Question

Use the following data on ethanol-chloroform solutions at 35°C to calculate the activity coefficients of ethanol and chloroform based on the deviations from Raoult’s Law. Explain the meaning of your answers.

c EtOH, liq	0	0.2	0.4	0.6	0.8	1.0
Y EtOH, vap	0.000	0.138	0.186	0.255	0.425	1.000
Ptot (kPa)	39.35	40.56	38.69	34.39	25.36	13.70

Can someone help me with this please!? Thank you!!

Answer 1

Partial pressure of component in a mixture = total pressure x mole fraction of component

At 35 oC

P*(EtOH) = 13.705 kP

P*(CHCl3) = 39.34 kPa

activity coefficient = Partial pressure of component/(P* x mole fraction of component)

Table completed:

c[EtOH]       E[CHCl3]       Ptot[kPa]     P[EtOH]        P[CHCl3]        Y[EtOH]     Y[CHCl3]

   0                     0                39.35           0                       0                    0                0

  0.2               0.138             40.56         8.112             5.597               2.96           0.178

  0.4               0.186             38.69        15.476            7.196               2.82           0.305

  0.6               0.255             34.39        20.634            8.769               2.51           0.557

  0.8               0.425             25.36        20.288           10.778              1.85           1.370

  1. 0                1.0               13.70         13.70            13.700               1.0               0

   Here,

Y[EtOH] and Y[CHCl3] are acitivty coefficients for EtOH and CHCl3

the results show activity coefficient of liquid is greater than the vapor CHCl3 at any given time in the mixture.