65212: CHEMISTRY 2

Spring, Semester 2

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Chemistry 2NotesContents

1 Introduction to criminal law

1.1 Crime1.2 Structure of criminal law1.3 NSW criminal law1.4 Common law and code states1.5 Jurisdiction1.6 Legal definitions1.7 Burden of proof 1.8 Standard of proof1.9 The golden thread

2 General principles and introduction to elements of crime

2.1 Elements of criminal liability2.2 Legal capacity2.3 Actus reus2.4 Mens rea2.5 Subjective and objective standard2.6 Strict and absolute liability2.7 Temporal coincidence

3 Homicide

3.1 Actus reus of homicide3.2 Concept of human being3.3 Requirement of voluntariness3.4 Voluntariness and the burden of proof3.5 Causation3.6 Responsibility for omissions

4 Murder

4.1 Actus reus of murder4.2

2

Strong Weak

Acids HCl (hydrochloric)

HBr (hydrobromic)

HI (hyrdroiodic)

Almost all other acids are weak

e.g. HA = HF, HNO2, HClO, H2S,

H2CO3, NH4+

1 Acid-base chemistry

1.1 Bronsted-Lowry acids and bases

Proton transfer reactions

An H+ ion is a proton with no surrounding valence electron

Define acids and bases by ability to transfer protons:

(a) Acid – donates proton

(b) Base - accepts proton

E.g. HCl dissolves in water. HCl is Bronsted-Lowry acid and donates a proton to

H2O, a Bronsted-Lowry base.

HCl (g) + H2O (l) H3O+ (aq) + Cl- (aq)

Does not have to be in aqueous solution e.g. HCl and NH3. Ammonia is base because

adding it to water increases concentration of OH-; accepts proton from H2O.

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)

Conjugate acid-base pairs

An acid and a base such as HX and X- that differ only in the presence or absence of a

proton e.g. OH- is conjugate base of H2O

Acid always has one more H

Strengths of acids and bases

The more easily an acid gives up a proton, the less easily its conjugate base accepts a

proton

The stronger an acid, the weaker its conjugate base

Types of acids:

(a) Strong acid:

Dissociate 100% (transfer all their protons to water); bases do not accept protons

– negligible base. [H+] is conc. of solution.

(b) Weak acid:

Partially dissociate; bases react partly with water to give small conc. of OH- –

weak base. [H+] < conc. of solution.

(c) Negligible acid:

Contain hydrogen but have no acidic behaviours; bases dissociate 100% to give

OH- and a cation – strong base.

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Strong Weak

Acids HCl (hydrochloric)

HBr (hydrobromic)

HI (hyrdroiodic)

Almost all other acids are weak

e.g. HA = HF, HNO2, HClO, H2S,

H2CO3, NH4+

Ka = [H + ] [A - ] [HA]

Strengths of acids and position of equilibrium

If the base in the forward reaction is a stronger base than the conjugate base of HX,

equilibrium will lie to right (when HX is a strong acid).

e.g. HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)

If base in forward reaction is a weaker base than the conjugate base of HX, equilibrium

will lie to left (when HX is a weak acid).

e.g. CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3OO- (aq)

Position of equilibrium favours transfer of proton to stronger base.

Extent of dissociation (acid strength) measured by equilibrium constant Ka

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pH = - log [H+] pOH = - log [OH-] pH + pOH = 14.00

Bigger Ka - stronger acid (Ka >1)

Smaller Ka – weaker acid (Ka < 1)

1.2 Autoionisation of water

Water as acid or base

Water can act as Bronsted-Lowry acid or base

In presence of acid, water acts as base (vice versa)

Water molecule can donate proton to another water molecule:

H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

Ion product of water

The equilibrium constant for water is:

Kw = [H3O+] [OH-]

Can also be written H2O (l) H+ (aq) + OH- (aq)

Kw = [H+] [OH-] = 1.0 x 10-14 (at 25°C)

Applicable to water and aqueous solutions (in water)

Can use to calculate [H+] if [OH-] known (vice versa)

If [H+] = [OH-] substance is neutral

[H+] > [OH-] substance is acidic

[H+] < [OH-] substance is basic

1.3 pH and pOH

[H+] in terms of pH

Acidic when [H+] > 1.0 x 10 -7 M

e.g. [H+] > 1.0 x 10 -3 M

pH = - log (1.0 x 10 -3)

= 3

Basic when [OH-] > 1.0 x 10 -7 M

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e.g. 0.10 M solution of formic acid (HCOOH) has pH 2.38 at 25°C.

(a) Calculate Ka for formic acid at this temperature.

1. Write equation for equilibrium reaction. Ionisation equilibrium for formic acid is:

HCOOH (aq) H+ (aq) + HCOO- (aq)

2. The equilibrium constant expression is:

Ka = [H + ] [HCOO - ]

[HCOOH]

3. From the measured pH, can calculate [H+]:

pH = -log [H+]

log [H+] = -2.38

[H+] = 10-2.38 = 4.2 x 10-3 M

4. Can determine the concentrations of the species involved in equilibrium.

HCOOH H+ HCOO-

Initial 0.10 M 0 0

Change - 4.2 x 10-3 M + 4.2 x 10-3 M + 4.2 x 10-3 M

Equilbrium 0.10 - 4.2 x 10-3 M 4.2 x 10-3 M 4.2 x 10-3 M

0.10 - 4.2 x 10-3 M is approximately 0.10 M

5. Insert equilibrium concentrations into Ka.

Ka = (4.2 x 10 -3 ) (4.2 x 10 -3 ) = 1.8 x 10-4

0.10

6. Check:

Answer reasonable as Ka for a weak acid is between 10 -3 and 10 -10

(b) What percentage of the acid is ionised in this 0.10 M solution?

1. % acid that ionises given by concentration of H+ or HCOO- at equilibrium / initial acid

concentration, x 100%

Percent ionisation = (4.2 x 10 -3 ) x 100% = 4.2 %

0.10

Calculating Ka from pH

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Calculating pH from Ka

For weak polyprotic acids (more than one proton to donate) – both dissociations will be

weak so no need to consider the contribution to pH of the second or third etc

dissociation

H2SO4 is an exception as first dissocation is strong

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e.g. Determine the pH of a 0.0020 M Solution of carbonic acid (pKa1 = 6.35, pKa2 = 10.33)

1. Write the ionisation equilibrium

H2CO3 (aq) H+ (aq) + HCO3- (aq)

2. Calculate what Ka equals

First dissociation is weak, second is much weaker so can ignore.

Ka1 = 10-6.35 = 4.5 x 10-7

3. Tabulate the concentrations of species involved in equilibrium reaction, where x equals

[H+]

H2CO3 H+ HCO3-

Initial 0.002 0 0

Change - x + x + x

Equilbrium 0.002 - x x x

4. Insert equilibrium concentrations into Ka.

Ka = x 2 = 4.5 x 10-7

0.002 – x

Neglect x as it is small compared to 0.002

5. Solve equation for x

x = √(0.002 x 4.5 x 10-7)

= 3.0 x 10-5

[H+] = 3.0 x 10-5

pH = -log (3.0 x 10-5) = 4.5

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9

e.g. H2PO4- and HPO4

2-

For phosphoric acid, H3PO4, pKa values are 2.12, 7.21, 12.38

pKa (H3PO4) = 2.12

pKb (H2PO4-) = 14 – 2.12 = 11.88

pKa (H2PO4-) = 7.21

(H2PO4-) is acidic pKa < pKb

pKb (HPO42-) = 14 - 7.21 = 6.79

pKa (HPO42-) = 12.38

(HPO42-) is basic pKb < pKa

1.4 Hydrolysis of ions

Hydrolysis: when a cation or anion reacts with water to change its pH

Hydrolysis of cations

Most cations act as acid in aqueous solution – undergo acid hydrolysis

Exceptions: group 1 and some group 2 metal ions – pH neutral so no effect on pH

Cations make solutions acidic

Acid-dissociation constants for hydrolysis increase with increasing charge and

decreasing radius of ion e.g. Cu2+ which has smaller charge and larger radius that Fe3+

forms less acidic solutions

E.g. Fe(H2O)63+ (aq) Fe (H2O)5(OH)2+ (aq) + H+ (aq)

Fe (H2O)5(OH)2+ (aq) Fe (H2O)4(OH)2+ (aq) + H+ (aq)

Hydrolysis of anions

The anion of a weak acid (HA) is a weak base – hydrolyse to produce OH-

A- + H2O HA + OH-

Exceptions: OH-, HSO4-, Cl-, Br -, I-, NO3

-, ClO4- most neutral so don’t affect pH,

HSO4- is a weak acid not base

e.g. CH3COO- + H2O CH3COOH + OH-

For anions that could be acidic or basic look at pKa and pKb and see which is smaller,

or which K is bigger

pKa < pKb then acidic

pKb < pKa then basic

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pH of salt solutions

Given the formula of salts able to see if acidic, basic or neutral in water

E.g. NaF:

Na+ is neutral (group 1 metal ion)

F- is the anion of a weak acid – basic

So basic

E.g. KCN:

K+ is neutral (group 1 metal ion)

CN- is anion of a weak acid – basic

So basic

E.g. NH4Cl:

NH4+ is an ammonium ion – acidic

Cl- is anion of strong acid – neutral

So acidic

E.g. NaNO3:

Both ions are neutral so will be pH 7

If cation is acidic and anion is basic need to look at pKa and pKb – smallest pK

(biggest K) wins

e.g. NH4F:

pKa (NH4+) = 9.24 pKb (F-) = 10.83

pKa < pKb so salt is slightly acidic in water

1.5 Buffer solutions

Buffer: solution that resists change in pH when small amounts of acid / base added to it

Maintain pH at safe levels in body

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pH = pKa + log10 [A - ] pH = pKa + log10 [base] [HA] [acid]

e.g. If pH 9 buffer needed, choose NH4+/ NH3 because pKa (NH4

+) is 9.25.

What concentrations of the two species are needed? What approximation is being

made in this type of calculation? – concentrations in HH eqn are meant to be

equilibrium concentrations but we usually put in initial concentrations.

pH = pKa + log10 [NH3] [NH4

+]

log10 [NH3] = pH - pKa = 9 – 9.25 = 0.25 [NH4

+]

[NH3] = 10 -0.25 = 0.562

[NH4+]

So – any concentrations in the ratio 0.562 will work – higher concentrations will be

better (more buffer capacity)

e.g. Calculate the pH of a buffer which is 0.05M in acetic acid and 0.07M in

sodium acetate. pKa (CH3COOH) = 4.76

pH = pKa + log10 [CH3COO - ] [CH3COOH]

pH = 4.76 + log10 (0.07 / 0.05) = 4.91

Mixture of weak acid (HA) and its conjugate weak base (A-) in similar conc.

acid: consumes added H+ base: consumes added OH-

Best buffering occurs at pH values near pKa of conjugate acid

Different pairs given buffering at different pH values – can prepare buffers of any pH

e.g. for buffering near pH 4-5 acetate buffer (CH3COOH / CH3COO-) as pKa of

CH3OOH is 4.76

Reactions: CH3COO- + H+ → CH3COOH or CH3COOH + OH- → CH3COO- + H2O

Henderson-Hasselbalch equation

Determine concentration of conjugate acid and base for desired buffering pH (vice

versa)

Best buffer solutions have ration of [A-] / [HA] close to 1

CH3COOH

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e.g. How much will the pH change if 10mL of 0.1 M NaOH is added to 20mL of a

solution containing 0.005 mol of NaH2PO4 and 0.006 mol of Na2HPO4? pKa

(H2PO4-) = 7.21

Before: pH = pKa + log10 [Na2HPO4] [NaH2PO4]

pH = 7.21 + log10 [0.006 / 0.02] = 7.29 [0.005/0.02]

After: pH = 7.21 + log10 [0.007 / 0.02] = 7.45 [0.004/0.02]

Buffer capacity

Amount of added acid or base a buffer can tolerate before pH changes significantly

Determined by actual concentrations of conjugate acid and base that make up buffer

I.e. can only add so much strong acid before base in buffer runs out (vice versa)

1.6 Titrations

Titration: incremental addition of reactant of known concentration to another of unknown

concentration, in order to determine unknown.

