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SCH4U- Atomic Structure Review
1) Describe the contributions of the following scientists to our current understanding of atomic structure and the
quantum theory:
a) Bohr: energy levels
b) DeBroglie: wave properties of matter
c) Heisenberg: uncertainty principle
d) Schrodinger: wave equation for orbitals
e) Pauli: magnetic spin property
2) Based on the quantum mechanical model of the atom:
a) What is an "orbital"'?
A region in which an electron with a certain set of quantum numbers has a certain probability of being found.
b) What is the maximum electron carrying capacity of the following energy levels: n=2, n=5
n=2: 8
n=5: 50
c) What orbital sub-levels would be found in the following energy levels: n=2, n=5
n=2: s, p (l=0,1)
n=5: s,p,d,f,g (l=0,1,2,3,4)
d) How would the number of electrons determined in 2(b) be distributed amongst the orbital sub-levels in 2(c)
(i.e.: number of e- per sub level)
n=2: 2 in s, 6 in p
n=5: 2 in s, 6 in p, 10 in d, 14 in f, 18 in g
3) a) Write the ground state electron configuration of a neutral atom of nitrogen.
1s2 2s2 2p3
2p
2s
1s
|
||
||
|
|
b) Assign quantum numbers to each or the electrons in your nitrogen atom.
n
m
ml
ms
2p
2
1
1
+.5
2p
2
1
0
+.5
2p
2
1
-1
+.5
2s
2
0
0
-.5
2s
2
0
0
+.5
1s
1
0
0
-.5
1s
1
0
0
+.5
SCH4U- Bonding and Molecular Shapes Review
1) Use Lewis diagrams to draw the following molecules or ions. Once completed, use the VSEPR theory to predict
the shape of each. For each molecule, decide whether it is polar or non-polar.
a) PCl3
b) SiF4
c) CO32d) H2S
e) PO43trig pyram
tetrahed
planar trig
bent
tetrahed
3 single
4 single
1 double
1 single
4 single
1 lone pair
2 single
2 lone pair
(1 co-ord)
polar
non-polar
non-polar
non-polar
non-polar charged
charged
f) O3
bent
1 single
1 double
1 lone pair
non-polar
(1 co-ord)
g) CO
straight
2 double
non-polar
h) NO31planar trig
1 double
1 single
(1 co-ord)
non-polar charged
2) Methane and fluoromethane are two very similar molecules. Despite their similarities, there is an 86/C difference
between their boiling points. Which of these two substances would have the higher boiling point? Explain your
answer.
CH3F, has a higher boiling point because it is polar and has greater London forces.
3) The element iodine exists as solid crystals composed of I2 molecules. A chemist wishing to dissolve iodine has a
choice of two solvents; water and tetrachloromethane. Which of these two solvents would be the best choice?
Explain your answer.
Tetrachloromethane, since I2 is non-polar and so is CCl4.
4) Consider a molecule of ethanal.
a) What type of orbital hybridization is each atom in the molecule using for bonding?
#1-carbon is 2sp2, 2p
#2-carbon is 2sp3
oxygen is probably not hybridized, but cannot know because it has only one sigma bond.
b) What geometric arrangement would each of these orbitals take?
#1-carbon is trigonal planar
#2-carbon is tetrahedral
c) Describe the orbital overlap that accounts for each bond and specify the type of bond formed (sigma, pi)
#1-carbon has a sigma bond to carbon 2, a sigma bond to oxygen, and a pi bond to oxygen
d) What are the bond angles around each carbon atom in this molecule?
#1-carbon: 120o
#2-carbon: 109.5o
5) Consider the data in the following table for substances labeled 'A" to “G"
___________________________________________________________________________________________
Substance
Melting point (/C)
Boiling point (/C)
Solubility in H2O
Electrical Conductivity
as; solid/liquid/aq. sol.
A (CH3OH) mol
-94
65
SOLUBLE
NO/NO/NO
B (quartz) netw 1610
2230
INSOLUBLE
NO/NO/NA
C (S)
mol
119
445
INSOLUBLE
NO/NO/NA
D (Al) metal
660
2467
INSOLUBLE
YES/YES/NA
E (KCl) ionic
770
1500
Slightly SOLUBLE
NO/YES/slightly
F (CH4) mol.
