chemistry203 finalexam -...

Chemistry 203 – Final Exam 1

(C) Pavel SedachLearnfaster.ca

Questions? Suggestions? Email me and I will try to find time to reply. [email protected]

All the best!

Lecture Slides Booklet Solutions Test Solutions Extra notes and advice Advice on which exams and questions to write Corrections

Available Saturday, December 15th at: http://learnfaster.ca/blog/chem203final/

Thursday

4:00 PM to 5:30 Equilibrium through Acids and Bases

5:30 to 6:00 Break

6:00 to 7:30 Acids and Bases and Ksp

7:30 to 7:40 Break

7:40 to 9:00 PM Equilibrium and Thermodynamics

Saturday

4:00 PM to 5:30 Electrochemistry

5:30 to 6:00 Break

6:00 to 7:30 Kinetics

7:30 to 7:40 Break

7:40 to 9:00 PM Review

The larger goal of this course is to teach you that Equilibrium IS Electrochemistry 2


A reaction is spontaneous in the forward direction if K is bigger than Q, right?

ΔG0 = −RT ln K = −nFE0

Well that means the reaction shifts to the right, so the Δ𝐺 is negative (Spontaneous)!

This also means that the voltage is Positive (also Spontaneous)

The Climax of this course is in understanding that all 3 (Gibbs Free Energy, K and Voltage) are the same.

Acid and Bases


3

Water ionizes itself (autoionization)


4

2 H2O l ⇌ H3O aq+ + OH aq

−

Kwater = Kw = H3O aq+ OH aq

− = 1.00 × 10−14 (𝑎𝑡 298 𝐾)

In neutral water, H3O aq+ = OH aq

−

Where does Neutral pH come from?

1.00 × 10−14 = H3O aq+ OH aq

− = x2

x =

pH of neutral Water


5

At 90 ℃, the Kw of water is 9.00 × 10−12, what is the pH of neutral water at that temperature?

If the pH of neutral water at 2 ℃ is 8.255, what is Kw?

Is the dissociation of water endothermic or exothermic?

pH (potential of hydrogen) and other “p” Scales


6

pH is a measure of acidity. The lower the pH the more acidic the solution.

The letter “p” is a mathematical symbol like ×,÷,− or +.

p means log(x), for instance pH = −log[H3O+], pOH = −log[OH−], pKa = − log Ka

H3O aq+ pH = −log(H3O aq

+ )

10−14 M 14

10−10 M 10

10−7 = 0.0000001 M 7

10−4 = 0.0001 M 4

100 = 1 M 0

10+1 = 10 M 1

Commercial HClis 13.6 M, so its pH is even lower!

Question – what is p90?

H3O aq+ OH aq

−

14 = pH + pOHpH pOH

Kw = H3O aq+ OH aq

−

−log H3O aq+

− log OH aq−

10−pOH10−pH

Strong Acids –ON FORMULA SHEET!


7

Covalent compounds that ionize 100%,

No K value because they ionize 100%

HCl, HBr, HI, H2SO4, HClO4, HNO3

What is the pOH of an aqueous solution of HI aq with an initial concentration of 1.0 × 10−3 M?

What is the pH of 6.0 × 10−9 M HBr aq ?

a) 7.00

b) 8.22

c) 6.98

d) 6.20

Acid = H3O+

if HClO4 = 1 mol/L, H3O+ = 1 mol/L

hydrochloric acid, hydrobromic acid, hydroiodic acid, sulfuric acid, perchloric acid, nitric acid

Strong Bases –ON FORMULA SHEET!


8

Ionic compounds, Dissociate ~100% to give OH− in solution.

1st column metal hydroxide: NaOH → Na+ + OH−

NaOH = OH−

If NaOH = 1 mol/L, then OH− = 1 mol/L

What is the pH of an aqueous solution of LiOH(aq) with an initial concentration of 130.0 × 10−7 M?

2nd column metal hydroxide: Mg OH 2 Mg2+ + 2 OH−

2 × Mg OH 2 = OH−

If Mg OH 2 = 0.0001 mol/L, then OH− = 0.0002 mol/L (times 2!)

With 2+ metals, you may have to perform a Ksp (value of Ksp is provided) calculation to determine acidity/basicity as they are

not very soluble.

Acidity and Basicity

© Pavel SedachLearnfaster.ca

9

The Conjugate of Strong Acid is a Neutral Salt (exception is H2SO4)

Strong AcidsIonize 100%

Strong Bases are Metal Hydroxides – their conjugate acid is water

Weak Acids Ionize Partially Because they Produce Weak Base

The Conjugate of a Weak Acid is a Weak Base

The conjugate of a strong acid is a

______________

⇌⇌⇌⇌⇌⇌

Relationship Between Acidity and Basicity(C) Pavel Sedach Learnfaster.ca 10

CH3COOH aq + H2O l ⇄ H3O aq+ + CH3COO aq

− , Ka = 1.8 × 10−5

CH3COO aq− + H2O l ⇄ OH aq

− + CH3COOH aq , Kb = 5.5 × 10−10

Net Reaction 2 H2O l ⇌ H3O aq+ + OH aq

− , Kw = 1.0 × 10−14

∴ Ka × Kb = Kw

Conjugate Acid Strength

Ka

Conjugate Base Strength

Kb

Acid/Base Strength

+ ×

Strong Acids ionize 100% their conjugate is neutral! Example: I−, Br−, ClO4−, NO3

−

HCl + H2O → H3O+ + Cl−

Weak acids/bases do not ionize 100% because on the other side there is a weak conjugate A/B

