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A-Level · Chemistry

AC3 Energetics and Thermodynamics

AQA 7405 · Calculator allowed · about 130 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question is about standard enthalpy changes and the direction of energy transfer in exothermic and endothermic reactions.

(a)Define what is meant by the standard enthalpy change of formation of a compound.(2)

(b)State two standard conditions that must apply when quoting a standard enthalpy change.(2)

(c)Hydrogen reacts with oxygen: 2H2(g) + O2(g) -> 2H2O(l), delta H = -572 kJ/mol. State, with a reason, whether this reaction is exothermic or endothermic.(2)

(d)Using ideas about bond breaking and bond making, explain why this reaction is exothermic overall.(2)

(Total for Question 1 is 8 marks)

Methane reacts with chlorine in a substitution reaction: CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g). Mean bond enthalpies (kJ/mol): C-H = +413, Cl-Cl = +242, C-Cl = +346, H-Cl = +431.

(a)State what is meant by the term mean bond enthalpy.(2)

(b)Calculate the enthalpy change for this reaction using the mean bond enthalpies given. Show your working.(3)

(c)Explain why a value calculated using mean bond enthalpies may differ from the actual (experimentally measured) enthalpy change for this reaction.(2)

(d)State one other limitation of using mean bond enthalpies, in terms of the physical states of the reactants and products.(1)

(Total for Question 2 is 8 marks)

Standard enthalpies of combustion (kJ/mol): delta Hc[C(s)] = -393, delta Hc[H2(g)] = -286, delta Hc[CH4(g)] = -890.

(a)Write a balanced equation, including state symbols, to represent the standard enthalpy of formation of methane.(2)

(b)Using a Hess's law cycle and the enthalpies of combustion given, calculate the standard enthalpy of formation of methane. Show your working.(3)

(c)The accepted literature value for the standard enthalpy of formation of methane is -74.8 kJ/mol. Comment on the agreement between your calculated value and this literature value.(1)

(d)State Hess's law.(2)

(Total for Question 3 is 8 marks)

Standard entropies, S degree, at 298 K: N2(g) = 192 J/(K mol), H2(g) = 131 J/(K mol), NH3(g) = 193 J/(K mol). Reaction: N2(g) + 3H2(g) -> 2NH3(g).

(a)State what is meant by the term entropy.(1)

(b)Predict, with a reason, whether you expect the entropy change for this reaction to be positive or negative, before doing any calculation.(2)

(c)Calculate the standard entropy change, delta S, for this reaction.(3)

(Total for Question 4 is 6 marks)

A student determines the enthalpy of combustion of ethanol using a simple calorimetry method, as in the required practical for measuring an enthalpy change. A spirit burner containing ethanol is weighed, then used to heat 150 g of water in a copper calorimeter; the maximum temperature reached is recorded, and the spirit burner is reweighed.
Data:
mass of water = 150 g
initial temperature of water = 18.5 degC
final (maximum) temperature of water = 40.5 degC
initial mass of spirit burner + ethanol = 78.23 g
final mass of spirit burner + ethanol = 77.39 g
specific heat capacity of water, c = 4.18 J/(g K)
Mr(ethanol, C2H5OH) = 46.0

(a)Calculate the temperature rise of the water.(1)

(b)Calculate the mass of ethanol burned during the experiment.(1)

(c)Calculate the energy transferred to the water, q = m x c x delta T, giving your answer in kJ.(2)

(d)Calculate the amount, in mol, of ethanol burned.(1)

(e)Calculate the experimental enthalpy of combustion of ethanol, delta Hc, in kJ/mol. Give your answer to 3 significant figures.(2)

(f)The literature value for the standard enthalpy of combustion of ethanol is -1367 kJ/mol. Calculate the percentage difference between the experimental value and the literature value.(2)

(g)Suggest two reasons why the experimental value is much less exothermic than the literature value.(2)

(Total for Question 5 is 11 marks)

Enthalpy of hydration of Na+(g) ion = -406 kJ/mol. Lattice dissociation enthalpy of NaCl(s) = +771 kJ/mol. Enthalpy of solution of NaCl(s) = +3.9 kJ/mol.

