0.6 Equilibrium and the second law of thermodynamics  (Page 9/11)

We can actually determine the conditions under which this is true. Since $\Delta (G)=\Delta (H)-T\Delta (S)$ , then when $\Delta (G)=0$ , $\Delta (H)=T\Delta (S)$ . We already know that $\Delta (H)=44.0\mathrm{kJ}$ for the evaporation of one mole of water. Therefore, the pressure of water vapor at which $\Delta (G)=0$ at 25°C is the pressure at which $\Delta (S)=\frac{\Delta (H)}{T}=147.6\frac{J}{K}$ for a single mole of water evaporating. This is larger than the value of $\Delta (S)$ for one mole and 1.00 atm pressure of water vapor, which as we calculated was $118.9\frac{J}{K}$ . Evidently, $\Delta (S)$ for evaporation changes as the pressure of the water vapor changes. We therefore need to understand why the entropy of the water vapordepends on the pressure of the water vapor.

Recall that 1 mole of water vapor occupies a much smaller volume at 1.00 atm of pressure than it does at theconsiderably lower vapor pressure of 23.8 torr. In the larger volume at lower pressure, the water molecules have a much largerspace to move in, and therefore the number of microstates for the water molecules must be larger in a larger volume. Therefore, theentropy of one mole of water vapor is larger in a larger volume at lower pressure. The entropy change for evaporation of one mole ofwater is thus greater when the evaporation occurs to a lower pressure. With a greater entropy change to offset the entropy lossof the surroundings, it is possible for the evaporation to be spontaneous at lower pressure. And this is exactly what weobserve.

To find out how much the entropy of a gas changes as we decrease the pressure, we assume that the number ofmicrostates $W$ for the gas molecule is proportional to the volume $V$ . This would make sense, because the larger the volume, the more places thereare for the molecules to be. Since the entropy is given by $S=k\ln W$ , then $S$ must also be proportional to $\ln V$ . Therefore, we can say that

We are interested in the variation of $S$ with pressure, and we remember from Boyle's law that, for a fixedtemperature, volume is inversely related to pressure. Thus, we find that

For water vapor, we know that the entropy at 1.00 atm pressure is $188.8\frac{J}{K}$ for one mole. We can use this and the equation above to determine the entropy at any other pressure. For a pressure of $23.8\mathrm{torr}=0.0313\mathrm{atm}$ , this equation gives that $S(23.8\mathrm{torr})$ ) is $217.6\frac{J}{K}$ for one mole of water vapor. Therefore, at this pressure, the $\Delta (S)$ for evaporation of one mole of water vapor is $217.6\frac{J}{K}-69.9\frac{J}{K}=147.6\frac{J}{K}$ . We can use this to calculate that for evaporation of one mole ofwater at 25°C and water vapor pressure of 23.8 torr is $\Delta (G)=\Delta (H)-T\Delta (S)=44.0\mathrm{kJ}-298.15K147.6\frac{J}{K}=0.00\mathrm{kJ}$ . This is the condition we expected for equilibrium.

We can conclude that the evaporation of water when no vapor is present initially is a spontaneous process with $\Delta (G)< 0$ , and the evaporation continues until the water vapor has reached itsthe equilibrium vapor pressure, at which point $\Delta (G)=0$ .

Thermodynamic description of reaction equilibrium

Having developed a thermodynamic understanding of phase equilibrium, it proves to be even more useful to examinethe thermodynamic description of reaction equilibrium to understand why the reactants and products come to equilibrium at the specificvalues that are observed.