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SOS Exam-AID
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CHM 1311
Topics in this Exam-AID 1
The Fundamentals Balancing Chemical Equations Stoichiometry % Composition by mass + Determining
Chemical Equations Trivial things profs might ask to trip you up
(molality, mole fraction, %w/w) Molecular Geometry
Topics in this Exam-AID 2 Thermochemistry
Enthalpy Heats of Reaction and Heats of Formation Calorimetry
Entropy Free Energy, ΔG Equilibrium and Equilibrium Constants for:
Gases Solutions Acids/Bases Solubility
Topics in this Exam-AID 3
Electrochemistry Redox Reactions Cell potentials
Kinetics Rate equations Rate-Determining Steps
Quantum Numbers
Balancing Chemical Equations 1
Unbalanced:
Ca(OH)2 + H
3PO
4 = Ca
3(PO
4)
2 + H
2O
Balancing the atom which appears the fewest times on each side
Then balance the other atoms It can help to make a table
Balancing Chemical Equations 2
Balanced:
3 Ca(OH)2 + 2 H
3PO
4 → Ca
3(PO
4)
2 + 6 H
2O
It helps to know what the products of your reactions are!
Unbalanced:
C6H
5COOH + O
2 → ??
Balancing Chemical Equations 3
C6H
5COOH + O
2 → H
2O + CO
2
Balance the atom that shows up the fewest times on each side
C6H
5COOH + O
2 → 3 H
2O + 7 CO
2
LS RSC 7 7H 6 6O 4 17
C6H
5COOH + 15/2 O
2 → 3 H
2O + 7 CO
2
Stoichiometry
Ratios and Recipes
2 cups Baking Mix + 1 Cup Chocolate chips → 2 (terrible) Chocolate Chip Cupcakes
2 moles H2
+ 1 mole O2 → 2 moles H
2O
2:1:2 ratio
Stoichiometry
0.005 mols H2
+ Y moles O2 → Z moles H
2O
2 moles H2
+ 1 mole O2
→ 2 moles H2O
Because the ratios are the same, you can divide the equations in order to figure out the number of moles you need
% Composition by Mass 1
Given a chemical formula:
C6H
12O
6
Pretend you have one mole of the molecule, and multiple moles of the component atoms
1 mole C6H
12O
6
= 6 moles C, 12 moles H, 6 moles O
% Composition by Mass 2
How much do 6 moles of C weigh?
6 moles * 12.011g/mol = 72.066g How much do 12 moles of H weigh?
How much do 6 moles of O weigh?
Composition by mass 3
How much would one mole of the substance weigh?
The molar mass of C6H
12O
6 is 180.16g/mol
How much do each of the components weigh?
6 moles of C weigh (6 * 12.011), or 72.066g/mol
Take percentages 72.066/180.16 = 40.0%
Determining Chemical Formulas Steps
Assume one mole of molecule Use molar ratios to determine how many
moles of atoms you have Convert from moles to mass
To determine formula, do the opposite Assume 100g, find mass of each element Convert from mass to moles Use the number of moles to create molar
ratios to make your formula
Determining Chemical Formulas
You have a substance made of :
40% Ca, 12% C, and 48% O, molar mass of 100.0869 g/mol
Step One: Assume 100g of substance, find mass of each element
40% Ca * 100g = 40g 12% C * 100g = 12g 48% O * 100g = 48g
Determining Chemical Formulas 2
Step Two: Convert from mass to moles Remember: moles = mass / molar mass
40g Ca / 40.078g/mol = 0.998 mole Ca 12g C / 12.011g/mol = 0.999 mole C 48g O = ??? moles O
Determining Chemical Formulas 3 Step Three: Use the number of moles to create
molar ratios Take the smallest number of moles you have
and divide into all the other numbers 0.998 mole Ca, 0.999 mole C, 3.00 moles O
0.998 0.998 0.998
= 1 mole Ca, 1 mole C, 3 moles O If you don't use the smallest number...
0.998 mole Ca, 0.999 mole C, 3.00 moles O 3 3 3
= 0.33 mole Ca? 0.33 mole C? 1 mole O?
Step 4: Use the simplest formula to find the true formula
Find the molar mass of the simplest formula The molecular weight of the true compound
will be provided Make it match by multiplying the number of
atoms by an integer Why multiply?
