0.6 Equilibrium and the second law of thermodynamics  (Page 10/11)

Recall that $\Delta (G)=\Delta (H)-T\Delta (S)< 0$ for a spontaneous process, and $\Delta (G)=\Delta (H)-T\Delta (S)=0$ at equilibrium. From these relations, we would predict that most (but not all) exothermic processes with $\Delta (H)< 0$ are spontaneous, because all such processes increase the entropy of the surroundings when they occur. Similarly, we would predict thatmost (but not all) processes with $\Delta (S)> 0$ are spontaneous.

We try applying these conclusions to synthesis of ammonia

at 298K, for which we find that $\Delta (S^{°})=-198\frac{J}{\mathrm{mol}K}$ . Note that $\Delta (S^{°})< 0$ because the reaction reduces the total number of gas molecules during ammonia synthesis, thus reducing $W$ , the number of ways of arranging the atoms in these molecules. $\Delta (S^{°})< 0$ suggests that [link] should not occur at all. However, $\Delta (H^{°})=-92.2\frac{\mathrm{kJ}}{\mathrm{mol}}$ . Overall, we find that $\Delta (G^{°})=-33.0\frac{\mathrm{kJ}}{\mathrm{mol}}$ at 298K, which according to [link] suggests that [link] is spontaneous.

Given this analysis, we are now pressed to ask, if [link] is predicted to be spontaneous, why does the reaction come to equilibrium withoutfully consuming all of the reactants? The answer lies in a more careful examination of the values given: $\Delta (S^{°})$ , $\Delta (H^{°})$ , and $\Delta (G^{°})$ 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 atmpressure. Thus, the fact that $\Delta (G^{°})< 0$ for [link] at standard conditions means that, if all three gases are present at 1 atm pressure, thereaction will spontaneously produce an increase in the amount of $N{H}_3$ . Note that this will reduce the pressure of the $N_2$ and $H_2$ and increase the pressure of the $N{H}_3$ . This changes the value of $\Delta (S)$ and thus of $\Delta (G)$ , because as we already know the entropies of all three gases dependon their pressures. As the pressure of $N{H}_3$ increases, its entropy decreases, and as the pressures of the reactants gases decrease, their entropies increase. The result isthat $\Delta (S)$ becomes increasingly negative. The reaction creates more $N{H}_3$ until the value of $\Delta (S)$ is sufficiently negative that $\Delta (G)=\Delta (H)-T\Delta (S)=0$ .

From this analysis, we can say by looking at $\Delta (S^{°})$ , $\Delta (H^{°})$ , and $\Delta (G^{°})$ that, since $\Delta (G^{°})< 0$ for [link] , reaction equilibrium results in production of more product and less reactant than atstandard conditions. Moreover, the more negative $\Delta (G^{°})$ is, the more strongly favored are the products over the reactants at equilibrium. By contrast, the more positive $\Delta (G^{°})$ is, the more strongly favored are the reactants over the products at equilibrium.

Thermodynamic description of the equilibrium constant

Thermodynamics can also provide a quantitative understanding of the equilibrium constant. Recall that thecondition for equilibrium is that $\Delta (G)=0$ . As noted before, $\Delta (G)$ depends on the pressures of the gases in the reaction mixture, because $\Delta (S)$ depends on these pressures. Though we will not prove it here, it can be shown by application of [link] to a reaction that the relationship between $\Delta (G)$ and the pressures of the gases is given by the following equation:

(Recall again that the superscript ° refers to standard pressure of 1 atm. $\Delta (G^{°})$ is the difference between the free energies of the products and reactants when all gases are at 1 atm pressure.) In this equation, $Q$ is a quotient of partial pressures of the gases in the reaction mixture.In this quotient, each product gas appears in the numerator with an exponent equal to its stoichiometic coefficient, and each reactantgas appears in the denominator also with its corresponding exponent. For example, for the reaction