Download Mixing Ideal Mixtures Ideal Mixtures Colligative Properties Van t`Hoff

Mixing
• ∆G = ∆H - T∆S
• BIG IDEA: mixing of a solute with a solvent has a positive
(favorable) entropic component:
∆Smix = –R (x ln x + y ln y); 0.7R for 0.5:0.5
per mole of mixture
x and y are molar fractions < 1; ln x and ln y < 0; ∆Smix > 0
• WRONG INTERPRETATION: Dissolving any solute in any
solvent in any concentration is spontaneous and favorable.
• WHY NOT? Intermolecular interactions make ∆H and more
T∆S.
Ideal Mixtures
=
=
• ∆Hmix = 0 ; ∆Gmix = - T∆
∆Smix
• Components of ideal mixtures spontaneously and
favorably mix in all proportions.
• Usually chemically similar components.
⇔
Ideal Mixtures
• Problem: Toluene and benzene are similar non-polar
chemicals that mix in all proportions to form an ideal solution.
Estimate the total ∆H, ∆S, and ∆G of mixing 0.5 mol toluene
with 1.5 mol benzene at T = 300 K.
• Solution:
For an ideal solution, ∆Hmix = 0 ; ∆Gmix = –T∆Smix
Molar fractions of components x = 0.25 ; y = 0.75
∆Smix = –R (x ln x + y ln y) (per mole)
∆Smix = – 2 cal / (mol K) × (– 0.56) = 1.12 cal / (mol K) (per mole)
For two moles of the resulting mixture, ∆Smix = 2.24 cal / K
∆Gmix = –T∆Smix = – 300 K × 2.24 = – 672 cal
• Answer: ∆Hmix = 0 ; ∆Smix = 2.24 cal / K ; ∆Gmix = – 672 cal
Van t’Hoff Factor
• Problem: When the pharmaceutical formulation of
drug X, [X]2Ca2+ is dissolved in water, 30% of the
molecules dissociated into three ions, 30% into two
ions, and 40% did not dissociate. Calculate the van't
Hoff factor for the solution.
• Solution:
3×0.3 + 2×0.3 + 1×0.4 = 0.9+0.6+0.4 = 1.9
OR: Every 100 molecules will produce 3×30 + 2×30 +
1×40 = 190 particles.
• Answer: the van t’Hoff factor is 1.9 (to be used for
calculation of all colligative properties)
Colligative Properties
• At low concentrations of the solute…
• Adding solute lowers µ of water:
∆µwater = RT ln (1 – xsolute) ~ –RT xsolute
• Makes the liquid state more favorable.
• Consequently:
Equilibrium water vapor pressure decreases
Freezing point depresses
Boiling point elevates
Semipermeable membrane ⇒ osmosis
• All effects are proportional to molar fraction or
(os)molality of the solute (think dissociation!)…
• … and do not depend on the nature of the solute.
Major Equations
• Equilibrium water vapor pressure decreases:
∆Pw = xsolute Pw∗
Raoul’s Law, non-volatile solutes, x = molar fraction
• Freezing point lowers: ∆Tf = Kf xsolute
Kf = 1.86 K⋅kg / mol for water, x is molality
• Boiling point elevates: ∆Tb = Kb xsolute
Kb = 0.512 K⋅kg / mol for water, x is molality
• Semipermeable membrane ⇒ osmotic pressure:
∆PosmV = ∆nB•RT or ∆Posm = ∆[B]•RT
R ~ 0.082 L⋅atm / (K⋅mol); RT ~ 24.6 L⋅atm / mol
What if the solute is volatile?
vapor
vapor
water
Semipermeable
membrane
water
water
ice
Pw < Pw ∗
•
•
•
•
Tb > Tb ∗
Tf < Tf ∗
P > P∗
Raoul’s Law
• Problem: The pure water vapor pressure at 47°C is 0.1 bar.
Estimate the vapor pressure when 32 g of NaCl (MW = 58
g/mol) is added to 1 L of water. Assume that the salt is not
volatile, but dissociates completely.
• Solution:
32 g = 32/58 ~ 0.55 mol of salt
Dissolved AND DISSOCIATED in 1L = 55 mol of water
Molar fraction of the solute ~ 2 × 0.55 / 55 = 0.02.
Raoult's Law, ∆Pw = xsolute Pw∗ = 0.02 * 0.1 = 0.002 bar
Pw = Pw∗ – ∆Pw = 0.1 - 0.002 = 0.098 bar
• Answer: The new vapor pressure is ~ 0.098 bar.
• Bonus: now estimate the vapor pressure when another 16 g
of NaCl is added to the solution.
• Solution: Adding 32 g NaCl decreases VP by 0.002; adding
48 g should decrease it by 0.003. The new VP is 0.097 bar.
Osmotic pressure
• Problem: The blood glucose level of a diabetic
patient is approximately 0.198 g/dL (dL = 0.1L).
Given that glucose MW = 180.16 g/mol, calculate
the osmotic pressure created by glucose.
• Solution:
0.198 g/dL = 1.98 g/L ~ 11 mM = osmolarity (no
dissociation occurs, so van t’Hoff factor = 1)
∆Posm = 11 mOsm/L x 25 L atm/mol = 0.275 atm = 209
mmHg
• Answer:
209 mmHg
E.g. any gas, ethyl alcohol…
Solute vapor pressure is proportional to concentration
Henry’s Law: Psolute = xsolute ⋅ Kx
Henry’s Law constant Kx – specific values for each
solvent and solute; depend on interpretation of xsolute
(molar fraction, molarity, molality…)
• In a mixture of gases, Dalton’s Law:
Pgas = xgas ⋅ Ptotal; xgas is molar fraction
Tb and Tf changes
• Problem: After dissolving a certain amount of a nonionizable drug in water, the boiling temperature of
the solution, Tb, was elevated by 0.3°C. What will
happen if we add the same number of moles of
NaCl to this drug solution?
• Solution:
NaCl completely dissociates
The total osmolarity of the solution will triple
So will the Tb increase
• Answer: Tb will be further elevated by 0.6°C (total
increase of 0.9°C)
Henry’s Law
• Problem: The Henry's law constant K for CO2
dissolved in water is K = 1.6 × 103 atm at 300K (for
use with Pgas [atm] = K [atm] xaq, where xaq is the
molar fraction of the solute). If the partial pressure of
CO2 in air is 0.038 atm under STP, estimate the
concentration, in mol/L, of dissolved CO2 in blood
plasma that is in equilibrium with air.
• Solution:
x = P / K = 0.038 / (1.6×103)
x ~ 0.2375 × 10-4.
[CO2 ]aq = x × 55.5 mol/L = 1.3× 10-3 M = 1.3 mM
• Answer: 1.3 mM