Determining Chemical Formulas 4
You have an empirical formula CH2O, mm =
30.03g/mol
The formula of the true compound is 60.06g/mol
What is the true formula???
30.03 * 2 = 60.06
CH2O * 2 = C
2H
4O
2
Determining Chemical Formulas 5
Trivial Things
Molality:
= Moles of Solute / Mass of Solvent (not solution)
Mole Fraction [A]:
= Moles of A / Moles of A + Moles of B
%m/w (or w/w):
=(mass of solute / mass of solution) * 100%
Wrap-Up!
We discussed Balancing chemical equations Stoichiometry % composition by mass Determining chemical formula
Any questions?
Molecular Geometry
Sadly, molecular geometry is mostly memorization
Especially the names of the configurations Key Tricks to drawing molecules with proper
geometry: Figure out the electron arrangement first Electrons repel, and a tetrahedron provides
the most space for four pairs of electrons Memorize all the names though =/
Molecular Geometry
Why do Reactions Happen?
Thermochemistry and Kinetics! Thermochemistry can be broken into
Enthalpy (ΔH) and Entropy (S) The interaction between the two
form Free Energy (ΔG) Free Energy, Enthalpy and
Entropy also affect equilibrium (K)
Thermochemistry - Enthalpy
Change in Enthalpy (ΔH)
= Change in Internal Energy (ΔU)
[+ Work done/by on the system(W)]
Most times, work = 0 , ΔH = ΔU, except for gases
ΔH = ΔU + W
If so, change in internal energy is
ΔH = ΔU = q
Where q is heat energy transferred
All about q
q is all about heat transfer
If q is negative, heat is given off If q is positive, heat is absorbed
Reactions that give off heat usually happen Reactions that take up heat usually don't How do you figure out what q is?
Calorimetry
Experimentally: q = mc ΔT c tells you how much heat the material holds
per gram
Check the units of c, usually in J/g*K but sometimes in J/mol*K
If so, q =nC ΔT If you are given a calorimetry question, c will
be given to you Except for water. It's 4.18 J/gK
Hess's Law
Finding ΔHo is a pain. Luckily, Hess's Law states you can find ΔH
using any combination of reactions as long as they add up to the reaction you want
C(s) + O2 → CO2(g) ΔH = ??
C(s) + ½O2(g) → CO(g) ΔH = -110.5 kJ
CO(g) + ½O2(g) → CO2(g) ΔH = -283.0 kJ
Think of it like a building Shamelessly ripped from UOttawa, Pell/Mayer 2009
Hess's Law
Taken one step further, you can also take the ΔH of formation of the molecules in the equation
ΔH of formation is ΔHf
ΔHf is the energy it takes to form the molecule from the constituent elements
ΔHf of all products - ΔHf of all reactants = ΔH for reaction
Hess's Law
CH4(g) ΔHf= -74.87 kJ
O2 (g) ΔHf= 0 kJ
H2O(g) ΔHf= -241.83 kJ
CO2(g) ΔHf=-393.509 kJ
CH4(g) + 2O2(g) → 2H2O(g) + CO2(g)
ΔH=ΔHf (H2O(g) + CO2(g)) - ΔHf (CH4(g) + O2(g))
ΔH: Endothermic or Exothermic?
When ΔH is negative, the reaction is exothermic
Heat is given off to the atmosphere Good for the reaction!
When ΔH is positive, the reaction is endothermic
Heat is absorbed from the atmosphere Not so good.
When does Work Matter?
Remember, W = ΔPV When you have a gas that expands, it cools
off and does work If you just measured the temperature change,
you would get a false reading Therefore, ΔH = ΔU = q + w
=q + PΔV
Entropy
Entropy is a measure of disorder in the universe
Having more things will make them more disordered
Having a molecule that can twist and turn makes it more disordered
Gases are more disordered than solids
Enthalpy, Entropy and Free Energy
ΔGo is free energy
ΔGo = ΔHo - TΔSo
ΔSo is entropy (J/molK) ΔS and ΔH don't change with T
ΔGo is for T = 298K, standard conditions, 1 atm
Just like ΔH, negative is good!
