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RAFFLES INSTITUTION Year 5 H2 CHEMISTRY 2011

Tutorial 6a – Chemical Energetics I

Prepared by: Mrs Joan Siaw

Self-Check Questions

1.a) By writing relevant thermochemical equations, define the following terms.

(i) standard enthalpy change of reaction

(ii) standard enthalpy change of formation of CH3OH(l)

(iii) standard enthalpy change of combustion of C3H8(g)

(iv) standard enthalpy change of neutralisation

(v) standard enthalpy change of atomization of fluorine

(vi) standard enthalpy change of solution of MgCl2(s)

(vii) standard enthalpy change of hydration of Ca2+(g)

(viii) bond energy of the C−O bond

(ix) bond dissociation energy

(x) first Ionisation energy

(xi) second Ionisation energy

(xii) first electron affinity

(xiii) second electron affinity of oxygen

(xiv) lattice energy of Na2O(s)

b) The enthalpy change of combustion of butane, C4H8, is −3000 kJ mol−1. Calculate the mass of

water at 20 °C that could be brought to the boiling point by burning this 1.2 dm3 of gaseous butane at room temperature and pressure. Assume that 80% of the heat from combustion of butane was absorbed by the water.

c) Nitrogen and steam undergo a reaction to form ammonia and oxygen as given.

2

1N2(g) +

2

3H2O(g) → NH3(g) +

4

3O2(g)

Determine the enthalpy change for the above reaction using appropriate bond energy information from the Data Booklet.

d)

(i)

(ii)

(iii)

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Practice Questions

2. a) Define Hess’ Law and explain why it plays such an important role in thermochemistry.

b) Use the information given below to construct an energy cycle and calculate the standard

enthalpy change of formation ∆Hfθ of propane. [−104 kJ mol−1]

c) 5.60 dm3 of a mixture of propane and butane (measured at s.t.p) on complete combustion evolved 654 kJ of heat. Calculate the percentage composition of the mixture by volume.

[% comp of propane = 39.7 %; % comp of butane = 60.3 % ]

Given : ∆Hcθ [C3H8(g)] = −2220 kJ mol−1, ∆Hf

θ [H2O(l)] = −285.9 kJ mol−1, ∆Hfθ [CO2(g)] = −393.5

kJ mol−1 and ∆Hcθ [C4H10 (g)] = −2877 kJ mol−1.

3. a) When 2.76 g of potassium carbonate was added to 30.0 cm3 of approximately 2 mol dm−3 hydrochloric acid, the temperature rose by 5.20 oC. (i) Write an equation for this reaction.

(ii) Calculate the enthalpy change for this reaction per mole of potassium carbonate. Assume

that the specific heat capacity and density of all solutions are 4.18 J g−1 K−1 and 1.00 g cm−3

respectively. [− 32.7 kJ mol−1]

(iii) Explain why the hydrochloric acid need only be approximately 2 mol dm−3.

b) When 2.00 g of potassium hydrogencarbonate was added to 30.0 cm3 of the same hydrochloric acid, the temperature fell by 3.70 oC. (i) Write an equation for this reaction.

(ii) Calculate the enthalpy change for this reaction per mole of potassium hydrogencarbonate.

[+23.2 kJ mol−1]

c) When potassium hydrogencarbonate is heated, it decomposes into potassium carbonate, carbon dioxide and water. By applying Hess’ Law and your results in (a) and (b), calculate the enthalpy

change for the decomposition of potassium hydrogencarbonate. [+39.6 kJ mol−1]

4. To determine the enthalpy change of neutralisation of sodium hydroxide with sulfuric acid, 50.0 cm3

of 0.400 mol dm−3 sodium hydroxide was titrated thermometrically with 0.500 mol dm−3 sulfuric acid. The results of the titration were given below:

Vol of H2SO4 used / cm3 6.00 12.00 18.00 24.00 30.00

Temperature rise / oC 1.40 2.60 3.50 3.60 3.30

a) Define the term enthalpy change of neutralisation.

b) Plot a graph of temperature rise (oC) against volume of acid used (cm3). Account for the shape of the graph.

c) Calculate the enthalpy change of neutralisation of NaOH with sulfuric acid. State the

assumptions made in your calculations. [−56.3 kJ mol−1]

d) A similar experiment carried out with hydrogen cyanide, HCN(aq), and aqueous ammonia gave a

value of −5.4 kJ mol−1 for the enthalpy change of neutralisation. The corresponding value for

sodium hydroxide with hydrochloric acid was −57.0 kJ mol−1. Suggest a possible explanation for the difference in these two values.