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e.g. 20mL of 0.1 M CH3COOH + NaOH (0.1M)

Moles of CH3COO- product = moles of CH3COOH reactant

= c x v = 0.1 M x 0.02 L = 0.0002 mol

Final volume = original + added

= 20 mL + 20 mL = 0.04L

[CH3COO-] = n/v = 0.0002 / 0.04 = 0.005 M

Equivalence pt. PH – weak base calculation:

e.g. What i pH of 0.005M solution of CH3COO-?

Dissociation (hydrolysis) reaction:

CH3COO- + H2O CH3COOH + OH-

0.005 – x x x

pKa (CH3COOH) = 4.76 (given), so pKb (CH3COO-) = 14 – 4.76 = 9.24

Kb = 10-9.24 = 5.75 x 10 – 10

Kb = [CH3COOH][OH - ] = x 2 = 5.75 x 10-10 [CH3COO-] 0.005 - x

x2 = 0.005 x 5.75 x 10-10

x = 1.70 x 10-6 = [OH-]

pOH = -log10 (1.70 x 10-6) = 5.77

Buffers made by titrating the conjugate acid with small volume of NaOH or the base

with HCl until the desired pH is obtained.

Plot pH vs volume of base added

Equivalence point: equal number of moles of OH- and H+ - very steep rise in pH around

the equivalence point

Strong acid + strong base

e.g. HCl and KOH

Equivalence point = pH 7

Weak acid + strong base

e.g. CH3COOH + NaOH

Equivalence point = pH > 7 because solution contains conjugate base (CH3COO-) of

weak acid and neutral Na+ ions (spectators)

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e.g. 20mL of 0.1 M CH3COOH + NaOH (0.1M)

Moles of CH3COO- product = moles of CH3COOH reactant

= c x v = 0.1 M x 0.02 L = 0.0002 mol

Final volume = original + added

= 20 mL + 20 mL = 0.04L

[CH3COO-] = n/v = 0.0002 / 0.04 = 0.005 M

Equivalence pt. PH – weak base calculation:

e.g. What i pH of 0.005M solution of CH3COO-?

Dissociation (hydrolysis) reaction:

CH3COO- + H2O CH3COOH + OH-

0.005 – x x x

pKa (CH3COOH) = 4.76 (given), so pKb (CH3COO-) = 14 – 4.76 = 9.24

Kb = 10-9.24 = 5.75 x 10 – 10

Kb = [CH3COOH][OH - ] = x 2 = 5.75 x 10-10 [CH3COO-] 0.005 - x

x2 = 0.005 x 5.75 x 10-10

x = 1.70 x 10-6 = [OH-]

pOH = -log10 (1.70 x 10-6) = 5.77

Initial pH >1

NB: initial pH determined by the pKa of the weak acid

Around the mid-point of the titration, half of the acid is gone, replaced by conjugate

base – solution in the flask is a buffer

pH of solution = pKa of weak acid i.e. pKa can be read off graph

Strong acid + weak base

E.g. HCl + NH3

Equivalence point = pH < 7

Initial pH < 13

NB: initial pH calculated by weak base calculation

At midpoint pH = pKa of conjugate acid (NH4+)

1.7 Indicators in acid/base titrations

Indicator is weak acid or base that undergoes dramatic colour change over narrow pH

range

Endpoint of titration corresponds to change of indicator colour

Equivalence point it point at which an equivalent number of moles of one reactant has

been added to the other – not the same as endpoint!

Choosing indicator

Indicator endpoint should be close as possible to equivalence point

Good indicator changes colour in steepest part of titration curve

pH of indicator

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pH = pKa (Ind) + log10 [Ind - ] [HInd]

Indicators have differently coloured base (Ind-) and acid (Hind) forms

e.g. methyl red HInd (red, pKa = 5.1), Ind- (yellow)

Can use equ. to work out pH range of colour change for indicator

Colour change will occur between pH = pKa (Ind) – 1 and pH = pKa (Ind) + 1

Choose indicator with pKa (Ind) close to pH of equiv point

NB: don’t confuse pKa of indicator with pKa of acid!!

Applying this

Many diff indicators can be used for SASB titrations because steep section of curve is

so long (6-7pH units) – just because doesn’t change colour at pH = 7 doesn’t mean it

is unsuitable

Have to choose more carefully for SAWB and WASB

1.8 Practice questions

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1. Estimate the pH of the equivalence pt. when 25mL of 0.100 M ammonia

(NH3) solution is titrated with 0.200 M HCl. Use pKb = 4.76.

Hint:

Vol HCl required:

[NH4+] =

NH4+ H+ + NH3

... – x x x

Solve for x:

pH = Ans: 5.20

2. Calculate the change in pH on addition of 0.040 mol of sodium hydroxide

to one litre of an 0.05 M NH3/ 0.10 M NH4Cl buffer solution. Assume

there is no volume change on addition of the sodium hydroxide. Ka for NH+

= 6.3 x 10-10 at this temp.

Before:

pH = pKa + log10 [NH3]/[NH4+]

After:

OH- reacts with NH4+ to form more NH3, so moles (and conc) of NH4

+

drops to 0.06 and moles (and conc) of NH3 rises to 0.09

pH = Ans: 9.38

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18

E.g. if 0.012 mol of Ca(OH)2 dissolves per litre of water, then s = 0.012 M

Then [Ca2+] = 0.012 M, [OH-] = 2 x 0.012 = 0.024 M

E.g. for reaction AgCl (s) Ag+ (aq) + Cl- (aq)

Ksp = [Ag + ] [Cl - ] = [Ag+] [Cl-] [AgCl]

Solubility, s, of AgCl is 1.34 x 10-5 M

So [Ag+] = [Cl-] = 1.34 x 10-5 M

But Ksp = [Ag+] [Cl-] = s2 = 1.8 x 10-10

2 Solubility equilibria

2.1 Insoluble salts

All salts are soluble to some degree

Equilibrium pt. is saturated solution – no more solute can dissolve

Some undissolved solid must be present for equilibrium

Salts which are sparingly soluble in water:

Phosphates (PO43-)

Carbonates (CO32-)

Sulfides (S2-)

Hydroxides (OH-)

Some halides (Cl-, Br-, I-)

Sulfates (SO42-)

Exceptions Na+, K+, NH4+ salts

2.2 The solubility of a salt

Solubility of a salt (s), is the amount dissolved in a saturated solution / unit V

Measured in g L-1 or mol L-1

Varies with temp

When calcium hydroxide dissolves, dissociates according to:

Ca(OH)2 (s) Ca2+ (aq) + 2OH- (aq)

Because of 1:1 ratio, [Ca2+] same as s, molar solubility of Ca(OH)2

Solubility (s) of a salt not to be confused with solubility product Ksp

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Problem questions

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1. The solubility product for calcium phosphate, Ca3(PO4)2 is 1.2 x 10-29 at 25° C. What concentration for calcium ions will exist in a saturated solution of this salt at this temperature? What is the molar solubility of this compound?

Ca3(PO4)2 3Ca2+ + 2PO43-

Equilib conc: 3x 2x (if had old table)

Ksp = [Ca2+]3 [PO43-]2 = [3x]3 [2x]2 = 108x5 = 1.2 x 10-29

x = 6.44 x 10-7M

[Ca2+] = 3x = 1.93 x 10-6M

So, solubility of compound is a third of [Ca2+] - molar ratio 1:3 i.e. it is x, so solubility of Ca3(PO4)2 is 6.44 x 10-7M

e.g. Will a precipitate form if 10mL of 1.0 x 10-2 M NaCl and 10mL of 1.0 x 10-4 M

AgNO3 are mixed? Assume that the final volume of the solution is 20mL. For

AgCl, Ksp = 1.7 x 10-10 at this temp.

Finding out whether a precipitate of AgCl will form.

NB: both ions are diluted upon mixing

2.3 The ionic product

Reaction quotient for ionic dissolution

Determined by current concentrations (non-equilibrium)

Qsp = [Ag+] [Cl-]

If Q > K system is supersaturated... precipitation occurs until Q = K

Q = K, system is at equilibrium

Q < K, more solid can be dissolved until Q = K

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e.g. Will a precipitate form if 10mL of 1.0 x 10-2 M NaCl and 10mL of 1.0 x 10-4 M

AgNO3 are mixed? Assume that the final volume of the solution is 20mL. For

AgCl, Ksp = 1.7 x 10-10 at this temp.

Finding out whether a precipitate of AgCl will form.

NB: both ions are diluted upon mixing

1. Determine the solubility of AgCl in pure water. Ksp = 1.8 x 10-10

AgCl (s) Ag+ (aq) + Cl- (aq)

s = [Ag+] = [Cl-]

Ksp = [Ag+] [Cl-] = s2 = 1.8 x 10-10

s = 1.34 x 10-5M

2. Determine the solubility of AgCl in 0.1 M NaCl (common ion – Cl-).

Compare this result with the answer to part 1.

Let solubility, s = [Ag+] = x

AgCl (s) Ag+ (aq) + Cl- (aq)

Equilib. Conc: x x + 0.1

Ksp = [Ag+] [Cl-] = x (x + 0.1) = 1.8 x 10-10

Neglect x in brackets, so: 0.1x = 1.8 x 10-10

x = 1.8 x 10-9 M

2.4 The common ion effect

Reaction has reached equilibrium and more ions already in solution are added

For example adding AgNO3 to AgCl – Ag+ is common ion

The added Ag+ drives equilibrium to the left – reduces solubility of AgCl (Le

Chatelier’s Principle)

AgCl (s) Ag+ (aq) + Cl- (aq)

← added Ag+

Common ions reduce the solubility of ionic solutes i.e. decreases the solubility of the

original substance

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1. Determine the solubility of AgCl in pure water. Ksp = 1.8 x 10-10

AgCl (s) Ag+ (aq) + Cl- (aq)

s = [Ag+] = [Cl-]

Ksp = [Ag+] [Cl-] = s2 = 1.8 x 10-10

s = 1.34 x 10-5M

2. Determine the solubility of AgCl in 0.1 M NaCl (common ion – Cl-).

Compare this result with the answer to part 1.