-182
-164
INSOLUBLE
NO/NO/NA
G (CaO) ionic 2614
2850
SOLUBLE
NO/YES/YE5
____________________________________________________________________________________________
Classify each substance as to the type of solid it represents: molecular, metallic, ionic, network
Match the letter for each sample to one of the following substances (one letter per substance) and explain your
reasoning: quartz, CaO, Al, methanol, Sulfur, methane, KCl
SCH4U- Organic Chemistry Review
1) Draw the structural formula for the
following compounds:
a) N,N-dimethyl-2-pentamine
b) 3-ethylcyclohexane
should be ethylcyclohexane
c) methylpropanal
d) 2-chloro-3-methyl-2-hexene
e) 2-ethoxypropane
f) propylbenzoate
g) N-ethylpropanamide
ethyl propanamide
h) 2,3-difluorobutanoic acid
2) Name the compounds pictured here:
a) 2,5-dimethyl-2-heptene
b) 1-ethyl-3-methylbenzene
c) N-ethyl-N-methylaminopropane
d) 2-chloro-3-methylpentanoic acid
f) 2,3-dimethylcyclopentane
g) 2-methyl-3-heptyne
h) 4-methyl-2-hexanone
i) N-methylethyl propanamide
j) ethoxypropane
k) ethyl butanoate
3) Complete the following chemical reactions by naming and drawing the molecular formula the organic product(s).
Assume all required conditions are available for each reaction,
a) 3-methyl-1-butyne + 2Cl2
---> 1,1,2,2-tetrachlorobutane
b) methane + excess O2
---> CO2 + 2H2O
c) 2-methyl-3-pentanol + KMnO4
---> 2-methyl-3-pentanone
d) methylpropene + HBr
---> 2-bromomethylpropane
e) 2-methylbutanal + KMnO4
---> 2-methylbutanoic acid
f) cyclopentene + H2O (w/ H2SO4)
---> 1-hydroxycyclopentane
g) dimethyl-l,3-propandiol + KMnO4
---> dimethylpropanedioic acid
h) ethanoic acid + phenol
---> phenyl ethanoate
4) Consider the following organic compounds:
All exhibit London Dispersion Forces (LDF)
Highest Boiling Point
proponoic acid LDF, dipole-dipole, H-bonding
soluble
2-propanol
LDF, dipole-dipole, H-bonding
soluble
propanone
LDF, dipole-dipole
soluble
2-chloropropane LDF and weak dipole-dipole (delta EN is 0.5)
slightly soluble
propane
LDF only
insoluble
Lowest Boiling Point
a) For each compound, identify the type of intermolecular forces acting between molecules.
b) Rate the solubility of each compound in water as: insoluble, slightly soluble or very soluble
c) Arrange the compounds in order of increasing boiling point.
Generally speaking,
order of increasing boiling point: hydrocarbons, ethers, alkyl halides, aldehydes, ketones, alcohols, carboxylic acids
the more polar functional groups, the higher the boiling point and solubility
the greater the molar mass, the higher the boiling point
the more central the functional group, the greater the solubility and the lower boiling point (all other things equal).
SCH4U- Energy and Rates Review
1) When 13.4g of ammonium chloride dissolve into 2.00x102 g of water the temperature changes from 20.0/C to
15.3/C. Determine the molar enthalpy of solution of ammonium chloride and write a balanced thermochemical
equation for the reaction. Sketch an enthalpy diagram for this reaction. (Ans: +16.7kJ/mol)
molar mass of NH4Cl
mN = 14.01+(3x1.01)+35.45 = 52.49 g/mol
moles of NH4Cl
n = m/mN
= 13.4 g / 52.39 g/mol
= 0.256 mol
enthalpy change for this process
= msurrCp)T
qsurr
= 2.00x102g x 4.18 J/oC x (15.3 oC - 20.0 oC)
= -3929.2 J
= -3.93 kJ
)H = qsys
= -qsurr
= 3.93 kJ
)Hrxn = )H/n
= 3.93 kJ / 0.256 mol
= 15.36 kJ/mol
2) The following chemical reaction occurs in a car battery:
Pb(s) + PbO2(s) + 2H2SO4(aq) –> 2PbSO4(s) + 2H2O(l)
Given the following enthalpy of formation values, calculate the enthalpy change for the above reaction as written
(also known as the enthalpy of reaction)
PbO2(s)
H2SO4(aq)
PbSO4(s)
H2O(l)
)H o f
-277.0 kJ/mol
-909.3 kJ/mol
-902.1 kJ/mol
-285.5 kJ/mol
)H = E (n x )Hof)prod - E (n x )Hof)react
= ((2 x -902.1) + (2 x -285.5)) - ((1 x 0)+(1 x -277.0)+(2 x -909.3))
= (-1804.2 + -571.0) - (0 + -277.0 + -1818.6)
= -2375.2 - -2095.6
= -279.6 kJ
b) If it takes 39.5 kJ to start your car, what mass of lead (II) sulphate is produced when starting the car using the
battery reaction described in 'a'?
