a2 17 ans (1) hbuh

TOPIC 17 ANSWERS & MARK SCHEMES

QUESTIONSHEET 1

ENTHALPY OF ATOMISATION

a) The standard enthalpy of atomisation of an element is the energy required (½)to produce 1 mole (½)of gaseous atoms (½)from the element in its standard state / most stable state under standard conditions (½)The standard enthalpy of atomisation of a compound is the energy required (½)to convert 1 mole (½)of the gaseous compound (½)into gaseous atoms under standard conditions (½)

b) Enthalpy of atomisation of hydrogen = ½ H—H bond dissociation enthalpy (1)

c) (i) 920/2 = 460 kJ mol-1 (1)

(ii) Data book value includes values for the O—H bond in other molecules (1)

d) (i) C(s) + 2H2(g) → CH

4(g) (1 mark for formulae & balance; 1 for state symbols)

(ii) ∆H for bond breaking: 715 + (4 × 218) = +1587 kJ mol-1 (1)∆H for bond making: -4 ∆H

be (C—H) (1)

∆Hêf (CH

4) = -74.9 = + 1587 - 4∆H

be (C—H) (1)

∴ ∆Hbe

(C—H) = (1587 + 74.9)/4= 415 kJ mol-1 (1)

Or

Or accept an enthalpy diagram (2)By Hess’s law,-74.9 + 4∆H

be (C⎯H) = +715 + 872 (1)

∴ ∆Hbe

(C⎯H) = (715 + 872 + 74.9)/4 = 415 kJ mol-1 (1)

(2) Deduct 1 mark for each error

C(g)+715 kJ mol-1

C(g)

+2H2(g) 4H(g)

CH4(g)

4∆Hbe

(C−H)

4(+218) kJ mol-1

-74.9 kJ mol-1


QUESTIONSHEET 2

LATTICE ENTHALPY

a) Enthalpy / heat required to separate the gaseous ions (1)in 1 mole of solid (1)to an infinite distance from one another (1)

Or enthalpy / heat released when gaseous ions (1)infinitely far apart (1)join together to give 1 mole of solid (1)

b) (i) Change Decreases (1)Explanation Ionic radii increase (1)

Hence charge density / charge : radius ratio of the cations decreases (1)

(ii) Melting point Decrease (1)Solubility in water Increase (1)

c) (i)

By Hess’s law, ∆Hêsoln

= +2352 – 1275 – 1060 (1)= +17 kJ mol-1 (1)

(ii) ∆Hêsoln

for BaF2(s) is endothermic while the value of ∆Hê

soln for BaCl

2(s) is exothermic (1)

∴ dissolving of BaF2(s) is thermodynamically less favourable (1)

Ba2+(g) + 2F-(g) + aq (½)

Ba2+(aq) + 2F-(aq) (½)

Ba2+(aq) + 2F-(g) + aq (½)

Ba2+(F-)2 (s) + aq (½)

∆Hêsoln

(½)

2 ×∆Hêhyd

∆Hêhyd

-1275 kJ mol-1 (½)

2 × (-530) = -1060 kJ mol-1 (½)

∆Hêlattice +2352 kJ mol-1 (½)


QUESTIONSHEET 3

a) Born-Haber cycle

BORN-HABER CYCLES I

Lattice enthalpy of lithium fluorideBy Hess’s law, 159.5 + 520.0 + 79.0 – 334.0 + ∆Hê

lattice = -616.0 (1)

∴ ∆Hêlattice

= -616.0 – 159.5 – 520.0 – 79.0 + 334.0= -1040.5 kJ mol-1 (1)

b) Calculation is based on the assumption that bonding is purely ionic (1)In reality, bonding has some covalent character (1)

Li+(g) + e− + F(g) (1)

Li+(g) + e− + ½F2(g) (1)

Li(g) + ½F2(g) (1)

Li(s) + ½F2(g) (1)

Li+(g) + F−(g) (1)

Li+F-(s) (1)

∆Hêlattice

(1)

+79.0 kJ mol-1 (1) -334.0 kJ mol-1 (1)

+520.0 kJ mol-1 (1)

+159.5 kJ mol-1 (1)

-616.0 kJ mol-1 (1)


QUESTIONSHEET 4

BORN-HABER CYCLES II

(ii) ∆Hêf (MgCl) = +150 + 736 + 121 + (-364) + (-756) (1)

= -113 kJ mol-1 (1)