Let solubility, s = [Ag+] = x

AgCl (s) Ag+ (aq) + Cl- (aq)

Equilib. Conc: x x + 0.1

Ksp = [Ag+] [Cl-] = x (x + 0.1) = 1.8 x 10-10

Neglect x in brackets, so: 0.1x = 1.8 x 10-10

x = 1.8 x 10-9 M

e.g. Calculate the maximum pH required to prevent Cu(OH)2 precipitation from a

0.1M solution of Cu(NO3)2. Ksp of Cu(OH)2 = 2.2 x 10-20

Cu(OH)2 (s) Cu2+ + 2OH-

Ksp = [Cu2+] [OH-]2 = 2.2 x 10-20

For equilibrium, [OH-] = √( Ksp / [Cu2+]) = 6.49 x 10-10 M ???

pOH = 9.33, pH = 4.67: above this pH, [OH-] will be high enough for precipitation.

2.5 Dependence of solubility on pH

Most metal hydroxides e.g. Cu(OH)2 are “insoluble” – sparingly soluble

The solubility of many metal ions will depend on pH as it is related to [OH-]

Can use Ksp to decide whether a metal cation will be soluble at given pH

Metals usually more soluble in acid

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2.6 Salts with basic anions

If sparingly soluble salt has anion which is weak base – addition of H+ will increase its

solubility

E.g. adding H+ to CaF2 will move reaction to the right – more CaF2 will dissolve

CaF2 Ca2+ + 2F-

→ added H+

3 Chemical kinetics

The study of the rates of chemical reactions and the mechanisms by which they occur.

3.1 Reaction rate

Rate: change in concentration per unit time

mol L-1 s-1

Reaction rate positive: expressed in terms of products (concentrations increasing with

time)

Reaction rate negative: expressed in terms of reactants (concentrations decreasing with

time)

Example of data set for rate of decomposition reaction:

Factors that determine reaction rate

Concentration (surface area)

24

Average rate = change in concentration time interval

NB: positive if products, negative if reactants

Temperature

Physical state

Molecular structure / size

Catalysts

3.2 Average rate

25

Instantaneous rate = slope of tangent to curve at time t

= - d [R] or d [P] dt dt

NB: positive if products, negative if reactants

Can read values off graph to determine

e.g. Using graph below:

Average rate = - (6.66 x 10 -3 ) – (7.12 x 10 -3 )

600 -500

= 4.6 x 10-6 mol L-1 min -1

3.3 Instantaneous rate

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For aA + bB → cC + dD

- 1 d [A] = - 1 d [B] = 1 d [C] = 1 d [D] a dt b dt c dt d dt

E.g. What are the relative rates of change in concentration of products and reactants

in the following decomposition reaction?

2NOCl (g) → 2NO (g) + Cl2 (g)

- 1 d [NOCl] = 1 d [NO] = d [Cl 2] 2 dt 2 dt dt

Instantaneous rate more useful than average rate

Rate of reaction usually means instantaneous rate

3.4 Reaction rates and stoichiometry

Relates to molar ratio So for H2 (g) + I2 (g) → 2HI (g):

HI must be produces at twice the rate (in moles per second) that H2 or I2 is disappearing d [HI] = 2 x - d [H 2]

dt dt

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3.5 Rate laws

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- d [R] = k [R1]a [R2]b

dt

k is rate constant

R1, R2... are reactants

a, b... are orders of the reaction with respect to R1, R2...– determined by experiment

a + b + ... is overall order of the reaction

If the exponent is 1 – rate first order, 2- second order etc. Exponents show how rate affected by concentrations of reactants In some cases change in concentration of reactant does not affect rate – zero order –

concentration of component does not appear in rate law NB: Coefficients not relevant to rate law!! k large: reaction is fast (k > 10 sec-1) k small: reaction is slow (k < 10-3 sec-1)

Applying reaction order If reaction order for reactant is 1 – rate proportional to conc. of that reactant

E.g. for A + B → products rate = k [A][B]2

Doubling the concentration of A will double rate of reaction If reaction order for reactant is 2 – rate depends on concentration of that reactant

squaredE.g. rate = k [A][B]2

Doubling the concentration of B will quadruple rate of reaction Units of k change with overall order:

3.6 Determining rate laws experimentally

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E.g. the reaction 2I- + S2O82- → 2SO4

2- + I2 is assumed to have rate law of the form:

- d [I - ] = k [I-]a [S2O82-]b

dt

Determine the vales of k, a, and b using the data collected in the three experiments:

Determine the vales of k, a, and b using the data collected in the three experiments:

So – compare experiments 2 and 3 to obtain a (with respect to I-) – [S2O82-] constant

Rate 2 = k [I - ] a [S 2O82- ] b = ([I-](2)/ [I-](3))a

Rate 3 k [I-]a [S2O82-]b

i.e. 1.5 x 10 -4 = (6.0x 10-2 / 2.0 x 10-2)a

5.0 x 10-5

There are two methods to find values for k, a, b, etc.:

1. Initial rate method

2. Integrated rate law method

3.7 Initial rate method

Reaction repeated in series of experiments

Initial concentration of one reactant at a time is altered between experiments

Initial rate of reaction determined each time

Calculations – arithmetic method

1. Pick a pair of experiments

2. Find the reactant whose concentration does change between experiments

3. Determine ratio (n) of reactant’s concs. in the two experiments

4. If the order of this reactant is a – the corresponding ratio of rates is n a

5. Solve for a

Graphical method

Can determine a using a graph.

Since rate = d [I - ] = k [I-]a[S2O82-]b and [S2O8

2-]b is fixed in the experiment can write: dt

So plot log10 d [I - ] vs. log10 [I-] dt

Will be a straight line with slope = a

log10 d [I - ] = a log10 [I-] + constant dt

30

E.g. the reaction 2I- + S2O82- → 2SO4

2- + I2 is assumed to have rate law of the form:

- d [I - ] = k [I-]a [S2O82-]b

dt

Determine the vales of k, a, and b using the data collected in the three experiments:

Determine the vales of k, a, and b using the data collected in the three experiments:

So – compare experiments 2 and 3 to obtain a (with respect to I-) – [S2O82-] constant

Rate 2 = k [I - ] a [S 2O82- ] b = ([I-](2)/ [I-](3))a

Rate 3 k [I-]a [S2O82-]b

i.e. 1.5 x 10 -4 = (6.0x 10-2 / 2.0 x 10-2)a

5.0 x 10-5

Having found a = 1, b = 1, rearrange rate law to find k.

Rate = k [I-]a [S2O82-]b

k = rate / ([I-]a [S2O82-]b)

Substitute vales for rate, [I-] and [S2O82-] from any one experiment in the table to

find k = 0.125 L mol-1 s-1

NB: second order reaction overall as a + b = 2

t1/2 = ln 2 k

NB: if we know t, we know k (vice versa)

3.8 Integrated rate law method (rate and conc. vs. time)

Aim to find a way of producing a straight line plot of the data - linear plot required depends

on the order of the reaction.

Integrated first order rate law

If reaction A → B + C is first order in A, rate law is:- d [A] = k [A] dt

Rearranged as: [A]t = [A]0 e-kt

A plot of reactant conc. vs. time shows exponential decrease

Plot of ln[A] vs. time gives a straight line, slope = -k

Half-life (t1/2 ) of first order reactions Half life of first order reaction: time taken for concentration to decrease by half

For first order reaction – half-life is a constant and can be measured from any point in

the reaction

Integrated second order rate law If reaction 2A → B + C is second order in A, rate law is:

31

E.g. NO2 + CO → NO + CO2 is found to have rate = k [NO2]2 [CO]0

Proposing a mechanism that is consistent with this rate law – 2-step mechanism:

Here NO3 is an intermediate – produced then consumed.

As step 2 is must faster than step 1 (k2 >k1), step 1 is rate determining.

Rate of overall reaction = rate of step 1. For step 1 rate = k [NO2]2 – rate law

predicted by this mechanism agrees with the one observed experimentally.

- d [A] = k [A]2

dt Rearranged as: 1 - 1 = kt

[A]t [A]0

Plot of 1 vs. time gives a straight line, slope = k [A]

Distinguishing first and second order reactions

Compare graphs to see which is linear

E.g. 2 NOBr (g) → 2 NO (g) + Br2 (g)

4 Reaction mechanisms

Rate laws are used to propose mechanisms for chemical reactions.

Mechanisms: series of steps that describes the molecular events taking place in reaction.

4.1 Steps in mechanism

Rate determining (or limiting) step (RDS)

Slowest step in mechanism

As reaction cannot go faster than slowest step RDS is also rate of overall reaction

32

E.g. NO2 + CO → NO + CO2 is found to have rate = k [NO2]2 [CO]0

Proposing a mechanism that is consistent with this rate law – 2-step mechanism:

Here NO3 is an intermediate – produced then consumed.

As step 2 is must faster than step 1 (k2 >k1), step 1 is rate determining.

Rate of overall reaction = rate of step 1. For step 1 rate = k [NO2]2 – rate law

predicted by this mechanism agrees with the one observed experimentally.

E.g. (CH3)3CCl + OH- → (CH3)3COH + Cl-

Two different mechanisms proposed – one unimolecular and one bimolecular

Unimolecular mechanism:

(CH3)3CCl → (CH3)3C+ + Cl- slow, RDS

(CH3)3C+ + OH- → (CH3)3COH fast

Rate law predicted:

- d [(CH 3)3CCl] = k [(CH3)3CCl] dt

Bimolecular mechanism:

(CH3)3CCl + OH- → [(CH3)3CCl]- slow, RDS

[(CH3)3CCl]- → (CH3)3COH + Cl- fast

Rate law predicted:

- d [(CH 3)3CCl] = k [(CH3)3CCl][OH-] dt

Elementary step

One step reaction

Rate law can be determined from stoichiometry (unlike for overall reaction!!)

E.g. for elementary step 2A + B → C rate = k [A]2 [B]

Number of reactant molecules called molecularity

Reactions usually unimolecular or bimolecular (1 or 2 reactant molecules in RDS)

33

When kinetics experiments are performed, the rate law was found to be:

rate = k [(CH3)3CCl] i.e. the unimolecular mechanism is supported

4.2 Dependence of reaction rates on temp

Reaction rates increase by 2x with every 10° increase in temp (T)

Rate and k increase exponentially with T

E.g.

T (°C) k (s-1)

189.7

198.9

230.3

251.2

2.52 x 10-5

5.25 x 10-5

6.30 x 10-4

3.16 x 10-3

4.3 Arrhenius equation – rate and temp

Explains exponential relationship between rate and temp by proposing: reactant molecules

need a minimum energy (activation energy) to react.

A – pre-exponential factor (same units as k)

Ea – activation energy (J mol-1)

T – temperature (K)

R – ideal gas constant ( = 8.3145 J K-1 mol-1)

34

k = A e – (Ea/RT)

35

NB: reaction rates decrease as Ea increases

Fraction of collisions As temp increases, a greater fraction of collisions occurs with sufficient energy (> Ea)

to react. The fraction (f) of molecules that has energy equal to or greater than Ea i.e. with

sufficient energy to react, is:

f = e-Ea / RT

f increased by:

- decreasing activation energy

- increasing temp

Summary

Collision between reactant molecules necessary for reaction

Not all collisions lead to reaction, only the fraction f with energy greater than Ea

A is the number of correctly-oriented collisions per second (molecules must make

contact at right pts)

Using the Arrhenius equation

If the rate constant, k, is measured at different temps – can determine Ea

Plot ln k vs. 1/T.