CH3COOH aq + H2O l ⇄ H3O aq+ + CH3COO aq

−

Calculating the pH of aWeak Acid Solution


11

What is the pH of a solution of 0.0017 moles of H2S aq (hydrosulfuric acid) added to 10 mL water? (Ka = 8.9 × 108)

H2S aq + H2O l⇄ H3O aq

+ + HS aq−

Initial

Change

Equilibrium

Ka =HS aq

− H3O aq+

H2S aq= 8.9 × 108

_________ ________________

= 8.9 × 108

_________ ________________

= 8.9 × 108 pH = −log H3O aq+

Calculating the pH of aWeak Base Solution


12

What is the pH of 0.1 mol/L CH3COO aq− ? (acetate ion) (Ka CH3COOH aq

= 1.8 × 10−5)

CH3COO aq−

+ H2O l⇄ OH aq

− + CH3COOH aq

Initial

Change

Equilibrium

Kb =Kw

Ka=

10−14

1.8 × 10−5= ______________ =

CH3COOH aq OH aq−

CH3COO aq−

_________ ________________

= ______________

_________ ________________

= ______________

pOH = − log OH aq−

Review


13

What is the pH of 0.25 M NaHCO3 aq (sodium hydrogen carbonate)? (KaH2CO3 aq= 4.5 × 107) (KaHCO3 aq

− = 4.7 × 1011)

+ H2O l⇄ aq +

Initial

Change

Equilibrium

Amphiprotic species act as an Acid or a Base


14

Will H2PO4− behave as an acid or base in neutral water?

Species Ka Value Kb Value

H3PO4 aq 7.1 × 10−3

H2PO4 aq− 6.3 × 10−8

HPO4 aq2− 4.2 × 10−13

PO4 aq3− Not an Acid

Review


15

Can we find the pH of a potassium citrate (KC3H5O COOH 2COO aq ) solution given the following information?

Species Ka Kb =10−14

Ka

HC3H5O COOH 3+

C3H5O COOH 3 7.4 × 10−4

C3H5O COOH 2COO aq− 1.35 × 10−11

C3H5OCOOH COO 2 aq2− 4.0 × 10−7

C3H5O COO 3 aq3− 2.5 × 10−8

Neutral, Acidic or Basic?


16

Neutral Compounds

The conjugate of most strong acids is NEUTRAL

Exception = ?

HCl – Cl−

HBr – Br−

HI – I−

HClO4 – ClO4−

HNO3 – NO3−

Alkali and Alkali Earth Metals are neutral: Na+? Cs+? Sr2+? Fe3+? Be2+?

Acidic Compounds (usually POSITIVE CHARGE)

Transition Metals (Fe3+, V3+, Cr3+ and others with unfilled valencies)and Al3+,

Biggest giveaway? A 𝐊𝐚 is provided for a metal!

These metals undergo a “Lewis Acid” = Electron Acceptor reaction in water: Al aq3+ + 3H2O l ⇌ Al OH 3(aq) + 3 H aq

+

If it contains a −COOH or − OH or − SH or − NH+ (the PLUS is a big give away!), they’re usually acidic.

Especially if there is a POSITIVE NITROGEN.

Basic Compounds (usually NEGATIVE CHARGE)

−COO− or − O− or − S− or − N (bases have a negative charge OR a neutral or negative NITROGEN)

Weak A/B + Strong A/B Reactions


17

The reaction of weak acids/bases with strong bases/acids forms the conjugate species:

NaOH(aq) + CH3COOH aq → CH3COONa aq + H2O l

Result: __________ solution

HCl(aq) + KCN(aq) → HCN(aq) + KCl(aq)Result: __________ solution

HCl(aq) + NH3(aq) → NH4Cl aqResult: __________ solution

Is a Salt Basic or Acidic?


18

Split the salt into its cation and anion

Ka H2CO3= 4.5 × 10−7

Ka HCO3− = 4.7 × 10−11

If a salt contains a cation that contributes to H3O aq+ production, as well as an anion that contributes to OH aq

− production, the

influence of that salt on pH is dependent on the relative values of Ka for the cation and Kb for the anion.

(Kb CN− = 1.6 × 10−5, Ka(NH4+) = 5.6 × 10−10)

Ka > Kb, the solution is acidic

Ka < Kb, the solution is basic

Ka = Kb, the solution is neutral

LiHCO3(aq)

NH4CN aq

Strong Acid/Base Titration:


19

HCl(aq) + NaOH aq ⇄ H2O l + NaCl aq

Equivalence point: whenmoles acid equal moles base

To visualize the equivalence point, a chemical indicatoris used. An indicator is a weak acid or base, whoseconjugate is a different color.

Indicators are added in small amount and will change color over a specific pH range – this pH range must overlap the equivalence point.

The end point is when the indicator changes color

∴ choosing the correct indicator is important

pH 7

0

14

mL of base added

pH = 7mol HCl = mol NaOH at the equivalence point

HCl is the only species present

NaOH is present in excess

pH usu. 0 to 2the indicator chosenMUST overlap the equivalence point

Titration of aWeak Acid with a Strong Base


20

CH3COOH aq + NaOH aq → H2O l + NaCH3COO aq

Volume (mL) of Strong Base

pH 7

0

14

CH3COOH aq

CH3COOH aq /CH3COO aq−

CH3COO aq−

NaOH aq

If we are titrating a weak acid with a strong base, the pH at equivalence will be greater than 7 because the conjugate base is produced.

If we are titrating a weak base with a strong acid, the pH at equivalence will be less than 7 because theconjugate acid is produced.

Polyprotic Species


21

1 molar phosphoric acid is triprotic because it has three ionizable hydrogen atoms (protons).