Finding ΔGo, ΔHo,
ΔSo
If ΔGo isn't given, you can calculate it from ΔH
and ΔS
You can also calculate it from ΔGof from a table
of ΔGo by using Hess's law
Finding ΔGo, ΔHo,
ΔSo
ΔHfo
(kJ/mol) ΔG
fo (kJ/mol) ΔSo
(J/molK)
CO(g)
-110.5 -137.2 197.7
Cl2(g)
0 0 223.1
COCl2(g)
-218.8 -204.6 283.5
ΔGo = ΔGfo (products – reactants)
= -204.6 + 137.2 = -67.4 kJ
ΔGo = ΔHfo (products – reactants) - T ΔSo (products -
reactants) = -218.8kJ + 110.5kJ + (298) (283.5-223.1-197.7)J/molK = -108.3 kJ + 40915.4J = -108.3 kJ + 40.91 kJ = -67.4 kJ Pirated from Waterloo
Chemistry Pages
Finding ΔGo, ΔHo,
ΔSo
H2(g)
+ N2(g)
↔ 2NH3(g)
ΔHf for NH
3(g) is -46.1kJ/mol
ΔS for H2(g)
is 130.7J/molK, for N2(g)
is 191.56J/molK, and for NH
3(g) is 192.77J/molK
What is ΔGo , assuming standard conditions?
ΔGo and ΔG?
ΔGo is great! And useless by itself
It is only for T = 298K ΔG (no o) is the value for the actual condition ΔGo lets you get ΔG!
ΔGo = ΔHo - T ΔSo
ΔG = ΔH - T ΔS
ΔG and Reaction Favourability
If ΔG = negative, reaction likely to happen If ΔG = positive, reaction will not happen; it might happen in the reverse direction
By plugging values into ΔG = ΔH - T ΔS, you
can find ΔG = 0 At ΔG = 0, the reaction is in equilibrium
Decomposition in Equilibrium
2AlCl3 (s) ↔ 2Al(s) + 3 Cl2 (g)
At what temperature will the decomposition of AlCl
3(s) be in equilibrium?
Al(s) Aluminum solid 0 28.3 0
Cl2 (g) Chlorine gas 0 223.08 0
AlCl3 (s)Aluminum Chloride
-705.63 109.29 -630.0
Enthalpy (kJ/mol)
Free Energy (kJ/mol)
Entropy(J/molK)
Cl2 (g) 0 223.08 0
Al (s) 0 28.3 0
AlCl3 -706.63 109.29 -630.0
Decomposition in Equilibrium
ΔH = ΔHf (products) – ΔH
f (reactants)
= 2 (-706.63) – 0
= -1413.26 kJ/molΔS = ΔS
(products) – ΔS
(reactants)
= 2(-630.0) – 0 = -1260 J/molK ΔG = ΔH - T
ΔS
0 = -1413.26 + (T) 1.26 T = 1121 K
Equilibrium Constant: K
K is the equilibrium constant; it is temperature dependent
Given balanced formula
aA + bB ↔ cC + dD K = [C]c [D]d
[A]a [B]b
In the case of gases Kp = p[C]c p[D]d
p [A]a p[B]b
K > 1 is product favoured
K< 1 is reactant favoured
K
Keq
/Kc= K for equation (c for concentration)
Ksp
= K for dissolved substances (sp for solubility product)
Ka /K
b = K for an acid / K for a base
ALL Ks are calculated the same way! Do not include solids/pure liquids in the calculation.
K = [C]c [D]d
[A]a [B]b Except for gases, which * can * be calculated with
pressure
Kp and Gases
For gases, Kp = p[C]c p[D]d
p [A]a p[B]b
Pressures are just easier to use, especially since
Total Pressure = Sum of the individuals pressures of the gases
Kp = K
c (RT)n
P = RTn/V
Examples of K
H2SO
4(aq) + H
2O
(l) → HSO
4-(aq)
+ H3O+
(aq)
Ka = [HSO
4-] [H
3O+] / [H
2SO
4]
2SO2(g)
+ O2(g)
→ 2SO3(g)
Kp = p[SO
3]2 / p[SO
2]2 [O
2]
CaCO3(s)
→ Ca2+(aq)
+ CO3
2-(aq)
Ksp
= [Ca] [CO3]
K, ΔG, and Equilibrium
When ΔG = 0, the reaction was at equilibrium
How can this be related to K? ΔG = ΔGo + RT ln K ΔGo = -RT ln K
ΔGo = -600 kJ/mol
T = 200K T = 1200K RlnK = 3 RlnK = 0.5
Q and Lechatalier's Principle
Lechatalier's principle is like a seesaw
Reactants ↔ Products
Where ΔG is experimental, while ΔGo is for standard conditions
Q is experimental, while K is for theoretical equilibrium
Q
If the experimental conditions = thereoretical equilibrium, Q = K
I2(g)
+ H2(g)
↔ 2HI(g)
at some temperature/ pressure
For this T and P, Kc = 1
Therefore 1 = [HI]2 / [H2] [I
2]
You have 10 moles of HI, two moles of H, and two moles of I in the vessel. Which way will the reaction go?
