Carbon tetrachloride (CCl4) and benzene (C6H6) form ideal solutions. Consider an equimolar solution of CCl4 and C6H6 at 25°C. The vapor above this solutionis collected and condensed. Using the following data, determine the composition in mole fraction of the condensed vapor.

Question

Substance ∆Gf°
C6H6 (l) 124.50 kJ/mol
C6H6 (g) 129.66 kJ/mol
CCl4 (l) -65.21 kJ/mol
CCl4 (g) -60.59 kJ/mol

Answer 1

i was hoping someone else would answer

Answer 2

4.04

Answer 3

To solve the problem, you need to find the relative vapor pressures for the two species. Since you're given the delta g values you can calculate the delta g for the reaction of the liquid going to vapor
C6H6(l)->C6H6(g)
Delta g= 129.66-124.5= 5.16
Do the same for CCl4 you'll get 4.62
Then use delta g= -RTln(K)
And solve for K
For example, for C6H6
5160=-8.314*298*ln(k)
K= .125
Do same for CCl4 and you'll get
K= 0.155
Since k indicates how much each reaction is going to the right, it is proportional to vapor pressure. Therefore, the relative vapor pressure of CCl4 to C6H6 is K(CCl4)/K(C6H6)= 0.155/0.125=1.24
So when vapor pressure C6H6 is 1, the vapor pressure of CCl4 is 1.24. Use the vapor pressure/tot pressure to get mole fraction in vapor:
Mole fraction C6H6= 1(0.5)/(1.24(.5)+1(.5))= 0.446
Where .5 is mole fraction in solution because they are equimolar
Then
Mole fraction CCl4= 1-.446= .554