CHEM123 Midterm Review February 27, 2011

CHEM123 Midterm ReviewFebruary 27, 2011

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CHEM123 Midterm

• Chp 12: Liquids, Solids, Intermolecular Forces• Chp 14: Chemical Kinetics• Chp 15: Chemical Equilibrium

Tips for CHEM123

• Heavily calculation focused - do lots of HARD problems

• Watch out for units!!! (i.e. Ideal gas constant, radius of atoms)

Chapter 12

Topics include:•Phase diagrams and transitions•Clausius-Clapeyron equation•Intermolecular forces•Solids & Cubic packing•Ionic solids & Interstitial sites•Born-Haber Cycle

Phase Transitions

•Enthalpy all should be measured at fixed T and P.

Phase Diagrams • Phase diagrams show regions of

uniform phases for different P and T– And regions where phase

equilibria exist.• As P increases for constant TA. Pure gasB. Solid gas interface (sublimation)

(normal sublimation T because P = 1atm).

C. Pure solid• As T increases for constant PD solid liquid interface (melting)E. Pure liquid• As P decreases with constant TF. Liquid gas interface (boiling)G. Gas

Carbon dioxide

A

B

CD

E

F

G

Triple point

Petrucci Fig 13-19

Phase Diagram - water

• Many different solid structures– Polymorphism

• Slope for S/L is (-)– MP ↓as P↑– Liquid denser than Solid

Water

Liquid Properties

• Vapour pressure (VP): pressure given off material when system reaches equilibrium – Rate of evaporation = Rate of Condensation

• Normal Boiling Point :Temp at which VP = 1 atm• Standard Boiling Point :Temp at which VP = 1 bar• Surface Tension: Energy to increase surface area of liquid• Viscosity (μ): liquid’s resistance to flow

↑Intermolecular attraction = ↑ BP, ↑ γ, ↑ μ, ↓VP

Clausius-Clapeyron Eq.

• Clausius-Clapeyron equation:

• Simple ratio for P means P2 and P1 units just have to match• T must be in Kelvin• Hvap assumed to be constant, R is 8.314 J/ mol K∙

211

2 11ln

TTR

H

P

P vap

* NoteThe C- C equation can also be applied to other phase transitions, provided that the proper heat of phase transition is used.

Problem1:

• Ans: 0.0445 atm

Intermolecular forces

Intermolecular forces: forces of interaction among molecules

Intermolecular forces strength examplevan der Waalsor London dispersion ~ 0.1 kJ mol-1 Ar, S8, CH4

Dipole-dipole ~ 10 kJ mol-1 CO, NO2

hydrogen bonding ~20-40 kJ mol-1 H2O, HF, NH3 ionic attraction ~100-1000 kJ mol-1 NaCl, KBr

Intramolecular: covalent bonding ~100-2000 kJ mol-1 diamond

Explain each term and correlate properties such as hardness, m.p., b.p., H, of materials with intermolecular forces in them.

Intermolecular forces

1. Dipole-Dipole: - polar molecules (dipole moments μ)2. London Dispersion Forces:- All molecules (polarizability α)- ↑ with ↑ molecular weight3. Hydrogen Bonding: - H and (O, N or F)- Strongest of the 34. Ionic

Intermolecular forces

Types of intermolecular forces

Problem2:

• Rank from lowest boiling point to highest– NaF, HF, H2

– Acetaldehyde CH3CHO, Acetic acid anhydride (CH3COO)2O, Acetone CH3COCH3, Ethanol C2H5OH, Alkane (CH3CH2CH3)

ans: H2<HF<NaFCH3CH2CH3< CH3CHO < CH2COCH3< C2H5OH < (CH3COO)2O

Heating Curves

Enthalpy of vaporization: ΔHvap

Heat Capacity: Csolid, liquid, gas

Enthalpy of fusion: ΔHfus

Solid Structures

• Amorphous• Crystalline

– Nature of bonding: Ionic, Network covalent, Molecular, Metalic – Geometry/packing: CUBIC, trigonal, tetragonal, hexagonal,

monoclinic, triclinic, orthorhombic

Cubic packing (1 type of atom)

Simple Body Centred Face Centred

N cell 1/8 * 8 = 1 (1/8 * 8) + 1 = 2 1+ (4 * ½) + 1= 4

Coordination# 6 8 12

Packing Efficiency(density)