(iii) The formation of MgCl2 is more exothermic than MgCl (1)

Hence MgCl2 is more stable / at a lower energy level than MgCl (1)

b) Lattice enthalpy of MgCl (1)

c) Second ionisation energy of magnesium (1)Lattice enthalpy of MgCl

2 (1)

d) Greater or less than ∆Hêlattice

(MgCl2) Less (1)

Explanation Mg+ has a smaller charge than Mg2+ (1) and is larger (1)Or Mg+ has a lower surface charge density than Mg2+ (2)

a) (i)

Mg+(g) + e− + Cl(g) (½)

+121 kJ mol-1 (½)-364 kJ mol-1 (½)

Mg+(g) + e− + ½Cl2(g) (½)

Mg+(g) + Cl−(g) (½)

Mg(g) + ½Cl2(g) (½)

Mg(s) + ½Cl2(g) (½)

MgCl(s) (½)

+150 kJ mol-1 (½)

∆Hêf (MgCl) (½)

+736 kJ mol-1 (½)

-756 kJ mol-1 (½)


QUESTIONSHEET 5

ENTROPY AND FREE ENERGY CHANGE

a) ∆Gê Standard free energy change / free energy change at 298 K and 1 atm (½)kJ mol-1 (½)

T Absolute temperature / temperature in kelvins (½)K (½)

∆Hê Standard enthalpy change / enthalpy change at 298 K and 1 atm (½)kJ mol-1 (½)

∆Sê Standard entropy change / entropy change at 298 K and 1 atm (½)J K-1 mol-1 (½)

b) (i) Decrease (1)O

2 gas (disordered) → MgO solid (ordered) (1)

(ii) Increase (1)Mg solid (ordered) → H

2 gas (disordered) (1)

(iii) Remain approximately the same (1)Reactants and products are all gaseous (½)No change in the number of gas molecules (½)

(iv) Increase (1)Increased number of gaseous molecules (1)

c) (i) Total entropy of products = (2 × 213.6) + (3 × 69.9) = 636.9 J K-1 mol-1 (1)Total entropy of reactants = (160.7) + (0) = 160.7 J K-1 mol-1 (1)∴ ∆Sê = 636.9 – 160.7 = 476.2 J K-1 mol-1 (1)

(ii) ∆Gê = -1367.3 – (298 × 476.2 × 10-3) (1)= -1509.2 kJ mol-1 (1)


QUESTIONSHEET 6

ENTROPY AND STATE

a) (i) more ordered/less random structures (1)so fewer energy quanta available (1)

(ii) sodium 2Na(s) + Cl

2(g) → 2NaCl(s) (1)

72-(51 + 223/2) (1) = -90.5 (1)

hydrogen H

2(g) + Cl

2(g) →2HCl(g) (1)

187-(130+223) (1) 2 = 10.5 (1)

(iii) NaCl : product more ordered structure (1)so loss of entropy (1)but entropy of surroundings increases (1)

thenas exothermic (1)

orso overall loss of free energy (1)

b) (i) HF 7.5 × 1000/293 (1) = 25.6 (1)

HCl 16.2 × 1000/188 (1) = 86.2 (1)

(ii) ∆G = 0 (1)∆H = T/∆S (1)

(iii) HF retains order in gas state (1)giving lower change in randomness (1)due to strong H bonding (1)


QUESTIONSHEET 7

TEST QUESTION I

a) Step 1 Enthalpy of atomisation / sublimation of sodium (1)

Step 2 (First) ionisation energy of sodium (1)

Step 3 Hydration enthalpy of the sodium ion (1)

Step 4 Dissociation enthalpy of water (1)

Step 5 Negative(½) hydration enthalpy of the proton / hydrogen ion (½)

Step 6 Negative (½) ionisation energy of hydrogen (½)

Step 7 Negative (½) enthalpy of atomisation of hydrogen / half the bond dissociation enthalpy of the H—Hbond (½)

(ii) By Hess’s law,-183.7 = +109 + 500 + ∆Hê

x + 1090 – 406 – 1316 – 218 (1)

∴ ∆Hêx

= -183.7 – 109 –500 – 1090 + 406 + 1316 + 218= +57.3 kJ mol-1 (1)

c) Predicted value -57.3 kJ mol-1 (1)Explanation Because neutralisation / H+(aq) + OH-(aq) → H

2O(l)

is the reverse of the dissociation of water (1)