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5. a) By referring to CH4, explain what you understand by the term ‘bond energy’.

b) The standard heat change of formation of the following four hydrocarbons are given below:

Hydrocarbon C2H6 (g) C2H4 (g) C2H2 (g) C6H6 (l)

∆Hfθ / kJ mol−1 −84.7 +52.3 +227 +82.9

Using bond energy data from the data booklet and given that the standard enthalpy change of

atomisation of graphite is +715 kJ mol−1 and the standard enthalpy change of vaporization of

benzene is +34.7 kJ mol−1, calculate the C−C bond energies of the four hydrocarbons and comment on their values.

[BE(C−C) for C2H6 = 363 kJ mol−1; C2H4 = 610 kJ mol−1; C2H2 = 819 kJ mol−1;

C6H6 = 503 kJ mol−1]

6. One of the most important uses of alkanes is for fuels. In some countries, where crude oil is either scarce or expensive, biofuels such as ethanol are increasingly being used for fuels instead of hydrocarbons. a) Use the bond energies given in Data Booklet to calculate a value for the enthalpy change of

combustion of octane, C8H18. [−4090 kJ mol−1] b) The accurate experimental enthalpy changes of combustion of three hydrocarbons are given in

the table below.

Alkane Formula ∆∆∆∆Hc/kJ mol−1

Heptane C7H16 −4817

Octane C8H18 −5470 Nonane C9H20 −6125

(i) Suggest a reason for the discrepancy between ∆Hc for octane you calculated in (a) and that given in the table.

(ii) Suggest what the regular increase in the values of ∆Hc given in the table represents.

c) The enthalpy change of combustion of enthanol is −1367 kJmol−1 and the densities of enthanol and octane are 0.79 g cm−3 and 0.70 g cm−3 respectively. Calculate the heat produced by the

complete combustion of 1.0 dm3 of each fuel. [2.35 × 104 kJ and 3.36 × 104 kJ] N2005/III/3 or

7. a) Using the data provided, construct a Born−Haber cycle for magnesium chloride, MgCl2, and from

it determine the first electron affinity of chlorine. [1st EA of Cl = −348 kJmol−1]

∆H / kJ mol−1 Enthalpy change of atomisation of chlorine +122 Enthalpy change of atomisation of magnesium +148 First ionisation energy of magnesium +738 Second ionisation energy of magnesium +1451

Lattice energy of magnesium chloride −2526

Enthalpy change of formation of magnesium chloride −641

b) The theoretically calculated value for the lattice energy of magnesium chloride is –2326 kJ mol−1. Explain the difference between the theoretically calculated value and the experimental value given in the data in (a), in terms of bonding of magnesium chloride.

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8. a) Describe, in terms of the energetics involved, the process of dissolution of an ionic solid in water.

b) Given that the lattice energy of lithium chloride is −843 kJ mol−1 and its hydration energy is

−883 kJ mol−1, draw an energy level diagram and use it to calculate the enthalpy change of

solution of lithium chloride. [−40 kJ mol−1] Specific heat capacity of water = 4.20 J g−1 K−1

c) 50.0 cm3 of water was placed in a thin, well-lagged calorimeter of negligible heat capacity. The

temperature of the water was 18.2 oC. 3.00 g of potassium chloride, KCl, was added and the

mixture well-stirred until all the crystals had dissolved. The minimum temperature reached was

14.9 oC. Calculate the enthalpy change of solution of potassium chloride. [+17.2 kJ mol−1] d) Comment on the results obtained in (b) and (c).