(a)Define enthalpy of hydration.(2)

(b)Construct a Hess's law cycle linking the enthalpy of solution, the lattice dissociation enthalpy and the two enthalpies of hydration for NaCl, and use it to calculate the enthalpy of hydration of the chloride ion, Cl-(g). Show your working.(4)

(c)The small positive value for the enthalpy of solution of NaCl shows that dissolving NaCl is only very slightly endothermic overall. Explain what this tells you about the relative magnitudes of the lattice dissociation enthalpy and the total enthalpy of hydration of the ions.(2)

(Total for Question 6 is 8 marks)

A student reacts 25.0 cm3 of 1.00 mol/dm3 hydrochloric acid with 25.0 cm3 of 1.00 mol/dm3 sodium hydroxide solution in an insulated polystyrene cup, as in the required practical for measuring an enthalpy change. Initial temperature of both solutions = 19.2 degC. Maximum temperature reached = 26.0 degC. Assume the density of the solution is 1.00 g/cm3 and its specific heat capacity is 4.18 J/(g K).

(a)Calculate the total mass of solution in the calorimeter.(1)

(b)Calculate the heat energy released in the reaction.(2)

(c)Calculate the amount, in mol, of HCl (and of NaOH) that reacted.(1)

(d)Calculate the enthalpy change of neutralisation, delta Hneut, in kJ/mol.(2)

(Total for Question 7 is 6 marks)

In the experiment described in Question 7, the student then repeats the experiment using 25.0 cm3 of 1.00 mol/dm3 ethanoic acid (a weak acid) in place of the hydrochloric acid, keeping the volume and concentration of sodium hydroxide, and all other conditions, the same. The temperature rise recorded is smaller than in the original experiment with hydrochloric acid, giving a less exothermic enthalpy of neutralisation.

Explain, in terms of the energy changes involved, why the enthalpy of neutralisation for ethanoic acid is less exothermic (smaller in magnitude) than for hydrochloric acid.

(Total for Question 8 is 6 marks)

The table gives lattice enthalpies (magnitude; all values are exothermic) for four sodium halides:
NaF = -918 kJ/mol
NaCl = -780 kJ/mol
NaBr = -742 kJ/mol
NaI = -705 kJ/mol

(a)Describe the trend in lattice enthalpy shown by these values as the halide ion changes from F- to I-.(1)

(b)Explain this trend in terms of ionic radius and electrostatic attraction.(3)

(c)Magnesium oxide, MgO, has a much greater lattice enthalpy (-3791 kJ/mol) than sodium chloride (-780 kJ/mol), even though the ions in MgO are a similar size to those in NaCl. Explain why.(2)

(Total for Question 9 is 6 marks)

The table gives enthalpy data needed to construct a Born-Haber cycle for magnesium chloride, MgCl2.
Enthalpy of formation of MgCl2(s), delta Hf = -641 kJ/mol
Enthalpy of atomisation of Mg(s), delta Hat = +148 kJ/mol
First ionisation energy of Mg, IE1 = +738 kJ/mol
Second ionisation energy of Mg, IE2 = +1451 kJ/mol
Enthalpy of atomisation of chlorine (per mole of Cl atoms), delta Hat = +122 kJ/mol
Electron affinity of Cl, EA = -349 kJ/mol

(a)Define the term first ionisation energy.(2)

(b)Explain why the enthalpy of atomisation of chlorine is given as a value per mole of Cl atoms, rather than per mole of Cl2 molecules.(1)

(c)Using the data given, construct a Born-Haber cycle and calculate the lattice enthalpy of formation of magnesium chloride, MgCl2. Show all steps in your working.(4)

(d)The theoretical (calculated) lattice enthalpy of MgCl2, based on a perfect ionic model, is -2326 kJ/mol. Compare this with your Born-Haber (experimental) value from part (c), and suggest what this comparison indicates about the bonding in MgCl2.(2)

(Total for Question 10 is 9 marks)

For the Haber process, N2(g) + 3H2(g) -> 2NH3(g), delta H = -92.0 kJ/mol and delta S = -199 J/(K mol) (as calculated in Question 4(c)). Use delta G = delta H - T x delta S to answer the following.