Since there are two moles of lead (II) sulphate produced in this reaction, the molar enthalpy for this reaction with
respect to Lead (II) sulphate is: )Hrxn = )H/n
= -279.6 kJ / 2 mol
= -139.8 kJ/mol
To calculate the number of moles of Lead (II) sulfate produced for a specific enthalpy change, use the equation:
)H = n )Hrxn
n = )H/)Hrxn
= -39.5 kJ / -139.8 kJ/mol (-39.5 kJ since the battery needs to release
-39.5 kJ to the car’s starter motor)
= 0.282 mol
molar mass of Lead (II) sulfate
mN = 207.20 + 32.06 + (4 x 16.00)
= 303.26 g/mol
mass of Lead (II) sulfate produced m = n x mN
= 0.282 mol x 303.26 g/mol
= 85.6 g
3) The combustion of ethane is illustrated by the following equation:
2C2H6(g) + 7O2(g) –> 4CO2(g) + 6H2O(g)
a) Show how the following equations can be combined using Hess’ Law to determine the enthalpy of combustion
of Ethane. (Ans: -3028 kJ) (HS says -2973.8 kJ)
)H1 = -1409 kJ
1) C2H4(g) + 3O2(g) –> 2CO2(g) + 2H2O(g)
)H2 = -136.7 kJ
2) C2H4(g) + H2(g) –> C2H6(g)
3) H2(g) + ½O2(g) –> H2O(g)
)H3 = -214.6 kJ
2 x 1) = 4) 2C2H4(g) + 6O2(g) –> 4CO2(g) + 4H2O(g)
-2 x 2) = 5) 2C2H6(g) –> 2C2H4(g) + 2H2(g)
2 x 3) = 6) 2H2(g) + O2(g) –> 2H2O(g)
4) + 5) + 6) =
)H4 = -2818 kJ
)H5 = +273.4 kJ
)H6 = -429.2 kJ
2C2H4(g) + 6O2(g) + 2C2H6(g) + 2H2(g) + O2(g) –> 4CO2(g) + 4H2O(g) + 2C2H4(g) + 2H2(g) + 2H2O(g)
2C2H6(g) + 7O2(g) –> 4CO2(g) + 6H2O(g)
)H
= )H4 +)H5 +)H6
= -2818 + 273.4 - 429.2
=-2974 kJ
b) If you wish to burn ethane as fuel to heat 550.0 g of water from 12.5 /C to 100.0 /C, how many moles of ethane
will be require? (Ans: 0.135 moles)
Given that the )H from the equation above is for 2 moles of ethane, the molar enthalpy of combustion for ethane is:
)Hrxn = )H/n = -2974 kJ / 2 mol = -1487 kJ/mol
To heat the water (we will call it the surroundings), the energy needed is calculated by the calorimetry equation
qsurr = m Cp )T = 550.0 g x 4.18 J/g oC x (100 oC - 12.5 oC) = 201163 J = 201 kJ
)H = -qsurr = -201 kJ
To calculate moles of ethane required, use the equation:
n = )H / )Hrxn ()Hrxn = )H/n rearranged)
= -201 kJ / -1487 kJ/mol
= 0.135 mol
4) From your text on pages. 414-416 answer
13: a) rate doubles, b) rate drops by half, c) no change (reactants are not gases).
14: a) a series of steps, since there are 6 moles of reactants, see back of text for b)
15: a) first, b) first, c) second, d) r=k[HI][O2], e) k = r/([HI][O2]) = 0.0042/(0.01x0.01) = 42, f) 2
g) the coefficients do not match the exponents in the rate law expression
16: a) 2NO(g) + 2H2(g) —> N2(g) + 2H2O(g)
b) Anything that is not in the overall balanced equation.
c) r = k[NO(g)]2
18: Consider the physical states and past experience with precipitation reactions; reaction ‘a’ seems likely.
SCH4U- Introduction to Equilibrium
page 457
1: a) left, b) right, c) no change, d) left, e) right
2: a) F2 line spikes up followed by F2 and Br2 lines sloping down and the BrF line sloping up
b) HOCl spikes down followed by sloping up while the other two lines slop down
In both cases, all the lines become horizontal again.
3: a) It would speed it up.
b) No effect.