9. Hydrazine is used as rocket fuel and to prepare gas precursors used in air bags. Approximately 260

thousand tonnes of hydrazine are manufactured annually.

Liquid hydrazine undergoes combustion according to the following equation:

N2H4(l) + O2(g) → N2(g) + 2H2O(l)

A chemist conducted an experiment to determine the standard enthalpy change of combustion of hydrazine. In the experiment, 0.210 g of hydrazine was burnt as fuel to heat up a beaker containing 200 cm3 of water. The temperature of water rose by 4 °C. You may assume the process has 80 % efficiency.

a) Calculate the standard enthalpy change of combustion of hydrazine.[−6.37 x 102 kJ mol-1]

b) Given the following data: Enthalpy change of formation of steam = −242 kJ mol-1 Enthalpy change of vapourisation of water = +44 kJ mol-1

and using the value you have calculated in (a), draw an appropriate energy cycle to determine the standard enthalpy of formation of hydrazine. [+ 65 kJmol-1]

c) The standard enthalpy change of formation of hydrazine gas is +235 kJ mol-1. i) Using appropriate data from the Data Booklet, draw an energy level diagram to calculate the

average bond energy of N−H bond in hydrazine. ii) Suggest a reason for the difference in the N-H bond energy value obtained from (c)(i) with

the value given in the Data Booklet. [368 kJ mol-1] MJC08 Prelim/III

Challenging Questions (Optional)

10. a) Using the information below and the Data Booklet, construct a Born−Haber diagram and calculate

the lattice energy of Cu(I) oxide (Cu2O). [−3232 kJ mol−1]

Energy change / kJ mol−1

Atomisation energy of Cu 339 First electron affinity of O − 141

Second electron affinity of O 791 Heat of formation of Cu(I) oxide − 166

b) The lattice energy of Cu(II) oxide is − 4143 kJ mol−1. Explain why this value is different from the

one calculated in (a).

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c) Using the information in (a) and (b), draw the appropriate energy cycle and calculate the enthalpy change of reaction for:

Cu2O(s) � CuO(s) + Cu(s)

Based on the value obtained, predict whether the reaction will proceed under standard conditions.

[−35 kJ mol−1] d) Consider the following reaction:

Cu2O(s) + H2SO4(aq) � CuSO4(aq) + Cu(s) + H2O(l)

When 2.86 g of solid Cu2O is added to 60.0 cm3 of dilute sulfuric acid of approximately 2 mol dm-3, the temperature of the solution was raised by 8.9 K. Calculate the enthalpy change

for the above reaction, stating the assumptions made. [−112 kJ mol−1] e) Hence, calculate the enthalpy change of neutralisation for the reaction between H2SO4(aq) and

CuO(s). [−77 kJ mol−1]

Suggested Answers to Self−−−−Check Questions 1a) Refer to lecture notes.

Please give specific definition when asked for the enthalpy change for a specific substance. Example :

(ii) Standard enthalpy change of formation of CH3OH(l) is the energy change when 1 mole of CH3OH(l) is formed from its constituents elements, C(s), O2(g) and H2(g) at 298 K and 1 atm.

C(s) + ½ O2(g) + 2H2(g) ���� CH3OH(l) ∆∆∆∆Hfθθθθ [CH3OH(l)]

b) No. of moles of butane = 1.2/24 = 0.0500

Heat released by reaction = ∆Hc(butane) x nbutane = 3000x103x0.0500 = 150000 J Heat absorbed by water = (80/100) x heat released by reaction (80% efficiency) = 80/100 x 150000 = 120000 J

120000 = m x 4.18 x (100 – 20)

Mass of water heated, m = 359 g

c) 2

1N2(g) +

2

3H2O(g) → NH3(g) +

4

3O2(g)

∆Hr = [2

1BE(N≡N) + 3BE(O-H)] – [3BE(N-H) +

4

3BE(O=O)]

= [2

1(994) + 3(460)] – [3(390) +

4

3(496)] = +335 kJ mol−−−−1

d) (i) C, (ii) B, (iii) D

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Suggested Solutions to Tutorial 6a: Chemical Energetics I 1a) Refer to lecture notes.

