In the Haber process, the chemical reaction is as follows:

$ \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3 $

The change in enthalpy ($\Delta \text{H}^\circ$) is negative, indicating that the reaction is exothermic.

The change in entropy ($\Delta \text{S}^\circ$) is also negative because the number of moles of gas decreases during the reaction.

Key points to consider:

As temperature increases, the term $-\frac{\Delta H_R^0}{T}$ decreases.

The free energy change is given by:

$ \Delta \text{G}^\circ = -\text{RT} \ln \text{K}_{\text{eq}} $

Rearranging gives:

$ \text{R} \ln \text{K}_{\text{eq}} = -\frac{\Delta \text{G}^0}{T} $

For an exothermic reaction, as temperature increases, the equilibrium constant ($ \text{K}_{\text{eq}} $) decreases.

Both $\Delta \text{H}^\circ$ and $\Delta \text{S}^\circ$ are almost constant with respect to temperature.