H3PO4 aq + H2O l ⇄ H3O aq+ + H2PO4 aq

− , Ka1 =H2PO4 aq

− H3O aq+

H3PO4 aq= 7.1 × 10−3

H2PO4 aq− + H2O l ⇄ H3O aq

+ + HPO4 aq2− , Ka2 =

HPO4 aq2− H3O aq

+

H2PO4 aq− = 6.3 × 10−8

HPO4 aq2− + H2O l ⇄ H3O aq

+ + PO4 aq3− , Ka3 =

PO4 aq3− H3O aq

+

HPO4 aq2− = 4.2 × 10−13

For all polyprotic acids Ka1 > Ka2 > Ka3

Acid dissociation decreases when acid concentration increases in solution.

For a polyprotic acid, the pH of the solution is ~ equal to the [𝐇𝟑𝐎 𝐚𝐪+ ] produced from only the first ionization step.

Ideal Polyprotic Titration


22

triprotic acid

H3PO4 aq + H2O l ⇄ H3O aq+ + H2PO4 aq

−

diprotic acid

H2PO4 aq− + H2O l ⇄ H3O aq

+ + HPO4 aq2−

also amphoteric

monoprotic acid

HPO4 aq2− + H2O l ⇄ H3O aq

+ + PO4 aq3−

also amphoteric base only

Buffer – amixture of a weak acid and it’s conjugate base


23

Buffer solutions resist changes to pH from the addition of acid, base, or water.

Let’s look at how a CH3COOH aq /CH3COO aq− buffer does this:

CH3COO aq− + HCl(aq)CH3COOH aq + Cl(aq)

−

CH3COOH aq + NaOH aq NaCH3COO aq + H2O l

The pH of a buffer solution is given by the HendersonHasselbalch Equation:


24

pH = pKa + logA−

HA to get a specific pH, use the buffer whose pKa is closest to pH!

Are the following solutions buffers? What pKa do we use to calculate pH?

HCN/NaCN? HPO42− / H2PO4

−

HNO3/KNO3? H2PO4− / PO4

3−

The addition of which of the following two components could make a buffer solution?


25

*There are two answers

a) 0.1 mol HCl and 0.1 mol KOH

b) 0.1 mol CH3COONa and 0.1 mol KCN

c) 0.1 mol HCN and 0.5 mol NaCN

d) 0.1 mol HNO3 and 0.5 mol CH3COOH

e) 0.05 mol HNO3 and 0.1 mol CH3COO−

Titration – Initial pH


261.00 L of 0.10 mol/L HOCl aq (Ka HOCl aq = 4.0 × 10−8) is titrated with 0.10 M NaOH aq What is the pH before any NaOH aq is added?

Step 1. Stoichiometry.

HOCl aq + NaOH aq NaOCl aq + H2O l

Initial

Change

Final

Step 2. Identify the system and do an appropriate calculation

3 options –Weak Acid/Base Strong Acid/Base Buffer

0

14

pH 7

mL of NaOH aq

1.0 L of 0.10 mol/L HOCl aq (Ka HOCl aq = 4.0 × 10−8) is titrated with 0.10 M NaOH aq What is the pH when 0.25 L of NaOH aq is added?



Initial

Change

Final



pH before Equivalence Point


27

0

14

pH 7

mL of NaOH aq

pH at ½ Equivalence Point





Initial

Change

Final



0

14

pH 7

mL of NaOH aq

pH at Equivalence Point





Initial

Change

Final



0

14

pH 7

mL of NaOH aq

pH after Equivalence Point

(C) Pavel Sedach Learnfaster.ca




Initial

Change

Final



0

14

pH 7

mL of NaOH aq

0 0.25 L 0.5 L 1.0 L 1.5 LVolume (mL) of Strong Base

Review


31

The pH of the solution at the midpoint is equal to the pKa of the weak acid because A− = HA and log Base

Acid= log(1) = 0

This means if you have a titration curve (created with a pH meter), you can find the half equivalence point and therefore the pKa,Ka and finally the identity of an unknown acid.

pH 7

0

14NaOH aq

4.2

6.92 7.40

10.05

12.3

Indicators


What is the best indicator for a titration with an equivalence point at pH=7.5?

Indicator pH RangeColor Change aspH Increases

Ka

cresol red0.0 − 1.07.0 − 8.8

red to yellowyellow to red

~3 × 10−1

3.5 × 10−9

thymol blue1.2 − 2.88.0 − 9.6

red to yellowyellow to blue

2.2 × 10−2

6.3 × 10−10

bromocresol green 3.8 − 5.4 yellow to blue 1.3 × 10−5

chlorophenol red 5.2 − 6.8 yellow to red 5.6 × 10−7

phenolphthalein 8.2 − 10.0 colorless to pink 3.2 × 10−10

alizarin yellow R 10.1 − 12.0 yellow to red 6.9 × 10−12

32

Choosing the best buffer


If we want a buffer of pH = 8.00, what should we add to a solution of NaHSO3?

pH = pKa + logA−

HAAcid Conjugate Base Ka pKa

H2SO3 HSO3− 1.4 × 10−2

HSO3− SO3

2− 6.3 × 10−8

33

Does the above buffer have a larger acid (ability to resist acid) or base (ability to resist base) capacity?

Can we use the base table to create buffers?


34

Acid Ka Base Kb pKb pKa

Bicarbonate HCO3− 4.8 × 10−11 Acetate CH3COO− 5.6 × 10−10

Carbonic acid H2CO3 4.2 × 10−7 Aniline C6H5NH2 3.8 × 10−10

Hydrazoic acid HN3 1.9 × 10−5 Hydrazine N2H4 8.5 × 10−7

What is the pH of a 1:1 H2N3+/HN3 mixture?

What is the pH of a 1:1 N2H4/N2H5+ solution?

*For the base table, you can only make a buffer with the conjugate acid.

