Question

6. At 60°C, pure substance A has a vapor pressure of 380 mmHg and pure substance B of 140 mmHg. The mole fraction XB=0.15 is in the range of validity of Henry's law. The partial pressure of B at XB=0.1 is 20 mmHg. (a) What is the vapor partial pressure PB at XB=0.15? Answer: PB=30 mmHg. (b) What would PB be at XB=0.15 if the solution were ideal? Answer: PB=21 mmHg. (c) What is the activity coefficient of B, γB, at XB=0.15, using pure liquid B as the standard state for B? Answer: 1.428. (d) What is the molar free energy change ∆GM at XB=0.15 after mixing A and B? Answer: ∆GM= -1022.78j. (e) Is the attraction between A and B stronger or weaker than the attraction between A and A, and B and B? Explain your answer.

Answer 1

a) Given P_A^o = 380 mm Hg ; P_B^o = 140 mm Hg
By using Dalton' s law , P_total * X_B = P_B
   P_total ^* 0.1 = 20 mm Hg
   P_total = 200 mm Hg
at X_B = 0.15 ; P_B = 200 * 0.15 = 30 mm Hg

b) If the solution were ideal, just use the Raoult's law to solve
   P_B = X_B * P_B^o
   P_B = 0.15 * 140 mm Hg = 21 mm Hg

c) at X_B = 0.15 ;
   γ_B = P_B / P_B^o = 200 / 140 = 1.428571

d) molar mixing free energy is = ΔG_m = RT (X_A ln X_A + X_B ln X_B )
X_A = 1 - 0.15 = 0.85
  ΔG_m = 8.314 * (60+273.15) [ 0.85 * ln(0.85) + 0.15* ln(0.15) ] = -1170.823 J

e) as , γ_B > 1 the interactions between B-B repel each other, as the pure vapor pressure is increases in A
   the interactions between A-A will be stronger than B-B. Pure V.P exerted by A and B will be more than
Individual A and B itself so, the attractive forces between A-B will be strongest as its activity coeffcient
   decreases , which is usually less than zero.
NOTE:
If the interactions attract each other then γ<1, If they repel one another then γ>1
If there are no interactions then γ=1