0.23 Free energy and thermodynamic equilibrium  (Page 5/6)

We can conclude that the evaporation of water when no vapor is present initially is a spontaneous process with ΔG<0, and the evaporation continues until the water vapor has reached the equilibrium vapor pressure, at which point ΔG = 0. Equivalently, the condensation of water vapor when the pressure is 1 atm is spontaneous with ∆G<0 until enough water vapor has condensed to reach the equilibrium vapor pressure, at which point again ∆G = 0. Thermodynamics tells us exactly what the conditions are for phase equilibrium!

Observation 2: thermodynamic description of reaction equilibrium

Now that we have developed a thermodynamic understanding of phase equilibrium, it is even more useful to examine the thermodynamic description of reaction equilibrium. We want to understand why the reactants and products come to equilibrium at the specific concentrations and pressures that are observed.

Recall that ΔG = ΔH – TΔS<0 for a spontaneous process, and ΔG = ΔH – TΔS = 0 at equilibrium. From these relations, we would predict that most (but not all) exothermic processes with ΔH<0 are spontaneous, because all such processes increase the entropy of the surroundings when they occur. Similarly, we would predict that most (but not all) processes with ΔS>0 are spontaneous.

We try applying these conclusions to synthesis of ammonia

N ₂ (g) + 3 H ₂ (g) → 2 NH ₃ (g)

at 298 K, for which we find that ΔS˚ = 198 J/mol·K. Note that ΔS˚<0 because the reaction reduces the total number of gas molecules during the ammonia synthesis, thus reducing W , the number of ways of arranging the atoms in these molecules. ΔS˚<0 suggests that N ₂ and H ₂ should not react at all to produce NH ₃ . On the other hand, ΔH˚ = -92.2 kJ/mol from which we can calculate that ΔG˚ = -33.0 kJ/mol at 298 K. This suggests that N ₂ and H ₂ should react completely to form NH ₃ . But in reality, we learned in the Concept Development Study of reaction equilibrium that this reaction comes to equilibrium, with N ₂ , H ₂ , and NH ₃ all present. We determined the equilibrium constant

${\text{K}}_{\text{p}}=\frac{\text{P}{\left(\text{N}{\text{H}}_3\right)}^2}{\text{P}\left({\text{N}}_2\right)·\text{P}{\left({\text{H}}_2\right)}^3}=6\times {10}^5$

Why does the reaction come to equilibrium without fully consuming all of the reactants?

The answer, just as before, lies in a more careful interpretation and application of the values given: ΔS˚, ΔH˚, and ΔG˚ are the values for this reaction at standard conditions, which means that all of the gases in the reactants and products are taken to be at 1 atm pressure. Thus, the fact that ΔG˚<0 for Reaction (6) at standard conditions means that, if all three gases are present at 1 atm pressure, the reaction will spontaneously proceed to increase the amount of NH ₃ .

From this analysis, we can say that, since ΔG˚<0 for Reaction (6), reaction equilibrium results in the production of more product and less reactant than at standard pressures. This is consistent with our previous measurement that K _p = 6×10 ⁵ , a number much, much larger than 1. As a result, we can say that Reaction (6) is “thermodynamically favorable,” meaning that thermodynamics accurately predicts that the equilibrium favors products over reactants.