VCE Chemistry Unit 4

Key Knowledge

• energy profile diagrams and the use of ΔH notation including: activation energy; alternative reaction pathways for catalysed reactions; and deduction of ΔH for an overall reaction given energy profiles or ΔH of two related reactions

Chemical energy

• The chemical energy of a substance is made up of its potential energy and kinetic energy.

• Potential energy – attraction between nuclei and electrons –Repulsion between electrons –Repulsion between nuclei

• Kinetic energy - movement of particles –Movement of electrons –Vibrations of and rotations around

bonds. –Actual movement in the substance

(gases vs liquids and solids)

Enthalpy

• The chemical energy of a substance cannot be directly measured, but it can be determined theoretically.

• It is known as enthalpy or heat content and has the symbol H.

• However, a change in enthalpy, 'H, for a chemical reaction can be determined.

Measuring enthalpy • The change in temperature, 'T, gives an

indication of change in enthalpy. y If temperature increases during the

course of a reaction, then heat energy is being released to the surroundings by the reaction

ySo enthalpy of the products is less than

enthalpy of reactants.

Energy profile diagram Exothermic reaction

• We can visualise this change using an energy profile diagram:

• If temperature decreases during the course of a reaction, then heat energy is being absorbed from the surroundings by the reaction

• So enthalpy of the products is greater

than enthalpy of reactants.

Energy profile diagram endothermic reaction

• To visualise the decrease in enthalpy when a temperature decrease occurs we can also use an energy profile diagram:

Definition of 'H

'H = H(products) – H(reactants) • So if the temp increases,

H(products) < H(reactants), So 'H is negative

This is an exothermic reaction • If the temp decreases

H(products) > H(reactants), So 'H is positive

This is an endothermic reaction

• We can recognise an exothermic reaction by the increase in temperature of the surroundings

• eg Combustion reactions, neutralization reactions

• Endothermic reactions are much less common. The temperature of the surroundings decreases

• eg chemical ‘cold packs’

Activation energy

• The activation energy is the amount of energy required to break the bonds of the reactants.

• It is the amount of energy needed to start a reaction.

• eg a spark to start a fire, or to burn fuel in an engine.

Reversing the reaction

• If we were to turn products back into reactants, then the activation energy could still be determined from the energy profile diagram:

Catalysts and activation energy

• A catalyst increases the rate of a reaction. • It does this by creating an alternative

energy pathway for the reaction. • This pathway has a lower activation energy than the original reaction.

Sample exam question – energy profile diagrams

The reaction A + B → C involves a two step process. A + B → X: 'H positive and X → C; 'H negative Which one of the diagrams below best represents the energy changes during the course of the reaction?

VCAA June 2007 Q19

Key Knowledge

• energy profile diagrams and the use of ΔH notation including: (activation energy; alternative reaction pathways for catalysed reactions;) and deduction of ΔH for an overall reaction given energy profiles or ΔH of two related reactions

Thermochemical equations • More information about a reaction can be

included in a chemical equation by adding the value of the change in enthalpy.

• This creates a thermochemical equation • eg combustion of methane: CH4(g) + 2O2(g) o CO2(g) + 2H2O(l) ; 'H = -890 kJ mol-1

y In a thermochemical equation, the change in enthalpy is directly related to the actual amounts of reactants and products in the equation.

yThis is because 'H is calculated from the

enthalpy absorbed when reactant bonds are broken and the enthalpy released when product bonds are made.

yBond breaking is endothermic

yBond making is exothermic

• The more reactants and products there are, the more bonds there will be broken and made and the greater the difference between H(products) and H(reactants), 'H, there will be.

• Remember that 'H = H(products) – H(reactants)

Enthalpy for x mol of reactants

Enthalpy for 2x mol of reactants

Energy put in to break reactant bonds

+200 kJ +400 kJ

Energy released when product bonds are formed

-600 kJ -1200 kJ

'H -400 kJ -800 kJ

• So while CH4(g) + 2O2(g) o CO2(g) + 2H2O(l) ; 'H = -890 kJ mol-1 When the amounts are doubled: 2CH4(g) + 4O2(g) o 2CO2(g) + 4H2O(l) ; 'H = -1780 kJ mol-1 and halved: ½CH4(g) + O2(g) o ½CO2(g) + H2O(l) ;'H = -445 kJ mol-1

Also the reverse reaction is endothermic CO2(g) + 2H2O(l) o CH4(g) + 2O2(g) ; 'H = +890 kJ mol-1

Calculating 'H

• The thermochemical equation for a reaction can be used to calculate energy released when any amount of reactants react together.

• Ratios are the key here: 'H is the enthalpy of reaction in which n1 mole of reactant reacts E is the energy released/absorbed by n2 mole of reactant

ΔH E=n n

Example – calculating 'H

• Find 'H for the reaction 2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g) Given that the combustion of 0.2 mol of methanol releases 145 kJ of energy.

ΔH E=n n

1452 0.200

145 2 14500.200

Hess’ Law

• This law states that the enthalpy difference for a reaction will be the same, no matter what pathway the reaction takes.

• This leads us to being able to determine 'H for two related reactions.

• So, if A → B ; but 'H1 is unknown, we can use

A → C ; 'H2 and C → B ; 'H3 to work out 'H1, since 'H2 + 'H3 = 'H1

Example – using two related equations to find 'H

Calculate the enthalpy of reaction for the formation of NO2(g) according to the equation:

N2(g) + 2O2(g) → 2NO2(g) 'H1 = ? given the following thermochemical equations: N2(g) + O2(g) → 2NO(g) 'H2 = +180 kJ mol-1

2NO2(g) → 2NO(g) + O2(g) 'H3 = +112 kJ mol-1

Solution: Reverse 2nd equation: 2NO(g) + O2(g) → 2NO2(g) 'H4 = -112 kJ mol-1 Then add to 1st equation N2(g) + O2(g) → 2NO(g) 'H2 = +180 kJ mol-1

When these are added together (and cancelled out) you get the required equation, so you can also add the 'H values. 2NO(g) + O2(g) + N2(g) + O2(g) → 2NO2(g) + 2NO(g) N2(g) + 2O2(g) → 2NO2(g) ; 'H1 = 'H2 + 'H4 = +180 +(-112) = +68 kJ mol-1

Sample exam question - 'H The energy diagram below relates to the following two reactions.

The enthalpy change for the reaction NO2(g) → ½ N2O4(g) will be A. +58 kJ mol-1

B. +29 kJ mol-1 C. -58 kJ mol-1 D. -29 kJ mol-1

VCAA Nov 2010 Q 15

Key Knowledge

• collision theory and factors that affect the rate of a reaction including temperature, pressure, concentration and use of catalysts, excluding: a formal treatment of the Maxwell-Boltzmann distribution, reaction mechanisms and rate laws

Collision Theory

• For a reaction to occur, the molecules must collide.

• In a successful reaction, the molecules collide

1. With the correct orientation 2. With enough energy to achieve the

activation energy (overcome the ‘activation energy barrier’).

Colliding with correct orientation

Achieving the activation energy

Maxwell-Boltzmann distribution

• While this does not have to be known formally, it is useful to see another representation of meeting the activation energy

Rate of reaction

• The rate of reaction is equal to the change in concentration of reactants, or products per unit time.

• The greater the proportion of collisions that are successful, the greater the rate will be..

Typical concentration-time graphs

Factors affecting the rate of a reaction

Factor Affect of increase Collisions

Temperature Increase in average kinetic energy of particles

Increase in number and energy of collisions

Concentration More particles available to react Greater number of

collisions

Surface area More particles at surface, therefore available to react

Greater number of collisions

Pressure More particles available in a given volume

Greater number of collisions

Catalyst Activation energy is lowered, so greater proportion of particles have enough energy to react

Increased proportion of successful collisions

Summary

• To increase the rate of a reaction – Increase the temperature – Increase concentration of an aqueous solution – Increase pressure of gases – Make particles of solids smaller – Use a catalyst (if possible)

Catalysts

• A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy.

• With a lower Ea, a greater proportion of particles have enough energy to react.

Sample exam question – rates of reaction

A 2.0 g piece of Mg ribbon was added to a known volume of 2.0 M HCl. The volume of H2(g) produced during the reaction was measured and recorded. The graph below shows the result of this experiment. (b) In a second experiment, 2.0 g of Mg powder was added to the same volume of 2.0 M HCl as used in the first experiment. On the axes above, sketch the expected graph of volume of hydrogen against time for this second experiment. Give an explanation for the shape of your graph.

VCAA Nov 2008 Q1

Key Knowledge

• equilibrium: representation of reversible and non-reversible reactions: homogeneous equilibria and the equilibrium law (equilibrium expressions restricted to use of concentrations), Le Chatelier’s Principle and factors which affect the position of equilibrium

Reversible reactions

• Changes of state: Eg H2O(l) ⇌H2O(g)

‘⇌’ indicates a reversible reaction – When heat is added, the liquid water

evaporates, – When heat is removed (ie cooled), the reverse

reaction can occur – gaseous water condenses.