Q
Q = [HI]2 / [H2] [I
2]
=[10]2 / [2] [2]
Q = 100/4 = 25
if Q = K, the reaction would be at equilibrium
Q > K, the reaction will go towards reactants
Q < K, the reaction will go towards products
ΔG= ΔGo + RT ln Q
Q in a Ksp
Problem
PbCl2 is highly insoluble in water. The K
sp of PbCl
2 is
1.6 * 10−5. PbCl
2 is added to 0.1L of water at 298K,
1atm pressure.
a) How much PbCl2 dissolves?
b) What is ΔGo?
0.058g of NaCl are added to the solution.
c) Find Q.
d) What is ΔG right when the NaCl is added?
e) How much PbCl2 is left dissolved?
Q in a Ksp
Problem
a) How much PbCl2 dissolves?
Ksp
= 1.6 * 10−5
PbCl2(s) ↔ Pb2+(aq) + 2Cl-(aq)
I / 0 0C / +x +2xE / x 2x
Ksp
= (x) (2x)2
1.6* 10-5 = 4x3
x = 0.0159Therefore 0.00159 moles of PbCl
2 dissolve
Keep in mind that ICE tables are for concentration! Therefore the concentration of the ions is x and 2x, not actual amount.
Q in a Ksp
Problem
b) What is ΔGo?
ΔG= ΔGo + RT ln QΔG = 0At equilibrium right now, Q = K
sp = 1.6 * 10−5
T= 298 KR= 8.314Therefore, ΔGo
= - RT ln 1.6 * 10 -5
ΔGo = 27359 kJ/mol
Q in a Ksp
Problem
0.0058g of NaCl are added to the solution.
c) Find Q.mm
NaCl = 58.44g/mol
Therefore 0.001 mols are added to the solution
PbCl2(s) ↔ Pb2+(aq) + 2Cl-
(aq)
I / 0.00159 (0.00318+0.001) 0.1 0.1
Q = 0.0159 * (0.0418)2
= 2.8 *10-5
Q in a Ksp
Problem
d) What is ΔG right when the NaCl is added?
ΔG= ΔGo + RT ln QQ = 2.8 *10-5
ΔGo = 27359 kJ/molΔG = 27359 + 8.314 * 298 * ln 2.8 *10-5
ΔG = 1385 kJ/mol
Q in a Ksp
Problem
e) How much PbCl2 is left dissolved?
Ksp
= 1.6 * 10−5
PbCl2(s) ↔ Pb2+(aq) +
2Cl-(aq)
I / 0.0159 0.0418C / -x -2xE / 0.0159- x 0.0418 -2x
Ksp = (0.0159-x) (0.0418-2x)2
. . .
Ksp = (0.0159-x) (0.0418-2x)2
Is not exactly solvable with the math we know
Using a simplification
Ax3 + Bx2 + Cx + D = Ax3 + D
X = 0.0057
0.0159- x 0.0418 -2x= 0.0102 M Pb2+ 0.0304M Cl-
= 0.00102 Pb
Van't Hoff + Clausius-ClapeyronVan't Hoff:
Clausius-Clapeyron:
ln k=− H o
RT S o
R
lnk 1
k 2
= H o
R 1
T 1
− 1T 2
Note: H is kJ/mol, while S is J/molK
This equation allows you to find one K if given another K
Using Van't Hoff and Clausius-Clapeyron
You have the reaction:
CaCO3 ↔ CaO + CO
2
At 400o C, K = 3.6*10-6
At 500o C, K = 2.2 * 10-4
What is ΔHo, ΔSo, ΔGo, and K at 500o C?
