Unit 2 Equilibria, energetics and elements (F325) Energy UNIT 2 Module 2... · 1 | P a g e Unit 2 Equilibria, energetics and elements (F325) Energy Topics covered in this module:

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Unit 2 Equilibria, energetics and elements (F325) Energy

Topics covered in this module:

1. Lattice enthalpy

2. Constructing Born-Haber cycles

3. Born-Haber cycle calculations

4. Further examples of Born-Haber cycles

5. Enthalpy change of solution

6. Understanding hydration and lattice enthalpies

7. Entropy

8. Free energy

9. Redox

10. Cells and half cells

11. Cell potentials

12. The feasibility of reactions

13. Storage and fuel cells

14. Hydrogen for the future

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1. Lattice enthalpy

Unit 2 – Equilibria, energetics and elements

Energy Done

Explain and use the term lattice enthalpy.

Key areas to concentrate on…

You will need to be able to explain what lattice enthalpy means and to construct Born-Haber cycles based upon information given – state symbols will be relevant here!

You may be asked about some of the AS material covered in AS Unit F322: Module 2 - Alcohols, halogenoalkanes and analysis including:

o Q = mcΔT

o standard enthalpy change of reaction

o standard enthalpy change of combustion

o standard enthalpy change of formation

o average bond enthalpy.

Energy cycles involving Hess’ law may also feature.

You need to understand new terms such as:

lattice enthalpy.

When writing an atomisation equation remember that only one mole of atoms is formed. Students frequently get this equation wrong in exams!

Learn all key definitions for lattice enthalpy, Hess’ Law, standard enthalpy change of formation, enthalpy change of atomisation, first and second ionisation energy, and first and second electron affinity.

State symbols are critical and it is important to write all state symbols in your equations.

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2. Constructing Born-Haber cycles


Energy Done

Use the lattice enthalpy of a simple ionic solid and relevant energy term to construct Born-Haber cycles.




o Q = mcΔT







lattice enthalpy.

When drawing a Born-Haber cycle it is important to make sure that you have sufficient space to complete your diagram. Start with the elements in their standard states a quarter of the way up from the bottom of the space.

It is important to show all the equations with the correct state symbols in the exam. Marks will be lost if state symbols are missed out.

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3. Born-Haber cycle calculations


Energy Done

Construct Born-Haber cycles.

Carry out related calculations.




o Q = mcΔT







lattice enthalpy.

Remember that exothermic energy changes have a negative value and the arrows go downwards. Endothermic changes have a positive value and the arrows go upwards.

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4. Further examples of Born-Haber cycles


Energy Done

Construct Born-Haber cycles.

Carry out related calculations.




o Q = mcΔT







lattice enthalpy.

Three common mistakes made in Born-Haber cycle calculations are:

Not doubling the enthalpy change of atomisation when you require two moles of gaseous atoms e.g. the two moles of chlorine atoms required to make CaCl2.

Not doubling the electron affinity when needing two moles of negative ions e.g. the two moles of electrons required to make the two moles of chlorine ions in CaCl2.

Not representing electron affinities correctly e.g. in the formation of copper (II) oxide the first electron affinity is exothermic, but the second electron affinity is endothermic.

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Specimen paper questions

The table below shows the enthalpy changes needed to calculate the lattice enthalpy of calcium oxide, CaO.

process enthalpy change/ kJ mol–1

first ionisation energy of calcium +590

second ionisation energy of calcium +1150

first electron affinity of oxygen –141

second electron affinity of oxygen + 791

enthalpy change of formation of calcium oxide –635

enthalpy change of atomisation of calcium +178

enthalpy change of atomisation of oxygen +248

(a) (i) Explain why the second ionisation energy of calcium is more endothermic than the first ionisation energy of calcium.

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[2]

(ii) Suggest why the second electron affinity of oxygen is positive.

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[2]

[Turn over]

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(b) Complete the Born–Haber cycle for calcium oxide below.

Use the data in the table to calculate the lattice enthalpy of calcium oxide.

lattice enthalpy = ............................ kJ mol–1

[5]

(c) The lattice enthalpies of calcium oxide and magnesium oxide are different.

Comment on this difference.

In your answer you should make clear how the sizes of the lattice enthalpies are related to any supporting evidence.

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[3]

[Total 12 marks]

energy/ kJ mol–1

Ca(g) + 0.5 O (g)2

Ca(s) + 0.5 O (g)2

CaO(s)

Ca(s) + O(g)

You will

know how

to do this

part after

covering

section 6,

‘Understan

ding

hydration

and

lattice

enthalpies

’

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Past paper questions

The table below shows the enthalpy changes needed to calculate the enthalpy change of formation of calcium oxide.

process enthalpy change/kJ mol–1

lattice enthalpy for calcium oxide –3459

first ionisation energy for calcium +590

second ionisation energy for calcium +1150

first electron affinity for oxygen –141

second electron affinity for oxygen +798

enthalpy change of atomisation for oxygen +249

enthalpy change of atomisation for calcium +178

(a) (i) Explain why the first ionisation energy of calcium is endothermic.

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[1]

(ii) Explain why the first electron affinity for oxygen is exothermic.

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[1]

[Turn over]

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(b) (i) Draw a Born-Haber cycle for calcium oxide.

Include

• correct formulae and state symbols • energy changes in kJ.

[3]

(ii) Use your Born-Haber cycle in (i) to calculate the enthalpy change of formation for calcium oxide.

enthalpy change of formation = .........................................

[2]

(iii) The lattice enthalpy for iron(II) oxide is –3920 kJ mol–1.

Suggest a reason for the difference in lattice enthalpy between calcium oxide and iron(II) oxide.

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[1]

[Total 8 marks]

You will

know how

to do this

part after

covering

section 6,

‘Understan

ding

hydration

and

lattice

enthalpies

’

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Examiner’s comments

Candidates of all abilities were able to draw a recognisable Born–Haber cycle but frequently candidates found parts (a) and (c) very demanding.

