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20 B Week V Chapters 11 and 18Colligative Properties and Chemical Kinetics
• Dissolution reactions and Arrhenius type Acid/Base rxns
• Colligative properties, Vapor Pressure Lowering, Boiling Point Elevation, Freezing Point Depression and Osmotic Pressure,
• The Rates of Chemical Reactions A-->B
Rate=d[A]/dt= -d[B]/dt. units molL-1s-1
A is the Reactant and B is the product of the reaction
• Rate Laws at early times =k[A]n
k is the rate constant and n is the reaction order
• Elementary reactions: single step reactions
Unimolecular(1st), Bimolecular(2nd), Termolecular
Fig. 10-21, p. 463
Fig. 10-16, p. 459
The Rate of evaporation vs time as the vapor pressure approaches Equilibrium Evaporation rate equals the condensation Rate
All the Macroscopic Properties, P, V, and Tare only defined at Equilibrium.Which means PV=nRTand the vdW Eq. canonly be use under Equilibrium conditions
C2H4(l) C2H4(g)Evaporation
Equilibrium P,V and T well defined
Table 10-3, p. 460
H2O P-T Phase Diagram
Super heatedH2O liquidWill spontaneouslyvaporize
Super cooled H2O liquid willSpontaneously freeze
Equilibrium Vapor Pressure
In both spontaneous processesthe system will go to the Equilibrium State (Phase) and Pressure
Fructose C6H12O6
Hydrated Fructose C6H12O6
H-bonds
Dissolution of a Nonvolatile((zero partial pressure) solute, e.g., sugars or salts. Heterogeneous Phase equilibrium of a 2 component solution
The dissolution reaction with water as a solvent: A(s)A(aq) A is a molecular solid and the molecular units, or monomers, do not dissociate in solution. Like fructose with lots of OH ( hydroxyl groups) for forming H-bonds and which makes the monomer more stable in aq soln
Electro Staticpotential
Table 10-3, p. 460
H2O P-T Phase Diagram
Super heatedH2O liquidWill spontaneouslyvaporize
Super cooled H2O liquid willSpontaneously freeze
Equilibrium Vapor Pressure
In both spontaneous processesthe system will go to the Equilibrium State (Phase) and Pressure
Fig. 11-10, p. 491
Solvent vapor pressure versus Mole fraction X1 =n1/(n1 + n2)
Ideal SolutionRaoult’s Law Ignores intermolecular Forces (interactions)
Nonideal or Real solnsWhere intermolecularForces are alwayspresent
Raoult’s LawP1=X1P°
1
P°1=vapor pressure
of pure solventP1 = solvent vaporPressure
Fig. 10-16, p. 459
The Rate of evaporation vs time as the vapor pressure approaches Equilibrium Evaporation rate equals the condensation Rate
All the Macroscopic Properties, P, V, and Tare only defined at Equilibrium.Which means PV=nRTand the vdW Eq. canonly be use under Equilibrium conditions
C2H4(l) C2H4(g)Evaporation
Equilibrium P,V and T well defined
Fig. 11-10, p. 491
Solvent vapor pressure versus Mole fraction X1 =n1/(n1 + n2)
Raoult’s LawP1=X1P°
1 basis of all4 colligative PropertiesVapor Pressure LoweringBoling Point ElevationFreezing point depressionAnd Osmotic pressure
Table 10-3, p. 460
H2O P-T Phase Diagram
Super heatedH2O liquidWill spontaneouslyvaporize
Super cooled H2O liquid willSpontaneously freeze
Equilibrium Vapor Pressure
In both spontaneous processesthe system will go to the Equilibrium State (Phase) and Pressure
Vapor Pressure Lowering in a two component heterogeneous soln
Raoult’s Law P1=X1P°1
Can also be written as P1=P1 - P°
1 = X1 P°1 - P°
1 = -(1- X1) P°1 =- X2P°
1
Implies Vapor Pressure Lowering since P1< 0When the solute is added.
Fig. 11-10, p. 491
Solvent vapor pressure versus Mole fraction X1 =n1/(n1 + n2)
Raoult’s LawP1=X1P°
1 basis of all4 colligative PropertiesVapor Pressure LoweringBoling Point ElevationFreezing point depressionAnd Osmotic pressure
Attractive Forces
Table 10-3, p. 460
Pure H2O P-T Phase Diagram
Equilibrium Vapor Pressure
Vapor PressureLowering After dissolution
Fig. 11-11, p. 493
Boiling Point Elevation =T’b = Tb + X2P°1/S
Tb = boiling ptT’b elevatedboiling ptVapor pressureLoweringWith added solventP1= - X2P°
1
Tb=T’b – Tb
(1/S)X2
S=-P/Tb
Fig. 11-11, p. 493
Boiling Point Elevation
Tb = boiling pointT’b = elevated boiling pt.
Vapor pressure LoweringWith added solvent
P1= - X2P°1
S= -P/Tb
Solve for Tb
Tb=T’b – Tb=(1/S)X2
Fig. 10-6, p. 450
In Solutions, for example when NaCl(s) is dissolved in H2O(l).
+ H2O
NaCl(s) + H2O(l) Na+(aq) +Cl-(aq)
(aq) means an aqueous solution, where water is the solvent,major component.The solute is NaCl, which is dissolved, minor componentWater molecules solvates the ions the Cation (Na+) and the Anion (Cl-). The forces at play here areIon dipole forces
Dissolution of a polar solid by a polar solid by a polar liquidA non-polar liquid e.g., benzene, would not dissolve NaCl?