Please give specific definition when asked for the enthalpy change for a specific substance. Example :

(ii) Standard enthalpy change of formation of CH3OH(l) is the enthalpy change when 1 mole of CH3OH(l) is formed from its constituents elements, C(s), O2(g) and H2(g) at 298 K and 1 atm.

C(s) + ½ O2(g) + 2H2(g) ���� CH3OH(l) ∆∆∆∆Hfθθθθ [CH3OH(l)]

b) No. of moles of butane = 1.2/24 = 0.05

Heat released by reaction = ∆Hc(butane) x nbutane = 3000x103x0.05 = 150000 J Heat absorbed by water = (80/100) x heat released by reaction (80% efficiency) = 80/100 x 150000 = 120000 J

120000 = m x 4.18 x (100 – 20)

Mass of water heated, m = 359 g

c) 2

1N2(g) +

2

3H2O(g) → NH3(g) +

4

3O2(g)

∆Hrxn = [2

1BE(N≡N) + 3BE(O-H)] – [3BE(N-H) +

4

3BE(O=O)]

= [2

1(994) + 3(460)] – [3(390) +

4

3(496)] = 335 kJ mol−−−−1

d) (i) C, (ii) B, (iii) D

2a) Hess’ Law states that the enthalpy change of a reaction is determined by the initial and final

states of the system and is independent of the pathways taken. Beauty of it is allowing one to

calculate ∆Hr of a reaction that is difficult or impossible to perform.

b) ∆Hfo

3C(s) + 4H2(g) C3H8(g)

3∆Hfo[CO2(g)] 4∆Hf

o[H2O(l)]

∆Hco[C3H8(g)]

3 CO2(g) + 4 H2O (l)

∆Hfo [C3H8(g)] = 3 ∆Hf

o [CO2(g)] + 4 ∆Hfo [H2O(l)] – ∆Hc

o [C3H8(g)]

= −−−−104 kJ mol−−−−1 (to 3 sf)

c) Enthalpy, ∆H = heat change per mol

Actual heat change = ∆H x n, where n is the no of mol of compound burnt (reactant) Total amt = 5.6 / 22.4 = 0.25 mol Let the amt of propane be n mol, then amt of butane is (0.25 – n) mol.

Total heat change = {n.∆Hco [C3H8(g)] + (0.25-n).∆Hco [C4H8(g)]} = −654 J

% composition of propane = 39.7 % (to 3 sf) % composition of butane = 60.3 % (to 3 sf)

+ 5 O2 (g) + 3 O2 (g) + 2 O2 (g)

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3a) (i) K2CO3(s) + 2HCl(aq) → 2KCl(aq) + CO2(g) + H2O(l)

(ii) Heat released = mc∆T = 30.0 x 4.18 x 5.20 = 652.1 J Mr of K2CO3 = 2(39.1) + 12.0 + 3(16.0) = 138.2 Amt of K2CO3 = 2.76 / Mr =2.76 / 138.2 = 0.01997 mol

Amt of HCl = 30/1000 x 2 = 0.06 mol; Hence, HCl is in excess.

Enthalpy change of reaction, ∆H1 = −−−−652.1/0.01997

=−32650 J mol−1 (to 4 sf) = −−−−32.7 kJ mol−−−−1 (to 3 sf)

(iii) The only requirement is that the H+ be in excess. A metal carbonate reacts effectively with a dilute acid.

b) (i) KHCO3(s) + HCl(aq) → KCl(aq) + CO2(g) + H2O(l)

(ii) Heat absorbed = 30.0 x 4.18 x 3.70 = 464.0 J Mr of KHCO3 = 39.1 + 1.0 + 12.0 + 3(16.0) = 100.1

Amt of KHCO3 = 2.00 / Mr = 0.01998 mol Amt of HCl = 30 / 1000 x 2 = 0.06 mol; Hence, HCl is in excess.