(a) The majority of candidates were unable to explain the direction of energy transfer in first ionisation energy and the first electron affinity. It was not sufficient to state that first ionisation energy required energy and first electron affinity released energy. To gain credit candidates had to refer to the attraction between electrons and the nucleus as well as electron loss in part (i) and electron gain (ii). There was evidence that a significant proportion of candidates did not read the question sufficiently well and answered about first and second ionisation energy.

(b) The majority of candidates drew energy level diagrams but other correct cycles were given full credit. The most common errors were the

• use of the wrong formulae e.g. O+(g)

• wrong stoichiometry e.g. Ca(s) + O2(g) rather than Ca(s) + ½O2(g),

• wrong labelling of energy changes

The correct value for the enthalpy of formation was –635 kJ mol–1. The unit was required for full credit. An error carried forward mark was allowed from incorrect cycles.

In part (iii) only a small proportion of candidates realised that Fe2+ must have a

smaller ionic radius than Ca2+. Credit was not given to references to the wrong type of particle for example Fe has a smaller atomic radius.

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The table below shows the enthalpy changes needed to construct a Born-Haber cycle for sodium oxide, Na2O.

process enthalpy change / kJ mol–1

first ionisation energy of sodium +495

first electron affinity of oxygen –141

second electron affinity of oxygen +791

enthalpy change of formation for sodium oxide –416

enthalpy change of atomisation for sodium +109

enthalpy change of atomisation for oxygen +247

(a) Use the table of enthalpy changes to complete the Born-Haber cycle by putting in the correct numerical values on the appropriate dotted line.

[4]

2Na (g) + O (g)+ 2–

lattice

enthalpy of

sodium oxide

2Na (g) + O(g) + 2e+ –

H = ..........kJ

ΔH = ..........kJ

ΔH = ..........kJ

2Na (g) + O (g) + e

2Na

+ – –

+(g) + 12

1

O (g) + 2e

2Na(g) +

2–

2O (g)2

H = ..........kJ

H = ..........kJ

H = ..........kJ

2Na(s) + 12O2(g)

Na O(s)2

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(b) Use the Born-Haber cycle to calculate the lattice enthalpy of sodium oxide.

lattice enthalpy = ....................................kJ mol–1

[2]

(c) Which one of the following compounds has the most exothermic lattice enthalpy?

• calcium bromide • calcium chloride • potassium bromide • potassium chloride

Explain your answer in terms of the ions present.

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[4]

[Total 10 marks]

You will

know how

to do this

part after

covering

section 6,

‘Understan

ding

hydration

and

lattice

enthalpies

’

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This question focussed on lattice enthalpy and the Born-Haber cycle. Candidates from the whole of the ability range were able to access the questions and even the weakest candidates often scored at least four marks.

(a) Many candidates were able to complete the Born-Haber cycle using the appropriate enthalpy changes. A significant proportion only scored two marks since they did not double the first ionisation energy of sodium and the enthalpy change of atomisation of sodium.

(b) Candidates gained full marks for the correct lattice enthalpy of –2521 kJ mol–1 or for a value that was an error carried forward from their enthalpy changes in the Born-

Haber cycle in (a). The most frequent error that was carried forward was –1917 kJ mol-1. A common misconception was to use the negative value for the enthalpy change of formation i.e. +416 rather than –416. Good answers were exemplified by clear working out that allowed the Examiner to see any error carried forward.

(c) Most candidates selected calcium chloride in Q.1(c) and then gave an explanation based on the difference between anionic radius and cationic charge. Many candidates referred to the wrong particles or did not specify a type of particle at all in Q.1(c). Typical incorrect statements referred to the charge of calcium rather than the calcium ion or referred to the difference in the atomic radii rather than the ionic radii. The use of the wrong particle was only penalised once in this question. Common misconceptions included using the difference in electronegativity of the halogens and the reactivity of the metals to explain differences in lattice enthalpies. Other candidates referred to polarisation which was ignored.

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The table below shows the enthalpy changes needed to calculate the lattice

enthalpy of calcium chloride, CaCl2.

process enthalpy change / kJ mol–1

first ionisation energy of calcium +590

second ionisation energy of calcium +1150

electron affinity of chlorine –348

enthalpy change of formation for calcium chloride –796

enthalpy change of atomisation for calcium +178

enthalpy change of atomisation for chlorine +122

(a) The Born-Haber cycle below can be used to calculate the lattice enthalpy for calcium chloride.

(i) Use the table of enthalpy changes to complete the Born-Haber cycle by putting in the correct numerical values on the appropriate dotted line.

[3]

[Turn over]

Ca +2 (g) + 2Cl(g) + 2e

(g) + Cl (g) + 2e

Ca

2–

+(g) + Cl (g) + e

Ca(g) + C

2–

l (g)2

Ca(s) + Cl (g)2

CaCl (s)2

Ca +2 (g) + 2Cl (g)–

H = .................... kJ mol –1

H = .................... kJ mol –1

H = .................... kJ mol –1

H = .................... kJ mol –1

H = .................... kJ mol –1

lattice enthalpy ofcalcium chloride

H = .................... kJ mol –1

–

Ca2+

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(ii) Use the Born-Haber cycle to calculate the lattice enthalpy of calcium chloride.

answer ........................... kJ mol–1

[2]

(iii) Describe how, and explain why, the lattice enthalpy of magnesium fluoride differs from that of calcium chloride.

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[3]

(b) Explain why the first ionisation energy of calcium is less positive than the second ionisation energy.

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[2]

[Total 10 marks]

You will

know how

to do this

part after

covering

section 6,

‘Understan

ding

hydration

and

lattice

enthalpies

’

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This question focused on the Born-Haber cycle and lattice enthalpy.