Incomplete reactions

• Some reactions can be reversed • As a result these reactions appear to be

incomplete – it is impossible to achieve a 100% yield.

• A reaction may be reversed if the reactants and products stay in contact with one another.

• The reaction between copper sulfate and water is reversible:

CuSO4(s) + 5H2O(l) ⇌ CuSO4.5H2O white blue

• Rechargeable batteries involve reversible

reactions. – The forward reaction releases energy while the

backward reaction requires energy.

Reaching equilibrium

• When two reactants are mixed, they react to make products

• As soon as products are made, there is a chance that they may collide and make reactants.

• As the concentration of products increases, the chance of the backward reaction occurring increases.

y As the concentration of products increases, the rate of the backward reaction gradually increases.

y Similarly, the rate of the forward reaction is decreasing because the concentration of reactants is decreasing.

y Finally the rate of the forward reaction becomes equal to the rate of the backward reaction – this is called equilibrium.

Dynamic equilibrium

• At equilibrium, the reaction has not stopped. • So, it is termed dynamic (still proceeding) • The macroscopic properties of the reaction

become constant eg – Colour – pH – Temperature – Gas pressure

y Only a closed system can reach equilibrium y A closed system is a reaction in a sealed container

– nothing can enter or escape. y If the products can leave the reaction vessel, the

reaction cannot be reversed. y Eg The reaction between CaCO3 and HCl does not

reach equilibrium because the CO2 escapes

Homogeneous equilibria

• This is when all the reactants and products are in the same state.

• All the reactants and products could be gaseous, or in aqueous solution or liquid (organic reactions such as esterification)

• The opposite of homogeneous equilibria is heterogeneous equilibria.

Sample exam questions - equilibria

Phosphorus(V) chloride decomposes to form PCl3 and Cl2 according to the equation Some gaseous PCl5 is placed in an empty container. When equilibrium is reached, the mass of the gas mixture, compared to the initial mass of PCl5, is A. halved B. unchanged C. one and a half times greater D. doubled.

VCAA Nov 2008 Q10

2. When 2-propanol reacts to form an equilibrium mixture with propanone and hydrogen, which one of the graphs below best represents how the rates of the forward and back reactions change over time?

VCAA June 2007 Q7

Key Knowledge

The equilibrium law

• A system at equilibrium can be identified by a constant.

• This is called the equilibrium constant, Kc.

• The equilibrium constant is temperature dependent

• ie it only changes if the temperature changes.

The equilibrium constant expression

• For a general reaction aA + bB + … ⇌ pP + qQ + …

• The equilibrium constant, K, is expressed as:

[ ] [ ] ...[ ] [ ] ....

a bP QKA B

examples

• For the equation PCl5(g) ⇌ PCl3(g) + Cl2(g) The equilibrium constant expression is • For the equation CO(g) + 2H2(g) ⇌ CH3OH(g) The equilibrium constant expression is

[PCl ][Cl ]=[PCl ]

[ ][ ][ ]CH OHKCO H

• The equilibrium constant that we will study this year is a concentration fraction

• Concentration fractions can be calculated for

the reaction at any time, not just when it is at equilibrium, but these are then not called the equilibrium constant! (they are just concentration fraction)

Using the value of K to determine whether equilibrium has been reached

• When the reaction is approaching equilibrium, the value of the concentration fraction will change with time.

• As the reaction proceeds, more products are being made and the value of the fraction will increase (numerator is increasing)

• When the reaction has reached equilibrium, the

concentration fraction (equilibrium constant) will remain constant.

A point to note

• The direction that is defined as ‘forward’ is from left to right as the equation has been written.

• If the equation is written around the other way, then the equilibrium constant is rewritten (it is the inverse)

What does the value of the equilibrium constant mean?

• The larger the equilibrium constant, the larger the ratio of products to reactants

• When the reaction has almost gone to completion, it will have a very large value of K

• When the reaction hardly proceeds (ie mostly reactants are present), the value of K will be very small.

Calculations

Calculate the equilibrium constant for the reaction:

PCl3(g) + Cl2 ⇌PCl5(g) at 20qC

Given that the equilibrium concentrations of the species involved are : [PCl3] = 1.50 × 10-3 M, [Cl2] = 8.25 × 10-2 M, [PCl5] = 1.67 M

Solution

• K =

= 1.34 × 104 M-1

[PCl ][PCl ][Cl ]

1.671.50 10 8.25 10� �u u u

Example 2 • Calculate the equilibrium concentration of

Ag+(aq) ions, given the following: Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2

+(aq) K = 1.60 × 104 M2 at 25 qC, [NH3] = 5.00 × 10-3 M and [Ag(NH3)2

+] = 0.401 M Solution K = [Ag+] = = 1.00 M

[Ag(NH ) ][Ag ][NH ]

0.4011.60 10 (5.00 10 )�u u u

Example 3

• 4.00 mol of HI was admitted to an evacuated 2.0 L reaction vessel and allowed to reach equilibrium at constant temperature according to the equation:

H2(g) + I2(g) ⇌ 2HI(g) At equilibrium 0.250 mol of I2 had formed (a) Determine the equilibrium concentrations of all species (b) Determine the equilibrium constant for this reaction

2.00 L

[HI][H ][I ]

Example 4

• 1.800 mol of ammonia is introduced into an evacuated 5.00 L vessel and allowed to come to equilibrium. The concentration of Nitrogen at equilibrium was determined to be 0.0435M.

• Calculate the equilibrium constant for the reaction shown below, given that the temperature remained constant at 160 qC throughout the experiment.

N2(g) + 3H2(g) ⇌ 2NH3(g)

solution

5.00 L

Sample exam question - The Equilibrium law

The concentrations of reactants and products were studied for the following reaction H2(g) + F2(g) ⇌ 2HF(g); K = 313 at 25 qC In an experiment, the initial concentrations of the gases were [H2] = 0.0200 M, [F2] = 0.0100M and [HF] = 0.400M. When the reaction reaches equilibrium at 25 qC, the concentration of HF will be A. 0.400 M B. 0.420 M C. Between 0.400M and 0.420 M D. Less than 0.400 M

Key Knowledge

Changing the equilibrium position of a reaction

• When a closed system is at equilibrium, products and reactants are both present in the mixture.

• “When a change is made to a system at

equilibrium, the system will adjust to partially oppose the change”

• This is called Le Chatelier’s principle

Changes that could be made to a system at equilibrium

1. Add or remove reactants 2. Add or remove products 3. Decrease volume of a gaseous system 4. Increase volume of a gaseous system 5. Increase volume of an aqueous system 6. Increase the temperature 7. Decrease the temperature 8. Add a catalyst

Adding a reactant • The addition of one reactant will temporarily

increase the concentration of that reactant. • Since all reactants are still present in the system,

the system will act to oppose the change by making the concentration of the reactant decrease – by reacting it with the other reactant

• More products will be made • The position of equilibrium has moved to the

right • OR ‘there is a net forward reaction’

Removing a reactant

• Concentration of that reactant is temporarily decreased

• The system will act to oppose the change by making the concentration of the reactant increase – by making products turn into reactants (backward reaction)

• More reactants will be made • The position of equilibrium has moved to the left • OR ‘there is a net backward reaction’

Adding a product

• This has exactly the same effect as removing a reactant

• The position of equilibrium moves to the left

Removing a product • This has exactly the same effect as adding a

reactant • The position of equilibrium moves to the right

Gaseous reactions Decreasing volume of system

• The concentration of all reactants and products is temporarily increased (same amounts in smaller volume)

• The system will act to oppose the change by making the concentration of particles decrease

• What follows is dependant on the particular reaction:

1. There are more reactant particles than product particles

eg N2 + 3H2 ⇌ 2NH3

4 particles 2 particles To decrease the concentration of particles

overall, the system must convert reactants to products

The position of equilibrium moves to the right

2. There are equal numbers of reactant particles and product particles

eg I2 + H2 ⇌ 2HI

2 particles 2 particles There is nothing the system can do to

decrease the concentration of particles overall,

The position of equilibrium does not move

3. There are less reactant particles than product particles

eg N2O4 ⇌ 2NO2

1 particle 2 particles To decrease the concentration of particles

overall, the system must convert products to reactants

The position of equilibrium moves to the left

Increasing the volume of a system

• The concentration of all reactants and products is temporarily decreased (same amounts in smaller volume)

• The system will act to oppose the change by making the concentration of particles increase

• The position of equilibrium will move depending on the equation (as before)

Increasing the volume of an aqueous system ( by adding water)

• This is the same as increasing the volume of a gaseous system.

• The concentration of all reactants and products will initially be decreased, so the system will act to oppose the change – in which ever manner matches the stoichiometry of the particular reaction.