Enthalpy change of reaction, ∆H2 = +464.0 /0.01998

= +23220 J mol−1 (to 4 sf) = +23.2 kJ mol−−−−1 (to 3 sf) c) 2KHCO3(s) K2CO3(s) + H2O (l) + CO2(g) heat +2HCl (aq) +2HCl (aq)

2∆H2 ∆H1 2KCl (aq) + 2H2O(l) + 2CO2(g)

By Hess’ Law,

2 ∆Hr = 2∆H2 − ∆H1 = +79.1 kJ mol−1

∴Heat change for the decomposition of 1 mole KHCO3 = +39.6 kJ mol−−−−1 (to 3 sf)

4a) Enthalpy change of neutralisation is the enthalpy change when an amount of acid or alkali is neutralised to form 1 mole of water, the reaction being carried out using dilute aqueous solutions.

b) Temp (20.0,3.85) rise/oC Volume of acid /cm3

• Initially, temperature rose as heat of neutralisation is released.

• Maximum temperature rise signified end of the neutralisation reaction.

• Further addition of acid resulted in drop in temperature as addition of room temperature acid to the salt solution cooled it down.

c) Heat released = (50.0 + 20.0) x 4.18 x 3.85 = 1126.5 J

∆Hneut = −1126.5 / (0.0500 x 0.400) = −56.3 kJ mol−1 (to 3 sf)

Assumptions: 1. No heat loss to surroundings

2. Density of solution ≅ density of water 3. Specific heat capacity of soln. approximates that of water.

2∆Hr

Points plotted traces out a curve. But we extrapolate straight lines to find the maximum temperature that can be reached (heat is lost to the surroundings through the

heating and cooling portions of the heat change)

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d) HCN is a weak acid and energy is required to ionise it to produce one mole of H+ ions before neutralisation. NH3 is a weak base and energy is required to ionise it to produce one mole of OH- ions before neutralisation can take place. Heat of ionisation of the weak acid is endothermic and

that resulted in a less exothermic heat of neutralisation of the weak acid−weak base as

compared to that of a strong acid−strong base. 5a) The bond energy of a C-H bond is the average energy absorbed when 1 mole of C-H bonds of

methane are broken. The successive breaking of each of the 4 C-H bonds has a different bond dissociation energy.

Thus CH4(g) → C(g) + 4H(g) ; ∆H /4 = bond energy BE(C-C) + 6BE(C-H) b) C2H6(g) 2C(g) + 6H(g)

∆Hfo

= −84.7 kJ mol−1 2 ∆Hatmo [C (g)]

+ 3 BE(H-H)

2C(s) + 3H2(g)

∆Hfθ = energy consumed in bond breaking + energy released in bond formation

−84.7 = 2 ∆atmo [C(s)] + 3 BE(H−H) − {BE(C−C) + 6 BE(C−H)}

BE(C−C) = + 363 kJ mol−−−−1 (to 3 sf)

Hydrocarbon C2H6 C2H4 C2H2 C6H6

BE(C-C) / kJmol-1 363 610 819 503

Note: Regardless of the multiplicity of the bond, the BE(C-C) should not be divided by the bond

order as the bond energy of the σ and π-bonds are different. Comments:

• Data booklet: BE[H-H] = +436 kJ mol−1 and BE[C-H] = +410 kJ mol−1

• For calculation involving benzene, C6H6(l) needs to be vaporized. Enthalpy change of

vaporization is +34.7 kJ mol−−−−1

• Bond strength: C≡C > C=C > C-C

• C-C bond in benzene is intermediate in bond length and strength between C=C and C-C bonds due to resonance.

• Bond energy of C=C is not twice that of C-C ⇒ π-bond is weaker than a σ-bond.

6a) C8H18 + 25/2 O2 → 8CO2 + 9H2O

∆Hc [C8H18] = Σ[bond energy broken in reactants] − Σ[bond energy formed in products]

= [7BE(C−C) + 18BE(C−H) + 25/2BE(O=O)] − [8BE(C=O) + 18BE(H−O)]

= (7 x 350 + 18 x 410 + 25/2 x 496) − (16 x 740 + 18 x 460)

= −4090 kJ mol−1

b)(i) ∆Hc assumes that all substances are in their standard states eg H2O formed is in the liquid state at

298 K. However, bond energy involves breaking bonds (including O−H bonds in H2O) in the

gaseous phase. Thus ∆Hc calculated from bond energy data is different from that obtained experimentally. Alternative Answers: 1. Bond energy data are average bond energies and may differ from bond energies of actual

bonds broken and formed in the combustion reaction. OR

2. Can also state that octane is liquid at standard conditions. Hence actual ∆Hc is different from experimental value as bond energy involves breaking bonds in gaseous phase.