Taking logs of equ.

ln k = ln A - Ea

RT

Plot is straight line with slope = - E a

RT

4.4 Reaction profiles

36

The activation energy is a barrier that must be overcome for reaction to occur

Energetic changes during reaction represented by ‘reaction profile’

Plot of potential energy vs. extent of reaction

Exothermic reaction

37

Ea – energy difference between reactants and top of barrier

∆H – amount of energy absorbed or released by the reaction (diff between reactants

and products)

Exothermic – energy released by reaction (energy of reactants greater than energy of

products i.e. ∆H is negative)

Endothermic – energy absorbed by reaction (energy of reactants less than energy of

products i.e. ∆H is positive)

4.5 Catalysts

A catalyst:

(a) Increases rate of reaction w/out being consumed

(b) Is reactant and a product in reaction mechanism

(c) Reduces Ea by changing mechanism of reaction

Catalyst may be heterogeneous (different phase to reactants) or homogeneous (same

phase).

Phase refers to states (solid, liquid, gas) and different liquids such as water and oil.

Heterogeneous e.g. solid catalyst with reactants as liquids or gases.

38

4.6 Enzymes

Enzymes are biological catalysts that:

(a) Perform chemical reactions on substrate molecules

(b) Are very specific (lock and key model)

(c) Enable reactions that are normally impossible (too slow)

(d) Are proteins

4.7 Relationship between kinetics and equilibrium

At equilibrium, rates of the forward and reverse reactions are equal. So for following

reaction at equilibrium:

aA + bB cC + dD

We may write:

forward rate = reverse rate

kf [A]a [B]b = kr [C]c [D]d

∴ [C]c [D]d / [A]a [B]b = kf / kr = Kc

i.e. the equilibrium constant is the ratio of the forward and reverse rate constants

5 Organic chemistry: part 1

5.1 Isomers

Isomers (same parts): two or more compounds with the same molecular formula

39

5.2 Constitutional (structural) isomers

Atoms connected in different order

Example: C8H8O3

5.3 Stereoisomers

Atoms connected in same order but arranged differently in space.

Non-superimposable.

5.4 Enantiomers (optical isomers)

Non-superimposable mirror images of each other.

Pairs of stereoisomers that are chiral – mirror image actually a different molecule.

Have identical physical and chemical properties.

E.g. human hands.

40

Chirality

Chiral molecules: non-superimposable mirror images – have chiral centre or sterocentre –

lacks internal plane of symmetry!

Achiral molecules: superimposable on mirror images (same molecule) – possess plane of

symmetry

Cahn-Ingold-Prelog Rules (R, S sequence priority rules)

Information about molecule’s absolute configuration

Method

1. Locate the centre of chirality

2. Assign priority to each substituent from 1 (highest) to 4 (lowest) based on atomic

number

3. Orient the molecule so that the group of lowest priority points away – don’t use

priority group 4 for working out direction!

4. Read the remaining groups from highest to lowest

Clockwise = R

Anticlockwise = S

Chirality centre: an atom that has 4 different groups bonded to it

41

e.g. Assign the absolute configuration as R or S to the following:

1. Locate the chirality centre.

2. Assign priorities, note for 2 and 3, if two atoms have same priority move along

chain till there is difference – here O has higher priority than H.

3. Read the remaining groups from highest to lowest.

Clockwise - R

e.g. Assign the absolute configuration as R or S to the following:

1. Locate the chirality centre.

2. Assign priorities and point lowest priority group away

3. Read the remaining groups from highest to lowest.

Clockwise - R

42

5.5 Diastereoisomers

Stereoisomers that are not enantiomers

Molecules with two or more stereocentres

Cis/trans isomers

Have different chemical and physical properties e.g. melting points

Cis/ trans isomers

Cis = same side of bond

Trans = different side of bond (diagonal)

Meso isomers

An achiral (has internal plane of symmetry – same molecules) molecule possessing

two or more stereocentres

Stereocentre: molecule with at least 3 different attachments

43

HOW TO DO R S PRIORITISING FOR LONGER MOLECULES!!

44

5.6 Unsaturated hydrocarbons

Unsaturated hydrocarbons: contain one or more double or triple bonds

Carbons aren’t attached to 4 separate molecules

Alkenes

At least one C-C double bond.

Alkynes

At least one C-C triple bond.

Carbon in triple bond has linear geometry.

5.7 Aromatic hydrocarbons (arenes)

Aromatic hydrocarbons: contain one or more benzene rings.

Not alkenes

Molecules have planar ring structure – all atoms in common plane, symmetrical

45

Examples:

Benzene geometry

The C atoms in benzene are sp2 hybridised with 120° bond angles (sp2 if three things

attached to central atom, sp3 if four things attached etc. )

Unhybridised p orbitals are at 90° to the plane of the ring.

Benzene has delocalised π bonds – double bonds move in structure i.e. it is

resonating

e.g.

Can represent this by drawing two resonance structures or by a circle in middle of

structure. (Hydrogens are assumed)

5.8 Aromaticy

Aromatic: cyclic, conjugated molecule with stability (due to delocalisation) greater than that

of localized structure

* Conjugated molecule: has double or triple bonds separated by single bond

46

Huckel’s Rules for Aromaticity

1. Molecule must have a ring with at least one unhybridised p orbital on each atom

2. Molecule must be planar

3. The number of π-electrons must equal [4n+2] where n is an integer (0, 1, 2)

π-electrons is number of electrons in double bonds (2 in each)

Anti-aromatic: number of π-electrons = 4n

Non-aromatic: sp3 hybrids – 4 attachments to central atom

Naming arenes

E.g. Benzene We know every atom in ring is sp2 hybrid Molecule is planar Check if number of π-electrons equals 4n+2 or 4n.

(no. of π-electrons = 6)

4n+2 = 6, n = 1 4n = 6, n =3/2

π-electrons equals 4n+2 as n must be whole number.

Therefore we know it is aromatic.

E.g. Cyclopentadiene

Atom at top is sp3 hybrid (has 4 attachments) – therefore non-aromatic.

E.g. Furan

If you can identify sp2 hybridised atom attached to another atom with lone pairs – that atom is also sp2 hybridised

Molecule is planar Check if number of π-electrons equals 4n+2 or 4n.

(no. of π-electrons = 6)

4n+2 = 4, n = 1 4n = 6, n =3/2 Therefore aromatic.

47

Names of benzene-containing compounds derived from benzene – common names

retained in some instances

Names end in –benzene

If benzene group has an OH attached, name ends in –phenol

Label as though the first group is 1

When benzene ring has two substituents – has 3 constitutional isomers

5.9 Reactivity of benzene

Benzene reacts differently to alkenes due to resonance

48

Aromatic compounds undergo substitution reactions – electrophilic aromatic

substitution

Hydrogen replaced by a substituent (Lewis acid); often needs catalyst

Two step process:

(1)

(2)

Examples:

5.10 Reactivity of alkenes

49

Naming alkenes

1. Find and name longest carbon chain containing C=C bind

2. No the chain from the end closest to C=C bond

3. List position of double bond(s) using smaller number

4. List each substituent and its position

e.g.

Reactions due to C=C double bound

1. Hydrogenation

2. Hydration

3. Halogenation

4. Polymerisation

5. Hydroxylation

6. Hydrohalogenation

Hydrogenation

Alkenes react readily with H2 at room temp in presence of a metal catalyst (Ni, Pd, Pt)

Reduction of alkene to corresponding alkane

e.g. ethene ethane

Saturates compound

Exothermic – releases energy

Carbon skeleton preserved

50

Hydration

Addition reaction

Adding water to double bond to produce alcohol

Catalysed by strong acid

Markovnikov’s Rule – when you add a molecule HX to carbon carbon double bond,

hydrogen joins carbon atom that has more H atoms attached to it

Ethene Ethanol

Halogenation

Addition of Br2 and Cl2

One atom bonds to each side of the double bond (trans addition)

Does not need light/ heat as catalyst

Colour change indicates reaction

brown/ orange clear colourless

Reaction used for:

- testing presence of double bond

- distinguishing between alkanes and alkene/alyne

51

Hydroxylation

Alkenes react with KMnO4 (potassium permanganate)

Gives 1,2 – diols

diol: An alcohol containing two hydroxyl groups -OH

Adds OH to each end of C=C bond

Reaction used as test for double bond

Polymerisation

C=C double bonds react with each other

Form long chains of singly bonded carbons (more stable)

Alkene molecules become polymer chain of repeating units

Polymer: large molecule composed of repeating structural units

52

Hydrohalogenation

Addition of hydrohalic acids (composed of hydrogen and halogen – group 17) to

alkenes

Yields corresponding haloalkanes (single bonds)

Example using symmetrical alkene

Example using asymmetrical alkene

Two structural isomers possible

Markovnikov’s Rule states H adds to C of double bond which is already bonded to

more H atoms – H adds to C(1) and Br adds to C(2)

5.11 Alcohols

C(1) is bonded to 2 H atomsC(2) is bonded to 1 H atoms

Markovnikov’s Rule

In the addition of an acid to a double bond, the hydrogen attaches to the carbon already

bonded to the greatest number of hydrogens.

53

General formula R-OH where R is a carbon group

Naming alcohols Main chain must contain –OH group (doesn’t have to be at end of chain)

1. Name the longest C chain which contains the –OH group

2. Number the chain from the end which gives OH the smallest number

OH group gives suffix –ol

Two OH groups gives suffix – diol

3. List any other substituents and their positions

Classification

Determines reactions

Primary (1°) alcohol:

Carbon atom bearing OH group also bonded to 2 hydrogen atoms

Secondary (2°) alcohol:

Carbon atom bearing OH group also bonded to 1 H

Tertiary (3°) alcohol:

Carbon atom bearing OH group not bonded to a H atom

54

Properties

Difference in electronegativity between O and H allows alcohols to hydrogen bond

Compared to alkanes:

o More soluble in water

o Higher boiling and melting pt

Soluble in both H-bond forming and non H-bond forming solvents

Phenols

Phenols: class of alcohol with direct attachment of OH group to benzene ring

Phenols more acidic than alcohols

Negative charge can be delocalised!

Reactions

Substitution

(alcohol to

haloalkanes)

Glue in pic

Nucleophiles (love +) have a pair of electrons which are used

to form a bond to the electrophile (love -)

55

Leaving group departs to make room for nucleophile

Nucleophile converts lone pair to a bond and becomes positive

by +1

Bond from C to leaving group collapses into a lone pair on the

leaving group when becomes negative by -1

Nucleophile is lewis base, electrophile is lewis acid

β– Elimination

(haloalkane to

alkene)

Glue in pic

A base as catalyst can remove the leaving group and an

adjacent H-creating a π (double) bond.

Competition between β– elimination and nucleophilic substitution of alcohols

Glue in pic

Reactant is a strong base – elimination

Reactant is nucleophile but poor base – substitution

Result is dependant on:

1. Accessibility of α-carbon (first carbon that attaches to functional group) bearing the

halide (group 17 metal + another element)

o More accessible – substitution favoured

2. Size of the alkoxide (conjugate base of alcohol)

o Bigger the alkoxide – elimination favoured

3. Accessibility ofβ- hydrogen (hydrogen that attaches to second carbon)

56

o More accessible – elimination favoured

Nucleophilic substitution reactions of alkyl halides

* alkyl halides: single bonded coumpound with C, H, and a halogen (Cl, Br, I, Fl)

E.g. predict the major product from the following reaction:

Using rules:

Therefore the elimination reaction is favoured.

