Question

The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ.

H₂O(l)⇌ H₂O(g) AG°₂₉₈ = 8.58 kJ

a) Is the evaporation of water under standard thermodynamic conditions spontaneous?

b) Determine the equilibrium constant, Kp, for this physical process.

c) By calculating AG, determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water is 0.011 atm.

d) If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of PH₂O in air?

Answer 1

The evaporation of water at 298 K is nonspontaneous under standard thermodynamic conditions. The equilibrium constant, Kp, for the process is 0.031. Wet laundry can only dry when the value of PH₂O in the air is less than Kp or less than 0.031 atm.

Step-by-step explanation:

a) No, the evaporation of water under standard thermodynamic conditions is nonspontaneous.

b) The equilibrium constant, Kp, for the physical process is 0.031.

c) By calculating AG, we can determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, PH₂O, is 0.011 atm.

d) If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, the value of PH₂O in the air must be less than Kp or less than 0.031 atm.