Question

N2(g) + 3 H₂(g) ⇌ 2 NH₃ (g)

K = 5.6 x 105 at 298 K
∆H°ᵣₓₙ = -91.8 kJ/molrxn

The synthesis of NH₃ is represented by the equation above. Based on the equilibrium constant, K, and ∆H°ᵣₓₙ given above, which of the following can best be used to justify that the reaction is thermodynamically favorable at 298 K and constant pressure?

A. ∆G° = - RT In K > 0 because K >>1
B. ∆G° = - RT In K < 0 because K >>1
C. ∆G° = ∆H° - T∆S° < Obecause ∆H° < 0 and ∆S > 0
D. ∆G° = ∆H°-T∆S°> O because ∆H° <0 and ∆S < 0

Answer 1

The synthesis of ammonia is thermodynamically favorable at 298 K due to a large equilibrium constant (K) and a negative change in enthalpy, leading to a negative Gibbs free energy change.

Step-by-step explanation:

The synthesis of ammonia (NH₃) via the reaction N2(g) + 3 H2(g) ⇌ 2 NH3 (g) is thermodynamically analyzed using the provided equilibrium constant (K) and change in enthalpy (ΔH°rxn).

Given that the reaction has a large equilibrium constant (K = 5.6 x 105 at 298 K), and a negative change in enthalpy (ΔH°rxn = -91.8 kJ/molrxn), we can determine the spontaneity of the reaction by calculating the Gibbs free energy change (ΔG°). The relationship between ΔG°, K, and T (absolute temperature) is given by the equation ΔG° = - RT ln K, where R is the universal gas constant and T is the temperature in kelvins.

Because K is much greater than 1, ln K will be positive, and, since ΔH°rxn is negative, we can deduce that TΔS° must also be negative to result in a negative ΔG° (ΔG° = ΔH° - TΔS°). Therefore, the correct justification is: B. ΔG° = - RT ln K < 0 because K >> 1, indicating that the reaction is thermodynamically favorable at 298 K and constant pressure.