57

Alkyl halides react with nucleophiles and bases

Nucleophiles replace the halide in C-X bonds of alkyl halides

5.12 Carboxylic acids

Contain the functional group –COOH

carboxylic acid carboxylate ion

Naming carboxylic acids

1. Name the longest C chain which contains the –COOH group

2. Number the chain so that the C of the –COOH group is #1

3. Give suffix –oic acid

4. List any other substituents and their positions

Acidity and dissociation in water

Weak acids

Do not dissociate completely in water

Any carboxylic acid that dissolves in water behaves in this way:

58

% dissociation depends on pH

Knowing the pKa - calculate % dissociation using Hendersen-Hasselbalch equation:

pH = pKa + log [conjugate base]

conjugate acid

Reaction with base

Carboxylic acid + base

carboxylate ion + water

When converted into salts – they are more stable, soluble compounds

Resonance

Carboxylic acids are more acidic than alcohols

This is due to resonance in carboxylate ions

The negative charge on the O atom is delocalised and spread between both O’s

weaker attraction for H+ - weaker bases

This makes the carboxylic acid a stronger acid – can donate H+ more easily

e.g. In what form is acetic acid blood plasma (pH 7.4)?

pH - pKa = log [CH3CO2- ]

CH3COOH

7.4 – 4.8 = 2.6

[CH3CO2- ] = 102.6 = 400

CH3COOH

1/400 x 100 = 0.3% CH3COOH100 – 0.3 = 99.7% CH3CO2

-

59

Trends in acid strength

Electronegative groups (such as F or Cl) near the –COOH group increases acidity

This is because the negative charge is shared out amongst electronegatives in the ion

– weaker attraction for H+ - weaker base – carboxylate is stronger acid

NB: smaller pKa is more acidic!!

Separating acid from neutral compounds (experiment)

Separating compounds

1. Place acid and neutral compound in separatory funnel with an organic solvent and

water

o Two compounds have similar polarities and both dissolve in organic layer (of

moderate polarity)

2. Separate the compounds by converting the water insoluble acid into a water soluble

salt by reaction with aqueous base (e.g NaOH) – moves into aqueous layer

o The other compound will not react with NaOH and remains dissolved in

organic layer

60

3. Use separatory funnel to separate aqueous and organic layer – drain one out.

Recovering the compounds in pure form

4. Dry organic layer and evaporate solvent

5. Recrystallise the acid by adding strong acid to aqueous layer – neutralises strong base

6. Filter to isolate the solid then dry

7. Measure the mass of each and melting points to test purity

5.13 Amines

Weak organic bases related to ammonia

One or more H atoms replaced by hydrocarbon groups

Amine Ammonia

Naming amines

1. Name the longest C chain

2. Number the chain so that it gives the NH2 group smallest number

3. Give suffix –amine

4. List any other substituents and their positions

Other rules

If not the principal functional group give prefix – amino

Compound with multiple amino groups – diamine, triamine, tetraamine

61

Reactions

Lone pair on N accepts H+

Amine Ammonia

With Water

Forms basic solution CH3NH2 + H2O → CH3NH3+ + OH- NH3 + H2O → NH4

+ + OH-

With Acid

Forms salts of

ammonium ion

CH3NH2 + H3O → CH3NH3+ + H2O NH3 + H3O → NH4

+ + H2O

Acidity and dissociation in water

Weak bases

Do not dissociate completely in water

% dissociation depends on pH

Knowing the pKa - calculate % dissociation using Hendersen-Hasselbalch equation:

pH = pKa + log [conjugate base]

conjugate acid

Classification

e.g. In what form is methyl amine (pKa 10.6) at pH 7.4?

pH - pKa = log [CH3NH2]

CH3NH3+

7.4 – 10.6 = -3.2

[CH3NH2] = 10-3.2 = 0.00063

CH3NH3+

0.00063/1 x 100 = 0.06% CH3NH2

100 – 0.06 = 99.94% CH3NH3+

62

How many carbons bonded to N

Primary (1°):

1 carbon bonded to N

Secondary (2°):

2 carbons bonded to N

Tertiary (3°):

3 carbons bonded to N

Quaternary (4°):

4 carbons bonded to N

1°, 2°, 3° ammonia ions react with OH- to form amine (donate H+ to form water)

4° ammonia ions not able to donate H+ as there is no H on the N atom

Physical properties of amines

1. Boiling points: alcohol > amine > alkane

o Increases with number of lone H’s – available for H bonding

2. More soluble than alcohols

o Amines with fewer than 6C’s / N are 100% soluble

3. Less soluble than ammonium salts (ionic) in water

4. Have fishy odour

Separating base from neutral compounds (experiment)

Separating compounds

1. Place base and neutral compound in separatory funnel with an organic solvent and

water

o Two compounds have similar polarities and both dissolve in organic layer (of

moderate polarity)

63

2. Separate the compounds by converting the water insoluble base into a water soluble

salt by reaction with dilute acid (e.g HCl) – moves into aqueous layer

o The other compound will not react with HCl and remains dissolved in organic

layer

3. Use separatory funnel to separate aqueous and organic layer – drain one out.

Recovering the compounds in pure form

4. Dry organic layer and evaporate solvent

5. Recrystallise the base by adding base to aqueous layer – neutralises acid

6. Filter to isolate the solid then dry

7. Measure the mass of each and melting points to test purity

Reactions

Reaction with a haloalkane (alkyl halide)

Ammonia and 1°, 2°, 3° amines

Reaction with a ketone or aldehyde → imine

Ammonia and 1° amines

+ +

+ OH-

+ H2O

64

Reaction with a carboxylic acid → amide

Synthesis of amines

Reduction of imines

5.14 Esters

Close relatives of carboxylic acid (just no H)

Identifying esters

Naming esters

65

Useful to think of ester containing an acid part and an alcohol part

First component of name – alkyl (alcohol)

1. Root name based on longest chain containing –OH group

2. Chain numbered so as to give –OH lowest possible number

Second component of name –oate (carboxylic acid)

1. Root name based on longest chain including –COOH group

2. Since –COOH group is at the end of the chain it must be C1

3. Suffice -anoate

Complete name: alkyl alkanoate

Example:

Solubility

Methyl acetate 24% soluble

Water solubility decreases with increasing chain length

Longer chains dissolve well in organic solvents e.g. dimethyl ethyl

Synthesis of esters

1. Condensation reaction of carboxylic acid and alcohol

Condensation: 2 molecules form one 1 molecule with water as byproduct

66

2. Reaction of acid chlorides and alcohol

Cl and H from alcohol bond

No water formed

3. Reaction of acid anhydrides and alcohol

First half of anhydride (without centre O) and OH from alcohol bond

No water formed

Reactions

Hydrolysis

Forms carboxylic acid and alcohol (reverse of synthesis)

o Ester formation in equilibrium with ester hydrolysis (double arrows)

Ester splits at O-C, H added to first half, OH added to second half

67

Hydrolysis promoted by a base – soap production!

5.15 Amides

Carboxylic acids, esters and amides are interconverted by condensation and

hydrolysis reactions.

Naming amides

Useful to think of amide containing an acid part and an amine part

1. Name the longest C chain with amide group

2. Number so amide group is smallest

3. Give suffix –amide

4. List any other substituents and their positions

5. N precedes nitrogen substituent(s)

68

Synthesis of amides

1. Condensation reaction of carboxylic acid and amine

2. Reaction of acid chlorides and amine

Physical properties

Due to strong hydrogen bonds!!

1. Amides have higher mp and bp than corresponding carboxylic acids

69

2. Amides more water soluble than corresponding carboxylic acids

Reactions

Hydrolysis

Amides hydrolyse to acid and amine

o Does not occur at room temp

o Promoted by boiling with strong acid or base

Hydrolysis with base

Hydrolysis with acid

Difference between amides and amines

70

1. Amides have C=O double bond

2. Amides are not basic

o Lone pair of electrons on N is delocalised

5.16 Amino acids

Building blocks of peptides and proteins

20 α amino acids – amine + carboxylic acid group on SAME CARBON

o Differ in side chain

o Classified by polar, non-polar, acidic and basic sidechains

Polar Non-polar

Acidic and Basic

71

Properties

Apart from glycine, they are chiral! (glycine has two H’s)

S configuration (clockwise)

Reactions

1. Acid-base equilibria

Have acidic –COOH and a basic – NH2 group so equilibrium possible

Amino acids are ionic at any pH

Position of equilibrium depends on pH of solution

Isoelectric point: pH at which amino acid has no net electrical charge

pI = ½ (pka1 + pka2)

pH < pI, charge positive

pH > pI, charge negative

72

2. Side chain reactions

Similar calculations to above

Work out isoelectric point

73

3. Peptide bond formation

Peptides are amino acids joined together

Amino acids bonded by condensation reaction (water byproduct)

OH comes off end of first one H comes off start of next one ect.

5.17 Peptides and proteins

Peptide: has two or more amino acids joined by peptide bonds

Sequence of amino acids determines biological activity

Proteins

Protein: typically has 50+ amino acids held in 3D shape

Held by intermolecular forces including H bonds

Sequence of amino acids determines 3D shape and function

Some proteins have extra components bound to polypeptide chain e.g. metal ions,

sugars

Structure (in water-soluble protein)

Non-polar side chains point inwards

Polar and ionic side chains point outwards – in contact with water

Levels of protein structure

Primary (1°) structure:

amino acid sequence

Secondary (2°) structure:

folding of chain due to H bonds between amide groups in polypeptide chain

Tertiary (3°) structure:

final folded shape of polypeptide chain

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Quaternary (4°) structure:

more than one polypeptide chain makes protein

Active sites

Enzymes and receptors rely on one or more active sites – specific 3D orientation of

amino acid side chains

May provide polar or non-polar space (lock and key model)

Protein folding

3D structures of proteins are limited

Small proteins can fold spontaneously into 3D structure

Many large proteins need chaperone to fold

chaperone: enzyme that guides folding

Can fold incorrectly – called prions – pass on conformations causing disease

Protein denaturation

Changes in enviro may cause protein to unfold

o H-bonding substances e.g. alcohols, phenols – H bonding

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o Detergents - hydrophobic interactions

o pH change – salt bridges b/ween side chains

o Reducing agents – salt bridges

o Heavy metal ions - -SH groups e.g. cysteine residue bound to Hg2+

Inactivates protein

Separation and detection

Separated by electrophoresis

Detected by UV absorption

5.18 Carbohydrates

Often formula: Cx(H2O)y

Family tree

Monosaccharides

Classified and named by whether it is aldose or ketose, hexose or pentose

Do not dissolve in organic solvents

High mp’s

Do dissolve in water – H bonding

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Aldose or ketose

Depends whether ketone or aldehyde group is present

e.g. triose (3 C sugar)

Pentose or hexose

Most monosaccharides have 5 or 6 carbons

5 C’s – pentose

6 C’s – hexose

e.g. glucose and fructose are 6 C sugars

D or L configuration

Whether OH group shown is drawn on right or left (bottom OH group)

When C=O is on top

Right – D

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L eft – L

Enantiomers – non-superimposable mirror images

Family tree of D-aldoses

Diastereoisomers

Differ in:

o Shape

o Polarity

o Mp, bp, density

o Solubility

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Hemiacetal groups

Sugars are predominately cyclic hemiacetals

The hemiacetal functional group is: (where O-R is not OH)

e.g. identifying hemiacetal groups:

Closed ring or open chain structure

79

Anomeric carbon in cyclic sugars

C bonded to 2 O’s in a cyclic sugar

Anomers of cyclic monosaccharides

α-anomer:

OH on bottom of ring (opposite side to CH2OH)

β-anomer:

OH on top of ring (same side as CH2OH)

Anomers are stereoisomers and enantiomers – differ in physical properties

Anomers come to equilibrium in water

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Chemical reactions of monosaccharides

1. Easily oxidised

o Oxidising agents used to test for presence of reducing sugar

o Reducing sugar – has free hemiacetal group – aldoses not ketoses

2. Undergo condensation to form disaccharides – catalysed by enzymes!

o Glycoside/acetal formation

3. Substitution reaction with alcohol

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Disaccharides

Linkage of two monosaccharides

Undergoes oxidation (i.e. is reducing) if there is free hemiacetal group

Undergo hydrolysis to form monosaccharides

o Glycosidic bond is cleaved

5.19 Nucleic acids

Nucleotides

Composed of sugar, phosphoric acid, and a nitrogen base

Two types of bases:

- purine base (2 rings)

- pyrimidine base (1 ring)

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Formation of sugar and base

Produces nucleoside

Nucleoside is missing no. 1 carbon in below picture

DNA and RNA

RNA less stable than DNA

Differ in their sugar

DNA: deoxyribose

RNA: ribose

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Differ in their bases

DNA: A-T, C-G

RNA: A-U, G-C

DNA forms double helix

Two strands wound around each other like spiral staircase

Pairs of bases supported by two sugar-phosphate backbones

One strand is the compliment of the other

There are 2 H bonds between A-T and 3 between G-C

DNA replication

Double helix unwound and chains separated

Nucleotides are added to build two new complementary strands

Results in two identical copies of original molecule

Transcription

From DNA to m-RNA

Ribonucleotides are added to the strands

Translation

From m-RNA to peptide sequence

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Dideoxynucleotides and DNA sequencing

In DNA replication, deoxyribonucleotides are added only to the 3’ ends of DNA

strands

Dideoxyribose has no –OH group in the 3’ position, so once a dideoxynucleotide is

incorporated chain cannot grow!

2’ and 3’ position OH’s missing!

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When dd-TTP is added, the chain stops growing

Produces mixture of all possible fragments ending in T

When dd-CTP is added, the chain stops growing

Produces mixture of all possible fragments ending in C

6 Thermochemistry

Thermodynamics: study of transformations of energy (e.g. heat into work)

Thermochemistry: branch of thermodynamics – examines heat released/absorbed by

chemical reactions

Applied to fuels, combustion, nutrition, biological metabolism etc.

6.1 Enthalpy of reactions

Enthalpy (H): total heat content of a system

Enthalpy change (∆H) = heat transferred to a system at constant pressure

(approximation of total energy change of system – different being work done)

State function - ∆H depends only on initial and final enthalpies, and not on

intermediate values

Use balanced chemical equations to show enthalpy change, called ‘thermochemical

equations’

Endothermic (∆H > 0)

Heat is transferred from surroundings to system

Heat absorbed

H2O (l) → H2O (g) ∆H = +44.0 kJ

Exothermic (∆H < 0)

∆H = ΣnHproducts – ΣmHreactants

n and m are coefficients of the products and reactants in balanced equ.

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Heat is transferred from system to surroundings

Heat released

H2O (g) → H2O (l) ∆H = - 44.0 kJ

∆H and quantity of material undergoing change

∆H is proportional to quantity of material undergoing change

2H2O (l) → 2H2O (g) ∆H = +88.0 kJ

NB: when process is reversed, ∆H has same magnitude but opposite sign

e.g. Given the following thermochemical equation:

H2 (g) + Cl2 (g) → 2HCl (g) ∆H = -185 kJ

Calculate ∆H when:

a) 1 mole of HCl is formed

∆H = -185 kJ /2 = -92.5 kJ

b) 1.00g of Cl2 reacts

n = m/Mn = 1.00g/70.9 g mol-1

n = 0.0141 mol

∆H = 0.0131 mol x -185 kJ mol-1 = -2.61 kJ

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6.2 Standard enthalpy of reaction

Set standard to compare enthalpies of different reactions

Standard enthalpy (∆H°) : enthalpy change when all reactants and products are in

standard states

Most enthalpy values for temp of 25°C (298 K)

e.g. The major source of aluminium in the world is bauxite (mostly aluminium oxide). Its thermal decomposition is represented by:

Al2O3 (s) → 2Al (s) + 3/2 O2 (g) ∆H = 1676 kJ

If Al is produced in this way, how many grams of Al can from when 1000 kJ of heat is transferred?

2 mole of Al forms when 1676 kJ of heat is absorbed.

Heat transferred for 1 mole: 1676 kJ / 2 mol = 838 kJ mol-1

Number of moles when 1000 kJ transferred: 1000 kJ / 838 kJ mol-1 = 1.193 mol

m = n x Mm = 1.193 mol x 26.98 g mol-1

m = 32.19 g Al

Standard states:

Liquids and solids (l) (s) – pure form at 1atm

Solutes in solution (aq) – 1M concentration

Gases (g) – partial pressure of 1atm

e.g. for the themochemical equation:

2H+ (aq) + Zn (s) → H2 (g) + Zn2+ (aq) ∆H° = -154 kJ

This means:

(a) Start with 2 moles of H+ (aq) in a 1M solution and 1M of zinc(b) Finish with one mole of H2 (g) at 1 atm pressure and one mole of Zn2+ (aq) in 1 a

1M solution(c) Enthalpy change -154 kJ

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6.3 Calorimetry

Calorimetry: measurement of heat flow to determine ∆H using a calorimeter.

Determining ∆H experimentally

Measure the heat flow accompanying a reaction at constant pressure

When heat flows in or out of substance, temp changes

6.4 Heat capacity (C)

Amount of heat required to raise the temperature of object/substance by 1°C (1 K)

Measured in JK-1 or J°C-1

Usually recorded on a per mole or per gram basis

Specific/ molar heat capacity

Specific heat capacity (Cs): amount of E needed to increase temp of kilogram by one

Kelvin – measured in JK-1 g-1 or J°C-1 g-1

Molar heat capacity same as specific heat capacity but units differ

Molar heat capacity (Cm): amount of E needed to increase temp of one mole by one

Kelvin – measured in JK-1 mol-1 or J°C-1 mol-1

Calculating heat transfer with temp change

Heat (q) is measured in joules; temp change (∆T)

Use relationships:

Cm = Cs x mmole

mmole is atomic weight (mass per mole)

q = nCm∆T ..... for n moles of substance

q = mCs∆T ..... for m grams of substance

q = C∆T ......... for an object such as a calorimeter

NB: ∆T = Tfinal - Tinitial

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Problem questions

6.5 Constant-pressure calorimetry (coffee cup)

Simple apparatus known as “coffee cup” calorimeter

Measures heat transfers at lab atm pressure

Consists of known mass of solution in insulated container equipped with a

thermometer and stirrer

Tinitial of solution measured – stirred as reaction takes place and Tfinal is measured

Heat gained by calorimeter = heat lost by reaction (opposite sign)

e.g. a 466 g sample of water is heated from 8.5°C to 74.6°C. Calculate the amount of

heat absorbed by the water. The specific heat capacity of water is 4.184 J°C-1g-1.

q = mCs∆T

= mCs (Tfinal – Tinitial )

= 4.66 g x 4.184 J°C-1g-1 x (74.60 – 8.50)°C

= 1.29 x 105 J

= 129 kJ

e.g. a 0.295 kg aluminium component at an initial temperature of 276 K absorbs 85.0

kJ of heat. What is the final temperature of the part? The specific heat capacity of

aluminium is 0.900 JK-1g-1.

q = mCs∆T

∆T = q/ mCs

Tfinal = q/ mCs + Tinitial

= 85.0 x 10-3 J / 0.900 J K-1g-1 + 276 K

= 596 K

NB: check units!!

qsoln = -qrxn

qsoln = (specific heat capacity of solution) x (grams of solution) x ∆T = -qrxn

qsoln: heat gained by solution qrxn: heat produced by reaction

90

e.g. when 50.0mL of 0.100 M AgNO3 and 50.0mL of 0.100 M HCl are mixed in a

constant-pressure calorimeter, the temperature of the mixture increases from 22.30°C

to 23.11°C. The temperature increase is caused by the following reaction:

AgNO3 (aq) + HCl (aq) → AgCl (s) + HNO3 (aq)

Calculate ∆H for this reaction expressed in kJ mol-1 of HCl, assuming that the

combined solution has a mass of 100.0 g and a specific heat capacity of 4.18 J°C-1 g-1.

q = mCs∆T

= 100 g x 4.18 J° C-1 g-1 x (23.11°C – 22.30°C)

= 338 J

Heat gained by solution is equal to heat released in reaction: q = -338 J

Number of moles of HCl

n = c x V

= 0.100 M x 50.0 x 10-3 L

= 5.00 x 10-3 mol

∆H = -388 J / 5.00 x 10-3 mol

= -6.76 x 104 J mol-1

= -67.6 k J mol-1

e.g. a 25.0 mL sample of 0.500 M HCl at 25.0°C is added to 25.0 mL of a 0.500 M

NaOH, also at 25.0°C, in a coffee-cup calorimeter. After mixing, the temperature of

the solution is found to have increased to 28.3°C. Calculate ∆H for the reaction

expressed in kJ mol-1 of HCl. The specific heat capacity of the solution is 4.1 J°C-1g-1.

Heat transferred during the reaction:

q = mCs∆T

= (25.0 g +25.0 g) x 4.18 J° C-1 g-1 x (28.3°C – 25.0°C)

= 690 J

Heat gained by solution is equal to heat released in reaction: q = - 690 J

Number of moles of HCl

n = c x V

= 0.500 M x 25.0 x 10-3 L

= 0.125 mol

∆H = -690 J / 0.0125 mol = -55.2 kJ mol-1

= -6.76 x 104 J mol-1

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6.6 Constant-volume calorimetry (bomb)

Apparatus known as bomb calorimeter

Measure heat released in combustion reaction

Electrically heated coil ignites a substance within a steel bomb containing O2

Heat released from the combustion reaction is transferred to rest of calorimeter and

temp is measured

qrxn = -Ccal x ∆T

Ccal: heat capacity of calorimeter

e.g. A 0.5865 g sample of lactic acid (C3H6O3) is burned in a bomb calorimeter, the

heat capacity of which is 4.812 kJ°C-1. The temperature increases from 23.10°C to

24.95°C. Calculate the enthalpy of combustion of lactic acid per gram and per mole.

qrxn = -Ccal x ∆T

∆T = 24.95°C – 23.10°C

= 1.85°C

Ccal = 4.812 kJ°C-1

qrxn = -Ccal x ∆T = - (4.812 kJ°C-1) (1.85°C) = -8.902 kJ

qrxn (per gram) = -8.902 kJ / 0.5865 g = -15.18 k J g-1

qrxn (per mole) = -8.902 kJ / 6.511 x 10-3 mol = -1.367 x 103 k J mol-1

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6.7 Foods and fuels

Fuel value: energy released when 1g of a substance is combusted

Most energy in our bodies comes from carbs (av. fuel value: 17 kJg-1) and fats (av.

fuel value: 38 kJg-1)

6.8 Hess’s law

If a reaction is carried out in steps, ∆H (enthalpy change) for overall reaction = sum

of enthalpy changes for individual steps

e.g. In a bomb calorimeter, a 0.332 g sample of benzoic acid (C7H6O2) is burned and

causes a temperature rise from 23.13°C to 24.61°C. The heat of combustion of

benzoic acid is -3227 kJ mol-1. Determine the heat capacity of the calorimeter.

qrxn = -Ccal x ∆T

∆T = 24.61°C – 23.13°C

= 1.48°C

qrxn (per mole) = -3227 kJ mol-1

qrxn = -3227 kJ mol-1 x (2.72 x 10-3 mol) = -8.78 kJ

-Ccal = qrxn / ∆T = -- 8.78 kJ / 1.48°C

Ccal = 5.93 kJ°C-1

∆Hoverall = ∆H1 + ∆H2 + ∆H3

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∆H for a reaction may be calculated from other known enthalpies – do not need to

determine experimentally.