Sample exam question – Le Chatelier’s Principle

• The following reaction systems are at equilibrium in separate sealed containers. The volumes of the containers are halved at constant temperature. Which reaction has the largest percentage change in the concentration fraction immediately after the volume change?

A. N2O4(g) ⇌ 2NO2(g) B. H2(g) + I2(g) ⇌ 2HI(g) C. 2CO2(g) ⇌ 2CO(g) + O2(g) D. CO(g) + 2H2(g) ⇌ CH3OH(g)

VCAA Nov 2010 Q7

Solution:

Key Knowledge

Changing the temperature

• This is the only change that can affect the value of Kc

• It is a permanent change, unlike those described earlier.

• The value of 'H influences the direction in which the equilibrium moves

Exothermic reaction eg N2 + 3H2 ⇌ 2NH3 'H = -92 kJ mol-1

• When the products are made, heat is also released (could think of it as ‘a product’)

• If the temperature is increased, it is like adding a product (adding heat), so the position of equilibrium moves to the left (to oppose the change)

• The extra heat energy is used in the backward reaction that occurs (but the change is not ‘undone’)

• If the temperature is decreased, it is like removing a product (removing heat), so the position of equilibrium moves to the right (to oppose the change)

• Extra heat energy is produced in the forward reaction that occurs.

• So… An exothermic reaction is favoured (more products made) by decreasing the temperature

Endothermic reaction eg NH4NO3(s)+ aq ⇌ NH4

+(aq)+ NO3-(aq) ;

'H = +25 kJ mol-1

• Heat can be considered to be a ‘reactant’ (it is used to make products)

• If the temperature is increased, it is like adding a reactant (adding heat), so the position of equilibrium moves to the right (to oppose the change)

• If the temperature is decreased, it is like removing a reactant(removing heat), so the position of equilibrium moves to the left (to oppose the change)

So… An endothermic reaction is favoured (more products made) by increasing the temperature

Using graphs to indicate these changes to an equilibrium system

• Graphs of concentration vs time are often used to track the response of a system at equilibrium to a change.

• These are typified by a horizontal line at the points where the system is at equilibrium.

Graph examples Addition of a product

• At time = 10 mins, some H2 was added. The system acted to oppose the change and used up the H2 while also making CH4 in a 3:1 ratio

Change in volume 2NO2 ⇌ N2O4

• The volume was decreased, so the concentration of

both reactant and product increased instantaneously.

[N2O4] continued to increase, while [NO2] decreased (in a 1:2 ratio) as the new position of equilibrium was reached.

Change in temperature • At time = 4 mins, the gradual change in all

concentrations suggests a change in temperature. Since the reaction is endothermic and the forward reaction is favoured, we can conclude that the temperature increased.

Catalysts and equilibrium • The addition of a catalyst does not

change the position of equilibrium in any way.

• It does, however, allow equilibrium to be reached more quickly, as it lowers the activation energy and increases the rate of both the forward and backward reactions.

Sample exam questions – temperature changes and graphs

The following gaseous equilibrium is established at high temperatures in the presence of a finely divided Ni catalyst. A particular reaction is carried out using equal amounts of CH4(g) and H2O(g). Which one of the following sets of changes in conditions would lead to the greatest increase in the proportion of the reactants converted to products?

VCAA Nov 2008 Q1

Volume of reaction vessel

Temperature

A. increased increased B. increased decreased C. decreased increased D. decreased decreased

2. Hydrogen iodide dissociates into its elements according to the following equation

2HI(g) ⇌ H2(g) + I2(g) 'H = +9 kJ mol-1

A mixture of H2(g), I2(g) and HI(g) rapidly comes to equilibrium in a 2.0 L container. After the reaction has been at equilibrium for 10 minutes, the volume of the container is suddenly reduced to 1.3 L at constant temperature. Which one of the following graphs best represents the effect of this decrease in volume on the concentration of the gases in the mixture?

Key Knowledge

• pH as a measure of strength of acids and bases; Kw, (Ka for weak acids)

The self-ionisation of water

• Water is both a weak acid and a weak base, so in any aqueous solution, water molecules react together:

H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq) • This is called the self-ionisation of water • The equilibrium constant for this reaction is

• The equilibrium constant for this reaction would be written as

• Since the concentration of water is vastly greater than that of H3O+ or OH-, it can be removed from the equation to become part of a new constant, the ionisation constant of water, Kw

Kw = [H3O+][OH-]

Temperature dependence of Kw • The temperature of the water must be stated

before a value can be assigned. At 25°C, Kw = 1.0 u 10-14 M2

• The self-ionisation of water is an endothermic

reaction H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq); 'H = +57 kJ mol-1

• So as the temperature increases, Kw increases and

as the temperature decreases, Kw decreases.

How Kw changes with temperature

Temperature (qC) Kw (M2) 0 1.14 × 10-15

5 1.85 × 10-15

15 4.51 × 10-15

25 1.00 × 10-14

35 2.09 × 10-14

45 4.01 × 10-14

55 7.29 × 10-14

• Recall that pH is a measure of the acidity of an aqueous solution.

• The lower the pH, the more acidic a solution is

• It is defined mathematically:

pH = -log10 [H3O+]

Finding pH from Kw

• Kw = [H3O+][OH-] • Since [H3O+] = [OH-] in pure water, then Kw = [H3O+]2

=1.0 u 10-14 M2 so [H3O+] = = 1.0 u 10-7 M at 25°C Since pH = –log10 [H3O+] = –log10 1.0 u 10-7 M = 7

141.0 10�u

Acid, base, neutral

• An acidic solution is defined as one in which [H3O+] > [OH-]

• In a neutral solution [H3O+] = [OH-] • And in a basic solution [H3O+] < [OH-] • At 25°C, an acidic solution has a pH < 7,

neutral solution has pH= 7 and basic solution has a pH > 7

• BUT ….

• pH is calculated only from [H3O+] • At 55°C, Kw = 7.29 u 10-14 M2

• So if [H3O+] = [OH-] in pure water (a neutral solution),

then [H3O+] = = 2.7 u 10-7

and pH = -log10(2.7 u 10-7 ) = 6.57 (not 7 )

147.29 10�u

• At 15°C, Kw = 4.51 u 10-15 M2

so [H3O+] = = 6.71 u 10-8

and pH = -log10(2.7 u 10-7 ) = 7.17 • In summary, the pH of a neutral solution

is less than 7 above 25°C and greater than 7 below 25°C.

154.51 10�u

Using Kw in other calculations

• Just as pH = -log10 [H3O+] , also pOH = -log10 [OH-] ‘p’ in front of a quantity means that -log10

should be taken. • So pKw = -log10 Kw • The value of Kw can be used to calculate

pH of a basic solution

y Kw = [H3O+][OH-] = 1.0 u 10-14 M2 at 25°C ySo [H3O+] = or [OH-] = yAlso pKw = pH + pOH = 14 (This may be useful in some circumstances)

1.0×10[OH ]

1.0×10[H O ]

Sample exam question • The value of the ionisation constant, Kw, of a sample of

pure water at different temperatures is shown in the graph below.

Which one of the following statements about the effect of increasing temperature on the pH and acidity of water is correct? A. The pH is always 7 and the water remains neutral B. The pH decreases and the water remains neutral C. The pH decreases and the water becomes acidic D. The pH increases and the water remains neutral.

Key Knowledge

• pH as a measure of strength of acids and bases; (Kw), Ka for weak acids

Acidity Constants

• When an acid-base reaction is considered as an equilibrium, an equilibrium constant can be written for it.

• For strong acids, the concentration of reactants is very tiny due to complete dissociation of the acid, however for weak acids, there is a significant concentration of both reactants and products.

Consider the equilibrium involving the dissociation of the weak acid, ethanoic acid: CH3COOH(aq) + H2O(l) ⇌ CH3COO-(aq) + H3O+(aq) •Given that the concentration of H2O is virtually constant, the equilibrium constant becomes an acidity constant when [H2O] is taken out •The acidity constant will therefore be:

+ -3 3

[H O ][CH COO ]K =[CH COOH]

What is the use of Ka? • The smaller the value of Ka is for an acid, the

weaker the acid is. • Table 12 in the data book gives you a range of Ka

values for some weak acids. • In that table, the ammonium ion, NH4

+, with a Ka of 5.6 × 10 -10 is the weakest acid

and hydrofluoric acid, HF with a Ka of 7.6 × 10 -4 is the strongest acid, although it is still very much a weak acid. • Stronger acids have larger Kas!

Table 12 on p11 of Data book

Calculating the pH of a weak acid • The expression for Ka and its value can be used

to find [H3O+] and hence the pH of any acid, but it is most useful for a weak acid.

• When a weak acid, HA, dissociates in water, we assume that the concentration of H3O+ formed is equal to the concentration of the conjugate base, A-, of the weak acid

HA + H2O ⇌ H3O+ + A-

weak acid conjugate base • We also assume that the concentration of HA

is effectively constant because it is such a weak acid.