52% 68% 74%

a and radius a= 2r a = 2.3 r a = 2.82 r

Unit cell

Closest-packed Structures

19

Density of a Crystalline Solid

=(ncell * (molar mass/avogadro))/a3

Closest-packed sphere structure: 2 typesSequence Type

ABAB… hcp (hexagonal closest packing) ABCABC... ccp (cubic closest packing) or fcc (face centred cubic)

Binary Ionic solids

20

• Vapour pressure (VP): pressure given off material when

Problem 3: A metal has a face-centered cubic lattice, a density of 11.4 g/mL, and an atomic radius of 175 pm. What is its relative atomic mass? [Avogadro number = 6.022e23] a. 207.2 g / mol b. b. 197.0 g / mol c. c. 107.9 g / mol d. d. 106.4 g / mol e. e. 72.6 g / mol

Ans: a

Problem 4: A

Ans: a

23

Chapter 14Chemical Kinetics

Chemical Kinetics

Topics include:

•Rates of chemical reactions•Differential & integrative rate laws: 0th, 1st, 2nd order•Effect of temperature •Measuring reaction rates•Reaction mechanisms, steady state approx.•Catalysis

Rates of reactions

• Measuring rates of reactions by measuring change in property that depends on concentration with time– Colour, density, acidity.

aA + bB cC + dD

dt

Ad

a

][1

dt

Bd

b

][1

dt

Cd

c

][1

Rate laws• Differential rate law: relate rate of reaction to concentration

of reactants

• Rate law are generally EMPIRICAL, and not derived from stoichiometry

• note: we use small “k” to denote rate constant, and big “K” to denote equilibrium constant

• Integrative rate laws: relate concentration with time

mn BAkdt

Ad

arate 1

n : order of reaction with respect to Am: order of reaction with respect to Bn + m: overall order of reactionk: reaction rate constant (function of temperature)

Rate Laws

Order Diff. rate law

Integrative rate law Units of k t1/2

0 r = k [A] = -kt + [A]0 mol.L-1s-1 [A]0/2k

1 r = k[A] ln [A] = - kt + ln [A]0 s-1 ln 2/k

2 r = k[A]2 [A]-1 = kt + [A]0-1 L.mol-1s-1 1/k[A]0

** Summary of rate laws**

Chemical Kinetics

• Arrhenius equation: Temp vs K:

• Determining rates of reaction: method of initial rates– Set up a series of reactions with varying concentrations of reactants. – keep [A] fixed and vary [B] => Measure initial rates.– Then keep [B] fixed and vary [A] => Measure initial rates. – Take ratios of initial rates with the reagent concentrations to determine n and m.– Once n and m are determined, rate law can be solved for k

211

2 11ln

TTR

E

k

k a

Ea - activation energyT - Temperature in KR = 8.314 J/mol K∙

A and Ea depends on specific rxn

Chemical Kinetics

PROBLEM 5 : Experiments were performed with different initial concentrations of A and B (no C) for the reaction A+ B C The initial reaction rates were determined and are given in the following table.

Determine the order of the reaction with respect to A and B and the value of the rate constant.• solution: k=0.844L2/(mol2 s) (see appendix)∙

exp't

[A]0 [mol/L]

[B]0 [mol/L]

-(d[A]/dt)0 [mol/(L s)]

1 0.30 1.00 0.0762 1.50 1.00 1.9003 1.50 0.25 0.475

Chemical Kinetics

PROBLEM 6 :

Reaction Mechanisms

• Reaction Mechanism must be consistent with overall reaction stoichiometry and the overall rate law

• Each step is termed “elementary process”– Unimolecular: A → products (dissociation)– Bimolecular: A + B → products (bimolecular collision) – Termolecular: A + B + C → products (very rare)

• Exponents for concentration are same as stoichiometric factors

• Intermediates are produced in one elementary step and consumed in another

• Rate determining step = slowest elementary step

Reaction Mechanisms

Transition state ≠ intermediates!