(a) Many candidates could score three marks but it was rare for candidates to score all five. The most common error was the failure to use 2 × the enthalpy change of atomisation of chlorine and 2 x the electron affinity for chlorine. The correct answer was

–2262 kJ mol–1. A small but significant proportion of candidates gave a positive lattice enthalpy. In order to score full marks for part (iii), candidates had to refer to the difference in ionic radii or the charge density of both the cations and the anions. Many candidates referred to the difference in atomic radii and this was not given credit. Another misconception was that compounds had ionic radii. The final mark was awarded for the connection between the change in lattice enthalpy of the two compounds and the strength of the electrostatic attraction between the ions. A significant proportion of candidates did not appreciate that the lattice enthalpy of magnesium fluoride was more exothermic than that of calcium chloride and referred to a bigger or greater lattice enthalpy with no regard for sign.

(b) In this part, a small but significant proportion of the candidates referred to the gain of electrons rather than the loss of electrons from a calcium atom. A common misconception was that the nuclear charge increased as electrons were lost, rather than the nuclear charge remained constant but attracted a smaller number of electrons.

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5. Enthalpy change of solution


Energy Done

Use the enthalpy change of solution of a simple ionic solid and relevant enthalpy change terms to construct Born-Haber cycles.

Carry out relevant calculations.




o Q = mcΔT







enthalpy change of solution

lattice enthalpy.

Lattice energy is an exothermic process and the sign is –ve. The process involved in breaking down the ionic lattice is endothermic and the sign is +ve.

The enthalpy change of solution can be either +ve or –ve.

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6. Understanding hydration and lattice enthalpies


Energy Done

Explain, in qualitative terms, the effect of ionic charge and ionic radius on lattice enthalpy and enthalpy change of hydration.




o Q = mcΔT







enthalpy change of solution

enthalpy change of hydration

lattice enthalpy.

Remember that lattice enthalpy is the enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions.

e.g. K+(g) + Cl-(g) KCl(s) ΔH = -711 kJ mol-1

It is important to state that ions affect the magnitude of the lattice enthalpy and enthalpy change of hydration. Sloppy wording such as ‘chlorine’ and ‘bromine’ instead of chloride and bromide must be avoided at all costs.

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In this question, one mark is available for the quality of spelling, punctuation and grammar.

The lattice enthalpy of magnesium chloride, MgCl2, can be determined using a

Born-Haber cycle and the following enthalpy changes.

name of process enthalpy change / kJ mol–1

enthalpy change of formation of MgCl2(s) –641

enthalpy change of atomisation of magnesium +148

first ionisation energy of magnesium +738

second ionisation energy of magnesium +1451

enthalpy change of atomisation of chlorine +123

electron affinity of chlorine –349

• Define, using an equation with MgCl2 as an example, what is meant by

the term lattice enthalpy.

• Construct a Born-Haber cycle for MgCl2, including state symbols, and

calculate the lattice enthalpy of MgCl2.

• Explain why the lattice enthalpy of NaBr is much less exothermic than

that of MgCl2.

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[11]

Quality of Written Communication [1]

[Total 12 marks]

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This question focussed on the Born-Haber cycle and lattice enthalpy.

The question also included one mark for the quality of spelling, punctuation and grammar. To be awarded this mark, candidates had to write at least two sentences that addressed the question set and had no significant errors of spelling, punctuation and grammar. A significant proportion of candidates could not be awarded this mark because they did not write any sentences and just carried out the calculation and the Born-Haber cycle.

Although many candidates answered the questions in the order of the bullet points others concentrated on the qualitative parts first and left the Born-Haber cycle and the calculation until the end.

Many candidates did not write the correct equation associated with the lattice enthalpy of magnesium chloride. The most frequent error was the omission of state symbols. A majority of candidates gave the correct definition for lattice enthalpy although two common misconceptions were that it was the energy required and that lattice enthalpy involves the gaseous elements rather than the gaseous ions.

Most candidates drew Born-Haber cycles as energy level diagrams and labelled the species and energy changes. One mark was awarded for all the correct formulae in the cycle and one for the correct state symbols. Many candidates were awarded one mark out of the two available.

The correct value for the lattice enthalpy was –2526 kJ mol–1 but many candidates made errors because they did not double the enthalpy of atomisation of chlorine or the electron affinity of chlorine. An error carried forward mark was allowed for these errors.

Candidates tended to use the terms atom, ion and molecule as though they meant the same thing when trying to explain why the lattice enthalpy of MgCl2 was more

exothermic than that of NaBr. Another similar misconception was to refer to the charge density of NaBr or MgCl2. Candidates were only penalised once within the

question for such errors. Good answers compared the ionic charges and the ionic radii of the ions involved and then made a comment about the electrostatic attraction between the ions.

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7. Entropy


Energy Done

Explain that entropy is a measure of the disorder of a system, and that a system becomes energetically more stable when it becomes more disordered.

Explain: the difference in entropy of a solid and a gas; the change when a solid lattice dissolves; the change in a reaction in which there is a change in the number of gaseous molecules.

Calculate the entropy change for a reaction given the entropies of reactants and products.


You must be able to:

carry out simple entropy calculations

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8. Free energy


Energy Done

Explain that the tendency of a process to take place depends on the temperature, T, the entropy change in the system, ΔS, and the enthalpy change, ΔH, with the surroundings.

Explain that the balance between entropy and enthalpy change is the free energy change, ΔG.

State and use the relationship ΔG = ΔH – TΔS.

Explain how endothermic reactions are able to take place spontaneously.


You must be able to:

carry out simple entropy calculations

understand the meaning of the term free energy change, ΔG, and the equation ΔG = ΔH – TΔS

understand the relationship between entropy and enthalpy.

The free energy change, ΔG, is the net driving force that determines whether a chemical reaction will happen.

In calculations, you need to get the units the same.

ΔH is usually given in kJ mol-1

ΔS is usually given in J K-1 mol-1

First get ΔS into kJ K-1 mol-1:

To convert J to kJ, divide by 1000.

Also remember that entropy is worked out using temperature in K:

To convert oC to K, add 273.