Na+(aq) + OH-(aq) + H(g)

Na+(g) + OH-(aq) + H+(g)

Na+(g) + OH-(aq) + H+(aq)

Na+(g) + H2O(l)

Na(g) + H2O(l)

Na(s) + H2O(l) Start

Na+(aq) + OH-(aq) + ½H2(g) Finish

Na+(aq) + OH-(aq) + H+(g)

∆Hêx

+1090 kJ mol-1-406 kJ mol-1

-1316 kJ mol-1

-218 kJ mol-1

-183.7 kJ mol-1

+109 kJ mol-1

+500 kJ mol-1

b) (i)

4 marks (Deduct ½ for each error or omission)


QUESTIONSHEET 8

TEST QUESTION II

a) Mg(s) + ½O2(g) → MgO(s) (2)

Award (1) for correctly balanced equation and (1) for correct state symbols.

b) (i)

(ii) ∆Hêf (MgO) = [+147] + [+249] + [+738] + [+1451] + [-142] + [+844] + [-3884] (1)

= -597 kJ mol-1 (1)

c) ∆Sê = [27.0] – [(32.7) + ½(205)] (1)= -108.2 J K-1 mol-1 (1)

d) (i) ∆Gê = ∆Hê - T∆Sê (1)= [-597 × 103] – [(298) × (-108.2)] (1)= -564 800 J mol-1

= -564.8 kJ mol-1 (1)

(ii) Reaction has high activation energy (1)Energy is required to form Mg2+ and O2- ions before lattice enthalpy can be released (1)∴ reaction is infinitely slow (1)Maximum 2 marks

Mg2+(g) + 2e− + O(g)

Mg2+(g) + O2−(g)

Mg2+(g) + e− + O−(g)

Mg+(g) + e− + O(g)

Mg(g) + O(g)

Mg(g) + ½O2(g)

Mg(s) + ½O2(g)

MgO(s)

-142 kJ mol-1

+844 kJ mol-1

+1451 kJ mol-1

+738 kJ mol-1

+249 kJ mol-1

+147 kJ mol-1

∆Hêf (MgO)

-3884 kJ mol-1

(4) Deduct 1 mark for each error.


QUESTIONSHEET 9

TEST QUESTION III

a) (i) The enthalpy change when 1 mole of gaseous ions (1)becomes hydrated by a large / infinitely large quantity of water (1)

(ii) The enthalpy change when 1 mole of the solid (1)is dissolved in a large quantity of water / to give an infinitely dilute solution (1)

b) (i)

(ii) ∆Hêsolution

= ∆Hêlattice

+ ∆Hêhyd(cation)

+ ∆Hêhyd(anion)

(1)

(iii) -155 = +2493 + ∆Hêhyd

(Mg2+(g)) + [-728] (1)∴∆Hê

hyd (Mg2+(g)) = [-155] + [-2493] + [+728]

= -1920 kJ mol-1 (1)

c) (i) Value becomes more negative (1)

(ii) Size of the metal cation increases (1)∴ charge density of the metal cation decreases (1)

(iii) Decrease in lattice enthalpy is much greater than decrease in hydration enthalpy (1)∴ ∆Hê

solution becomes more negative and solubility increases (1)

Mg2+(g) + 2Cl−(g)

Mg2+(aq) + 2Cl−(g)

∆Hêhyd

(Mg2+(g))

-728 kJ mol-1

Mg2+(aq) + 2Cl−(aq)

MgCl2(s)

+2493 kJ mol-1

-155 kJ mol-1

(3) Deduct 1 mark for each error.


QUESTIONSHEET 10

PRINCIPLES OF ELECTROCHEMICAL CELLS

a) (i) Oxidation Zn(s) → Zn2+(aq) + 2e- (1)

Reduction Cu2+(aq) + 2e- → Cu(s) (1)

(ii) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) (1)

(iii) Solution changes colour from blue to colourless (1)Pink / red solid is precipitated (1)

b) (i) Zinc rod becomes smaller (1)Copper rod becomes larger (1)Blue colour of CuSO

4(aq) fades (1)

(ii) With reaction in a cell, energy is released as electricity (1)With reaction in a beaker, energy is released as heat (1)

c) (i) For a redox reaction to occur, there must be oxidation at one half-cell (1)and reduction at the other (1)

Or electrons are released at one (1)and have to be accepted at the other (1)