Reaction MechanismsH2(g) + ICl(g) → HI (g) + HCl(g) (slow)

HI(g) + ICl(g) → I2(g) + HCl(g) (fast)

H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)

Rate of reaction: r = k[H2][ICl]

HI= Reaction intermediate

Chemical Kinetics

• Steady State Approximation– Assume concentration of any intermediate is constant

• Rate of formation = rate of consumption– Intermediate is classified as neither product or reactant

Chemical Kinetics

Problem 7:For the reaction X2 + Y + Z --> XY + XZ, the mechanism is believed to be

1. X2 + Y ---> XY + X k1, very slow

2. X + Z ---> XZ k2, very fast

What is the reactive intermediate? a. X2

b. Y c. Xd. Ze. XY Ans: c

Chemical Kinetics

Problem 7 con’t:For the reaction X2 + Y + Z --> XY + XZ, the mechanism is believed to be

1. X2 + Y ---> XY + X k1, very slow

2. X + Z ---> XZ k2, very fast

What is the concentration of the intermediate, according to the steady state approximation? • [XY][XZ] / [X2][Y][Z]

• k1[X2][Y] - k2[X][Z]

• k1[X2][Y] / k2[X][Z]

• k2[XZ] / k1[XY]

• k1[X2][Y] / k2[Z]

Ans: e

Chemical Kinetics

• Enzyme kinetics: steady state assumption• General reaction:

Rate of production of P][

]][[2

SK

SEk

M

o

ESE + S

ES E + P

1

21

k

kkKM

If [S] high then rate is zero order

If [S] is low then rate is first order

Chemical Kinetics

• Catalysis: increase rate of reaction via alternate reaction path with lower Ea.

• Homogenous catalyst: dissolved in reaction medium through out the reaction (H+ is the catalyst)

HCOOH → H2O + CO

Chemical Kinetics• Heterogeneous catalysts: solid that speed up the ration rates

of gas/liquid– Require interaction with an active site on catalyst surface

• 2 CO + 2NO 2CO2 + N2

Chapter 15 Chemical Equilibrium

Chemical Equilibrium

Topics include:

•The Equilibrium Process•Equilibrium Constants Kc, Kp•Manipulating K for Complex Reactions.•External Effects on Equilibriums

Equilibrium

• Vapour pressure, Chemical equilibrium, solubility equilibrium• Chemical equilibrium: forward rate = backward rate

• Equilibrium constant (Kc) is constant for a given temperature– Kc>1: products in higher amount– Kc<1: reactants in higher amount

• Magnitude of Kc does not indicate if a reaction happens OR how fast it will react

AaA + bB cC + dD

ba

dc

cBA

DCK

Equilibrium

• Manipulating Kc:

nA n B

2' cn

n

c KA

BK

A B A

BKc

B A c

c K

1

B

A'K

Equilibrium

• Kc for sequential reactions:

A B (1)

B C (2)

A

BKc

1

B

CKc

2

A C (3) 213 ccc KKA

CK Overall

Reaction

Equilibrium – Gases & Solids

• For gas: can also be defined in terms of partial pressure

• in general: R = 0.082 atm L/mol K• For solids: only concentration of gas/dissolved reactants is

considered

AaA + bB cC + dD

b

Ba

A

cC

dD

pPP

PPK

Kp = Kc (RT) ∆ngas

Equilibrium – Gases & Solids

PROBLEM 8:

(1)

(2)

Equilibrium - Q

• Reaction Quotient (Q): predict direction of change in Equilib.

• Q is defined for ANY concentration of reactants/products• K is defined ONLY when chemical equilibrium is reached

AaA + bB cC + dD

ba

dc

BA

DCQ

b

Ba

A

dD

cC

PP

PPQ

Equilibrium - Q

• Therefore, if Q = Kc chemical equilibrium

Q < Kc reactants are in excess.– System must change to reduce reactants and increase

productsQ > Kc products are in excess.– System must change to reduce products and increase

products.

Equilibrium

• Effects on equilibrium: Le Châtelier Principle:– Change in concentration of reactants/products– Change in volume – Change in pressure

• Adding inert gas (no change if constant V, change if constant P)

** the above will only change Q and disturb equilibrium **** Temperature will change K **

Equilibrium

• Van’t Hoff Equation: describes relationship between temperature and equilibrium constant, K

21

11ln

1

2

TTR

H

K

K

C

C

Kc1 is the equilibrium constant at the temperature T1.Kc2 is the equilibrium constant at the temperature T2.T1 and T2 in Kelvin.∆ H is the enthalpy of reaction (assumed constant).

Equilibrium

PROBLEM 9:

Ans: 0.133 mol/L (see appendix)

Thank you for supporting SOS, good luck on all your midterms!

AppendixPROBLEM 5 - Full solution

AppendixPROBLEM 9 - Full solution