Solution Equilibrium


35

Solubility Product When a solid dissolves in a liquid, equilibrium is reached betweendissolved ions and the solid


36

This is a normal equilibrium, K still equals Products

Reactants

Ca3(PO4)2 sH2O l

3 Ca aq2+ + 2 PO4 aq

3− , Ksp = Ca aq2+ 3

PO4 aq3− 2

= 2.07 × 10−33

Initial

Change

Equilibrium

The larger the value of Ksp, the greater the concentration of ions at equilibrium HOWEVER solubility is generally compared in

terms of molar solubility (moles of solute dissolved in liters of solvent) (i.e., the x value of the table above)

When setting up any equilibrium expression remember that solids and liquids are not included.

If Ksp < Qsp, SUPERSATURATED precipitation occurs

If Ksp > Qsp, UNSATURATED species remain in ionic dissolved form

If Ksp = Qsp, SATURATED dissolution and precipitation occur at the same rate and the solution is saturated

Molar Solubility


37

If the Ksp of calcium phosphate is 2.07 × 10−33 and its molar solubility is 1.14 × 10−7mol

L,

How many grams of the above salt can dissolve in 3.00 liters of water?

The Common Ion Effect


38

What is the molar solubility of calcium phosphate if it is added to a solution containing 0.01 M Li3PO4 aq ?

*Ksp lithium phosphate = 3.45 × 10−11, Ksp calcium phosphate = 2.07 × 10−33

Ca3(PO4)2 s ⇄ 3 Ca aq2+ 2 PO4 aq

3−

Initial

Change

Equilibrium

Review


39

What is the relative solubility of silver chloride AgCl, Ksp = 1.77 × 10−10 in

0.1 M NaOH 0.01 M HCl water 0.1 M Na2S 0.1 M KI

Additional data:

2 Ag aq+ + S aq

2− ⇌ Ag2S s , K = 1.23 × 1050

Ag aq+ + I aq

− ⇌ AgI s , K = 1.17 × 1016

Review


40

If a 2.50 × 10−2 M cadmium fluoride solution is mixed with 3.50 × 10−2 M barium nitrate will a precipitate occur? If so, what will it be?

compound Ksp

cadmium fluoride 6.44 × 10−3

barium nitrate 4.64 × 10−3

barium fluoride 1.84 × 10−7

calcium nitrate N/A (very soluble)

Relevant Ksps

Review


41

If 1.30 grams of lithium fluoride are dissolved in 1.00 liter of solution, how would you describe the solution? (Ksp = 1.85 ×10−3)

Review


42

Which of the following compounds is more soluble in pure water?

Yttrium fluoride (YF3, Ksp = 8.62 × 10−21) or silver bromide (AgBr, Ksp = 5.35 × 10−13)

Thermochemistry


43

Assuming the following reaction goes to completion in a closed vessel,


44

C5H12(l) + 8 O2(g) ⇌ 5 CO2 g + 6 H2O g

If we started with 15 atm of oxygen and excess C5H12(l), How will the pressure at the beginning of reaction compare to the total

pressure at the end of reaction?

If you then cool the container, what will happen to the pressure and why?

If the container is actually an adjustable piston, will it expand or compress and why? At the molecular level, what happens to the number of molecular collisions with the container?

Work and Isobaric, Closed Systems


45

Given the following:

C3H8 g + 5 O2 g → 3 CO2 g + 4 H2O l , ΔHrxn = −1.911 × 103 kJ

An unknown quantity of propene (C3H8 g ) is burned with excess oxygen (O2 g ) at 298K. When this experiment takes place,

there is an internal energy change of 800 kJ and + 60.2 kJ of work is involved. How many moles of propene were present in thisexperiment?

∆U = q + w

∆U = ∆H + −P∆V

6 CO2 g + 6 H2O g

C6H12O6 g + 6 O2 g

Reaction Coordinate Diagrams


46

6 CO2 g + 6 H2O g

Energy

Reaction Coordinate

∆H = +2800 kJ

Endothermic (energy is taken in) Exothermic (energy is released)

Energy

Reaction Coordinate

C6H12O6 g + 6 O2 g

∆H∆H = −2800 kJ

A → B ∆H = +100 kJB → C ∆H = −50kJA → C ∆H = +50kJ

Energy

Reaction Coordinate

∆H = +100 kJ𝐀

B

𝐂∆H = −50 kJ

∆H = +50 kJ

Enthalpy, Entropy and Free Energy


47

1 A + 2 B ⇌ 3 C

∆Hrxn° = 3 × ∆Hf C

° − 2 × ∆Hf 𝐵° + 1 × ∆Hf A

°

∆Hrxn° = 𝑛∆Hf(Products)

° −𝑛∆Hf Reactants°

∆Srxn° = 𝑛Sf(Products)° −𝑛Sf Reactants

°

∆Grxn° = 𝑛∆Gf(Products)° −𝑛∆Gf Reactants

°

∆Grxn = ∆Hrxn − T∆Srxn

Review


48

Under SATP, 2 moles of CaCO3 s decomposed completely into CaO(s) and CO2 g , given the following information, what was the

change in internal energy of the system?

CaCO3 s → CaO(s) + CO2 g

∆𝐻 = 𝑃 − 𝑅

∆U = q + w

∆Hf° kJmol

CaCO3 s 1207.6

CaO(s) 634.9

CO2 g 393.5

Review


49

The given chemical equations represent the combustion of ammonia and hydrogen.

4 NH3(g) + 3 O2 g → 6 H2O g + 2 N2 g , ∆H = −1516 kJ

2 H2(g) + O2 g → 2 H2O g , ∆H = −572 kJ

What is the enthalpy of the formation of ammonia?