4: the equilibrium reaction is exothermic as written and will shift right is temperature is decreased.
page 465
1: N2O4 <––> 2NO2
K=[NO2]2/[N2O4] = 0.87
2
a) Q = 2.15 /5.30 = 0.87 Q=K, thus the system is at equilibrium
b) Q = 1.552/0.80 = 3.0
Q<K, thus the equilibrium will shift right
2: Q = 1.5 x 3.2 / 2.0 = 2.4 Q<K, thus it is not at equilibrium
page 481
1: N2 + 3H2 —> 2NH3;
Q = (2.2x10 -4)2 / ((4.0x10 -3)3 (2.0x10 -4)); compare Q to K
SCH4U -Solubility Equilibrium Review
1) Consider the following salts:
AgI, AgNO3, AgCl, Ag2CO3, Ag2CrO4
a) List these salts in order of increasing solubility.
If X=solubility (mol/L) then for a salt with the formula AnBm Ksp = (nX)n(mX)m
The trick is to know when a subscript is part of a polyatomic ion (has no effect on the Ksp expression) and
when the subscript is telling you how many ions are in the formula. This is important because two salts with
the same Ksp can have very different solubilities if the exponents in the Ksp expressions are different.
Salt
AgI
AgCl
Ag2CO3
Ag2CrO4
AgNO3
Ksp
8.5x10 -17
1.8x10 -10
8.5x10 -12
1.1x10 -11
very large
Solubility Product Expression
Ksp = x2
Ksp = x2
Ksp = (2x)2 x = 4x3
Ksp = (2x)2 x = 4x3
Ksp = x2
x
(Ksp)½ = 9.2x10 -9 <— lowest solubility
(Ksp)½ = 1.3x10 -5
(Ksp/4)1/3 = 2.0x10 -4
(Ksp/4)1/3 = 2.2x10 -4
(Ksp)½ = large
b)Write the chemical equilibrium equation for the formation of an aqueous solution of the least soluble silver salt.
AgI(s) <—> Ag+(aq) + I -(aq)
Ksp = [Ag+(aq)][I -(aq)]
c) Determine the solubility in mol/L and in g/L for the least soluble silver salt in pure water.
(Ans: 9.2X10-9 mol/L and 2.2X10-6 g/L)
molar mass of AgI is mN = 107.87 g/mol + 126.90 g/mol = 234.77 g/mol
molar solubility = 9.2x10 -9 mol/L
mass solubility = CN x mN = 9.2x10 -9 mol/L x 234.77 g/mol = 2.2x10-6 g/L
d)Determine the solubility in mol/L for the least soluble silver salt in pure water in a 0.025 M solution containing a
sodium salt with the same negative ion. (Ans: 3.5X10-15 mol/L)
Ksp = (x)(x+0.025) = 0.025x on the understanding that x is really small compared to 0.025
x = Ksp / 0.025 = 3.4x10 -15 mol/L
2) You are given a 100.0 mL sample of 0.0010 M silver nitrate and 100.0 mL of solution made by dissolving
6.4X10-4 g of sodium carbonate (not sodium bicarbonate) in enough water to make 100.0 mL of solution. If the two
solutions are mixed at 25/C, will a precipitate of silver carbonate form? (Ans; Q = 8.0x10-12 )
Sodium carbonate solution
molar mass of sodium carbonate (Na2CO3) mN = (2x22.99) + 12.01 + (3x16.00) = 105.99 g/mol
Cm = 6.4x10 -4 g / 0.1000 L = 6.4x10 -3 g/L
Cn = Cm / mN = 6.4x10 -3 g/L / 105.99 g/mol = 5.9x10 -5 M
Silver nitrate solution
Cn = 0.0010 M
These two concentrations are premix concentrations.
We need post-mix, or initial, concentrations (which will be half premix, since the volume is doubled on mixing).
Initial concentrations are Ci = (Cpm x Vpm)/Vi
AgNO3
0.0010 x 0.1 L / 0.2 L
0.0005
Na2CO3
5.9x10 -5 x 0.1 L / 0.2 L
2.95x10 -5
Q = [Ag+]2[CO32-] = (0.0005)2 x 2.95x10 -5 = 7.4 x 10 -12
Ksp = 8.5x10 -12
Q<Ksp, therefore no precipitate occurs
SCH4U - Acid/Base Equilibrium Review
1) Determine the pH of the following: (Ans: a: pH 12.4
a) 0.025M KOH solution.
b) 0.075M HNO3 solution
b: pH 1.1)
Both substances are strong electrolytes.
a) [OH-] = [KOH] = 0.025 M; pOH = -log[OH-] = -log(0.025) = 1.6 pH = (14 - pOH) = 14 - 1.6 = 12.4
a) [H+] = [HNO3] = 0.075 M; pH = -log[H+] = -log(0.075) = 1.1
2) Nicotinic acid is a weak monoprotic organic add with the formula C6H4NCOOH.