Taken from UOttawa, Prof St-Amant,, past midterms
Acid-Base Equilibrium
How do Ka, K
b, K
w, pK
a, pK
b, pH,
and
pOH all
relate?
H2O
(l) + H
2O
(l) ↔ H
3O+
(aq) + OH-
(aq)
The K of this reaction is 1.0*10-14
Hence Kw
= 1.0*10-14
Acid-Base Equilibrium
Whenever you add acid/base to a solution, it reacts with the water to form H
3O+ or OH-
In neutral water, [H3O+] is 10-7 [OH-] is 10-7
pH= - log [H3O+] = 7
pOH = - log [OH-] = 7
pH + pOH = 14
Adding acid increases [H3O+] , so pH decreases
Adding base decreases [H3O+], so pH increases
Acid-Base Equilibrium
In the same vein, pKa + pK
b = 14
HA + H2O ↔ H
3O+ + A- K
a =
Some #
A- + H2O ↔ HA + OH- K
b = Some #
= H2O + H
2O ↔ H
3O+ + OH- K
w = 10-14
Ka * K
b = K
w
Henderson-Hasselbalch Equation Instead of using an ICE table, you can simplify with:
This assumes that [acid] > 100Ka
pH = pK a logA
HA
A Buffer Example
Calculate the pH of 1.00L of a buffer system composed of 0.90M acetic acid, CH
3COOH and
0.60M sodium acetate, NaCH3COO
a) initially b) after the addition of 0.1L of 0.5M NaOH
The Kb of acetate is 5.6*10-10
Taken from UOttawa, Prof St-Amant,, past midterms
A Buffer Example
First, find Ka
Ka * K
b = 10-14
Ka = 10-14/5.6*10-10
= 1.8 * 10-5
We can calculate the pH of the solution before the addition of NaOH, using the Henderson-Hasselbalch equation, since [acid] > K
a
A Buffer Example
pH = pKa + log [A-] / [HA]
= - log 1.8 *10-5 + log ( [0.6] / [0.9])
= 4.57
CH3COOH + H
2O ↔ CH
3COO + H
3O+
A Buffer Example
b) After the addition of 0.1L of 5M NaOH? (How many moles of NaOH is that?)
CH3COOH + H
2O ↔ CH
3COO + H
3O+
I (0.9 - 0.5)/1.1 / (0.6+0.5)/1.1 0
C -x / +x + x
E 0.36 -x / 1+x x
Step 1: Neutralize acid/baseStep 2 : Add conjugate base/acid to other sideStep 3: Recalculate concentrationsStep 4: Finish your ace table
A Buffer Example
E 0.36 -x / 1+x x
Ka = 1.8 * 10-5 = (1+x) (x) / (0.36-x)
0 = 6.5 *10-6 -1.000018 x -x2
X = 0.0000065 molspH [H
3O+] = - log 0.0000065
= 5.19
CH3COOH + H
2O ↔ CH
3COO + H
3O+
Electrochem – Redox Reactions
Redox reactions are composed of
Reduction and Oxidation LEO (goes) GER
Loss of Electrons is OxidationGain of Electrons is Reduction
Reduction must be paired with Oxidation When do you know if something is
oxidized/reduced?
Oxidation States
1. Check the Oxidation states!
Unbalanced:
C6H
5NO
2 + Sn → C
6H
5NH
2 + Sn2+
Redox Reactions
2. Begin to write out half-reactions:
Sn → Sn2+ + 2e-
C6H
5NO
2 + 6e-
→ C
6H
5NH
2
3. Balance atoms besides O and H
4. Balance charge with H-
Sn → Sn2+ + 2e-
C6H
5NO
2 + 6e- + 6H+
→ C
6H
5NH
2
Redox Reactions
5. Balance LS and RS with H2O:
Sn → Sn2+ + 2e-
C6H
5NO
2 + 6e- + 6H+
→ C
6H
5NH
2 + 2H
2O
6. Add your reactions together so e- cancels out
3Sn → 3Sn2+ + 6e-
C6H
5NO
2 + 6e- + 6H+
→ C
6H
5NH
2 + 2H
2O
Redox Example
Balance the following equation:
S2O
62-
(aq) + HClO
2(aq) →
SO
42-
(aq) + Cl
2(g)
Taken from UOttawa, Prof St-Amant,, past midterms
Cell Potentials
Oxidation potential + Reduction Potential
Li+(aq)
+ e- → Li(s)
Eo= - 3.04 VCu2+
(aq) + 2e- → Cu
(s) Eo = 0.34 V
2 Li (s)
→ 2Li+(aq)
+ 2e-
Eo= 3.04 V Cu2+
(aq) + 2e- → Cu
(s) Eo= 0.34 V
= 3.34 V
ΔGo = - nFEocell
ΔG = - nFEcell
E cell =E cello −
RTnF
ln Q
They do NOT change when you multiply the molar coefficients!