Could we have predicted the sign (negative or positive) of the enthalpy of formation of ammonia?

ΔS (Entropy) – AMeasure of Change in Disorder


50

When predicting entropy:

1. gases > liquids > solids.

2. A substance is more disordered at a higher temperature.

3. Gas in a large volume is more disordered than in a small volume.

4. A larger/more complex molecule has greater entropy than a smaller molecule.

5. An increase in the number of moles of products compared to reactants means an increase in entropy.

Statements about entropy to memorize/understand for Multiple Choice Theory Questions:

At 0 K, particles in a system have no thermal energy, no motion, and no disorder.

The entropy of any system at 0 K is zero.

The entropy of a perfect crystal at absolute zero (zero Kelvin) is zero.

Entropy decreases when the system becomes more ordered.

Predict the entropy change or sign of ΔS for the following reactions

CaO s + CO2 g ⇌ CaCO3 s

2 NH3 g ⇌ N2 g + 3 H2 g

ΔS (Entropy) – AMeasure of Change in Disorder


51

Which of the following is more disordered?

*The enthalpy of formation of elements (ΔHf0) is zero HOWEVER, the entropy (𝐒𝟎) of an element is never zero!!!!

What is the predicted sign of ΔS for the following reactions:

H2O l or H2O g

H3O aq+ or H2O l

FeCl3(s) or Fe CH3COO 3 s

NaCl l or NaCl s

CaCO3 s or CaO(s) + CO2 g

H2O g at 110 °C or H2O g at 210 °C

ΔS

H2O g ⇌ H2O l + or

4 NH3(g) + 3 O2 g → 6 H2O g + 2 N2 g + or

Introduction to Gibbs Free Energy


52

The entropy of the universe increases in any spontaneous reaction;

ΔSuniverse = −ΔGreaction

A negative Gibb’s free energy of reaction indicates the entropy of the Universe increases! This means the reaction is favorable!

Therefore a spontaneous reaction has a negative ΔG

When the change in free energy (ΔG) is negative, the reaction is spontaneous in the direction written.

When the change in free energy (ΔG) is positive, the reaction is nonspontaneous in the direction written BUT it is

spontaneous in the opposite direction (backward reaction is favored).

When ΔG = 0, the reaction is ________________

Spontaneity


53

In some cases, nonspontaneous reactions can be made spontaneous by changing the temperature. For example, water will not boil at room temperature, but it will above 100 ℃.

Depending on the sign of ΔS and ΔH, reactions may be spontaneous at all temperatures, no temperature, high temperatures, or low temperatures:

∆Grxn = ∆Hrxn − T∆Srxn

( ) = ( ) – ( )

( ) = ( ) – ( )

( ) = ( ) – ( )

( ) = ( ) – ( )

Use the following thermochemical data to calculate ∆S, ∆H and ∆G at 298K for:


54

C s,graphite + H2O g → CO g + H2 g ∆H𝑓° 𝑘J

molS°

JKmol

C s,graphite 0 6

H2O g 241 188

CO g 111 164

H2 g 0 130

Notice these numbers are NEVER zero! Evenfor elements.

Could we have used common sense to deduce this ∆G value (through deducing∆S and ∆H)?


55

C s,graphite + H2O g → CO g + H2 g

∆S = +100 J/K

∆H = +130kJ

∆G =+100kJ

At what temperature does the above reaction become spontaneous?

2

1

Path A (4 meters)

Path B (6 meters)State

State

State Functions (typically has a Delta, ∆ or Change) Path Functions

Internal energy (∆E or ∆U) Heat(q)

Enthalpy (∆H) Work (w)

Entropy (∆S) Distance (d)

Gibbs free energy (∆G) Frequency (f)

Pressure (P)

Temperature (T)

Volume (V)

State and Path Functions


56

State functions – where you are (a property of the system)

Example: Temperature

Path functions – how did you get there (did we add heat or work?)

Example: Heat or work

Everyone in the class is sitting in the same room (state function) even though everyone travelled different paths (path function) to get here.

Equilibrium and Thermochemistry


57

The change in Gibb’s free energy of a reaction ΔGrxn under nonstandard conditions depends

on the position of the reaction in its current state Q relative to the position of equilibrium Keq :

𝚫𝐆𝐫𝐱𝐧 = 𝐑𝐓𝐥𝐧𝐐

𝑲 𝒆𝒒

If Q = Keq, then ΔGrxn will be zero and the system is at equilibrium

𝚫𝐆𝐫𝐱𝐧 = 𝐑𝐓𝐥𝐧

If Q < Keq, then ΔGrxn will be negative – the forward reaction is favored until equilibrium is reached


If Q > Keq, then ΔGrxn will be positive – the reverse reaction is favored until equilibrium is reached


Read this over BUT don’t struggle – we’ll make sense of it through questions!


58

Q = 1 under standard conditions because all the concentrations are 1 mol/L and all pressures are 1 atm!