Table 11-2, p. 494
Boiling Point Elevation= Tb’ = Tb + X2P°1/SIn dilute a solutions n1>>n2
X2= n2/(n1 + n2)~ n2/n1=(m2/M2)/(m1/M1)
Tb’ = Tb + X2P°1/STb’-Tb =(1/S) (m2/M2)/(m1/M1)T=Kb(m2/M2)/(m1[1000gkg-1)T=Kbm ( m=molality of soln)m=(m2/M2)/(m1[1000gkg-1])
M1(gmol-1) is the molar mass of the solvent and M2 (gmol-1) of the solute ad
Since S and M1 are properties of the solvent then we can define Kb= M1/S(1000 gKg-1)
Tb = Kbm ( m=molality of soln)
m1 is the mass of the solutem2 is the mass of the solvent
Boiling Point Elevation (Tb= T’b-Tb)
Tb = Kbm ( m=molality of soln)
As an example: NaCl(s) dissolves completely in water.
NaCl(s) + H2O(l) Na+(aq) +Cl-(aq)
1.0 mole NaCl(s) produces 2 moles of ions in solnGiven 0.058 grams of NaCl(s) is dissolved in 10 grams of H20(l)What is the boiling point(T’b) at p=1 atm?
Tb=100°C for pure waterTb = Kbm ( m=molality of soln)
Kb(H2O)= 0.512 K kg mol-1
The molality m=(m2/M2)/(m1/[1000gkg-1]) m2 mass of solute; m1 mass od solvent
As an example: NaCl(s) dissolves completely in water.
NaCl(s) + H2O(l) Na+(aq) +Cl-(aq)
1.0 mole NaCl(s) produces 2 moles of ions in solnGiven 0.0584 grams of NaCl(s) is dissolved in 10 grams of H20(l)What is the boiling point(T’b) at p=1 atm?
Tb=100°C for pure waterTb = Kbm ( m=molality of soln); Kb(H2O)= 0.512 K kg mol-1
m=(m2/M2)/(m1[1000gkg-1])=n2/(m1/[1000gkg-1])
n2=(2)x(0.0584 g/ 58.4 gmol-1) = 2.0 x 10-3 molsm1 =10 g m =2x10-3 mols/0.01 kg = 0.2 mols kg-1
Tb = (0.512)x(0.2) K = 0.1024 KTb = 100 °C + 0.102 °C= 100.102 °C
Fig. 11-11, p. 493
Boiling Point Elevation
Tb = boiling pointT’b = elevated boiling pt.
Vapor pressure LoweringWith added solvent
Tb=T’b – Tb=(1/S)X2
0.102 °C
Fig. 11-12, p. 496
Freezing Point Depression
only consider cases where the pure solvent crystalizesfrom solution, e.g., iceCrystalizes from salt water and NaCl(s) does not
Tf = - Kfm ( m=molality of soln)Tf = T’f - Tf
Melting point(Tf) lowered to keep the vapor pressureover the pure solid and liquid solution the same at Equilibrium!P1= - X2P°
1
S= -P/Tf
Solved for Tf
Table 11-2, p. 494
Tf = - Kfm ( m=molality of soln)
Tf =
Example : again for 0.058 gmol-1 of NaCl in 10 g H2O(l) over H2O(s)The molality is m = 0.02 gkg-1 (grams of solute per kg of solvent)
Table 11-2, p. 494
Tf = - Kfm ( m=molality of soln)
Tf = - (1.86)x(0.02) K= - 0.0372 °C
T’f = 0 +(-0.037)°C=-0.037°C
Example : again for 0.058 gmol-1 of NaCl in 10 g H2O(l) over H2O(s)The molality is m = 0.02 gkg-1 (grams of solute per kg of solvent)
Fig. 11-12, p. 496
Freezing Point Depression
only consider cases where the pure solvent crystalizesfrom solution, e.g., iceCrystalizes from salt water and NaCl(s) does not
Tf = - Kfm ( m=molality of soln)Tf = T’f - Tf
Melting point(Tf) lowered to keep the vapor pressureover the pure solid and liquid solution the same at Equilibrium!P1= - X2P°
1
S= -P1/Tf (pure solvent)Solved for Tf
P1
Tf = 0.037 °C = 0.037 K
Fig. 11-13, p. 497
Freezing point depression (Tb) v.s. Molality of the solution
M(s)M(aq)
MX(s)M(aq) +X(aq)
MX2(s)M(aq) +2X(aq)x
MX3(s)M(aq) +3X(aq)x
} Closer to ideal soln
p. 499
Osmotic pressure forces water out of a carrot placed I n a salt soln Water doesn’t leave a carrot
placed in pure water
Osmotic Pressure π
Measuring PressureHg Barometer
mm
Pressure= Force/unit areaF=mg=Vg= hAg
FairFH
g
FHg = Fair Pair =Fair/A=FHg/A= hAg/A= hg:
Pair= hg
=mass/V
Mass Density
kg/m3
or kg/cm3
mm
Fig. 11-14, p. 498
Osmotic Pressure π=[solute]RTSolute molar concentration
Solute moleculesLowers the rate ofSolvent moleculesCrossing the MembraneFrom the solution
At Equilibrium the rate of The Solvent molecules Crossing the membrane from solution Is equal to the rate from the solvent
Recall that pressurein the tube P=ghso π=ghVan’t Hoff proposedπ=[solute]RTWhich is similar to PV=nRTfor an ideal gas. Note that is the soluteConcentration but the Solvent mass density (kg meter-3)!