1. Reaction in Haber’s process

   Haber’s process is: $N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

2. Thermodynamic changes involved

   For this reaction:

   • It is exothermic, so $\Delta H<0$
   • The number of gaseous moles decreases from $4$ to $2$, so disorder decreases. Hence, $\Delta S<0$

3. Use Gibbs free energy relation

   $\Delta G=\Delta H-T\Delta S$

   Since both $\Delta H$ and $\Delta S$ are negative, let $\Delta H=-a,\quad \Delta S=-b \qquad (a,b>0)$

   Then, $\Delta G=-a-T(-b)=-a+bT$

   So $\Delta G$ varies linearly with temperature with:

   • positive slope $=b$
   • negative intercept $=-a$

   Thus, as temperature increases, $\Delta G$ increases.

4. Interpretation for Haber’s process

   At low temperature, $\Delta G$ is more negative, so formation of ammonia is more thermodynamically favorable.

   Therefore the correct graph should show:

   • $\Delta H$ approximately constant and negative
   • $\Delta S$ approximately constant and negative
   • $\Delta G$ increasing linearly with $T$

5. Matching with options

   The option that represents this behavior is Option A.

Final Answer

$\boxed{A}$