To convert K to oC, subtract 273.

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Most metals can be extracted by reduction from compounds obtained from their naturally-occurring ores.

Metals such as calcium and magnesium are normally extracted by electrolysis but it is feasible that calcium oxide could be reduced by carbon as shown in the equation below.

CaO(s) + C(s) Ca(s) + CO(g)

Use the data in the table below to help you answer parts (i)–(iii) below.

CaO(s) C(s) Ca(s) CO(g)

Hfο /kJ mol–1 –635 0 0 –110

Sο/J K–1 mol–1 39.7 5.7 41.4 197.6

(i) Calculate the standard enthalpy change for the CaO reduction in the equation.

Hο = ............................................ kJ mol–1

[1]

(ii) Calculate the standard entropy change for the CaO reduction in the equation.

Sο = ......................................... J K–1 mol–1

[1]

[Turn over]

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(iii) Calculate the minimum temperature at which the carbon reduction in the equation is feasible.

minimum temperature = ...............................

[5]

[Total 7 marks]

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9. Redox


Energy Done

Explain the terms redox, oxidation number, half-reaction, oxidising agent and reducing agent for simple redox reactions.

Construct redox equations using relevant half-equations or oxidation numbers.

Interpret and make predictions for reactions involving electron transfer.


You will revisit the idea of redox – first learnt in AS Unit F321: Module 1 - Atoms and reactions. At A2 you should be able to use the terms:

redox

oxidation number

half-reaction

reducing agent

oxidising agent appropriately.

Never get in a muddle: use OILRIG

Oxidation Is Loss

Reduction Is Gain

If one species gains electrons (reduction), another species loses the same number of electrons (oxidation).

If one species decreases its oxidation number (reduction), then another species must increases its oxidation number (oxidation).

The total increase in oxidation number equals the total decrease in oxidation number.

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The carbonates and nitrates of Group 2 elements decompose when heated.

(a) Barium nitrate decomposes when heated to make barium oxide, nitrogen dioxide and oxygen.

2Ba(NO3)2(s) → 2BaO(s) + 4NO2(g) + O2(g)

(i) Use oxidation states to explain why this decomposition reaction involves both oxidation and reduction.

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[3]

(ii) Calculate the enthalpy change of reaction, ∆Hr, in kJ mol–1, for the

thermal decomposition of barium nitrate using the enthalpy changes of formation, ∆Hf, given in the table.

compound ∆Hf /kJ mol–1

Ba(NO3)2(s)

BaO(s)

NO2(g)

–992

–558

+33

answer ........................... kJ mol–1

[3]

[Turn over]

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(b) A student investigates the volume of gas formed when barium nitrate is heated.

The diagram shows the apparatus the student uses.

(i) A 1.31 g sample of barium nitrate is completely decomposed.

Use the equation above to calculate the volume, in cm3, of gas formed at room temperature and pressure.

1 mol of gas molecules occupies 24 000 cm3 at room temperature and pressure.

answer ......................... cm3

[3]

[Turn over]

barium nitrate

heat

100cm gas syringe3

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(ii) Suggest one problem that the student may encounter when carrying out the investigation.

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[1]

[Total 10 marks]


This question focussed on the thermal decomposition of carbonates and nitrates of Group 2 elements.

(a) In part (i) many candidates were unable to deduce the oxidation numbers of barium, nitrogen and oxygen in the substances given in the equation. In particular a ‘combined’ oxidation number, e.g. –12 for oxygen in Ba(NO3)2, was given. A common

misconception was to give barium different oxidation numbers in Ba(NO3)2 and BaO. The

majority of candidates could interpret oxidation and reduction in terms of changes in oxidation numbers and so obtained an error carried forward mark. Many candidates correctly calculated the enthalpy change of reaction in part (ii) as +1000 kJ; however a significant proportion gave the answer –1000 kJ, while others did not use the correct number of moles as shown in the equation. Many candidates did not show sufficient working out. A small proportion of the candidates just added the numbers up as shown in the table.

(b) A large proportion of candidates obtained the correct answer of 300 cm3 in part (i). Other candidates did not appreciate that they had to calculate the volume of both oxygen

and of nitrogen dioxide and gave an answer of 60 cm3 or 240 cm3. A common misconception was to use the Mr of Ba(NO3)2 as 522.

In part (ii) the majority of candidates realised that the syringe was too small.

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In order to obtain full marks in this question, you must show all your working clearly.

In its reactions, sulphuric acid, H2SO4, can behave as an acid, an oxidising

agent and as a dehydrating agent.

The displayed formula of pure sulphuric acid is shown below.

Concentrated sulphuric acid will readily oxidise halide ions to the halogen.

The equation below represents the unbalanced equation for the oxidation of iodide ions by sulphuric acid.

H+ + SO4

2– + I– → I2 + H2S + H2O

(i) Write the oxidation numbers of sulphur and iodine in the boxes above the equation.

[2]

(ii) Balance the equation above.

[1]

[Total 3 marks]


Most candidates were able to show correct oxidation numbers for iodine but those of sulphur proved to be more elusive. Only the very best candidates correctly constructed

the equation, the difficulty being the need for 4I2 and 8I– to account for all electrons.

Correct equations had usually been balanced using oxidation numbers.

O

O

S

O

O

H

H

33 | P a g e


NO2 reacts with oxygen and water to form nitric acid, HNO3. In the

atmosphere, this contributes to acid rain. Construct a balanced equation for this formation of nitric acid and use oxidation numbers to show that this is a redox reaction.

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[Total 2 marks]


Many candidates were able to construct a fully balanced equation for the formation of nitric acid although some candidates did not seem to realise that the species were all given in the opening sentence of the question. When it came to using oxidation numbers to show that it was a redox reaction, the answers were rather poor, certainly more so than in previous years. A large number of candidates were satisfied with showing only that nitrogen was oxidised from +4 to +5 and did not discuss the changes to oxygen. Of those, who did, the most common error was to assign an oxidation number of –6 to the oxygen in HNO3.