We can then relate the change in standard Gibb’s free energy of a reaction to the equilibrium constant::

𝚫𝐆𝐫𝐱𝐧 = 𝐑𝐓𝐥𝐧𝐐

𝐊 𝐞𝐪

ΔGrxn0 = RTln1

K eq= ΔGrxn0 = RTln K eq

−1

This becomes 𝚫𝐆𝐫𝐱𝐧𝟎 = −𝐑𝐓𝐥𝐧 𝑲 𝒆𝒒

If ΔGrxn0 < 0 then Keq > 1

If ΔGrxn0 = 0 then Keq = 1

If ΔGrxn0 > 0 then Keq < 1

ΔGrxn0 can be used to determine the spontaneity of a reaction under any set of conditions via:

𝚫𝐆𝐫𝐱𝐧 = 𝚫𝐆𝐫𝐱𝐧𝟎 + 𝐑𝐓𝐥𝐧 𝐐

Review


59

The normal boiling point is the temperature at which a substance will change from liquid to gas phase at 1 atm of pressure.During any phase change, a dynamic equilibrium is established. Calculate the normal boiling point of mercury, Hg l , using the

following data:

∆Hf° for Hg g 50 kJ/mol

S° of Hg l 74 J/(K mol)

S° of Hg g 174 J/(K mol)

a) 300K

b) 400K

c) 500K

d) 600K

Review Manipulating Equilibrium Constants


60

12N2 g + O2 g → NO2 g , K1 = 1.2 × 10−4

2 NO2 g → N2O4 g , K2 = 52

What is the equilibrium constant for:

N2O4 g → N2 g + 2O2 g

Direction of Reversible or Equilibrium Reactions (F2015 Q16) (C) Pavel Sedach Learnfaster.ca

61

Δ𝐺 𝐸𝑐𝑒𝑙𝑙 Q Direction of Reaction

< 0(−)

> 0(+)

> 1 spontaneous in forward direction

> 0(+)

< 0(−)

< 1 spontaneous in reverse direction

0 0 1 no net reaction: system at equilibrium

Negative ΔG0 = spontaneous in forward direction

Positive ΔG0 = spontaneous in reverse direction

(equilibrium shifts depending on excess of reactants or products)


ΔGabsolute value ofGibbs Free Energy

If we plot the absolute value of ΔG0 vs. the amount of reactants/products we get the following:

0Reactants Only

0Products Only

At the bottom of the curve, ΔG = 0, rxn. is at equilibrium.

Where the bottom is relative to a 50:50 ratio defines

whether the reaction prefers products or reactants at K.

In this case, the reaction prefers _____________

and so ΔG0 is ________ and K is ___________ than 1.

ΔG is a large negative number ΔG is a large

positive number

Electrochemistry


62

Oxidation Numbers


63

Oxidation Numbers are based on the number of electrons elements exchange to reach a stable configuration – this is dependent on their existing configuration and their electronegativity

OxidizingAgents

ReducingAgents

+1

+2

+3

123

NaCl HCl HH

Oxidation Numbers (ONs)


64General Rules

1. The ON of any pure element is zero.

Ex: N2, P4, S8, He

2. The oxidation number of a monatomic ion equals its charge.

Ex: N3−, Cl−, Fe3+, Cu2+ and Na+

Main rules:1. The most electronegative element in a compound keeps it’s charge according to the periodic table, any other elements vary2. All the ONs on individual elements add to the net charge of the compound.

HCl Cl2 ClO− ClO4−

What do electrons have to do with chemistry?


65

Electrons are exchanged because the products are _______________________under the given conditions.

What key thermodynamic quality is this linked to? ________________


In an electrical circuit, electricity is the movement of _______________.

Reactants

Na(g) +12Cl2(g)

Voltage

Reaction Coordinate/Reaction Progress

V ProductsNaCl(s)

Cl2 g + 2 e− ⇌ 2 Cl aq− +1.36

Na aq+ + e− ⇌ Na s −2.71

Standard Electrode Potentials – Essentially a Reactivity Series!


66

Reduction HalfReaction Electrical Potential 𝐸° (𝑉)

F2 g + 2 e− ⇌ 2 F aq− +2.87

MnO4(aq)− + 8 H aq

+ + 5 e− ⇌ Mn(aq)2+ + 4 H2O l +1.51

Au aq3+ + 3 e− ⇌ Au s +1.50

Cl2 g + 2 e− ⇌ 2 Cl aq− +1.36

2 NO3 aq− + 4 H aq

+ + 2 e− ⇌ N2O4 g + 2 H2O l +0.80

Ag aq+ + e− ⇌ Ag s +0.80

Fe aq3+ + e− ⇌ Fe aq

2+ +0.77

Cu aq2+ + 2 e− ⇌ Cu s +0.34

2 H aq+ + 2 e− ⇌ H2 g 0.00

Fe aq2+ + 2 e− ⇌ Fe s −0.45

Zn aq2+ + 2 e− ⇌ Zn s −0.76

Al aq3+ + 3 e− ⇌ Al s −1.66

Na aq+ + e− ⇌ Na s −2.71

Li aq+ + e− ⇌ Li s −3.04

Assign oxidation number and point out the Oxidizing and Reducing agents:

SnCl2 + PbCl4 ⇌ PbCl2 + SnCl4

Zn s + 2 HCl aq ⇌ ZnCl2(aq) + H2(g)

Review


67

What is the most spontaneous reaction on this redox table?What is its voltage?


68

What is the least spontaneous reduction reaction and what is its voltage?

Under what conditions are these voltages calculated?

Reduction HalfReaction Electrical Potential 𝐸° (𝑉)

Cl2 g + 2 e− ⇌ 2 Cl aq− +1.36

2 NO3 aq− + 4 H aq

+ + 2 e− ⇌ N2O4 g + 2 H2O l +0.80

Ag aq+ + e− ⇌ Ag s +0.80

2 H aq+ + 2 e− ⇌ H2 g 0.00

Zn aq2+ + 2 e− ⇌ Zn s −0.76

Al aq3+ + 3 e− ⇌ Al s −1.66

Li aq+ + e− ⇌ Li s −3.04

What is the strongest reducing agent? (*Table is randomized!)


69

What is the second strongest reducing agent?

What is the strongest oxidizing agent?