• This means that we can simplify the expression

• Since pH = -log[H3O+] we can find pH if we know [H3O+]

• [H3O+] =

• and pH = -log

a[H O ][A ]K =

[HA]+ 2

[H O ]K =[HA]

aK [HA]

Percentage ionisation

• Consider the dissociation of a weak acid in water

HA + H2O ⇌ H3O+ + A-

• The acid is being ionised, so we can calculate the percentage ionisation by using [A-] and [HA]

• % ionisation =

-[A ] 100[HA] 1

Dilution of strong and weak acids

• It is important to recognise that pH changes to a different extent when strong and weak acids are diluted.

• The dissociation of a strong acid is represented by the equation:

HA + H2O → H3O+ + A-

• When a solution of a strong acid is diluted 10 fold, [H3O+] is decreased by a factor of 10, so pH increases by 1 unit.

• For a weak acid, the dissociation is an equilibrium: HA + H2O ⇌ H3O+ + A-

• When a solution of a weak acid is diluted 10 fold, the water used for diluting is also a reactant, so according to Le Chatelier’s Principle, the system will adjust to partially oppose the change

• So the position of equilibrium moves to the right and more H3O+ is made.

• As a result the [H3O+] does not decrease as much as for a strong acid and so the pH increases by less than 1 unit.

Sample exam question- acid-base equilibria

The table below lists the pH of 0.10M solutions of four different acids at 25 qC (a) Which one of the four acids listed has the smallest Ka value?

VCAA Q3 Nov 2008

Acid pH I 1.0 II 3.0 III 0.7 IV 2.1

(b) Which acid must have more than one acidic hydrogen per molecule? Give a reason for your answer. (c) Using the concentration and the pH of acid IV, calculate the percentage ionization of acid IV in the 0.10M solution

(d) Calculate the value of the ratio [OH-] acid II/[OH-]acid I present in the solutions of acids II and I. (e) Samples of the solutions of acids I and IV are diluted by a factor of 10. The resulting change in pH units would be (Tick one of the boxes)

Greater for acid I than acid IV

Greater for acid IV than for acid I The same for both acids

(f) Methanoic acid is a weak monoprotic acid. (i) Calculate the concentration of a methanoic acid solution that will have the same pH as acid IV. (ii) The dissociation of methanoic acid in water is exothermic. If a solution of the acid is heated, will the pH of the solution increase, decrease or remain constant? Explain

Key Knowledge

• application of equilibrium and rate principles to the industrial production of one of ammonia, sulfuric acid, nitric acid: – factors affecting the production of the selected

chemical – waste management including generation,

treatment and reduction – health and safety considerations – uses of the selected chemical.

Assessed by one SAC (not exam)

Uses of sulfuric acid • Sulfuric acid is mostly used for making other

chemicals Major industrial uses

– Fertilizers – Paper manufacture, – Household detergents, – pigments, dyes and drugs. – car batteries (the electrolyte) – petroleum refining – metallurgical processes

Laboratory uses

• Strong diprotic acid H2SO4(l) + 2H2O(l) → 2H3O+ (aq) + SO4

2-(aq)

• Dehydrating agent – Used for removing water from gas mixtures – Can even remove water from compounds

• Strong oxidant

Production of sulfuric acid The Contact process

First stage (the burner)

• Unless SO2 is used, the first stage is to burn sulfur in air (O2)

Second stage (converter)

• Sulfur dioxide is oxidised to sulfur trioxide by oxygen

• This is an exothermic equilibrium, so the yield is increased by lower temperatures and a catalyst (in layers) helps increase the rate of reaction.

Third stage (absorption)

• Sulfur trioxide is absorbed into sulfuric acid • Reaction between sulfur trioxide and water does

make sulfuric acid, but the reaction is highly exothermic and results in a mist of acid being formed

• Absorption into sulfuric acid is less exothermic

H2SO4 is finally produced in two stages in the absorption tower:

1. Sulfur trioxide is absorbed into Sulfuric acid to make oleum

2. Oleum is mixed with water to make 98% sulfuric acid

Compromises related to the control of the equilibrium at stage 2

1. Increased rate of reaction is achieved by high temp

But it is an exothermic reaction – more products made at only moderate temperatures (400-450 qC)

So a catalyst is used 2. Excess of reactants will favour products, so

excess air (oxygen - cheap reactant) is used

3. Gaseous reaction - greater rate at high pressures

More products made at high pressures But high pressure vessels are very expensive Moderate pressures are used (1-2 atm) and

achieve a 99.5% conversion of SO2 to SO3

4. V2O5 catalyst is used - not the most effective

catalyst, but catalysts can be ‘poisoned’ and other catalysts are too expensive.

Large surface area is used to make it more effective.

Waste Management

• Use of SO2 (a waste from other industries) is environmentally and economically attractive.

• Emission of SO2 into atmosphere is strictly controlled, so conversion to SO3 must be maximised

• Double absorption process where gases are recycled back to the converter is used for this purpose

• Catalyst (vanadium(V) oxide) improved by adding caesium (‘doping’)

• Low air pollution due to H2SO4 because it has a high bpt (low vapour pressure)

• Little solid waste is produced in the process – spent catalyst goes to landfill

• Waste energy (exothermic processes) is used in other locations on site

Health and safety

Risk Controlled by Transportation of H2SO4

Minimising transport, industries are situated close together. Hazchem signs are used to warn of risks

H2SO4 is highly corrosive, can cause severe burns to skin and eyes, even blindness

To protect workers protective clothing is worn and work areas are well ventilated

Health and safety cont.

Risk Controlled by SO2 and SO3 in the air are major contributors to acid rain and the gases themselves are hazardous.

Strict guidelines are set for emission of gases. Authorities monitor levels of SO2 in the air and prosecute as needed.

Oleum is a highly corrosive oily liquid that produces sulfur trioxide fumes.

Safety clothing, gloves, eyewear is needed and SO3 fumes contained

Other safety measures

• In case of accidental spillage, it must be possible to trap the fumes and minimize risk to workers and the environment.

• Acid spills are contained with earth, clay or sand, then slowly diluted with water and finally neutralised with limestone (CaCO3) or sodium carbonate – otherwise direct neutralisation would release dangerous amounts of heat.

Sample exam question – H2SO4

A particular industrial process involves the following steps

VCAA Q6 Nov 2009

Would the rate of reaction 2 become higher, lower or remain unchanged?

Would the equilibrium yield of reaction 2 become higher, lower or remain unchanged?

Would the value of the equilibrium constant, K, of reaction 2 become higher, lower or remain unchanged?

The temperature of reaction 2 is lowered to 150°C. The pressure of reaction 2 is increased to 5 atm by pumping more B(g) and C(g) into the reaction vessel, at constant temperature.

a) It is possible to alter the temperature and pressure at which reaction 2 occurs.

b) Heat energy is released by reaction 2. Describe how the heat energy could be used within this industrial process.

c) i) Describe one waste management strategy, other than recycling heat, employed in the industrial production of your selected chemical.

ii) The following table includes a selection of HAZCHEM labels used to identify dangerous goods. (See next slide) iii) State two uses of your selected chemical.

Key Knowledge

Uses of ammonia, NH3

• Ammonia is a weak base, it is a polar molecule which can take part in hydrogen bonding and so is very soluble in water.

• Nitrogen is an element in demand by plants, so ammonia is commonly used to make fertilizers. eg ammonium nitrate: HNO3(aq) + NH3(g) → NH4NO3(aq)

• Other uses of ammonia include – Household cleaning (cloudy ammonia) – Commercial refrigeration – Used to make a number of economically

important chemicals such as • nitric acid; • fibres, such as nylon, • explosives, such as ammonium nitrate; • pharmaceuticals, such as sulfonamides.

Production of ammonia The Haber process

Raw materials Hydrogen, H2 – depends on the plant, but may

be obtained by steam reforming of natural gas, or by electrolysis of water (if energy is easily obtainable)

Nitrogen, N2 – is always obtained from the air The reaction: N2(g) + 3H2(g) ⇌ 2NH3(g) 'H = - 92 kJ mol-1

Flow chart of the Haber process

Compromises related to the control of the equilibrium

1. Excess of reactants will favour products, so excess nitrogen - cheap reactant is used.

2. Increased rate of reaction is achieved by high temp

But it is an exothermic reaction – more products made at only moderate temperatures (400-450 qC)

So a catalyst is used (porous Fe pellets)

3. Gaseous reaction - greater rate at high pressures

More products made at high pressures But high pressure vessels are very

expensive High pressures are used (200 atm) 4. Fe catalyst is used - not the most effective

catalyst, but catalysts can be ‘poisoned’ and other catalysts are too expensive. KOH is added to increase its efficiency.

The porous pellets have a surface area which also makes it more effective.

Waste Management

• If the H2 required to make NH3 is obtained from natural gas, there are waste management issues related to sulfur-containing compounds, nitrogen oxides, NOx and carbon monoxide.

• Carbon dioxide is also a waste and this can be sold to the food industry.

