0.16 Equilibrium and the second law of thermodynamics  (Page 8/11)

Thermodynamic description of phase equilibrium

As we recall, the entropy of vapor is much greater than the entropy of the corresponding amount of liquid. Alook back at [link] shows that, at 25°C, the entropy of one mole of liquid water is $69.9\frac{J}{K}$ , whereas the entropy of one mole of water vapor is $188.8\frac{J}{K}$ . Our first thought, based on our understanding of spontaneousprocesses and entropy, might well be that a mole of liquid water at 25°C should spontaneously convert into a mole of watervapor, since this process would greatly increase the entropy of the water. We know, however, that this does not happen. Liquid waterwill exist in a closed container at 25°C without spontaneously converting entirely to vapor. What have we leftout?

The answer, based on our discussion of free energy, is the energy associated with evaporation. The conversionof one mole of liquid water into one mole of water vapor results in absorption of $44.0\mathrm{kJ}$ of energy from the surroundings. Recall that this loss of energy from the surroundings results in a significant decrease in entropyof the surroundings. We can calculate the amount of entropy decrease in the surroundings from $\Delta (S_{\mathrm{surr}})=-\left(\frac{\Delta (H)}{T}\right)$ . At 25°C, this gives $\Delta (S_{\mathrm{surr}})=\frac{-44.0\mathrm{kJ}}{298.15K}=-147.57\frac{J}{K}$ for a single mole. This entropy decrease is greater than the entropy increase of the water, $188.8\frac{J}{K}-69.9\frac{J}{K}=118.9\frac{J}{K}$ . Therefore, the entropy of the universe decreases when one mole of liquid water converts to one mole of water vapor at 25°C.

We can repeat this calculation in terms of the free energy change: $$\Delta (G)=\Delta (H)-T\Delta (S)$$ $$\Delta (G)=44000\frac{J}{\mathrm{mol}}-298.15K118.9\frac{J}{K\mathrm{mol}}$$ $$\Delta (G)=8.55\frac{\mathrm{kJ}}{\mathrm{mol}}> 0$$ Since the free energy increases in the transformation of one mole of liquid water to one mole of water vapor, we predict that the transformation will not occurspontaneously. This is something of a relief, because we have correctly predicted that the mole of liquid water is stable at25°C relative to the mole of water vapor.

We are still faced with our perplexing question, however. Why does any water evaporate at 25°C? Howcan this be a spontaneous process?

The answer is that we have to be careful about interpreting our prediction. The entropy of one mole of water vaporat 25°C and 1.00 atm pressure is $188.8\frac{J}{K}$ . We should clarify our prediction to say that one mole of liquid water will not spontaneously evaporate to form one mole ofwater vapor at 25°C and 1.00 atm pressure. This prediction is in agreement with our observation, because we have found thatthe water vapor formed spontaneously above liquid water at 25°C has pressure 23.8 torr, well below 1.00 atm.

Assuming that our reasoning is correct, then the spontaneous evaporation of water at 25°C when no water vapor is present initially must have $\Delta (G)< 0$ . And, indeed, as water vapor forms and the pressure of the watervapor increases, evaporation must continue as long as $\Delta (G)< 0$ . Eventually, evaporation stops in a closed system when we reach thevapor pressure, so we must reach a point where $\Delta (G)$ is no longer less than zero, that is, evaporation stops when $\Delta (G)=0$ . This is the point where we have equilibrium between liquid andvapor.