Give the equation of the most spontaneous reaction:

HalfReaction Electrical Potential 𝐸° (𝑉)

Cl2 g + 2 e− ⇌ 2 Cl aq− +1.36

I2 g + 2 e− ⇌ 2 I aq− +0.535

Ag aq+ + e− ⇌ Ag s +0.80

Mg s ⇌ Mg aq2+ + 2 e− +2.37

Sn aq3+ + 2 e− ⇌ Sn aq

2+ +0.15

pH and pOH of Standard Cells


MnO4−

aq + 8 H aq+ + 5e− ⇌ Mn aq

2+ + 4 H2O l , 𝐸° = +1.51

I2 s + 2 e− ⇌ 2 I aq− , 𝐸° = +0.54

What is the spontaneous net reaction?

What is the voltage under standard conditions?

What is the concentration of H aq+ under standard conditions?

What is the pH under standard conditions?

70

pH and pOH of Standard Cells


2 MnO4−

aq + 16 H aq+ + 10 I aq

− ⇌ 2 Mn aq2+ + 8 H2O l + 5 I2 aq , 𝐸° = +0.97

If we increase [H+] to 10 mol/L, will E increase or decrease?

If the pH is raised to 4.0, what will be the new voltage of the cell?

a) +1.1 V

b) +1.3 V

c) +0.97 V

d) +0.59 V

What is the voltage of the cell at equilibrium?

As time passes, does the pH of the solution change?

What happens to the concentration of MnO4−

aq with time?

71

Consider the following reaction:

Cu aq2+ + 2 e− ⇌ Cu s , 𝐸° = +0.34

2 H2O l ⇌ O2 g + 4 H aq+ + 4 e−, 𝐸° = −1.23

What is the 𝐸°𝑐𝑒𝑙𝑙 for 2 Cu aq2+ + 2 H2O l ⇌ 2 Cu s + O2 g + 4 H aq

+ ?

a) +0.89 V

b) 0.89 V

c) +1.57 V

d) 1.57 V

Is the reaction spontaneous?

What is the oxidizing agent in the above reaction?

That is the reducing agent in the above reaction?

Review 72


8H+ + Cr2O7(aq)2− + 3 Cl aq

− ⇌ 2 Cr aq3+ + 3 Cl2 g + 4H2O(l)


73


What would happen to Q? To ΔG?

What would happen to 𝐸 if the pH is increased to 6?

What would happen to 𝐸 if the pH is lowered to 1?

What would happen to 𝐸 if the [Cr aq3+ ] is increased?

What would happen to 𝐸 if the PCl2 gwas changed to 2 atm?

Redox Reactions


74

MnO4− + I− ⇌ I2 + Mn2+Smart Split the reaction into two

African Balance All except H and O

Orangutans Balance O using H2O

Have Balance H using H+

Enormous Add electrons to balance charge

Brain If basic, add OH− forevery H+

Cells Combine the reactions to cancel electrons

MnO4− + I− ⇌ I2 + Mn2+

Your goal is to make sure of 3 things:

1. electrons lost = electrons gained2. Atoms balance3. Charges Balance

I consider this a good method because it is almost fool proof, gives you balanced half reactions before you get a net reaction and does not involve using oxidation numbers at all unless you want/need to

Redox Reactions


75

MnO4− + CH4 ⇌ HCOOH +Mn2+Smart Split the reaction into two







Redox Reactions – SELF STUDY SLIDE


76

Cr2O7(aq)2− + Br aq

− ⇌ Cr aq3+ + BrO aq

− , (acid solution)

Solution: 8 H+ + Cr2O7(aq)2− + 3 Br aq

− ⇌ 2 Cr aq3+ + 3 BrO aq

− + 4 H2O(l)

Smart Split the reaction into two







Kinetics Review


77

The following general factors affect reaction rates:


• Reactant concentrations collision frequency increases as concentration increases.

• Temperature speed of the molecules, thus collision frequency increases as temperature increases.

• Physical state of the reactants – influences collision frequency.

• Mechanism – reactions can occur at different rates for different pathways.

• Catalysts raises the reaction rate without being permanently consumed.

Temperature Catalysts

78

Kinetic Energy

Number of Molecules

Activation energy

low temperature

high temperature

Low concentration = Few collisions High concentration = more collisions

Kinetic EnergyNumber of Molecules

CatalyzedActivation energy

UnCatalyzedActivation energy

Rate = −Δ AaΔt

= −Δ BbΔt

= +Δ CcΔt

= +Δ DdΔt

If Δ SO4(aq)

2−

Δtis 2.40

mol

L s:

S2O8(aq)2− + 3 I aq

− → I3 aq + 2 SO4(aq)2−

What is Δ I3 aq

Δt?

a) 2.40 mol

L s

b) 1.20 mol

L s

c) +1.20 mol

L s

d) +2.40 mol

L s

What is the overall rate of reaction?

Reaction Rates aA + bB → cC + dD


79

Measuring Reaction Rates


80

As [reactant] decreases, rate also decreases. With fewer reactants, there is less opportunity for collisions.

H2O2 aq → H2O l +½ O2 g

H2O2 aq

time (minutes)

Instantaneous Rate

Initial Rate

Average Rate

0 1 2 3 4 5 6 7 8 9

What was the average rate of rxn. from 0 to 5 seconds? Provide units.

What was the instantaneous rate of formation of oxygen at 1.5 seconds?

Slope = 2.30

Slope = 2.10

Rates, Orders andMechanisms


81

NO2 g + NO2 g →𝑘1NO3 g + NO g , Slow

NO3 g + CO g →𝑘2NO2 g + CO2 g , Fast

Net: NO2 g + CO g → CO2 g + NO g , Rate = k

If the concentration of NO2 g is doubled, how would the rate change?