(ii) From heptane to nonane, there is an additional −CH2− group. Hence ∆Hc increases from heptane to nonane regularly as the increase represents the ∆Hc value for the combustion of the –CH2– group.

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c) Mass of 1 dm3 of ethanol = 0.79 x 1000 g = 790 g. Amount of ethanol in 1 dm3 = 790/46.0 = 17.17 mol. Heat produced by 1 dm3 of ethanol = 17.17 x 1367 kJ = 2.35 x 104 kJ Mass of 1 dm3 of octane = 0.70 x 1000 g = 700 g. Amount of ethanol in 1 dm3 = 700/114.0 = 6.140 mol. Heat produced by 1 dm3 of octane = 6.140 x 5470 kJ = 3.36 x 104 kJ

7a) Mg2+(g) + 2e- + 2Cl(g) Mg(g) + 2Cl(g)

2∆Hatom[Cl2(g)] Mg2+(g) + 2Cl-(g) Mg(g) + Cl2(g)

∆Hatom[Mg(s)] Mg(s) + Cl2(g) MgCl2(s)

By Hess’ Law,

∆Hf[MgCl2(s)] = ∆Hatom[Mg(s)] + 2∆Hatom[Cl2(g)] + 1st IE + 2nd IE + 2(1st EA) + LE

∴ 1st EA = ½ {∆Hf[MgCl2(s)]- ∆Hatom[Mg(s)] - 2∆Hatom[Cl2(g)] – 1st IE – 2nd IE – LE } = −348 kJmol-1 b) The calculated theoretical value for the LE is less exothermic than the experimental LE. This is

because theoretically, the ionic bond is assumed to be purely ionic. In reality, due to polarisation of the anion by the cation, the bond between the two oppositely charged ions is strengthened.

8a) When an ionic solid dissolves in water, two enthalpy terms are involved.

1. The ions must be separated from the ionic lattice. The energy required is the lattice dissociation enthalpy (-LE).

2. The specific ions interact with water molecules. The energy released (Σ∆Hhyd) as these attractive forces come into play is compensation for the energy required to dissociate the lattice.

∆Hsolnθ = – LE + Σ∆Hhyd

θ

(a positive quantity) (a negative quantity) b)

∆Hsoln [LiCl(s)] = ∆Hhyd [Li+ (g)] + ∆Hhyd[Cl− (g)] − lattice energy = −−−−40 kJ mol−−−−1

LE[MgCl2]

∆Hf[MgCl2(s)]

1st IE (Mg) + 2nd IE (Mg) 2{1st EA (Cl)}

LiCl(s)

Li +(aq) + Cl−(aq)

Li+(g) + Cl−(g)

LE [LiCl]

∆Hhyd [Li+ (g)] + ∆Hhyd[Cl− (g)]

∆Hsoln [LiCl(s)]

E / kJ mol−1

0

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c) Heat absorbed = mc∆T = 50.0 x 4.20 x 3.3 = 693 J

Enthalpy change of solution, ∆Hsoln [KCl(s)] = +693 / (3.00/74.6) = +17.2 kJ mol−−−−1 (to 3 sf)

d) Since LE is inversely proportional to interionic radius, the lattice energy of LiCl is slightly more exothermic than that of KCl. However, since hydration energy is proportional to charge density, Li+ will have a much more exothermic hydration enthalpy compared to K+. Therefore, the heat change of solution of LiCl is more exothermic that that of potassium chloride.

