sos exam-aid

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


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 → ??


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?


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


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


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


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


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)


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


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


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

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