sides

$\ln \left(K\right)=-\frac{\mathrm{Δ}G°}{RT}$

Differentiate both sides with respect to T

$\frac{d}{dT}\left(\ln \right(K\left)\right)=-\frac{1}{RT}\frac{d\mathrm{Δ}G°}{dT}+\frac{\mathrm{Δ}G°}{R{T}^2}\left(2\right)$

Change in free Gibbs energy

$dG=U-SdT+\mathrm{μ}dN$

At constant $N,dN=0$

$\frac{dG}{dT}=-S$

For standard Gibbs free energy, we have

$\frac{dG°}{dT}=-S°$

Taking difference, we get

$\frac{d\mathrm{Δ}G°}{dT}=-\mathrm{Δ}S°$

Substitute into (2), to get

$\frac{d}{dT}\left(\ln \right(K\left)\right)=\frac{\mathrm{Δ}S°}{RT}+\frac{\mathrm{Δ}G°}{R{T}^2}\left(3\right)$

The Gibbs free energy is:

$G=H-TS$

At constant temperature, the standard change

$\mathrm{Δ}G°=\mathrm{Δ}H°-T\mathrm{Δ}S°$

Substitute into (3)

$$\begin{gathered} \frac{d}{dT}\left(\ln \right(K\left)\right)=\frac{\mathrm{Δ}S°}{RT}+\frac{\mathrm{Δ}H°-T\mathrm{Δ}S°}{R{T}^2} \\ \frac{d}{dT}\left(\ln \right(K\left)\right)=\frac{\mathrm{Δ}S°}{RT}+\frac{\mathrm{Δ}H°}{R{T}^2}-\frac{\mathrm{Δ}S°}{RT} \\ \frac{d}{dT}\left(\ln \right(K\left)\right)=\frac{\mathrm{Δ}H°}{R{T}^2} \end{gathered}$$

Now that we have separated the variables, such as temperature in the RHS and natural logarithm in the LHS, we must integrate this equation.

$d\ln \left(K\right)=\frac{\mathrm{Δ}H°}{R{T}^2}dT$

Integrate both sides from $T_{1}to{T}_2$

$$\begin{gathered} {∫}_{T_1}^{T_2}d\ln \left(K\right)={∫}_{T_1}^{T_2}\frac{\mathrm{Δ}H°}{R{T}^2}dT \\ \left[\ln \right(K){]}_{T_1}^{T_2}=-{\left[\frac{\mathrm{Δ}H°}{RT°}\right]}_{T_1}^{T_2} \\ \ln \left(K\left(T_2\right)\right)-\ln \left(K\left(T_1\right)\right)=\frac{\mathrm{Δ}H°}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] \end{gathered}$$