Which of the following graphs correctly represents the variation of thermodynamic properties of Haber's process ?

The overall reaction in Haber’s process is

$$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$$

Step-1 Sign of $$\Delta H^{\circ}$$
From standard data the reaction is strongly exothermic, $$\Delta H^{\circ} \lt 0$$ and its magnitude does not change much with temperature in the range used industrially. Hence a horizontal line below the $$\Delta H^{\circ}=0$$ axis is the correct representation for enthalpy.

Step-2 Sign of $$\Delta S^{\circ}$$
Four gaseous moles (1 + 3) change to two moles. Disorder therefore decreases, so $$\Delta S^{\circ}\lt 0$$. A horizontal line below the $$\Delta S^{\circ}=0$$ axis is therefore correct for entropy.

Step-3 Temperature-dependence of $$\Delta G^{\circ}$$
The Gibbs free-energy change is given by $$\Delta G^{\circ}= \Delta H^{\circ}-T\Delta S^{\circ}$$. Substituting the signs determined above:

$$\Delta G^{\circ}= (\text{negative})-T(\text{negative})$$

$$\Rightarrow\;\Delta G^{\circ}= \Delta H^{\circ}+|\,\Delta S^{\circ}\,|\,T$$

The second term is positive and grows linearly with temperature. Therefore:

• at low temperature the negative $$\Delta H^{\circ}$$ term dominates and $$\Delta G^{\circ}\lt 0$$ (reaction spontaneous),
• as temperature rises, the straight-line graph of $$\Delta G^{\circ}$$ versus $$T$$ has positive slope and eventually crosses the $$\Delta G^{\circ}=0$$ axis, beyond which $$\Delta G^{\circ}\gt 0$$ (reaction non-spontaneous).

Step-4 Qualitative behaviour of the equilibrium constant $$K$$
$$\Delta G^{\circ} = -RT\ln K \; \Longrightarrow \; \ln K = -\dfrac{\Delta G^{\circ}}{RT}$$. Because $$\Delta G^{\circ}$$ increases with $$T$$, the value of $$K$$ (or $$\ln K$$) falls as temperature increases. Hence a plot of $$\ln K$$ against $$1/T$$ is a straight line with negative slope.

Step-5 Matching with the given graphs
Among the four sketches supplied in the question, only Option A simultaneously shows

• a horizontal negative line for $$\Delta H^{\circ}$$,
• a horizontal negative line for $$\Delta S^{\circ}$$,
• a straight line for $$\Delta G^{\circ}$$ rising with $$T$$ and intersecting the axis,
• and, if included, a downward straight line for $$\ln K$$ vs $$1/T$$.

No other option satisfies all these thermodynamic requirements.

Hence the correct choice is
Option A which is: the graph showing $$\Delta H^{\circ}\lt 0$$ (constant), $$\Delta S^{\circ}\lt 0$$ (constant), $$\Delta G^{\circ}$$ increasing linearly with $$T$$, crossing zero, and $$\ln K$$ decreasing with $$T$$.