Question

N₂(g)+3H₂(g)⇄2NH₃(g)

K=5.6×10⁵ at 298K
ΔH°rxn=−91.8kJ/molᵣₓₙ

The synthesis of NH₃ is represented by the equation above. Based on the equilibrium constant, K, and ΔH°rxn given above, which of the following can best be used to justify that the reaction is thermodynamically favorable at 298K and constant pressure?
A. ΔG°=−RT ln K>0 because K>>1
B. ΔG°=−RT ln K<0 because K>>1
C. ΔG°=ΔH°−TΔS°<0 because ΔH°<0 and ΔS°>0
D. ΔG°=ΔH°−TΔS°>0 because ΔH°<0 and ΔS°<0

Answer 1

Option B correctly justifies this by stating ΔG° = -RT ln K < 0 because K >> 1.

Step-by-step explanation:

The synthesis of ammonia (NH3) from nitrogen (N2) and hydrogen (H2) gases is represented by the chemical equation N2(g) + 3H2(g) ⇌ 2NH3(g). The given equilibrium constant, K, is very large (>1) at 298K, indicating that the reaction favors the production of ammonia at this temperature.

Additionally, the standard enthalpy change of the reaction (ΔH°rxn) is negative, meaning that the reaction is exothermic. To determine if the reaction is thermodynamically favorable, we can calculate the standard Gibbs free energy change (ΔG°) using the Gibbs free energy equation, ΔG° = ΔH° - TΔS°.

The correct justification for thermodynamic favorability at 298K and constant pressure is ΔG° = -RT ln K < 0 because K >> 1. For a reaction to be spontaneous, ΔG° must be negative. Therefore, option B is the correct choice, justifying the thermodynamic favorability of the reaction.