• Wastage of raw materials is minimised by almost complete conversion into ammonia by recycling unreacted gases back into the converter for further passes over the catalyst.

• Promoting energy efficiency is another way to minimise waste. This is done by – using heat released from exothermic

reactions to generate steam to drive turbines, compressors and other machines

– using waste heat to preheat reaction gas mixtures

– using cold gases from the refrigerator section of the plant to cool gas mixtures before compression

– computer control of all plant operations.

Health and safety Risk Controlled by Toxic gas An extreme irritant to the eyes, respiratory system and other parts of the body. Liquid ammonia can cause frost bite and severe burning.

Work areas must be well ventilated, and workers involved with liquid ammonia storage and transport wear impervious gloves, face shields, and rubber boots and aprons. Breathing apparatus must be available, and employees are drilled in emergency procedures

Ammonia spills can be dangerous to a community

Ammonia is allowed to disperse and areas are evacuated until safe

Other safety considerations

• Hydrogen is explosive, so plants where the H2 is produced must be carefully controlled

• Carbon monoxide (produced during hydrogen reforming) is toxic, so exposure to this must also be carefully monitored.

• Modern ammonia plants pay careful attention to operating, safety and maintenance routines.

Sample exam question – NH3

VCAA Q6 Nov 2009

Key Knowledge

Uses of nitric acid

• Nitrogen is an element in demand by plants, so nitric acid is commonly used to make fertilizers.

• Reaction between nitric acid and ammonia makes ammonium nitrate

HNO3(aq) + NH3(g) → NH4NO3(aq)

• Other uses of nitric acid include: – Explosives such as ammonium nitrate,

trinitrotoluene (TNT) and nitroglycerine – Production of Nitrate salts such as AgNO3

(silver plating and medicine), KNO3 (fertiliser), NaNO3(food additive), Sr(NO3)2 (fireworks)

– Purification of precious metals – Production of dyes and perfumes – Production of paints, pigments, dyes – As a nitrating agent in organic synthesis.

Production of nitric acid The Ostwald process

• Nitric acid is made from ammonia in a 3 step process

• NH3 →NO →NO2 →HNO3

Step 1 Catalytic conversion of ammonia to nitrogen(II) oxide

• Preheated air is mixed with ammonia and passed into a converter where oxygen oxidises the ammonia to nitrogen(II) oxide, NO

• This stage uses a catalyst made up of layers of a thin gauze woven from an alloy of 90% platinum and 10% rhodium

• High temperatures (820–930°C) and high pressures (about 11 atm)

Step 2 Oxidation of nitrogen(II) oxide

• Nitrogen(II) oxide, NO, is oxidised by oxygen to form nitrogen(IV) oxide, NO2

• This is an exothermic equilibrium, so the

yield is increased by lower temperatures. • The temperatures used are low: below 30 qC, since the rate of reaction actually decreases with a decrease in temperature (unusual)

Step 3 Absorption of nitrogen(IV) oxide

• Water is mixed with NO2 in an absorption tower

• Any NO formed in this reaction reacts with a stream of air which is also introduced into the tower. It is oxidised to nitrogen(IV) oxide, (as in step 2) which then reacts with water as above.

Points related to the control of the equilibrium at stage 2

1. Low temperatures are used to maximize equilibrium yield (does not reduce rate of reaction – actually increases it!)

2. This is a gaseous reaction - so has a greater rate at high pressures

The equilibrium has less product particles than reactants so high pressures favour the forward reaction

But high pressure vessels are very expensive, so the incoming gases are only sometimes compressed.

Waste Management

• The main emissions come from the absorption tower - mainly NO and NO2 (NOx) and trace amounts of nitric acid.

• Since these gases contribute to photochemical smog there is further treatment to reduce the levels that are emitted into the atmosphere.

• Treatment includes: – heating these gases with a fuel such as

natural gas, or hydrogen, over a catalyst, so that the NOx is reduced to N2.

– Or the absorption tower may be modified by increasing its size or operating pressure, or an additional absorption tower may be incorporated, this enables more NOx to be converted into nitric acid.

• Waste heat from the converter can be used to heat incoming gases or generate electricity.

Health and safety

Risk Controlled by

HNO3 is corrosive, can cause severe burns to skin and eyes

Full protective equipment and breathing apparatus must be available. Acid spills are isolated with sand or earth and carefully neutralised with Ca(OH)2 or Na2CO3

Fumes are harmful if inhaled

Careful monitoring in nitric acid plants for leaks and spills, employees must be familiar with appropriate procedures to handle them if they do occur.

Further safety considerations

Risk Controlled by Relatively low concentrations of NO2 may cause fluid in the lungs and excessive exposure may be fatal.

Levels of NO2 must be carefully monitored

The mixture of ammonia to air in the gas entering the converter could become explosive.

The ratio of ammonia to air in the gas entering the converter is continuously measured and controlled.

Sample exam question – HNO3

VCAA Q6 Nov 2009

Key Knowledge

• comparison of the renewability of energy sources including coal, petroleum, natural gas, nuclear fuels and biochemical fuels

Definition

• Fossil fuels are a non-renewable energy source – we use them at a greater rate than they are being supplied.

• Fossil fuels include coal, petroleum and natural gas

• They are carbon compounds which have been formed over millions of years from fossilized deposits of plants and animals.

Why do we need to use less fossil fuels?

yTheir supply is dwindling – we will run out

yBurning fossil fuels produces CO2 – greenhouse gas

yThey are needed for other chemical processes, such as production of polymers and paints

yThe use of non-renewable resources such as fossil fuels and uranium conflicts with the aims of sustainability

• Coal is used in Victoria to fuel electricity generating power stations.

• The coal is burnt, heating water to its boiling point. The steam then turns a turbine which operates the generator and electricity is made.

• Electricity can be made in a similar way by burning other fuels.

Crude oil - Petroleum • Crude oil is a mixture of a number of

different hydrocarbons (alkanes) • One part of this mixture is petroleum • Others include kerosene, diesel and

liquefied petroleum gas (LPG). • The parts of the mixture are separated

according to their boiling points using fractional distilllation.

Natural gas

• Made up mainly of methane but also includes other hydrocarbons such as ethane and propane

• Used for home heating and cooking • Some power stations (eg Newport) use

gas to generate electricity

Nuclear fuels yUranium is a nuclear fuel which is non-re-

newable. yAll the uranium we will ever have is in the

earth or being mined now. There is no way to make more uranium.

yUranium is mined in only a small number of countries

• Nuclear fission occurs when a large nucleus, such as a Uranium nucleus (235U) is bombarded by high energy neutrons.

• The nucleus splits and two new nuclei are formed while more neutrons and a large amount of energy are released.

• The large amount of energy from this nuclear reaction can be used to heat the water in a nuclear power plant to make steam, drive turbines and generate electricity.

Advantages and disadvantages of nuclear power

Advantages: yNo air pollution y Chance to conserve fossil fuels for other use y 1 kg of Uranium produces as much electrical

energy as 2500 tonne of Coal Disadvantages: y Accidental release of radioactive substances y Problem of storing wastes y Long construction time, limited working life.

Nuclear fusion

yOccurs in the Sun y Small nuclei come together to make a larger

nucleus and releasing lots of energy

yNot possible to control currently, but would be

very useful if it could be. y This could be classified as a renewable energy

source

Renewable energy sources

• Renewable energy sources are those that are constantly being replaced by natural processes

• Improvements in technology are making these more viable as time goes on.

• Biochemical fuels such as biodiesel and ethanol (see Unit 3) are renewable energy sources.

Sample exam question - fuels Most of Victoria’s electricity is generated by burning fossil fuels such as coal and natural gas. Alternative methods of generating electricity are currently being developed. (e) Biochemical fuels are an alternative fuel for generating

electricity (i) Name one biochemical fuel, other than methyl palmitate, and the raw material used in its production. (f) Some countries rely on nuclear fission for the large-

scale production of electricity. (i) state one disadvantage of using nuclear fission

VCAA Q4 Nov 2009

Key Knowledge

• application of calorimetry to measure energy changes in chemical reactions in solution calorimetry and bomb calorimetry, including calibration of a calorimeter and the effects of heat loss.

Specific heat capacity

• Specific heat capacity is a physical property of a substance.

• It is the amount of energy required to raise the temperature of 1 g of a substance by 1 °C.

• Water has a specific heat capacity of 4.18 J g-1 °C-1 (unusually high)

Finding energy change using specific heat capacity

• To calculate an energy change in a substance use

E = mc'T where m = mass of substance being heated(g) If the substance is water then = volume of

water (mL), since density of water is 1gmL-1

c = specific heat capacity of the substance 'T = change in temperature of the

substance(qC or K)

• This equation is most useful when a pure substance (with a measureable specific heat capacity) is being heated

eg calculate the energy required to heat 150 mL of water to boiling point if the initial water temperature is 15 qC.