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10. Cells and half cells


Energy Done

Describe simple half cells made from

i. metals or non-metals in contact with their ions in aqueous solution;

ii. ions of the same element in different oxidation states.

Describe how half cells can be combined to make an electrochemical cell.



redox

oxidation number

half-reaction

reducing agent


A way to earn some marks is by knowing the definition of standard electrode potential.

You will need to describe how standard electrode potentials can be calculated against the standard hydrogen electrode – both for metals and ions or pairs of ions.

You will need to know how to use values of electrode potentials in order to predict if reactions will occur.

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11. Cell potentials


Energy Done

Define the term standard electrode (redox) potential, Eο.

Describe how to measure standard electrode potentials using a standard hydrogen half cell.

Calculate a standard cell potential by combining two standard electrode potentials.



redox

oxidation number

half-reaction

reducing agent





Remember that the particles that carry charge in the electric current are:

electrons in the wire

ions in the salt bridge.

The sign of the standard electrode potential of a half cell gives the sign (‘polarity’) of the electrode compared with the hydrogen half cell.

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Use the standard electrode potentials in the table below to answer the questions that follow.

I Fe2+(aq) + 2e– Fe(s) Eο = –0.44 V

II V3+(aq) + e– V2+(aq) Eο = –0.26 V

III 2H+(aq) + 2e– H2(g) Eο = 0.00 V

IV O2(g) + 4H+(aq) + 4e– 2H2O(l) Eο = +0.40 V

An electrochemical cell was set up based on systems I and II.

(i) Write half-equations to show what has been oxidised and what has been reduced in this cell.

oxidation:

reduction:

[2]

(ii) Determine the cell potential of this cell.

Ecell = ......................................................... V

[1]

[Total 3 marks]

38 | P a g e


The standard electrode potential of Cu2+(aq) + 2e– Cu(s) is +0.34 V.

(a) Define the term standard electrode potential.

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

[3]

(b) Complete the diagram to show how the standard electrode potential of

Cu2+(aq) + 2e– Cu(s) could be measured.

[3]

[Total 6 marks]

copper rod

Cu2+(aq)

39 | P a g e


(a) Many candidates know this definition. Quite a large number did not include the actual standard conditions in the definition but these were then used in the diagram in (b). This was credited.

(b) Diagrams of the cell were usually of a good standard. Occasionally candidates forgot to draw in the liquid level and a few electrodes were not in contact with liquid, particularly for the hydrogen half-cell.

40 | P a g e


The standard electrode potentials for some redox systems involving vanadium are shown below. These are labelled A, B, C and D.

Eο/ V

A VO2+ + 2H+ + e–

VO2+ + H2O +1.00

B V3+ + e– V

2+ –0.26

C V2+ + 2e– V –1.20

D VO2+ + 2H+ + e– V

3+ + H2O +0.34

(a) Which of the vanadium species shown in A, B, C and D is the most powerful oxidising agent?

....................................................................................................................

[1]

(b) A student wishes to set up a cell with a standard cell potential of 0.60V.

(i) Which two of the redox systems, A, B, C or D, should he choose?

...........................................................................................................

[1]

(ii) Complete the labelling of the following diagram which shows the cell with a standard cell potential of 0.60V.

[4]

V

41 | P a g e

(iii) The emf of this cell is only 0.60 V under standard conditions. What do you understand by the expression standard conditions?

...........................................................................................................

...........................................................................................................

...........................................................................................................

[1]

[Total 7 marks]


(a) A majority of candidates misinterpreted the requirement of this part of the question. The common answer given was ‘A’ when the actual species which was the most

powerful oxidising agent, VO2+, was required.

(b) (i) A majority of candidates correctly chose B and D.

(ii) A common error here was to miss off the H+ ion for D and to use vanadium electrodes instead of platinum.

(iii) Standard conditions are well known.

42 | P a g e


The standard electrode potential of the Cl2/ Cl– half-cell may be measured

using the following apparatus.

(a) Suggest suitable labels for A, B, C and D.

A ................................................................................................................

B ................................................................................................................

C ................................................................................................................

D ................................................................................................................

[2]

(b) The half cell reactions involved are shown below.

Cl2 + e– Cl– Eο = +1.36 V

H+ + e– H2 Eο = 0.00V

(i) Use an arrow to show the direction of flow of electrons in the diagram of the apparatus. Explain your answer.

...........................................................................................................

...........................................................................................................

[2]

[Turn over]

2

1

C

H (g)2

B

A

D

Cl (aq)–

salt bridge

2

1

2

1

43 | P a g e

(ii) The values of Eο are measured under standard conditions. What are the standard conditions?

...........................................................................................................

...........................................................................................................

...........................................................................................................

[2]

(c) The half cell reaction for ClO3–/ Cl2 is shown below.

ClO3– + 6H+ + 5e– Cl2 + 3H2O Eο = +1.47 V

What does this tell you about the oxidising ability of ClO3– compared with Cl2?

Explain your answer.

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

[2]

[Total 8 marks]


(a) Some candidates were not awarded marks for giving appropriate state symbols for

H+ and Cl2

(b) (i) Most candidates gave the correct direction for the flow of electrons, although it was necessary to show them flowing through the wire and not the salt bridge.

(ii) A surprising number of candidates omitted to state that the concentration of the

solution should be 1 mol dm–3.

(c) Many candidates answered this correctly by comparing two electrode potentials

2

1

2

1

44 | P a g e


Chlorine gas may be prepared in the laboratory by reacting hydrochloric acid with potassium manganate(VII). The following standard electrode potentials relate to this reaction.

Cl2 + e– Cl–

Eο = +1.36 V

MnO4– + 8H+ + 5e–

Mn2+ + 4H2O Eο = +1.52 V


....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

[3]

(b) Determine the standard cell potential for a cell constructed from these two redox systems.

