How to solve this delta G problem? (gen chem 2)

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) ΔG°=8.58 kJ

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

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

(c) By calculating ∆G, 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, what must be the value of PH2O in the air?

Note: Start by writing the Kp expression, in terms of the pressures of the

species involved (applying our usual rules). Also consider how the term “standard” applies in

this context. In part (c), the partial pressure mentioned refers to that in the air above the water

sample. Note that all parts can be answered by mathematics, starting with the value of ΔGo for

the given process. But also try to relate this to the real-world behavior of a beaker – or puddle –

of water

Answer 1

Lots to unpack here.

H2O(l)⇌H2O(g) ΔG°=8.58 kJ

(a). Since dG^o is positive the reaction is NOT spontaneous under standard conditions

(b). @equilibrium dG^o = -RT lnK. Solving for K we have …

8.58 = -(0.008314)(298) lnK

lnK = -3.46

K = 0.0313 (note K is <1 consistent with the reaction being non spontaneous)

(c). dG = dG^o + RT lnQ = 8.58 + (0.008314)(298) * ln(0.011)

dG = 8.58 + (-11.17)

dG = -2.59 kJ/mole

Since dG is negative, the evaporation would be spontaneous under these conditions

(d). Solve the equation in (c) setting dG to zero and solving for Q. As soon as dG is negative, the evaporation will be spontaneous. 0 = 8.58+(.008314)(298)lnQ

0 = 8.58 + 2.48 lnQ

lnQ = -3.50

Q = 0.031

So pH2O would have to be 0.031 atm or less for the evaporation to be spontaneous at 298K.

Be sure to check all the math.