Question

Carbon tetrachloride (CCl₄) and benzene (C₆H₆) form ideal solutions. Consider an equimolar solution of CCl₄ and C₆H₆ at 25 ˚C. The vapor above the solution is collected and condensed. Using the following data, determine

the composition in mole fraction of the condensed vapor.

Substance ΔG˚f

C6H6(l) 124.50 kJ/mol

C6H6(g) 129.66 kJ/mol

CCl4(l) 265.21 kJ/mol

CCl4(g) 260.59 kJ/mol

Short answer

C6H6(l) mole fraction = 0.155

CCl4(l) mole fraction = 0.554

Step-by-step solution

Step 1:

We have to calculate ΔG⁰ from given data.

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)
ΔG = 129.66-124.5= 5.16
For CCl4 use the same equation then we get 4.62
Then use Δ = -RTln(K)
For K for C6H6
5160=-8.314*298*ln(k)
K= .125
Same for CCl4 K will be as follows
K= 0.155

Step 2:

K gives information about how much each reaction is going to the right, and 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= 0.554