Therefore, in following illustration any one of three enthalpy changes can be

calculated if other two are known.

Method for calculating unknown ∆H

1. Identify target reaction with unknown ∆H

2. Rearrange equations of known ∆H so that target reactants and products are on correct

sides with correct number of moles

Change sign of ∆H if equation is reversed

Adjust moles and ∆H by same factor

3. Add altered equations, cancelling substances not in target reaction (should be present

on both sides of reaction!).

E.g. Given the following combustion reactions:

(1) C (s) + O2 (g) → CO2 (g) ............∆H = -393.5 kJ(2) CO (g) + ½ O2 (g) → CO2 (g) .......∆H = -283.0 kJ

Determine the enthalpy of combustion of C to CO:C (s) + ½ O2 (g) → CO (g)

Reaction (1) has correct number of moles and reactants and products

Reaction (2) needs to be reversed to have CO (g) as a product

C (s) + O2 (g) → CO2 (g) ............∆H = -393.5 kJCO2 (g) → CO (g) + ½ O2 (g) .......∆H = 283.0 kJ

Add equations together and cancel out.C (s) + O2 (g) + CO2 (g) → CO2 (g) + CO (g) + ½ O2 (g)

C (s) + ½ O2 (g) → CO (g) ∆H = -110.5 kJ

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6.9 Enthalpy of formation

Enthalpy of formation ∆f H: enthalpy change associated with formation of compound from

constituent elements

Depends on temp, pressure and states of products and reactants

Standard enthalpy of formation: enthalpy change when one mole of the compound

is formed from its elements with all substances in their standard states.

- most stable form of element is used in reaction

e.g. for glucose at 25°C:

6C (graphite) + 6H2 (g) + 3O2 (g) → C6H12O6 (s) ∆H = -1268 kJ mol-1

NB: elemental source of carbon is graphite not diamond, O2 rather than O or O3

because it is stable form and standard pressure, H2 is most stable form of hydrogen

under standard conditions.

Unit: kJ mol-1

E.g. Calculate ∆H for the reaction:

2C (s) + H2 (g) → C2H2 (g) (acetylene)

Given the following thermochemical reactions:

(1) C2H2 (g) + 5/2 O2 (g) → 2CO2 (g) + H2O (l)........∆H = -1299.6 kJ(2) C (s) + O2 (g) → CO2 (g) .....................................∆H = -393.5 kJ(3) H2 (g) + ½ O2 (g) → H2O (l)................................∆H = -285.8 kJ

Reaction (1) needs to be reversed to make C2H2 a product.

Reaction (2) needs to be multiplied by 2 to get the right number of moles.

Reaction (3) left unchanged.

(1) 2CO2 (g) + H2O (l) → C2H2 (g) + 5/2 O2 (g) ........∆H = 1299.6 kJ(2) 2C (s) + 2O2 (g) → 2CO2 (g) .....................................∆H = -787.0 kJ(3) H2 (g) + ½ O2 (g) → H2O (l)................................∆H = -285.8 kJ

Add equations together and cancel out.2CO2 (g) + H2O (l) + 2C (s) + 2O2 (g) + H2 (g) + ½ O2 (g) → C2H2 (g) + 5/2 O2 (g) + 2CO2 (g) + H2O (l)

2C (s) + 2O2 (g) + H2 (g) → C2H2 (g) + 2O2 (g)

2C (s) + H2 (g) → C2H2 (g) ∆H = 226.8 kJ

95

Standard enthalpy of formation of most stable form of any element is zero – no

formation reaction needed when element in standard state. E.g. O2 (g) ∆f H° = 0 kJ

mol-1.

Standard enthalpies of formation tabulated and can be used to calculate enthalpy

change for any chemical reaction

∆rxn H ° = Σn∆fH° (products) - Σm∆fH° (reactants)

Standard enthalpy change of reaction = sum of standard enthalpies of formation of products – standard enthalpies of formation of reactants.

E.g. calculate the standard enthalpy of reaction for the following process:

2C2H6 (g) (ethane) + 7O2 (g) → 4CO2 (g) + 6H2O (g)

Use ∆f H° (kJ mol-1) for:

C2H6 (g) -84.7

CO2 (g) -393.5

H2O (g) -241.8

∆rxn H ° = Σn∆f H° (products) - Σm∆f H° (reactants)

= [4∆f H° (CO2) + 6∆f H° (H2O)] – [2∆f H° (C2H6) + 7∆f H° (O2)]

= [4 (-393.5) + 6 (-241.8)] – [2 (-84.7) + 7 (0)]

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Enthalpy of formation of H+ ions in solution = zero

E.g. calculate the enthalpy of reaction for the following process:

SiO2 (s) + 4HF (g) → SiF4 (g) + 2H2O (g)

Use ∆f H° (kJ mol-1) for:

SiF4 (g) -1614.9

H2O (g) -241.8

SiO2 (s) -910.9

HF (g) -273

∆rxn H ° = Σn∆f H° (products) - Σm∆f H° (reactants)

= [∆f H° (SiF4) + 2∆f H° (H2O)] – [∆f H° (SiO2) + 4∆f H° (HF)]

E.g. determine enthalpy of reaction - ∆rxn H ° for the following reaction:

2H+ (aq) + CO32- (aq) → CO2 (g) + H2O (l)

Use ∆f H° (kJ mol-1) for:

CO32- (aq) -677.1

CO2 (g) -393.5

H2O (l) -285.8

Recall that ∆f H° (kJ mol-1) for H+ = 0.

∆rxn H ° = Σn∆f H° (products) - Σm∆f H° (reactants)

= [∆f H° (CO2) + ∆f H° (H2O)] – [2 ∆f H° (H+) + ∆f H° (CO32-)]

= [(-393.5) + (-285.8)] – [ 2(0) + (-677.1)]

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6.10 Spontaneous processes

Spontaneous process: occurs of its own accord, without any ongoing outside intervention

Occurs in a definite direction – cannot be reversed.

e.g. brick falling or brick rising into a waiting hand

Processes that are spontaneous in one direction are non spontaneous in opposite

direction.

Temp and pressure determine whether process is spontaneous.

e.g. T > 0° ice melts spontaneously into liquid, T < 0° water freezes spontaneously to

ice. At T=0 the states are in equilibrium and neither conversion occurs spontaneously.

Speed not relevant.

E.g. determine the heat of formation - ∆f H ° of lead sulfate from the following reaction:

PbO2 (s) + SO42- (aq) + 4H+ (aq) + 2I- (aq) → PbSO4 (s) + 2H2O (l) + I2 (s)

∆H°= -194.4 kJ

Use ∆f H° (kJ mol-1) for:

PbO2 (s) -277.4

SO42- (aq) -909.3

I- (aq) -55.2

H2O (l) -285.8

Recall that ∆f H° (kJ mol-1) for H+ = 0.

∆rxn H ° = Σn∆f H° (products) - Σm∆f H° (reactants)

= [∆f H° (PbSO4) + 2∆f H° (H2O) + ∆f H° (I2)] – [∆f H° (PbO2) + ∆f H° (SO4

2-) + 4 ∆f H° (H+) ]

= [∆f H° (PbSO4) + 2(-285.8) + (-110.4)] – [ (-277.4) + (-909.3) + 4(0)]

∆f H° (PbSO4) = -649.1 kJ mol-1 ?? -918.9 ??

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6.11 Entropy

Entropy: thermodynamic function associated with number of different energy states or

spatial arrangements in which system may be found – randomness.

Helps us understand why processes are spontaneous or not.

State function – depends only on initial and final states of the system.

Absolute entropies: 0 for perfectly crystalline solids at 0 K

Determining whether entropy is positive or negative

Entropy is measure of disorder.

Order = low entropy Disorder = high entropy

Positive when creates disorder, negative when creates order.

E.g. positive – ice melting (solid to liquid), liquid vaporising (liquid to gas), sugar

dissolving in water (solid to liquid)

Scale of disorder (randomness)

Gases

Liquids

Solids

∴ Entropy ∆S increases in processes in which:

a) Gases are formed from either solids or liquids

b) Liquids or solutions are formed from solids

c) The number of gas molecules increases

∆S = Sfinal – Sinitial

E.g. predict whether ∆S is positive or negative for the following process:

H2O (l) → H2O (g)

Liquid to gas – randomness increases - ∆S is positive.

E.g. predict whether ∆S is positive or negative for the following process:

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6.12 Entropy changes in chemical reactions

Standard molar entropy (S°): entropy values of a substance in its standard state

Observations about S° values:

S° of elements at 298 K are not zero (unlike ∆H°)

S° values of gases are greater than those of liquids and solids

S° values increase with molar mass (M)

S° values increase with number of atoms in substance

Formula for entropy change

E.g. predict whether ∆S is positive or negative for the following process:

H2O (l) → H2O (g)

Liquid to gas – randomness increases - ∆S is positive.

E.g. predict whether ∆S is positive or negative for the following process:

∆S° = ΣnS° (products) - ΣmS°(reactants)

NB: n and m are coefficients in chemical equation

E.g. determine the standard entropy change ∆S° for the following reaction at 298K:

Al2O3 (s) + 3H2 (g) → 2Al (s) + 3H2O (g)

Use S° (JK-1 mol-1) for:

Al (s) 28.22

H2O (g) 188.83

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6.13 Gibbs free energy

Gibbs free energy: the energy associated with a chemical reaction that can be used to do

work.

Way to use ∆H enthalpy and ∆S entropy to predict whether a reaction at constant

temp and pressure will be spontaneous.

E.g. determine the standard entropy change ∆S° for the following reaction at 298K:

Al2O3 (s) + 3H2 (g) → 2Al (s) + 3H2O (g)

Use S° (JK-1 mol-1) for:

Al (s) 28.22

H2O (g) 188.83

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Gibbs free energy (G) of a system is defined as:

For a reaction at constant temp, determine the change in free energy of the system by:

Using sign of ∆G to determine spontaneity

∆G < 0: forward reaction is spontaneous

∆G = 0: system is at equilibrium (no tendency for reaction to occur in either direction)

∆G > 0: reverse reaction is spontaneous

Standard free energy changes (∆G°)

Free energy is a state function - only use final and initial values.