If the concentration of CO g is tripled, how would the rate change?

rate law

Method of Initial Rates


82

For the reaction: 2 l aq− + S2O8

2−aq → I2 aq + 2 SO4(aq)

2− the initial rates of reaction were determined as follows:

Calculate the rate law and the rate constant of the reaction.

l aq−

0(mol/L) S2O8

2−aq 0

(mol/L) Initial Rate (mol/L s)

0.60 0.10 500 × 10−5

0.30 0.10 250 × 10−5

0.60 0.05 250 × 10−5

Transition State Theory


83

CH3Cl + I− → CH3I + Cl−

This reaction proceeds through an activated complex. If you look below, this complex is a carbon with what looks to be five bonds – this is an extremely unstable species and requires lots of energy to occur “activation energy” (Ea).

A catalyst can lower this activation energy.

Raising the temperature can speed this reaction up by making sure more molecules have reached the critical energy needed for the reaction to proceed.

The Arrhenius Equation k = Ae−EaRT


Rate constant (k) is proportional to 𝒑 × 𝒛 × 𝒇

p – the fraction of collisions with proper alignment for breaking and forming new bonds.

z – the collision frequency, or the number of molecular collisions per unit time.

f – the fraction of collisions with sufficient kinetic energy to overcome the activation energy of the reaction (EK must be > Ea)

Explanation of p:

lnk2k1

=EaR

1T1

−1T2

84

k is the rate constant, A is the Arrhenius constant, Ea is the activation energy, R is the gas constant (8.314 J/molK) and T is the temperature in Kelvin.

Reaction Rates and Temperature


85

The rate constant of a first order reaction is 2.34 × 10−5𝑠−1 at 100 °C. If the activation energy is 20.6 kJ, what is the rate constant at a temperature of 500 °C?

0

5

10

15

20

0 25 50 75 100

0 Order Kinetics 1st Order Kinetics 2nd Order Kinetics


86

A t = −kt + A 0

[A]

ln[A]

1A

ln A t = −kt + ln A 0

1A t

= kt +1A 0time (s)

0

0.2

0.4

0.6

0.8

1

0 25 50 75 100

2.5

1.5

0.5 0 25 50 75 100

time (s)

time (s)

time (s)

time (s)

time (s)

[A]

ln[A]

1A

0

0.5

1

1.5

0 50 100 150 200

3

2

1

0

0 50 100 150 200

0

5

10

15

20

0 50 100 150 200

0

0.5

1

1.5

0 150 300 450 600 750

2.8

2.1

1.4

0.7

0

0 150 300 450 600 750

0

5

10

15

20

0 150 300 450 600 750time (s)

time (s)

time (s)

[A]

ln[A]

1A

Different Temperatures and Different Initial Concentrations


87

1A

timeWhich experiments were performed at the same temperature?What’s the difference?

Performed at different temperatures?

The graphs below describe the same reaction under different conditions.

The graphs below show ln[A] vs. time.

ln[A]

timeWhat is the relationship between the 3 lines?

Can you tell the order of reaction?

What extra information would be helpful?

B

A

C

Y

X

Z

What is a catalyst, an intermediate and a reactant/product?


A + C → D + E ∆H = +100 kJ

D + A → F + C ∆H = −50kJ

2A → E + F ∆H = +50kJ

Energy

Reaction Coordinate Diagram

Reaction Coordinate

∆H = +100 kJA + C

D + E + A

E + F + C∆H = −50 kJ

∆H = +50 kJ

88

Catalysts are used and regenerated

Intermediates are created and destroyed

Which of the two elementary steps is the rate determining step (RDS)?

1. Label any intermediates and any transition states on the diagram below:


89

H2SO4 + C5H11OH

HSO4− + C5H11

+ + H2O

H2SO4 + C5H10 + H2O

50 kJ 15 kJ10 kJ

30 kJ

2. Write the equation for the 1st step, 2nd step and the net reaction:

3. What is ΔH for the entire process?

4. What is the rate determining step?

5. What is the forward activation energy? The reverse activation energy?

Energy

Reaction Coordinate Diagram

Reaction Coordinate

6. Is there a catalyst?

7. Draw a dotted curve to show theuncatalyzed reaction if there is one.

8. Give the rate law for the reaction

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CHEM 203 Fall 2016 Final exam Page 6 of 15

16. For the reaction depicted above, which of the following statements is true?

a. ΔrG° < 0 and K > 1

b. ΔrG° < 0 and K < 1

c. ΔrG° > 0 and K > 1

d. ΔrG° > 0 and K < 1

17. You prepare an aqueous electrochemical cell with the following ion concentrations in solution:

[Sn4+] = 0.5 M [Pb2+] = 0.5 M [Sn2+] = 0.005 M [Pb4+] = 0.005 M

At these concentrations, which statement about the equilibrium reaction is true?

a. The forward reaction is spontaneous and the cell voltage (Ecell) is positive.

b. The forward reaction is spontaneous and the cell voltage (Ecell) is negative.

c. The reverse reaction is spontaneous and the cell voltage (Ecell) is positive.

d. The reverse reaction is spontaneous and the cell voltage (Ecell) is negative.

18. If this reaction is allowed to proceed in the spontaneous direction, what will we observe?

a. The magnitude of Ecell will decrease until it reaches E°cell

b. The magnitude of Ecell will increase until it reaches E°cell

c. The magnitude of Ecell will decrease until it reaches zero

d. The magnitude of Ecell will increase until it reaches zero

******************END OF MULITIPLE CHOICE QUESTIONS SECTION**************

The graph at the left describes the following equilibrium redox reaction:

Sn4+ (aq) + Pb2+ (aq) ⇌ Sn2+ (aq) + Pb4+ (aq)

Use the information in this equation and graph

to answer all of the next three questions

(Q16, Q17, Q18).

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