Kinetics
Thermodynamics is not the only thing that makes reactions go! Kinetics play a large role too
Kinetics tell you how fast reactions go Kinetically, some reactions proceed so slowly, they
don't proceed at all
Given A + B → C
Rate law = k [A]m [B]n
Rate laws are determined experimentally! The order of the reaction is the sum of the
coefficients
Reaction Rates
All reaction rates must be determined experimentally
You did this in one of your labs You can measure the initial rate of change of the
concentration of the reactants/problems
Experiment 1 2 3 4
[A]o
0.100 0.200 0.200 0.100
[B]o
0.100 0.100 0.300 0.100
[C]o
0.100 0.100 0.100 0.400
Rate 0.100 0.800 7.200 0.400
Integrated Rate Law
The whole point of the integrated rate law is to turn the rate law into something that can be modeled with a linear equation
From the linear equation, you can easily determine K, and figure out other points on the line
If 0th order: [A] = -kt + [A]o
If 1st order: ln [A] = -kt + ln [A]o
If 2nd order: 1/[A] = kt + 1/[A]o
Integrated Rate Laws
Time [H2O2]
ln [H2O2]
1/[H2O2]
0 1 0 1
120 0.91 -0.09 1.1
300 0.78 -0.25 1.28
600 0.59 -0.53 1.69
1200 0.37 -0.99 2.7
1800 0.22 -1.51 4.55
2400 0.13 -2.04 7.69
3000 0.082 -2.5 12.2
6000 0.05 -3 20
Attempt to graph [H2O
2] vs Time
Integrated Rate Laws
Time [H2O2]
ln [H2O2]
1/[H2O2]
0 1 0 1
120 0.91 -0.09 1.1
300 0.78 -0.25 1.28
600 0.59 -0.53 1.69
1200 0.37 -0.99 2.7
1800 0.22 -1.51 4.55
2400 0.13 -2.04 7.69
3000 0.082 -2.5 12.2
6000 0.05 -3 20
Attempt to graph 1/[H2O
2] vs Time
Integrated Rate Laws
Time [H2O2]
ln [H2O2]
1/[H2O2]
0 1 0 1
120 0.91 -0.09 1.1
300 0.78 -0.25 1.28
600 0.59 -0.53 1.69
1200 0.37 -0.99 2.7
1800 0.22 -1.51 4.55
2400 0.13 -2.04 7.69
3000 0.082 -2.5 12.2
6000 0.05 -3 20
Attempt to graph ln [H2O
2] vs Time
RDS
Chemical reactions are actually a series of complex reaction reactions involving many intermediates
The speed of the overall reaction is limited by the formation of the slowest intermediate
Think about an assembly line The rate law reflects the interactions of this slowest
step
If rate = k [A], molecule A must be in the slowest
step, and no other molecules If rate = k [A]2, two molecules of A are involved in
the slowest step, and no others
RDS and Catalysts
There are many reasons the RDS might be slow Usually, because it is not as thermodynamically
favourable This is reflected in the activation energy of the step
How much energy to stay in a reactive state Catalysts change the reactive state itself, or make it
easier to get into the reactive state, so the reaction can proceed
Quick Quantum Number Overview
Your Quantum numbers are n, l, mn, and m
l
n = “Principle,” what shell you are in S P D F
l = “Angular,” n-1
mn =
Magnetic, - “L” to +”L”
ms = “Spin,” +1/2 or -1/2
Quick Quantum Number Overview
Quick Quantum Number Overview