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

a) Spontaneous
b) Nonspontaneous
c) Equilibrium
d) Zero entropy change

1 Answer

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Final answer:

a) Spontaneous

The evaporation of water at 298 K is nonspontaneous under standard conditions with ΔG° of 8.58 kJ. However, when PH₂O is 0.011 atm, evaporation becomes spontaneous. For drying laundry, the PH₂O must be below 0.031 atm.

Step-by-step explanation:

The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ. According to thermodynamic principles, a process is considered spontaneous if the free energy change (ΔG) is negative.

Given that the ΔG for this process is positive, we can determine the spontaneity of the process as follows:

  • (a) Nonspontaneous because ΔG° > 0 under standard conditions.
  • (b) The equilibrium constant, Kp, for the evaporation of water is 0.031.
  • (c) When the partial pressure of water, PH₂O, is 0.011 atm, the calculation of ΔG indicates that the evaporation of water at 298 K is indeed spontaneous under these conditions.
  • (d) For laundry to dry, PH₂O in the air must be less than Kp or less than 0.031 atm, otherwise the evaporation of water would be nonspontaneous.
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