0.23 Free energy and thermodynamic equilibrium  (Page 3/6)

We next want to consider phase equilibrium between liquid and vapor. Our goal is to predict the temperatures and pressures at which equilibrium exists between liquid water and water vapor. Let’s start with the temperature at which liquid water and water vapor are at equilibrium when the pressure is 1 atm. We know the observed answer to this; it is the boiling point of water, 100 ºC.

Based on our reasoning, ∆S˚ - ∆H˚/T = 0 at the boiling point temperature at 1 atm pressure. Plugging in ∆Sº = 118.9 J/K·mol and ∆Hº = 44.0 kJ/mol, we can solve for T to find the equilibrium temperature at 370 K, which is very close to the observed boiling point of 373 K. (The slight error is that the values of ∆Sº and ∆Hº depend somewhat on the temperature, and we used values from data collected at 25 ºC. If we use the accurate values, we can predict the boiling point of water exactly.)

At other temperatures and pressures, the study of phase equilibrium involving vapor becomes more complicated because the entropy of a gas depends strongly on the pressure of the gas. This means that we cannot simply use the values of Sº , which are valid at 1 atm pressure. We’ll illustrate this by analyzing the thermodynamics of the evaporation of water. At equilibrium, this should lead us to predict the vapor pressure of water and, relatedly, the boiling point of water, but we have some work to do first.

As we recall, the entropy of one mole of vapor is much greater than the entropy of the same amount of liquid. A look back at Table 1 shows that, at 25 ºC, the entropy of one mole of liquid water is 69.9 J/K, whereas the entropy of one mole of water vapor at 1 atm is 188.8 J/K. This means that for the evaporation of one mole of liquid water at room temperature and 1 atm pressure, ∆ Sº>0. We might think that this means that a mole of liquid water at 25 ºC should spontaneously convert into a mole of water vapor, since this process would greatly increase the entropy of the water. We know, of course, that this does not happen. Learning from our previous study, we must also consider the energy associated with evaporation. The conversion of one mole of liquid water into one mole of water vapor results in the absorption of 44.0 kJ of energy from the surroundings. Recall that this loss of energy from the surroundings results in a significant decrease in the entropy of the surroundings. We combine both of these factors by calculating ∆S - ∆H/T for the evaporation of liquid water at 25 ºC:

∆S˚ - ∆H˚/T = 118.9 J/K·mol - (44.0 kJ/mol)/(298 K) = - 28.9 J/K·mol

We can repeat this calculation in terms of the free energy change:

ΔG˚ = ΔH˚ - 4ΔS˚

ΔG˚ = 44,000 J/mol – (298.15K)(118.9J/K mol)

ΔG˚ = 8.55 kJ/mol>0

How do we interpret these numbers? Since ∆S° - ∆H°/T<0 and ΔG˚>0, these calculations tell us that the evaporation of water at 25 ºC is not spontaneous. But note carefully that these calculations use the values of entropy and enthalpy changes valid for 1 atm pressure of the water vapor. So, a careful interpretation of these inequalities tells us that evaporation of water at 25 ˚C is not spontaneous when the pressure of the water vapor is 1 atm. We can also conclude from this result that the reverse process is spontaneous: the condensation of water vapor to liquid is spontaneous at 25 ˚C when the pressure of the water vapor is 1 atm. These predictions are all correct and match experimental observations. Although liquid water does have a vapor pressure at 25 ˚C, it is at 23.8 torr, very far below 1.00 atm. So water vapor at 1 atm pressure will condense spontaneously down to a low but non-zero pressure.