(ii) To complete the circuit (1)Contains saturated KNO

3(aq) / KCl(aq) (1)

as a gel or saturated on filter paper or in sintered glass salt bridge(1)

(iii) Electrons (1)which flow from Zn to Cu (1)

(iv) K+ ions to the Cu half-cell (1)NO

3- / Cl- ions to the Zn half-cell (1)

SO4

2- ions from the Cu half-cell to the Zn (1)


QUESTIONSHEET 11

ELECTRODE POTENTIALS AND CELL EMF

a) (i) Metal tends to ionise / form hydrated ions in solution (½)releasing electrons to the rod / giving the rod a negative charge (½)Reverse change also tends to occur (½)accepting electrons from rod / giving rod a positive charge (½)Equilibrium is established between the opposing reactions (1)

(ii) Enthalpy of atomisation (1)Ionisation energy / energies (1)Enthalpy of hydration (of ions) (1)

(iii) 298 K (1)1 M / 1 mol dm-3 (1)

(iv) Different reactivity (1)Highly reactive metals have a high tendency to ionise (½)∴ giving a high electron density on the metal rod / rod acquires a high –ve potential (½)Metals of low reactivity – converse argument (1)Maximum 2 marks

b) (i) Platinium (foil) (1)

(ii) 1M / 1 mol dm-3 (1)

c) (i) The voltage developed when the cell delivers zero current / the maximum voltage the cell can develop (1)

(ii) High resistance (1) voltmeter (1)in the external circuit (1)

(iii) Ecell = E+ – E - (1)Or E

RHS – E

LHS (applied to cell notation) (1)

Or Ereduction electrode

– Eoxidation electrode

(1)


QUESTIONSHEET 12

MEASUREMENT OF ELECTRODE POTENTIALS

a) (i) It is impossible to find the potential of a single electrode (1)because we can only measure a potential difference (1)Or because a voltmeter has two terminals and must be connected to two electrodes (1)

(ii)

Pt foil (1)

HCl(aq) or H2SO

4 (aq) (½)

[H]+ = 1 mol dm-3 (½)

at 298 K (1)

H2(g) (½) 1 atm (½)

b) (i) Reduction Fe3+(aq) + e- → Fe2+(aq) (1)

Oxidation Cu(s) → Cu2+(aq) + 2e- (1)

(ii) 2Fe3+(aq) + Cu(s) → 2Fe2+(aq) + Cu2+(aq) (1)

(iii) Ecell

= E+ – E

–

+0.43 = E+ – 0.34 (1)

E+ = + 0.77 V (1)

(iv) Ecell

= E+ – E

–

Ecell

= 0.43 – (– 0.37) (1)Ecell = 0.71 V (1)


QUESTIONSHEET 13

FACTORS INFLUENCING ELECTRODE POTENTIALS AND CELL EMF

a) (i) Pt(s) | H2(g)

, H+(aq) Ag+(aq) | Ag(s) (2)

(1for species in correct order, 1 for state symbols)

(ii) Ecell

= + 0.80 – 0.00 = + 0.80 V (1)

(iii) From hydrogen to silver (1)

(iv) H2(g) + 2Ag+(aq) → 2H+(aq) + 2Ag(s) (2)

(1 for species, 1 for balance & state symbols)

(v) pH decreases (1)because H+ ions are formed in the cell reaction (1)Or because [H+] increases (1)

b) (i) Effect on electrode potential Nil (1)Effect on e.m.f. Nil (1)

(ii) Effect on electrode potential More +ve / higher (1)Effect on e.m.f. Decreases (1)

(iii) Effect on electrode potential More +ve / higher (1)Effect on e.m.f. Decreases (1)

(iv) Effect on electrode potential More –ve / becomes –ve (1)Effect on e.m.f. Increases (1)

(v) Effect on electrode potential More +ve / higher (1)Effect on e.m.f. Increases (1)


QUESTIONSHEET 14

PREDICTION OF REDOX CHANGES FROM STANDARD ELECTRODE POTENTIALS

a) (i) I– (½) & S4O

62– (½)

(ii) Cl– (½) & SO4

2– (½)

(iii) None (1)

(iv) Fe3+ and I2 (½) & Cr3+ (½)

b) (i) Eêcell

=+ 0.77 – (+ 1.23) = – 0.46 V (1)Outcome No reaction (because Eê

cell is –ve) (1)