Question: what sort of acid to you think nicotinic acid is? If you said "carboxylic acid," give your self a gold star!
a) Write the equilibrium equation for nicotinic acid in water.
Ka = [H+][A-]/[HA]
(A- being C6H4NCOO-)
b) If a 0.050M solution of nicotinic acid has a pH of 3.1, determine the Ka of this acid. (Ans: 1.3x10-5)
HA <--->
H+
+
AI
0.050
0
0
Ka = [H+][A-]/[HA]
C
-x
+x
+x
= x2 / (0.050 - x)
E
0.050 - x
x
x
= (7.94 x 10-4)2 / (0.050 - 7.94 x 10-4)
= 1.28x10 -5
+
-pH
-3.1
-4
But, we know that [H ]eq = 10 = 10 = 7.94 x 10
3) Consider the following salts: KNO3, LiC6H5COO, N2H5Cl, LiCN, NH4Cl
a) Group the salts in to the categories: acidic, basic, neutral.
acidic: (conjugates of a weak base present) N2H5Cl, NH4Cl
basic: (conjugates of a weak acid present) LiC6H5COO, LiCN
neutral: (no weak conjugates present)
KNO3
b) Write an equilibrium Ka or Kb expression for all the non-neutral salts in the list.
N2H5+(aq) + Cl -(aq) + H20(l) <---> N2H4(aq) + H30+(aq) + Cl -(aq)
Ka = [N2H4(aq)][H30+(aq)]/[N2H5+(aq)]
NH4+(aq) + Cl -(aq) + H20(l) <---> NH3(aq) + H30+(aq) + Cl -(aq)
Ka = [NH3(aq)][H30+(aq)]/[NH4+(aq)]
Li+(aq) + C6H5COO -(aq) + H20(l) <---> Li+(aq) + C6H5COOH (aq) + OH -(aq)
Kb = [C6H5COOH (aq)][0H -(aq)]/[C6H5COO -(aq)]
Li+(aq) + CN -(aq) + H20(l) <---> Li+(aq) + HCN (aq) + OH -(aq)
Kb = [HCN(aq)][0H -(aq)]/[CN -(aq)]
4) What volume of 0.0150M HNO3, is needed to titrate 30.0 mL of 0.0250M NaOH to the equivalence point?
Since both the titrant and base are “monoprotic”, the number of moles of titrant (nt) must equal
the moles of base (ns)
ns = CsVs = 0.0250 M x 0.0300L = 0.000750 mol
nt = ns = 0.000750 mol
Vt = nt/Ct = 0.000750 mol / 0.0150 M = 0.050 L = 50.0 mL
5) Hydrosulphuric acid (H2S) is a weak acid with a Ka of 9.1x10-9
H2S + H2O --> H3O+ + HSa) Determine the pH af a 0.075M solution of this acid. (Ans: pH 4.6)
I
0.075
0
0
C
-x
+x
+x
E
0.075-x
x
x
Ka = [H3O+][HS -] / [H2S] = x2 / (0.075-x) = x2 / 0.075
x = (Ka x 0.075)1/2
= (9.1x10 -9 x 0.075)1/2
= 2.6x10-5 M
pH = -log(x) = 4.6
(0.075/Ka >> 100)
b) What is the percent dissociation of this acid? (Ans: 0.035%)
%dissoc. = 100% x [H+]eq/[H2S]i = 100% x 2.6x10-5 / 0.075 = 0.035%
6) Hydrocyanic acid is a weak acid with a Ka of 4.9x10-10
HCN + H2O --> H3O+ + CNa) What is the Kb of the cyanide ion? (Ans: Kb= 2.0x10-5)
KaKb=10 -14
Kb = 10 -14 / 4.9x10-10 = 2.0x10 -5
b) Determine the pH of a 0.15M NaCN solution. (Ans: pH 11.2)
CN - + H2O ---> HCN + OH Kb = x2/(0.15-x) = x2/0.15
(0.15/Kb >>100)
I
0.15
0
0
x = (2.0x10 -5 x 0.15)½ = 0.0017
C
-x
+x
+x
pOH = -log(0.0017) = 2.8
E
0.15-x
x
x
pH = 14 - pOH = 11.2