[1]

[Total 4 marks]


(a) This was generally well answered, but many candidates failed to state that the

standard conditions are 1 atmosphere pressure, 298 K and 1 mol dm–3.

(b) A majority of candidates scored this mark. A few omitted the unit and a few gave the obvious wrong answer of +2.88 V. There were also a few responses of –0.16 V.

2

1

45 | P a g e

12. The feasibility of reactions


Energy Done

Predict, using standard cell potentials, the feasibility of reactions.

Consider the limitations of predictions made using standard cell potentials, in terms of kinetics and concentration.



redox

oxidation number

half-reaction

reducing agent





Notice that a reaction takes place between reactants taken from different sides of the two relevant half-equations.

The equilibrium with the more negative Eο value goes to the left.

46 | P a g e


Some standard electrode potentials are shown below.

Eο/V

Ag+ + e– Ag + 0.80

Cl2 + e– Cl– + 1.36

Cu2+ + 2e– Cu + 0.34

Fe3+ + e– Fe2+ + 0.77

I2 + e– I– + 0.54


....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

[3]

(b) The diagram below shows an incomplete cell consisting of Cu/Cu2+ and

Ag/Ag+ half-cells.

(i) Complete and label the diagram to show how the cell potential of this cell could be measured.

[2]

(ii) On the diagram, show the direction of electron flow in the circuit if a current was allowed.

[1]

[Turn over]

2

1

2

1

Cu(s)

Ag (aq)+

47 | P a g e

(iii) Calculate the standard cell potential.

standard cell potential = ……………………V

[1]

(iv) Write the overall cell reaction.

...........................................................................................................

[1]

(c) Chlorine will oxidise Fe2+ to Fe3+ but iodine will not. Explain why, using the electrode potential data.

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

[2]

[Total 10 marks]


(a) This should have been an easy start to the paper but many candidates simply used the words ‘electrode potential’ in the definition, rather than the expected ‘voltage’ or ‘emf’. Standard conditions were often not given.

(b) This was generally well answered but a few candidates missed out the salt bridge and a number had electrons flowing from silver to copper. On the point of electron flow, candidates should indicate this on the wire between the electrodes. Too many candidates put an arrow somewhere between the wire and the salt bridge.

(c) This is a difficult concept. Good candidates had no problem with this but weaker candidates tended to discuss larger or smaller electrode potentials without referring to oxidation or reduction. There were too many statements such as ‘iodine is oxidised by iron’. Clearly weaker students do not fully understand the concept of electrode potential.

48 | P a g e


Chlorine gas may be prepared in the laboratory by reacting hydrochloric acid with potassium manganate(VII). The following standard electrode potentials relate to this reaction.

Cl2 + e– Cl–

Eο = +1.36 V

MnO4– + 8H+ + 5e–

Mn2+ + 4H2O Eο = +1.52 V

(a) Use the half-equations above to:

(i) construct an ionic equation for the reaction between hydrochloric acid and potassium manganate(VII);

...........................................................................................................

...........................................................................................................

...........................................................................................................

[2]

(ii) determine the oxidation numbers of chlorine and manganese before and after the reaction has taken place;

...........................................................................................................

...........................................................................................................

...........................................................................................................

[2]

(iii) state what is oxidised and what is reduced in this reaction.

...........................................................................................................

...........................................................................................................

...........................................................................................................

[2]

(b) If potassium manganate(VII) and very dilute hydrochloric acid are mixed, there is no visible reaction. Suggest why there is no visible reaction in this case.

....................................................................................................................

....................................................................................................................

[1]

[Total 7 marks]

2

1

49 | P a g e


(a) (i) More able candidates scored well here. Many weaker candidates showed chlorine on the left hand side and chloride on the right hand side. A few candidates left electrons in the final equation and were unable to correctly balance the equation.

(ii) This was generally well done. Candidates who placed chlorine on the left-hand side were awarded a mark here, carrying forward their error from part (i).

(iii) Many candidates failed to score here because they did not specify the actual species that was oxidised or reduced, often merely stating ‘chlorine is oxidised’ and ‘manganese is reduced’

(b) Many candidates commented that the concentration was too low without linking this to a slow reaction. A few suggested that the reaction would be slow owing to a small cell potential. Very few suggested that the reaction was no longer possible because conditions were not standard.

50 | P a g e

13. Storage and fuel cells


Energy Done

Apply the principles of electrode potentials to modern storage cells.

Explain that a fuel cell uses the energy from the reaction of a fuel with oxygen to create a voltage.

Explain the changes that take place at each electrode in a hydrogen-oxygen fuel cell.



redox

oxidation number

half-reaction

reducing agent





It is important to know the changes that can happen within fuel cells – e.g. in hydrogen-oxygen fuel cells.

You must also be aware of any economic, political and environmental issues.

In exams, you will not be expected to recall a specific storage cell. However, you might be expected to make predictions about a given cell. All relevant electrode potentials and other data will be supplied.

51 | P a g e

14. Hydrogen for the future


Energy Done

Outline the development of fuel cell vehicles (FCVs) that use hydrogen gas and hydrogen-rich fuels.

State the advantages of FCVs over conventional petrol- or diesel-powered vehicles.

Understand how hydrogen might be stored in FCVs.

Consider limitations of hydrogen fuels cells.

Comment about the contribution of the ‘hydrogen economy’ to future energy and discuss its limitations.



redox

oxidation number

half-reaction

reducing agent





It is important to know the changes that can happen within fuel cells – e.g. in hydrogen-oxygen fuel cells.

You must also be aware of any economic, political and environmental issues.

52 | P a g e


Use the standard electrode potentials in the table below to answer the questions that follow.

I Fe2+(aq) + 2e– Fe(s) Eο = –0.44 V

II V3+(aq) + e– V2+(aq) Eο = –0.26 V

III 2H+(aq) + 2e– H2(g) Eο = 0.00 V

IV O2(g) + 4H+(aq) + 4e– 2H2O(l) Eο = +0.40 V

An electrochemical fuel cell was set up based on systems III and IV.

