1. http://iweb.tntech.edu/snorthrup/chem1120/quiz1-18.html Quiz 1

2. CHAPTER 2: Thermodynamics description of mixtures

3. Partial molar quantities
Fundamental Equation of Chemical Thermodynamics (The Gibbs-Duhem Equation)
The thermodynamics of mixing (Enthalpy and Entropy of mixing)
Ideal solution and Raoult’s Law

6. Partial molar quantities molarity
MOLES, MASS, MOLARCONCENTRATION
molarity
C = n/V
C = m/V

7. Exercise What mass of glycine should be used to make 250 mL of a solution of molar concentration 0.15M NH2CH2COOH(aq)?

8. molality

9. Exercise Calculate the mole fraction of sucrose in an aqueous sample of molality 1.22 mol kg-1.

10. Exercise
Use the figure in next slide to calculate the density of a mixture of 20 g of water and 100 g of ethanol.
Solution: First calculate the mole fractions.
20 g H2O = 1.11 mol; 100 g EtOH = mol
xH2O = 0.34; xEtOH = 0.66
Then interpolate from the mixing curve (next slide):
Partial molar volume VH2O = 17.1 cm3 mol-1;
Partial molar volume VEtOH = 58.1 cm3 mol-1
Then plug the moles and partial molar volume
(1.11 mol)(17.1 cm3/mol) + (2.17 mol)(58.1 cm3/mol) =19.0 cm cm3 = 144 cm3
Finally, the total mass is divided by the total volume:
120 g/144 cm3 = 0.83 g/ cm3

11. Exercise: the Interpolation
read EtOH over here read water here

12. z Thermodynamics G VdP SdT - =
For other temperatures and pressures we can use the equation:
dG = VdP – SdT (ignoring DH for now)
where V = volume and S = entropy (both molar)
We can use this equation to calculate G for any phase at any T and P by integrating z G
VdP
SdT T P 2 1 - =
To integrate properly we must know how V and S vary with P and T (hence the calculus), but we shall simplify the math and assume V and S are constant
This simplifies our math considerably (but may lead to some errors)
We can check the validity of our assumptions by comparing to experiments or more rigorous calculations (performed by computer)

13. The Wider Significance of the Chemical Potential
(Classroom exercise)

14. The Gibbs-Duhem Equation
Gibbs-Duhem equation applies to ALL partial molar quantities:
X=V, μ,H,A,U,S, etc

15. Using The Gibbs-Duhem Equation
Aqueous solution of K2SO4:
Given the molar volume of water at 298K is mL/mol,
find the partial molar volume of water.

17. The Thermodynamics of Mixing
Is mixing ( composition change) spontaneous?

18. Mixing of two perfect gases or two liquids that form an ideal solution:

19. Exercise
Suppose the partial pressure of a perfect gas falls from 1.00 bar to 0.50 bar as it is consumed in a reaction at 25 oC. What is the change in chemical potential of the substance?
Solution: We want mJ,f - mJ,i But mJ,i = mJo so we want mJ ! mJo = RT ln (pJ/po) = (2.479 J/mol) (ln 0.50) = -1.7 kJ/mol

20. Exercise: at 25C, calculate the Gibbs energy change when the
partition is removed

21. Other Thermodynamic Mixing Functions
Entropy of mixing

22. Entropy of mixing of perfect gases

23. Other Thermodynamic Mixing Functions
Enthalpy of mixing
For perfect gases,
Understandable?

24. Ideal Solutions
Pure substance:
Vapor pressure
Solute:

25. Exercise
A solution is prepared by dissolving 1.5 mol C10H8 in 1.00 kg benzene. The v.p. of pure benzene is 94.6 torr at this temperature (25oC). What is the partial v.p. of benzene in the solution?
Solution: We can use Raoult’s law, but first we need to compute the mole fraction of benzene. MM benzene = 78.1 g/mol, so 1.00 kg = 12.8 mol.
xbenz = 12.8 mol / (12.8 mol mol) = 0.895
pbenz = xbenz p*benz = (0.895)(94.6 torr) = 84.7 torr

26. Exercise
By how much is the chemical potential of benzene reduced at 25oC by a solute that is present at a mole fraction of 0.10?
Solution: We want Dmbenz = mbenz ! m*benz
But mbenz = m*benz + RT ln xbenz so mbenz ! m*benz = RT ln xbenz
And if xsolute = 0.10, then xbenz = 0.90
Thus Dmbenz = (2.479 kJ/mol) (ln 0.90) = ! 0.26 kJ/mol

27. Exercise
What partial pressure of methane is needed to achieve 21 mg of methane in 100 g benzene at 25oC?
Solution: From Table 7.1, KB = 4.27 × 105 torr. mol CH4 = g / g mol!1 = mol. mol C6H6 = 100 g / 78.1 g mol!1 = 1.28 mol
xCH4 = mol / (1.28 mol + ~0 mol) =
pCH4 = KCH4 xCH4 = (4.27 × 105 torr)(0.0010) = 436 torr = 4.3 × 102 torr