9a) Amount of heat absorbed by water, Q = 200 x 4 x 4.18 = 3344 J

Amount of heat released by reaction, Q’ = Q / 0.8 = 4180 J No of moles of N2H4 = 6.56 x 10-3 Standard enthalpy change of combustion of hydrazine =−6.37 x 102 kJ mol-1

b)

N2 (g) + 2H2 (g) N2H4 (l) N2(g) + 2H2O (g) N2(g) + 2H2O (l)

By Hess’ law,

∆Hf (N2H4) = 2(-242) – (-637) – 2(+44) = + 65 kJmol-1

c) i)

By Hess Law, 160 + 4 x B.E(N-H) + 235 = 994 + 2(436) B.E (N-H) = 368 kJ mol-1

ii) The bond energy values obtained from the Data Booklet are average values and would differ from

the experimental values.

+O2 ∆Hc(N2H4)

+O2 2 x ∆Hf(H2O(g))

2 x ∆Hv(H2O)

∆Hf(N2H4)

N2H4 (g)

N2 (g) + 2H2 (g)

∆Hf (N2H4) = + 235 kJmol-1

B.E(N-N) + 4 x B.E(N-H) = + 65 kJmol

-1 B.E(N ≡ N) + 2x B.E(H-H)

= + 994 + 2(436) kJmol-1

2N (g) + 4H (g)

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10a) Born-Haber diagram 2Cu+(g) + O2−(g)

2Cu+(g) + O(g) + 2e−

1st EA [O]

= − 141 kJ mol-1

2Cu+(g) + O−(g) + e−

2nd EA [O] = +791 kJ mol-1

½ BE [O2] = ½(+496) kJ mol-1

2Cu+(g) + ½ O2(g) + 2e−

1st IE [Cu] x 2 = 2(+745) kJ mol-1

2Cu(g) + ½ O2(g)

LE [Cu2O]

∆Hatom [Cu(g)] x 2 = 2(+339) kJ mol-1

2Cu(s) + ½ O2(g)

∆Hf [Cu2O]

= − 166 kJ mol-1

Cu2O(s)

Lattice energy of Cu(I) oxide, Cu2O = −3232 kJ mol-1

(b) LE for CuO is more exothermic (− 4143 kJ mol-1) ⇒ stronger ionic bonds formed! for CuO,

• higher charge [Cu2+ vs Cu+] ⇒ q+ larger

• smaller cationic radius (r+) as e− clouds are attracted more strongly

to the nucleus ⇒ smaller interionic distance (r+ + r-)

(c) Cu2O(s) CuO(s) + Cu(s)

LE [Cu2O] = −3232 kJ mol-1 LE [CuO] = − 4143 kJ mol-1

2Cu+(g) + O2−−−−(g) Cu2+(g) + O2−−−−(g) + Cu(s)

2nd IE (Cu) 1st IE (Cu) ∆Hatom = +339 kJ mol-1 = +1960 kJ mol-1 = +745 kJ mol-1

Cu+(g) + Cu2+(g) + O2−−−−(g) +e−−−− Cu2+(g) + O2−−−−(g) + Cu(g)

∆H = −35 kJ mol-1

⇒ exothermic reaction ⇒ thermodynamically favourable ⇒ reaction probably spontaneous

(d) Heat evolved = mc∆T = 60 x 4.18 x 8.9 = 2243 J Amt of Cu2O (s) = 0.02 mol

Cu2O (s) + H2SO4 (aq)� CuSO4 (s) + Cu (s) + H2O (l)

Enthalpy change = Q/n = −−−−112 kJ mol-1 (to 3sf)

Assuming specific heat capacity of solution = 4.18 J g-1 K-1 Density of solution = density of water = 1.00 g cm-3

(e) Enthalpy change of neutralisation Heat change when 1 mole of water is formed when an acid neutralises a base (in dilute aqueous solution) at 1 atm and 298K

CuO (s) + H2SO4 (aq) � CuSO4 (aq) + H2O (l)

� � Cu2O (s) + H2SO4 (aq)

∆H = −−−−77 kJ mol-1

− Cu (s)

∆H = −112 kJ mol-1

− Cu (s)

∆H = −35 kJ mol-1

∆H

−+

−+

+=

rr

qqLE

∆H

E / kJ mol−1

0