(i) Construct an equation for the spontaneous cell reaction. Show your working.

[2]

(ii) Fuels cells based on systems such as III and IV are increasingly being used to generate energy.

Discuss two advantages and two disadvantages of using fuels cells for energy rather than using fossil fuels.

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

[4]

[Total 6 marks]

53 | P a g e

Sample student answers

A2 Unit F325: Equilibria, energetics and elements

Module 5: Energy

Question 1

Total marks: 15

(a) What is meant by the term lattice enthalpy?

Marks available: 2

Student answer:

(a) The enthalpy change when 1 mole of solid ionic compound is formed from its constituent ions in the gaseous state

Examiner comments:

(a) The change is exothermic, so don’t start off by writing ‘The energy required...’.

54 | P a g e

(b) Construct a Born–Haber cycle for magnesium chloride. Insert all the values from the information given below.

First ionisation energy of magnesium = +736 kJ mol–1

.

Second ionisation energy of magnesium = +1452 kJ mol–1

.

First electron affinity of chlorine = –349 kJ mol–1

.

Enthalpy of atomisation of magnesium = +148 kJ mol–1

.

Enthalpy of atomisation of chlorine = +122 kJ mol–1

.

Enthalpy of formation of magnesium chloride = –641 kJ mol–1

.

Marks available: 6

Student answer:

(b)

Examiner comments:

(b) Here are a few tips:

All the arrows must be in the correct direction.

As well as using minus (–) signs in front of exothermic values, it is a good idea to insert plus (+) signs in front of endothermic changes.

Check if multiples are needed such as with the Cl changes here.

55 | P a g e

(c) Calculate the lattice enthalpy of magnesium chloride.

Marks available: 2

Student answer:

(c) All values used are in kJ mol–1

: –641 = +148 + 736 + 1452 + (2 × +122) + (2 × –349) + LE LE = –641 – [+148 + 736 + 1452 + (2 × +122) + (2 × –349)]

= –2523 kJ mol–1

Examiner comments:

(c) Always show your working. Check the final answer by re-doing the calculation step by step, for example:

replace 2 × 122 by 244

calculate the value within the square brackets before doing a final subtraction.

(d) Define the term enthalpy of hydration.

Marks available: 2

Student answer:

(d) Enthalpy change when 1 mole of gaseous ions is dissolved in excess water

Examiner comments:

(d) As with most definitions, the term 1 mole will usually score a mark.

(e) Explain why 2.0 g of magnesium sulfate will dissolve in a test tube of water but 2.0 g of barium sulfate will not.

Marks available: 3

Student answer:

(e) Mg2+

ions are smaller than Ba2+

ions. This means they have a higher charge density. Although the lattice enthalpy of MgSO4 is more exothermic than BaSO4, meaning the solid lattice is held together more tightly, the enthalpy of hydration of Mg

2+ ions is even more

exothermic than Ba2+

ions. The increase in the magnitude of hydration is greater than the increase in magnitude of lattice enthalpy.

Examiner comments:

(e) Make sure you describe the correct particle, e.g. not Mg atoms or simply magnesium but Mg

2+ ions.

Try to use the phrase more exothermic lattice enthalpy rather than higher lattice enthalpy – this is because lattice enthalpy is always a negative value.

56 | P a g e

Module 5: Energy

Question 2

Total marks: 15

(a) Sulfuroxychloride and water react as shown by the equation given below: SOCl2(l) + H2O(l) SO2(g) + 2HCl(g) (i) Calculate the entropy change for the reaction:

ΔS SOCl2(l) = 218 J K–1

mol–1

ΔS H2O(l) = 70 J K–1

mol–1

ΔS SO2(g) = 248 J K–1

mol–1

ΔS HCl(g) = 187 J K–1

mol–1

(ii) The enthalpy change for the reaction is +47.1 kJ mol–1

which means it is endothermic.

Predict if this reaction is spontaneous at room temperature, 298 K.

Marks available: (i) 3 (ii) 3

Student answer:

(a) (i) [248 + (2 × 187)] – [218 + 70] = +334 J K–1

mol–1

(ii) ΔG = ΔH – TΔS

= +47100 – (298 × +334) = –52432 J mol–1

= –52.4 kJ mol–1

ΔG < 0 Therefore reaction will take place.

Examiner comments:

(a) (i) Showing your working is always useful since marks can still be allocated for the final answer, as allowances can be made for any mistakes as errors carried forward (ecf).

(ii) It is important to convert ΔH from kJ mol–1

to J mol–1

to match the units of ΔS. However,

the answer has to be written in the correct units: kJ mol–1

.

57 | P a g e

(b) Draw a diagram to show the standard hydrogen electrode and explain how you could use the standard hydrogen electrode to measure the standard electrode potential of Fe

3+(aq) + e

– Fe

2+(aq).

Marks available: 6

Student answer:

(b)

Examiner comments:

(b) In drawing this diagram marks awarded for:

Hydrogen electrode labelled.

Fe2+

/Fe3+

half-cell labelled.

Salt bridge and voltmeter labelled.

Inert platinum electrodes labelled.

Hydrogen at 1 atm.

Fe2+

, Fe3+

and H+ all at 1.00 mol dm

–3.

58 | P a g e

(c) A student believed aluminium would react with aqueous nickel ions after consulting standard electrode potentials. Demonstrate why he came to this conclusion and suggest one reason why he may be proved wrong. Al

3+ + 3e

– Al E = -1.67V

Ni2+

+ 2e– Ni E = -0.25V

Marks available: 3

Student answer:

(c) The Al3+

/Al equilibrium has the more negative (less positive) E value and so the equilibrium will move to the left and Al will react to form Al

3+ ions, supplying electrons as it

does so.

The Ni2+

/Ni equilibrium has the more positive (less negative) E value and so the equilibrium will move to the right and Ni

2+ ions will react to form Ni atoms, accepting

electrons as it does so. Thus the overall reaction should be 2Al + 3Ni

2+ → 2Al

3+ + 3Ni so the reaction is feasible.

