In the lab you chose to mix two substances. (Their moles have been calculated by

Question

In the lab you chose to mix two substances. (Their moles have been calculated by you previously in lab and are listed at the bottom). If it was a regular day, and the solution was treated as ideal, at 298 K, compute the partial pressures of both substances. Show what the total pressure is of the mixture and the Gibbs Energy. How would this differ it the solution was not treated as ideal(note for this part explain in short, no need to further calculate.) Show all work for credit.

For substance 1, the point it starts to boil is T*= 383K and its DeltaH Vap = 39kj/mol. For substance 2 its T* = 353K and DeltaHvap = 30.82 kj/mol.

substance 1 = 0.109 mol

substance 2 = 0.64

please use the formula p = p* X e^ (DeltaH/R)(1/T-T*)

Explanation / Answer

We do not have the vapor pressure of the gases at 298 K. To calculate partial pressure from Raoult's Law we need the vapor pressures at 298 K. We can do this using the Clausius Clapeyron equation:

ln(P1/P2) = (Hvap/ R) ( 1/T2 -   1/T1)

We know that vapor pressure of any liquid at the boiling point is equal to the pressure on the gas (Atmospheric pressure in this case). Take P1 = 1 atm and T1 = Boiling point, then we can calculate P2 at 298 K.

For the first component:

ln(P1/P2) = (Hvap/R)(1/T2 -   1/T1)

or, ln( 1 atm /P2) = (39000 J.mol-1) / (8.314 J.mol-1.K-1 )(1/298 -   1/383)

or, 1 atm /P2 = e3.49

or, P2 = 0.03 atm

Similarly, for the second component:

ln(P1/P2) = (Hvap/R)(1/T2 -   1/T1)

or, ln( 1 atm /P2) = (30820 J.mol-1) / (8.314 J.mol-1.K-1 )(1/298 -   1/353)

or, 1 atm /P2 = e1.94

or, P2 = 0.144 atm

Now we can use Raout's Law. Total no of moles = 0.109 + 0.64 = 0.749 moles.

Mol fraction of component 1 = 0.109/0.749 = 0.145

Mol fraction of component 2 = 1-0.145 = 0.855

Partial pressure of component 1 = 0.03 atm * 0.145 = 0.0044 atm

Partial pressure of component 2 = 0.144 atm * 0.855 = 0.123 atm

Total pressure = 0.0044 + 0.123 = 0.1274 atm

GIbb's Free energy of the mixture = nRT (x1lnx1 + x2lnx2) ; n= total moles, x = mole fration

= 0.749 moles * (8.314 J.mol-1.K-1) * 298 K (0.145 ln 0.145   + 0.855 ln 0.855)

= - 768.26 J

If the solutions are not ideal the above criteria (Gibb's Energy and Raoult's Law) will not be fulfilled. It will either show positive deviation or negative deviation from the Raoult's Law. If the inter-molecular attraction of the same species is higher than the inter-molecular attraction of different species, then positive deviation will occur. In case of positive deviation vapor pressure is always greater than it is in case of ideal solutions.

If the inter-molecular attraction of the same species is lower than the inter-molecular attraction of different species, then negative deviation will occur. In case of negative deviation vapor pressure is always lower than it is in case of ideal solutions.

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