chem123 midterm review february 27, 2011
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CHEM123 Midterm Review February 27, 2011. About me. No clue? NO PROBLEM!. [email protected]. CHEM123 Midterm. Chp 12: Liquids, Solids, Intermolecular Forces Chp 14: Chemical Kinetics Chp 15: Chemical Equilibrium. Tips for CHEM123. - PowerPoint PPT PresentationTRANSCRIPT
CHEM123 Midterm ReviewFebruary 27, 2011
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