Derive the van't Hoff equation,

$\frac{d\ln K}{dT}=\frac{\Delta H°}{R{T}^2}$

which gives the dependence of the equilibrium constant on temperature." Here $∆H°$is the enthalpy change of the reaction, for pure substances in their standard states (1 bar pressure for gases). Notice that if $∆H°$is positive (loosely speaking, if the reaction requires the absorption of heat), then higher temperature makes the reaction tend more to the right, as you might expect. Often you can neglect the temperature dependence of$∆H°$; solve the equation in this case to obtain

$\ln K\left(T_2\right)-\ln K\left(T_1\right)=\frac{\Delta H°}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Vant Hoff's equation:

The equilibrium constant in chemical reactions is given by:

$K=e^{-\Delta G°/RT}\left(1\right)$

Where

$\Delta G°$is the Gibbs free energy:

Taking log on both sides

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

Differentiate both sides with respect to T

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

Change in free Gibbs energy

$dG=U-SdT+\mu dN$

At constant $N,dN=0$

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

For standard Gibbs free energy, we have

$\frac{dG°}{dT}=-S°$

Taking difference, we get

$\frac{d\Delta G°}{dT}=-\Delta S°$

Substitute into (2), to get

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

The Gibbs free energy is:

$G=H-TS$

At constant temperature, the standard change

$\Delta G°=\Delta H°-T\Delta S°$

Substitute into (3)

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

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

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

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

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