E = mc'T = 150 u 4.18 u (100 – 15) = 53295 J = 53.3 kJ

Calorimetry • A calorimeter is an instrument that is used

to measure the change in temperature during a chemical reaction.

• Reactions in solution (eg neutralisation) are carried out in a solution calorimeter.

• Combustion reactions and other gaseous reactions are carried out in a bomb calorimeter.

A solution calorimeter

• The change in temperature of the water surrounding the reaction is measured to obtain 'Trxn • The calorimeter is insulated to prevent heat escaping (or entering).

A bomb calorimeter

The effects of heat loss in calorimetry

• When heat from a reaction is transferred to a container of water without any attempt to insulate the equipment, heat is lost to the surrounding air, the container and any other equipment that are in contact with the water. • This will result in a lower calculated value for

the enthalpy of reaction, since not all the energy generated is transferred to the water which is used for calculations: E = mc'T.

Calibration of calorimeters • A calorimeter is calibrated to determine

how the temperature of the water in the calorimeter responds to the input of a given amount of energy.

• A fixed amount of energy is put into the water of the calorimeter and the maximum temperature change (increase) is measured.

Electrical calibration

• In this case the fixed amount of energy is added directly into the water of the calorimeter using an electrical heater.

• Electricity is passed through the heater and energy is calculated using

E = VIt where V = voltage, I = current and t = time for which the current is passed.

Chemical calibration

• A reaction of known 'H can be used to calibrate a calorimeter.

• A known amount (in mol) of the reactants are allowed to react in the calorimeter and 'Tcalibration is measured.

• Energy added during calibration is calculated using 'H for the reaction and n(reactants)

E = n × 'H

Calculating and using a calibration factor

• Calibration gives us a calibration factor (cf) with units J qC-1

• cf =

• The calibration factor (cf)can then be used to find out how much energy has been given out or absorbed by a reaction using

Erxn = cf × 'Trxn

calibration

Energy addedΔT

Why is calibration needed?

• A calibration factor is specific to a particular calorimeter.

• Whereas using E = mc'T does not consider insulation and absorption of heat energy by other materials in the calorimeter, a calibration factor takes these into consideration, so gives a more accurate value for E and hence 'H

Calculations involving calorimetry

• Examples of actual calculations and applications of calorimetry will be covered in Calorimetry part 2

Formulas used so far

• Energy when no calibration has occurred: E = mc'T

• Energy added during electrical calibration E = VIt

• Energy added during chemical calibration E = n × 'H

• Calibration factor cf = • Energy of reaction Erxn = cf × 'Trxn

calibration

Energy addedΔT

Example – electrical calibration

A bomb calorimeter was calibrated by passing 1.82 A through the electric heater for 60.5 s at a potential difference of 5.50V. The temperature of the water in the calorimeter rose by 0.375 qC. Calculate the calibration factor of this calorimeter.

V = 5.50V I = 1.82 A t = 60.5 s 'T = 0.375 qC E = VIt = 5.50 × 1.82 × 60.5 = 605.605 J (keep in calculator) cf =

-1E 605.605 =1.61 kJ°CΔT 0.375

Heat of combustion and heat of neutralisation

• The heat of combustion is the energy released when a specified quantity

(1 mole, 1 g, 1 L) of a compound burns completely in oxygen. • Note that the equation for the combustion of

a compound may not involve one mole of the compound, so 'Hrxn may differ to 'Hc

• The heat of neutralisation is the energy released when 1 mole of an acid or a base is neutralised.

Example – chemical calibration

The heat of combustion of benzoic acid (C6H5COOH, M = 122.13) is -3227 kJ mol-1. In the calibration of a bomb calorimeter, 0.244g of benzoic acid was combusted and the change in temperature was 4.13 qC. Calculate the calibration factor of this calorimeter.

n(C6H5COOH) = Eadded = 'H × n = 3227 × 0.00200 = 6.45 kJ cf =

0.244 0.00200 mol122.13

-1addE 6.45 =1.56 kJ°CΔT 4.13

Finding 'H from calorimetry

• A calibrated calorimeter is used to measure the amount of energy released or absorbed by a chemical reaction (Erxn) occurring in that calorimeter.

• For this to occur, the amount of reactants must be known and the change in temperature during the reaction, 'Trxn

• The enthalpy of reaction ('H) must relate to the equation

• Recall that the enthalpy of reaction changes according to the amount of reactants in the equation eg:

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l); 'H = -3120 kJ mol-1

• Recall that Erxn = cf × 'Trxn

• To find 'H use where n1 is the amount of reactant in the equation and n2 is the amount used in the reaction for which the energy released was Erxn

ΔH E=n n

Example – finding 'H using a calibrated calorimeter

A bomb calorimeter with a calibration factor of 1.56 kJ qC-1 was used for the combustion of methane gas. Calculate 'H for the equation CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) if the temperature in the calorimeter rose from 20.187 to 20.900 qC when 1.25 × 10-3 mol of methane was burnt.

• 'Trxn = 20.900 – 20.187 = 0.713qC • n(CH4) = 1.25 × 10-3 mol • Erxn = cf × 'Trxn = 1.56 × 0.713 = 1.11 kJ For the reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) 'H = -890 kJ mol-1 (note the –ve sign!)

ΔH E=n n

ΔH 1.11= 890 kJ mol1 1.25×10

Sample exam questions - calorimetry

A foam cup calorimeter containing 100 mL of water is calibrated by passing an electric current through a small heater placed in the solution. Assuming that all measurements are accurate, which one of the following is the most likely calibration factor (in J qC-1) for the calorimeter and contents? A. 120 B. 240 C. 480 D. 960

VCAA Q14 Nov 2008

Potassium hydroxide and hydrochloric acid react in aqueous solution according to the following equation

KNO(aq) + HCl(aq) → KCl(aq) + H2O(l) A 50 mL solution containing 0.025 mol of KOH was mixed rapidly in an insulated vessel with a 50 mL solution containing 0.025 mol of HCl. The temperature increased by 3.5 qC. Assuming that the specific heat capacity of the solution is the same as that of the water, the enthalpy change, 'H, of this reaction , in kJ mol-1, is closest to A. -29 B. -59 C. -2.9 × 104

D. -5.9 × 104

VCAA Q10 Nov 2009 Also look at VCAA Nov 2010 Q4 (chemical calibration and 2009 Q4(a-d) (electrical calibration)

Key Knowledge

• use of the electrochemical series in predicting the products of redox reactions and deducing overall equations from redox half equations

• limitations of predictions made using the electrochemical series, including the determination of maximum cell voltage under standard conditions

Galvanic cells (revision of basics) • Zn(s) + CuSO4(aq) o Cu(s) + ZnSO4(aq)

'H –ve • This is a redox reaction – the zinc loses

electrons (reductant) and the Cu2+ gains electrons (oxidant)

• If the reductant and oxidant are separated from each other, the energy is released as electrical energy instead of heat energy, creating a galvanic cell.

The Daniell cell

Construction of a galvanic cell

• Each half cell contains an oxidant and a reductant (conjugate redox pair)

• The reductant may be the electrode OR an inert electrode is used and the reductant is in solution

• Reductants are usually metals or non-metal ions

Occasionally a transition metal ion may be a reductant.

• Oxidants are usually non-metals (gases or in solution) or metal ions.

• Oxidation occurs at the anode (the negative electrode)

• Reduction occurs at the cathode (the positive electrode)

• The salt bridge completes the circuit • Only ions move through the salt bridge.

Positive ions move away from the anode half cell and negative ions move towards the anode half cell.

Using the electrochemical series

• An electrochemical series reflects the relative ability of reductants to lose electrons

• The best reductants (bottom RHS) are those metals that react vigorously eg magnesium, sodium

• The best oxidants (top LHS) are those non-metals that react vigorously eg fluorine

Data book p4

• In the electrochemical series all the half cells are compared to the hydrogen half cell: H2/H+

• The E0 value (standard electrode potential) is the potential difference between the half cell and H2/H+

• These are measured at standard conditions: 25 °C, gas pressures of 1 atm, and solution

concentrations of 1 M

Using the electrochemical series with a galvanic cell

• When two half cells are connected, the cell with the better reductant will contain the negative electrode.

y The Cu electrode in the Cu/Cu2+ half cell is a stronger reductant than the reductant in the Cl2/Cl- half cell, (Cl-).

y So Cu is the negative electrode and the inert electrode in the Cl2/Cl- half cell is the positive electrode.

y Half equations: Anode: Cu(s) → Cu2+ (aq) + 2e-

Cathode: Cl2(g) + 2e- → 2Cl-(aq) y Overall equation: Cu(s) + Cl2(g) → Cu2+ (aq) + 2Cl-(aq)

Predicting the flow of electrons

yeg Ag/Ag+//Zn2+/Zn Ag+(aq) + e- o Ag(s) Eo = 0.80 V Zn2+(aq) + 2e- o Zn(s) Eo = - 0.76 V

Calculating the maximum cell voltage

yUnder standard conditions, the maximum cell voltage can be calculated using the standard electrode potentials (E0 values)

yCell voltage = Eo(oxidant ½ cell) – Eo(reductant ½ cell) Eg Under standard conditions, the voltage of Ag/Ag+//Zn2+/Zn would be 0.80 – (-0.76) = 1.56V

Predicting products of direct redox reactions

yA reductant can only react with oxidants that have a more positive Eo value than itself

y ie oxidants high on the left hand side of the electrochemical series only react with reductants that are lower on the right hand side.

yThis knowledge is used to predict whether a reaction will occur between a pair of reactants

example y If a piece of Pb was placed in FeCl3 predict

whether there would be a reaction, and if so write the equation for the reaction.