Examiner comments:

(c) There are other approaches to working this out, but it is always better to try to use what we refer to as ‘first principles’.

59 | P a g e

The Periodic Table of the Elements

Th

e P

eri

od

ic T

ab

le o

f th

e E

lem

en

ts

1

2

3

4

5

6

7

0

Ke

y

1.0

H

h

yd

roge

n

1

4.0

H

e

heliu

m

2

6.9

L

i lit

hiu

m

3

9.0

B

e

bery

lliu

m

4

rela

tive a

tom

ic m

ass

ato

mic

sym

bo

l n

am

e

ato

mic

(pro

ton)

num

ber

10.8

B

b

oro

n

5

12.0

C

ca

rbo

n

6

14.0

N

nitro

gen

7

16.0

O

oxygen

8

19.0

F

fluo

rin

e

9

20.2

N

e

neon

10

23.0

N

a

sodiu

m

11

24.3

M

g

ma

gnesiu

m

12

27

.0

Al

alu

min

ium

13

28.1

S

isili

co

n

14

31.0

P

p

hosp

horu

s

15

32.1

S

sulfur

16

35.5

C

l ch

lorine

17

39.9

A

rarg

on

18

39.1

K

pota

ssiu

m

19

40.1

C

a

calc

ium

20

45.0

S

c

sca

nd

ium

21

47.9

T

i

tita

niu

m

22

50.9

V

vana

diu

m

23

52.0

C

rchro

miu

m

24

54

.9

Mn

ma

ngane

se

25

55

.8

Fe

iro

n

26

58.9

C

ocoba

lt

27

58.7

N

inic

kel

28

63

.5

Cu

co

ppe

r

29

65.4

Z

n

zin

c

30

69

.7

Ga

ga

llium

31

72.6

G

ege

rma

niu

m

32

74.9

A

sars

enic

33

79.0

S

esele

niu

m

34

79.9

B

rb

rom

ine

35

83.8

K

rkry

pto

n

36

85.5

R

b

rubid

ium

37

87.6

S

r str

ontium

38

88.9

Y

ytt

rium

39

91.2

Z

r

zirco

niu

m

40

92.9

N

b

nio

biu

m

41

95.9

M

om

oly

bdenum

42

[98]

Tc

tech

ne

tiu

m

43

101

.1

Ru

ru

the

niu

m

44

102

.9

Rh

rhod

ium

45

10

6.4

P

d

palla

diu

m

46

10

7.9

A

g

silv

er

47

112.4

C

d

cad

miu

m

48

114

.8

Inin

diu

m

49

118

.7

Sn

tin

50

121.8

S

ban

tim

on

y

51

127

.6

Te

tellu

rium

52

126

.9

I io

din

e

53

131.3

X

exen

on

54

13

2.9

C

s

cae

siu

m

55

137

.3

Ba

b

arium

56

138

.9

La*

lan

thanum

57

178.5

H

f

ha

fniu

m

72

18

0.9

Ta

tan

talu

m

73

18

3.8

W

tu

ngste

n

74

186.2

R

e

rheniu

m

75

190

.2

Os

osm

ium

76

192

.2

Ir

irid

ium

77

19

5.1

P

t

pla

tinum

78

19

7.0

A

u

gold

79

20

0.6

H

g

merc

ury

80

20

4.4

T

l th

alli

um

81

20

7.2

P

ble

ad

82

209.0

B

ib

ism

uth

83

[209

] P

op

olo

niu

m

84

[210

] A

ta

sta

tine

85

[222

] R

nra

do

n

86

[22

3]

Fr

franciu

m

87

[226

] R

a

rad

ium

88

[227

] A

c*

actiniu

m

89

[261]

Rf

ruth

erf

ord

ium

104

[262]

Db

d

ubniu

m

105

[26

6]

Sg

seab

org

ium

10

6

[264]

Bh

b

oh

rium

107

[27

7]

Hs

ha

ssiu

m

10

8

[26

8]

Mt

me

itne

rium

10

9

[271]

Ds

darm

sta

dtium

11

0

[272

] R

g

roen

tge

niu

m

111

Ele

men

ts w

ith

ato

mic

nu

mbe

rs 1

12–

116 h

ave b

een

rep

ort

ed b

ut n

ot fu

lly

au

then

ticate

d

140.1

C

e

ce

rium

58

14

0.9

P

r pra

seo

dym

ium

59

14

4.2

N

dne

odym

ium

60

144.9

P

mpro

meth

ium

61

150

.4

Sm

sam

arium

62

152

.0

Eu

euro

piu

m

63

15

7.2

G

dga

do

liniu

m

64

15

8.9

T

b

terb

ium

65

16

2.5

D

ydysp

rosiu

m

66

16

4.9

H

oh

olm

ium

67

16

7.3

E

re

rbiu

m

68

168.9

T

mth

uliu

m

69

173

.0

Yb

ytt

erb

ium

70

175

.0

Lu

lute

tium

71

232.0

T

h

tho

rium

90

[231]

Pa

pro

tactiniu

m

91

23

8.1

U

ura

niu

m

92

[237]

Np

nep

tun

ium

93

[24

2]

Pu

plu

toniu

m

94

[24

3]

Am

am

ericiu

m

95

[247]

Cm

curi

um

96

[245

] B

k

be

rkeliu

m

97

[251]

Cf

ca

liforn

ium

98

[25

4]

Es

ein

ste

iniu

m

99

[253]

Fm

ferm

ium

100

[256

] M

dm

ende

levium

101

[254

] N

on

ob

eliu

m

102

[257

] L

rla

wre

nciu

m

103

• • • • • •

Unit 2 Equilibria, energetics and elements (F325) Energy UNIT 2 Module 2... · 1 | P a g e Unit 2 Equilibria, energetics and elements (F325) Energy Topics covered in this module:

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