20 B Week V Chapters 11 and 18 Colligative Properties and Chemical Kinetics Dissolution reactions...

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)!