Fe3+(aq) + e- ⇌ Fe2+(aq) Eo = +0.77 V Pb2+(aq) + 2e- ⇌ Pb(s) Eo = -0.13 V y The oxidant has a more positive Eo than the

reductant, so there will be a reaction 2Fe3+(aq) + Pb(s) o 2Fe2+(aq) + Pb2+(aq)

Limitations

• While the electrochemical series accurately predicts the possibility of a reaction, it gives no indication of the rate of that reaction.

• Eg hydrogen peroxide • Also the predictions only apply under

standard conditions, if the conditions are different the voltage may differ, or even the order of reactivity.

Sample exam question – electrochemical series

VCAA Q18 Nov 2008

Key Knowledge

• the chemical principles, half-equations and overall equations of simple primary and secondary galvanic cells

Primary cells y A primary cell cannot be recharged y They go flat when the cell reaction reaches

equilibrium. yMost primary cells utilize electrolytes that

are contained within absorbent or separating material (i.e. no free or liquid electrolyte), and so are called dry cells.

y Cells are named according to the electrolyte used y eg acid cells, alkaline cells, lithium ion cells

Zinc-carbon dry cell y Anode is zinc (this is the case) y Oxidation reaction: Zn(s) → Zn2+ (aq) + 2e-

y Cathode is a carbon rod in middle of cell y Reduction reaction: 2MnO2(s) + 2NH4

+(aq) + 2e- → Mn2O3(s) + 2NH3(aq) + H2O(l) y Voltage is 1.5V, but this will drop with prolonged

use. y Overall reaction Zn(s) + 2MnO2(s) + 2NH4

+(aq) → Zn2+(aq) + Mn2O3(s) + 2NH3(aq) + H2O(l)

• Electrolyte is a paste of powdered carbon and manganese dioxide.

Alkaline cells

yAnode – central steel rod yPowdered zinc around the rod reacts Zn(s) + 2OH-(aq) → Zn(OH)2(s) + 2e- yCathode – outer steel case yManganese dioxide is reduced 2MnO2(s) + H2O(l) + 2e- → Mn2O3(s) + 2OH-(aq)

Half equations and overall equations

y Just as the oxidation and a reduction half equation go together to give the overall equation for a cell, the overall equation can be used to derive the two half equations.

y eg The overall equation for the reaction in a zinc-silver button cell is Zn(s) + Ag2O(s) + H2O(l) → Zn(OH)2(s) + 2Ag(s)

Write the oxidation and reduction half equations:

Rechargeable cells and batteries

yA battery is a collection of cells joined in series.

eg car battery is made up of 6 cells yA rechargeable cell is also called a

secondary cell. yWhen a cell is recharged, the products are

converted back to reactants. yFor this to be possible they must remain

in contact with the electrodes.

Lead acid battery – a secondary cell

• This is the battery most commonly found in cars

During discharge (giving out electricity): y Anode reaction (oxidation) – at negative

electrode: Pb(s) + SO4

2-(aq) → PbSO4(s) + 2e-

y Cathode reaction (reduction) – at positive electrode: PbO2(s) + SO4

2-(aq) + 4H+(aq) + 2e- → PbSO4(s) + 2H2O(l)

y Overall equation: Pb(s) + PbO2(s) + 2SO4

2-(aq) + 4H+(aq) → 2PbSO4(s) + 2H2O(l) y Lead(II) sulfate is the product of both

reactions

During recharge (using electricity) y Reactions are reversed yNegative terminal is connected to the

negative of a power supply, so reduction occurs

PbSO4(s) + 2e- → Pb(s) + SO42-(aq)

yOxidation occurs at the positive terminal PbSO4(s) + 2H2O(l) → PbO2(s) + SO4

2-(aq) + 4H+(aq) + 2e- yOverall equation: 2PbSO4(s) + 2H2O(l) → Pb(s) + PbO2(s) + 2SO4

2-(aq) + 4H+(aq)

Sample exam question – primary and secondary cells

A vanadium redox battery is used to store electrical energy generated at a wind farm in Tasmania. The battery supplies electricity to the power grid as required through a control switch. The diagram shows the structure of a cell in a vanadium redox battery. The reactants are dissolved in an acidic solution, stored in large tanks and pumped through the cell. The cell is recharge using electricity generated by the wind turbines. A polymer membrane allows the movement of particular ions.

VCAA Q5 Nov 2009

The two relevant half equations for the vanadium redox battery are VO2

+(aq) + 2H+(aq) + e- ⇌ VO2+ (aq) + H2O(l) E0 = +1.004 V V3+(aq) + e- ⇌ V2+(aq) E0 = -0.255 V a)State the polarity of each electrode as the battery is discharged b)Write a balanced overall equation for the reaction that occurs when the cell is being recharged. c)Write a balanced overall equation to show why iron would be an unsuitable material to use as electrode B in the vanadium redox cell.

Solution

VO2+(aq) + 2H+(aq) + e- ⇌ VO2+ (aq) + H2O(l) E0 = +1.004 V

V3+(aq) + e- ⇌ V2+(aq) E0 = -0.255 V

a)V2+ is the reductant for the cell and VO2+ is

the oxidant. So electrode B is the negative and electrode A is the positive b)Discharge equation is V2+(aq) + VO2

+(aq) + 2H+(aq) → V3+(aq) + VO2+(aq) + H2O(l)

So recharge equation is V3+(aq) + VO2+(aq) + H2O(l) → V2+(aq) + VO2

+(aq) + 2H+(aq)

c) V3+(aq) + e- ⇌ V2+(aq) E0 = -0.255 V From data book: Fe2+ (aq) + 2e- ⇌ Fe(s) E0 = -0.44 V If an electrode made of Fe was placed in the V3+/V2+ solution, the Fe, being a stronger

reductant, would react with the V3+ rather than allowing the V2+ to act as the cell reductant.

Key Knowledge

• the chemical principles, half-equations and overall equations of fuel cells; advantages and disadvantages of fuel cells compared to conventional energy sources

What are Fuel cells?

• Reactants are supplied continuously • Use oxidation of a fuel to produce

electricity from chemical energy directly (compare to multiple steps in coal-fired power stations)

• Oxygen, O2, is always the oxidant • Fuel may be hydrogen, methane, methanol etc

Alkaline fuel cell

• At the anode (oxidation ? fuel reacts) H2(g) + 2OH-(aq)→ 2H2O(l) + 2e-

• Hydroxide ions come from the electrolyte (alkaline)

• At the cathode (reduction of oxygen) O2(g) + 2H2O(l) + 4e- → 4OH-

• Overall equation: O2(g) + 2H2(g) → 2H2O(l)

• We cannot tell from overall reaction whether it is alkaline or acidic electrolyte – need more information in question.

Acidic fuel cell

• eg phosphoric acid electrolyte • At anode (oxidation of fuel)

H2(g) o 2H+(aq) + 2e-

• At cathode (reduction of oxygen) O2(g) + 4H+(aq) + 4e- o 2H2O(l)

• Overall: O2(g) + 2H2(g) → 2H2O(l)

Uses of fuel cells • Space craft – as early as Apollo 11 (1969). Still

used in space shuttle • Trial of buses in Perth (2004-2007) – powered

by hydrogen fuel cells • Small scale electricity generation • Powering small appliances • Proposed for cars

Other points

• Low voltage, but cells can be connected in series as a ‘stack’

• Only byproducts are water and heat (both can be used)

• Electrodes behave as catalysts for the reaction – surface area is important.

Advantages and Disadvantages

• More efficient energy conversion

• Water as byproduct • Will continue as long

as fuel is available (no need to recharge)

• Variety of fuels • Useful for onsite

electricity generation • Quiet operation

• Expensive • Require constant fuel

supply • Difficult to produce,

store and distribute H2

• Fuel cells are very large (not v. portable)

• Generates DC (not AC) power

Differences between primary and secondary cells and fuel cells

Fuel cell Primary and secondary cells

Reactants are delivered continuously, so the supply of electricity is continuous.

Once the reactants have been consumed the device is either discarded or recharged.

Electrodes are porous and usually contain a catalyst

The anode may react as the reductant in the cell, or electrodes may be inert.