(ii) Eêcell

= + 1.33 – (+ 0.54) = + 0.79 V (1)Outcome Reaction goes to completion (because Eê

cell > 0.3 V) (1)

(iii) Eêcell

=+ 1.36 – (+ 1.23) = + 0.13 V (1)Outcome Dynamic equilibrium (because Eê

cell < 0.3 V) (1)

(iv) Eêcell

=+ 1.36 – (+ 1.20) = – 0.16 V (1)Outcome Dynamic equilibrium (because Eê

celll < 0.3 V) (1)

c) (i) Brown colour / iodine is expected (1)

from a comparison of Eê values / because Eêcell

is positive (1)Reason: high activation energy (1)Both reacting ions are negatively charged / repel each other (1)Maximum 3 marks

(ii) Comparison of Eê values / Eêcell

negative suggests no reaction (1)but hydrochloric acid is concentrated / [Cl–] is > 1M (1)

hence conditions are non-standard / Eê values do not give a reliable prediction (1)


QUESTIONSHEET 15

FUEL CELLS

a) Reactants / oxidising and reducing agents are fed in continuously / from outside the cell (1)Reaction products are removed continuously (1)

b) (i) H2(g) + 2OH–(aq) → 2H

2O(l) + 2e– (1)

(ii) O2(g) + 2H

2O(l) + 4e– → 4OH–(aq) (1)

(iii) 2H2(g) + O

2(g) → 2H

2O(l)

(1)

(iv) To increase surface area (1)to assist adsorption of gas (1)

c) (i) 1 kWh ≡ 4.3 × 106 / 96 500 = 44.56 mol electrons (1)2 mol e– ≡ 1 mol H

2

∴44.56 mol e– ≡ 22.28 mol H2 (1)

∴ V = 22.28 × 24 = 534.7 / 535 dm3 H2 (1)

(ii) n (H2) required in practice = (22.28 × 100) / 72 = 30.94 mol (1)

∴ m (H2) = 30.94 × 2 = 61.9 / 62 g H2 (1)

d) (i) The reaction product is water which is non-polluting (1)Because H

2(g) and O

2(g) can be obtained from water, there is no need to use non-renewable fossil fuels (1)

(ii) Storage as a hydride (1)or adsorbed on to a solid adsorbant (1)(Not “stored as compressed / liquefied gas”)


QUESTIONSHEET 16

TEST QUESTION IV

a) The e.m.f. of a cell (1)comprising a half-cell consisting of Pt foil (1)immersed in a solution of an aqueous halogen and halide ions (1)both at concentrations of 1.00 mol dm-3 (½) at 298 K (½)and a second half-cell which is a standard hydrogen electrode / SHE (1)

b) Oxidising power decreases as Group 7 is descended (1)

c) (i) Equation 2Fe2+(aq) + Cl2(aq) → 2Fe3+(aq) + 2Cl-(aq) (1)

e.m.f. = Eêreduction half-cell

- Eêoxidation

half-cell

= +1.36 – (+0.77) = +0.59 V (1)

Reaction is energetically feasible (½) because the e.m.f. is positive (½)

(ii) Equation 2Fe2+(aq) + I2(aq) → 2Fe3+(aq) + 2I-(aq) (1)

e.m.f. = +0.54 – (+0.77) = -0.23 V (1)

Reaction is not energetically feasible (½) because the e.m.f. is negative (½)

(iii) Equation 2Fe3+(aq) + 2I-(aq) → 2Fe2+(aq) + I2(aq) (1)

e.m.f. = +0.77 – (+0.54) = +0.23 V (1)

Reaction is energetically feasible (½) because the e.m.f. is positive (½)

In the whole of c) state symbols are necessary, but deduct 1 mark only if they are not shown.

d) FeI3 is an unstable compound / it would decompose into FeI

2 and I

2 (1)


QUESTIONSHEET 17

TEST QUESTION V

a) A = electrochemicalB = negativeC = positiveD = oxidantE = electrolyteF = catalysisG = carrier (7)

b) advantages renewable (1)non-polluting/produces only water (1)

disadvantages low energy yield per unit volume (1)not liquefiable under normal pressures (1)

c) (i) H2(g) → 2H+(aq) + 2e (1) equation (1) state subscripts

(ii) Increased cell e.m.f (1) Increased [Cu2+] moves (reversible) half-cell reaction to Cu (1) Increases e acceptance/+ half-cell potential (1)

(iii) 0.34V (1)

a2 17 ans (1) hbuh

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