Question

1: Discuss the vaporization of water in standard condition at 298 K in term of: -1A exothermicity and endothermicity -1B change of entropy of the system and the surrounding -1C spontaneity of the reaction. ------ 2- Explain how the temperature and the composition of the two phases (liquid and vapour) influence the spontaneity of the vaporization of water. Please show steps and give ample explanation. Thank you

Answer 1

1a) The standard enthalpy of vaporization of water is +44 kJ/mol. This implies the water vaporization process is strongly endothermic. The energy must be supplied to convert water from liquid to vapor phase.

1b) Vaporization process is always accompined by increase in entropy as it leads to increase in disorder. So, ΔS is positive.

1c) Standard Gibbs free energy (ΔG^o) of formation of liquid water at 298 K is −237.17 kJ/mol and that of water vapour is −228.57 kJ/mol. Therefore, H₂O(l)H₂O(g)  ΔG ^o =8.43 kJ/mol. Since ΔG is positive it not a spontaneous process at standard conditions. Note that the standard conditions for gases correspond to 1 bar of partial pressure and it is less than 1 bar at atmospheric conditions.

2) For a reaction to be spontaneous ΔG should be negative.   

Since ΔG = ΔH-TΔS and ΔS is positive for vaporization process, increase in temperature will make -TΔS more negative. So, increase in temperature has postive effect on the spontanetiy of the process.

Composition of liqiuid and vapor phase has significant effect on the spontanetiy of the process. The comosition is directly related to partial pressure of water.

It is realted to free energy accroding to the realtion, G = G^∘+ RT ln (p/p_∘).

If p/p_∘ =0 (100 % dry) then  ln p/p_∘ would be negative infinity. That makes ΔG highly negative and  the process  highly spontaneous.

Small p/p_∘ values ( general the atmosphere )would still lead to the ΔG of water vapor that is more negative than liquid water. So water evaporation is still spontaneous.