Tabulate standard free energies of formation for substances usually at 25°C

Standard conditions to note:

As with standard heats of formation ∆f H° - free energies of elements in most stable

form = 0.

Use standard free energies of formation to calculate the standard free energy change:

G = H – T S

T is absolute temperature (in degrees Kelvin).

∆G = ∆H – T ∆S

∆G is the fraction of total energy change that is free to do work. T is absolute temperature (in degrees Kelvin).

∆G° = Σn∆f G° (products) - Σm∆f G° (reactants)

m and n are coefficients in chemical equation.

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Gases 1 bar pressure

Solids Pure solid

Liquids Pure liquid

Substances in solution 1M concentration

6.14 Free energy and temperature

To estimate ∆G° at temps other than 25°C, use Gibbs-Helmholtz equation:

Spontaneity depends on the sign and magnitude of ∆H° and T∆S°

Often ∆H° and ∆S° have the same sign and spontaneity depends on temp.

E.g. determine the standard free energy change ∆G° for the reaction:

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

What does the sign of your answer tell you about the reaction?

Use ∆G° (kJ mol-1) for:

CO2 (g) -394.4

H2O (g) -228.6

CH4 (g) -50.8

∆G° = Σn∆f G° (products) - Σm∆f G° (reactants)

= [∆f G° (CO2) + 2 ∆f G°(H2O)] – [∆f G° (CH4) + 2∆f G° (O2)]

= [-394.4 + 2(-228.6)] – [-50.8 + 2(0)]

= -800.8 kJ

∆G° = ∆H° – T ∆S°

E.g. the production of ammonia via the Haber process involves the equilibrium:

N2 (g) + 2H2 (g) 2NH3 (g)

Assuming that ∆H° and ∆S° for this reaction do not change with temperature, determine the standard free energy change at 500°C. If ∆G° is -33.3 kJ for the Haber process at 25°C, comment on the effect of temperature on the equilibrium.

Use ∆fH° (kJ mol-1) for:

NH3 (g) -46.2

Use S° (JK-1 mol-1) for:

N2 (g) 191.5

H2 (g) 130.7

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∆rxn H ° = Σn∆f H° (products) - Σm∆f H° (reactants)

= [2∆f H° (NH3)] – [∆f H° (N2) + 3∆f H°(H2)]

= [2(-46.2)] – [0 + 3(0)]

= -92.4 kJ

∆S° = ΣnS° (products) - ΣmS°(reactants)

= [2S°(NH3)] – [S°(N2) + 3S°(H2)]

= [2(192.5)] – [191.5 – 3(130.7)]

= -198.6 JK-1 = 198.5 kJ K-1

500°C = 500 + 273 K = 773K

∆G° = ∆H° – T ∆S°

∆G° = -92.4 kJ – 773K(-198.6 x 10-3 kJ K-1)

= 61.1 kJ

As the temp increases from 25°C to 500°C the ∆G° changes from a negative value (-33.3 kJ) to a positive value (61.1 kJ). The spontaneous reaction will change from being a forward reaction to being the reverse reaction.

E.g. estimate the standard free energy change for the following reaction at 400K:

2SO2 (g) + O2 (g) → 2SO3 (g)

Use ∆fH° (kJ mol-1) for:

SO3 (g) -395.2

SO2 (g) -296.9

Note ∆fH° for O2 as standard state is 0 kJ mol-1.

Use S° (JK-1 mol-1) for:

SO3 (g) 256.2

SO2 (g) 248.5

O2 (g) 205.0

∆rxn H ° = Σn∆f H° (products) - Σm∆f H° (reactants)

= [2∆f H° (SO3)] – [2∆f H° (SO2) + ∆f H°(O2)] = -196.6 kJ 104

6.15 Free energy and the equilibrium constant

Calculating ∆G under non-standard conditions

Most chemical reactions occur under non-standard conditions

Free energy change for any conditions:

Larger K = more negative ∆G°Smaller K = more positive ∆G°

7 Electroochemistry

Electrochemistry: study of chemical reactions involving transfer of electrons (e-)

7.1 Redox reactions

Oxidation-reduction reactions

Not at equilibrium:∆G = ∆G° + RT ln Q

At equilibrium:0 = ∆G° + RT ln K∆G° = - RT ln K

R – ideal gas constant: 8.314 J mol-1 K-1

T – temperature in KelvinQ / K – reaction quotient / equilibrium constant (same formula)

E.g. calculate the equilibrium constant at 25°C for the Haber process:

N2 (g) + 3H2 (g) 2NH3 (g)

For this process ∆G° = -33.3 kJ mol-1

NB: ‘per mole’ means ‘per mole of the reaction as written’So ∆G° = -33.3 kJ mol-1 means per 1 mole N2, per 3 mole H2, per 2 mole NH3).

∆G° = - RT ln K

-33.3 kJ mol-1 = -8.314 J mol-1 K-1 x 298K ln K

K = 6.9 x 105

This is a large K which indicates that NH3 is favoured in the equilibrium mixture at 25°C (forward reaction).

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Involve the transfer of electrons between two chemicals

Definitions

Oxidation: loss of electrons by a substance

Oxidant/oxidising agent: causes something to be oxidised – it is reduced

Reduction: gain of electrons by substance

Reductant/ reducing agent: causes something to be reduced – it is oxidised

Oxidation numbers

Use to keep track of electron transfer

Do not always correspond to real charges on atom

Increase in oxidation number – oxidation

Decrease in oxidation number – reduction

Rules for assigning oxidation numbers:

1. Atom in elemental form – 0

2. Monatomic ion (one atom) – charge on ion

e.g. K+ has oxidation number +1

3. Non-metals have negative oxidation numbers, can be positive

Oxygen: -2 except peroxides H2O2: -1

Hydrogen: +1 when bonded to non-metals, -1 when bonded to metals

Fluorine: -1

4. Sum of oxidation numbers of all atoms in neutral compound is 0

5. The sum of oxidation numbers of all atoms in polyatomic ion equals charge of ion

e.g. H2O2

Oxidation number of each hydrogen is +1Oxidation number of each oxygen is -1

2(+1) + 2(-1) = 0

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Balancing redox equations – half equation method

Electrons lost in oxidation must equal electrons gained in reduction

The overall balanced equation can be written down by adding the half-equations to

eliminate electrons (simultaneous equations)

Multiply half-reactions by integers if necessary

Balancing rules:

1. Balance for ‘main element’ first

2. Balance for O by adding H2O to O deficient side

3. Balance for H by adding H+ to H deficient side

4. Balance for charge by adding e to more +ve side

E.g. Zn(s) + 2 H+ (aq) → Zn 2+

(aq) + H2 (g)

Oxidation half-equation:

Zn(s) → Zn 2+(aq) + 2e-

Reduction half-equation:

2 H+(aq) + 2e- → H2

Overall with cancelling:

Zn(s) + 2 H+ (aq) + 2e → Zn 2+

(aq) + H2 (g) + 2e

Zn(s) + 2 H+ (aq) → Zn 2+

(aq) + H2 (g)

7.2 Galvanic cells/voltaic cells

Use spontaneous reaction to generate electrical energy

Difference in chemical potential energy between reactants and products converted to

electrical energy

e.g. H3O+

Oxidation number of each hydrogen is +1Oxidation number of oxygen is -2

3(+1) + 1(-2) = +1 – equals net charge of ion

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How it works

Two half reactions are separated so that e- can be diverted through a wire

Two electrodes, anode (negative) and cathode (positive) conduct electricity

Oxidation occurs at anode

Reduction occurs at cathode

Electrodes inserted into electrolyte: ions involved in reaction

Circuit completed with salt bridge:

- cations migrate towards cathode

- anions migrate towards anode

Cell notation

Anode (oxidation half cell) written first

Single vertical lines represent phase boundaries

Double vertical lines represent salt bridge

Reactant then product on each side

e.g. for above cell Zn| Zn2+ || Cu2+ | Cu

7.3 Standard reduction potentials

Standard reduction potential (E°cell) is potential difference between metal electrodes,

measured in volts (V) when all substances present in standard states

i.e. solutions: 1M gases: 1 atm

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Measuring reduction potential of single electrode

Impossible to measure potential of half-cell single electrode

Measure half-cells relative to standard hydrogen electrode (SHE)

o Potential is 0

o Hydrogen gas bubbled over a Pt electrode in a 1M acid solution

o H+ (1M) | H2 (1 atm) | Pt

In cell constructed with half-cell of interest and SHE

SHE is anode SHE is cathode

Other half cell has standard reduction

potential of +E°cell

Ag+ (aq) + e- → Ag (s) E° = +0.80V

Other half cell has standard reduction

potential of -E°cell

Zn2+ (aq) + 2e- → Zn (s) E° = -0.76V

Tables of standard reduction potentials

Values tabulated with relevant half equations

Species at top left – most easily reduced

Species bottom right – most easily oxidised

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E.g. arrange the following species in order of decreasing strength as oxidising agents:

Fe3+, Br2, Cu2+

Fe3+ (aq) + e- → Fe2+ (aq) E° = +0.77V

Br2 (l) + 2e- → 2br- (aq) E° = +1.06V

Cu2+ (aq) + 2e- → Cu (aq) E° = +0.34V

Strongest oxidising agent is species on left with high highest potential so:

Br2 > Fe3+ > Cu2+

Predict cathode and anode in galvanic cells

Cathode – half-cell that site highest in table

Cathode (reduction) half-reaction will occur in direction appearing in table

Anode (oxidation) will occur in opposite direction appearing in table

Predict cell potential

Once cathode and anode identified, standard potential of galvanic cell given by:

E°cell = E°cathode - E°anode

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Predict spontaneity of redox reaction

E°cell > 0 spontaneous reaction

E°cell > 0 non-spontaneous reaction

7.4 Free energy and redox reactions

Relationship between cell potential and free energy

When reactants and products in standard states:

E > 0, ∆G < 0 spontaneous reaction

7.5 Nernst equation

Predict changing cell potential as concentrations of reactants and products change

Potential of cell drops as reactants converted to products

7.6 Corrosion

Oxidation of metals by substances in environment (mostly H2O and O2)

Electrochemical process

Corrosion of iron

∆G = - nFE

G: free energyn: electrons transferredF: Faraday’s constant (9.65 x 104 JV-1mol-1)E: reduction potential

∆G° = - nFE°

E = E° - RT ln Q

nF

E: reduction potential for reactionE°: standard reduction potentialR: universal gas constant (8.314 JK-1mol-1)T: temperature (usually 298K)F: Faraday’s constant (9.65 x 104 Cmol-1)Q: reaction quotient (pure solids not included)

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Iron is oxidised by O2 because standard reduction potential for Fe2+ is less than O2

Anode is iron at the centre of the drop, cathode is iron at outer edge of the drop

Overall reaction: 2Fe + O2 + 2H2O → 2Fe2+ + 4OH- → Fe(OH)2 (s)

This compound will oxidise further to product Fe2O3.3H2O (rust)

Salt accelerates corrosion

Prevention of corrosion

1. Protecting the surface from O2 and H2O (e.g. paint)

2. Cathodic protection

o Metal to be protected is made the cathode, and another metal is oxidised

‘sacrificial anode’

o Galvanised iron: iron coated with thin layer of zinc

o Standard reduction potential for Zn2+ is more negative than for Fe2+ so Zn

easier to oxidise than Fe

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