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

Question

The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ. H_2O(l) H_2O(g) DG degree _298 = 8.58 kJ Is the evaporation of water under standard thermodynamic conditions spontaneous? Determine the equilibrium constant. K_p. for this physical process. By calculating Delta G, determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water is 0.011 atm. If the evaporation of water were always nonspontancous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry. what must be the value of P_H_2O in the air?

Explanation / Answer

A)it is non spontaneous process because value of dG is positive

B) dG0 = - 2.303RTlog k from this formula we can determine the value of by taking antilog on both side

C) As the above formula suggest we can calulate the value of dG by assuming the atmosphere pressure equal to 1atm and applying these formula value of dG comes out to be positive. And hence this process is non spontaneous

D) In the laundry electromagnetic radiation from the sun provide necessary activation energy for dring of clothes. For drying process of in laundry the partial pressure of water at clothes should be more than atmospheric pressure.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.