All products of the reaction in a fuel cell are removed during operation

Products remain in the cell. In a primary cell their build up slows the operation of the cell. In secondary cell the products can be changed back to reactants.

Sample exam question – fuel cells A fuel cell that can provide power for buses is the phosphoric acid fuel cell, PAFC. The electrolyte is concentrated phosphoric acid and the reactants are hydrogen and oxygen gases. A simplified sketch of a phosphoric acid fuel cell is given below.

VCAA Q8 Nov 2008

a) Give the equation for the half reaction that takes place at the i) anode and ii) cathode

b) On the diagram of the fuel cell, draw an arrow to show the direction in which the H2PO4

- ion moves as the cell delivers an electrical current.

c) Describe one advantage and one disadvantage of such a fuel cell compared with a petrol-driven car engine.

Key Knowledge

• the chemical principles, half-equations and overall equations of simple electrolytic cells; comparison of electrolytic cells using molten and aqueous electrolytes, and inert and non-inert electrodes

• Electrolysis is used to force non-spontaneous reactions to occur by the addition of electrical energy.

• It is the opposite process to that occurring in a galvanic cell

Electrolysis of molten salts

• A molten ionic compound (salt) is made up on only one type of positive ion and one type of negative ion.

• The negative ion loses electrons (is oxidized) at the anode

• The positive ion gains electrons (is reduced) at the cathode

eg electrolysis of molten KBr

• When molten potassium bromide is electrolysed, the electrode reactions are:

Cathode: K+(l) + e- o K(l) Anode: 2Br-(l) o Br2(l) + 2e-

Aqueous electrolysis

• In an aqueous solution, water is present at both electrodes, in addition to the ions

• Water is able to act as an oxidant or a reductant

O2(g) + 4H+(aq) + 4e- ⇌ 2H2O(l) E0 = +1.23 V

2H2O(l) + 2e- ⇌ H2(g) + 2OH-(aq) E0 = -0.83 V

• Water is a stronger oxidant than Al3+, Mg2+, Na+ and Ca2+, so in aqueous solution, water will always react in preference to these ions.

• Water is a weak reductant, but it is a stronger reductant than Cl- and F-.

• This is most important in 1M solutions. If the conditions are not standard, Cl- may react in preference to the water anyway.

eg The electrolysis of aqueous copper(II) sulfate

1) With carbon electrodes • Carbon electrodes will conduct electricity but

will not react. (inert) • Consider the ions/polar molecules present in

solution and which electrodes they will be attracted to.

• Identify the strongest oxidant at the-ve electrode and the strongest reductant at the +ve electrode

+ve electrode -ve electrode

SO42- Cu2+

H2O H2O

Half equations: At Cathode: Cu2+ (aq) + 2e- → Cu (reduction) At Anode: 2H2O(l)→ O2(g) + 4H+(aq) + 4e-

(oxidation)

eg The electrolysis of aqueous copper(II) sulfate continued 2) With copper electrodes • The copper electrodes will conduct electricity

and may also react, since copper is a reductant.

• Consider the ions/polar molecules and electrodes present in solution and which electrodes they will be attracted to.

• Identify the strongest oxidant present at the-ve electrode and the strongest reductant at the +ve electrode

+ve electrode

-ve electrode

SO42- Cu2+

H2O H2O Cu Cu

Half equations: • At Cathode: Cu2+(aq) + 2e- → Cu(s) (reduction) • At Anode: Cu(s) → Cu2+(aq) + 2e- (oxidation)

Another example

• Predict the reactions and hence the products at each electrode if a solution of sodium nitrate is electrolysed using carbon electrodes

Species present: Note that the electrodes are inert, so cannot react.

+ve electrode -ve electrode NO3

H2O H2O

• Best reductant present at +ve (anode): H2O • Best oxidant present at -ve (cathode): H2O • Reaction at anode: 2 H2O(l) o O2(g) + 4H+(aq) + 4e-

• Reaction at cathode: 2 H2O(l) + 2e- o H2(g) + 2OH-(aq)

• Note that when these reactions occur: – gas is produced at both electrodes (twice as

much hydrogen as oxygen per mole of electrons through the circuit)

– the pH around the positive electrode decreases as H+ is produced and

– the pH around the negative electrode increases as OH- is produced.

y This can be detected experimentally using universal indicator and the gases can be collected and tested (pop test and glowing splint)

Sample exam question - electrolysis When comparing the electrolysis of molten NaF and that of a 1.0 M aqueous solution of NaF, which one of the following statements is correct? A. The product at the anodes is the same in both cells and

the product at the cathodes is the same in both cells. B. The product at the anodes is the same in both cells but

the products at the cathodes are different. C. The product at the cathodes is the same in both cells

but the products at the anodes are different. D. The products at the cathodes of the cells are different

and also the products at the anodes are different. VCAA Q16 Nov 2008

Key Knowledge

• application of Faraday’s laws in electrochemistry

Relating amount of product to charge passed through a circuit

• The amount of electricity that is needed to deposit one mole of a metal depends on the charge on the metal ion

eg Na+ + e- o Na Cu2+ + 2e- o Cu Cr3+ + 3e- o Cr • The greater the charge on the ion, the smaller

the amount of metal that can be deposited when one mole of electrons passes through the circuit.

Charge and current

• Electrons carry charge • One mole of electrons carries 96500 Coulomb of

charge (This is called Faraday’s constant, F) • Charge is calculated using the relationship

Q = I × t where Q is charge (coulombs), I is current (amps) and t is time (seconds) When current and/or time increases, the amount of charge (electrons) passing through the circuit increases.

Relating mole of electrons to charge

• Since one mole of electrons carries 96500 C of charge:

n(e) = (Note similarity to n = and n = )

Q It96500 F

Relating mole of element to charge

• We saw earlier that as the charge on an ion increases, the amount of metal that can be deposited by a fixed amount of electrons decreases.

• So, considering the half equation for the production of the element M:

Mx+ + xe- → M • n(M) = u n(e)

• OR n(element) =

1 It96500x

Using this formula • The mass of a metal that is deposited at a cathode

during electrolysis can be calculated using

m(element) = ×M(element)

• Where x = number of mole of electrons required to deposit one mole of the element (always write out the half equation)

and M(element) depends on the formula of the element eg Pb, or Cl2

1 It96500x

example

• eg Find (a) the mass of lead deposited and (b) the volume of oxygen gas produced at STP when molten PbO is electrolysed using a current of 5.00 A for 3.00 minutes. (a) From the formula, the reaction at the cathode is: Pb2+ + 2e- → Pb

m(element) = ×M(element)

m(Pb) =

1 It96500x

1 5.00 (3.00 60) 207 0.9652 96500

gu uu u

(b) The reaction at the anode is 2O2- → O2 + 4e- (note that equations for non-metals must show

the formation of a molecule – see electrochemical series for reminder)

• V(O2) = n × VM

1 5.00 (3.00 60) 22.44 965000.0522 L

= 52.2 mL

u u u u

Linking these formulas to fuel cells • Recall from the calorimetry section that E = VIt Now we know that Q = It and n(e) =

• So, substituting into the Energy equation: E = VQ and Q = n(e) × F so E = V × n(e) × F

example

A particular fuel cell operates at 0.92V. How much energy, in kJ, is delivered per mole of hydrogen used in this fuel cell? At anode: H2 → 2H+ + 2e-

• E = V× n(e) × F = 0.92 × 2 × 96500 = 177560 J = 1.8 × 102 kJ

Sample exam question – Faraday’s laws

A series of electrolysis experiments is conducted using the apparatus shown below. An electric charge of 0.030 faraday was passed through separate solutions of 1.0 M Cr(NO3)3, 1.0 M Cu(NO3)2 and 1.0 M AgNO3. In each case the corresponding metal was deposited on the negative electrode.

VCAA Q18 Nov 2011

The amount, in mol, of each metal deposited is Amount, in mol,

of chromium deposited

Amount, in mol, of copper

deposited

Amount, in mol, of silver

deposited

A. 0.030 0.030 0.030 B. 0.010 0.015 0.030 C. 0.090 0.060 0.030 D. 0.030 0.020 0.010

Sample exam question – fuel cell energy calculations

A ceramic fuel cell delivers a current of 0.500 A for 10.0 minutes at a potential of 0.600 V a) How much electrical energy, in joules, would

be provided by the cell? b) Calculate the charge, in coulomb, produced by

the cell. c) If this particular cell operated at 60.0%

efficiency, what amount of hydrogen gas, H2, in mole, would be consumed by the fuel cell?

VCAA Q5 Nov 2006

a) E = VIt = 0.600 × 0.500 × (10.0 × 60) = 180 J b) Q = It = 0.500 × (10.0 × 60) = 300 C

c) H2(g) → 2H+(aq) + 2e-

n(e-) = n(H2) = ½ n(e-) = ½ × 0.00311 = 0.00155 mol Operating at 60% efficiency:

n(H2) =

Q 300 0.00311 mol96500 96500

100 0.00155 0.00259 mol60

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