(a)Calculate delta G for this reaction at 298 K, in kJ/mol. Show your working.(3)

(b)State, with a reason, whether this reaction is thermodynamically feasible at 298 K.(1)

(c)Calculate the temperature above which this reaction is no longer thermodynamically feasible (i.e. the temperature at which delta G = 0). Give your answer in kelvin.(3)

(d)The Haber process is actually carried out industrially at a compromise temperature of around 450 degC (723 K), well above the temperature calculated in part (c). Suggest why, in terms of reaction rate and the position of equilibrium, an operating temperature above the calculated feasibility temperature might still be used industrially.(3)

(Total for Question 11 is 10 marks)

In the ethanol combustion calorimetry experiment described in Question 5, the temperature rise was found to be 22.0 degC (using a thermometer that could be read to the nearest 0.5 degC) and the mass of ethanol burned was found to be 0.84 g (using a balance that could be read to the nearest 0.01 g). The calculated enthalpy of combustion was delta Hc = -755 kJ/mol. Each quantity (temperature and mass) required two readings (an initial and a final reading) to obtain the value used.

(a)Calculate the total (absolute) uncertainty in the temperature rise, given that each individual thermometer reading has an uncertainty of ± 0.5 degC.(1)

(b)Calculate the percentage uncertainty in the temperature rise.(1)

(c)Calculate the percentage uncertainty in the mass of ethanol burned, given that each individual mass reading has an uncertainty of ± 0.01 g.(2)

(d)Hence estimate the total percentage uncertainty in the calculated enthalpy of combustion, and use this to give the value with its absolute uncertainty (in kJ/mol), to an appropriate precision.(3)

(Total for Question 12 is 7 marks)

Calcium carbonate decomposes on heating: CaCO3(s) -> CaO(s) + CO2(g). For this reaction, delta H = +178 kJ/mol and delta S = +161 J/(K mol).

(a)Explain why delta S is positive for this reaction.(1)

(b)Calculate the minimum temperature, in kelvin, at which this decomposition becomes thermodynamically feasible.(3)

(Total for Question 13 is 4 marks)

Industrially, calcium carbonate (limestone) is decomposed in a lime kiln at temperatures of around 1200 K. Using the data given in Question 13 for this reaction (delta H = +178 kJ/mol, delta S = +161 J/(K mol), and delta G = delta H - T x delta S), explain why this reaction does not occur spontaneously at room temperature (298 K) but does become thermodynamically feasible at the temperature used industrially.

(Total for Question 14 is 6 marks)

Mark scheme · AC3 Energetics and Thermodynamics

Question 1

(a) B1 the enthalpy change when 1 mole of a compound is formed from its constituent elements
(a) B1 with all reactants and products in their standard states, under standard conditions (100 kPa and a stated temperature, usually 298 K) oe
(a) Answer: The enthalpy change when 1 mole of a compound is formed from its constituent elements, with all substances in their standard states under standard conditions (100 kPa, a stated temperature, usually 298 K).
(b) B1 pressure of 100 kPa
(b) B1 a stated temperature (usually 298 K) oe; accept solutions at a concentration of 1 mol dm-3
(b) Answer: Pressure of 100 kPa; a stated temperature (usually 298 K). Solutions, if present, at a concentration of 1 mol dm-3.
(c) B1 exothermic
(c) B1 because delta H is negative, so energy is transferred to the surroundings (energy of products is less than energy of reactants) oe
(c) Answer: Exothermic, because delta H is negative (energy is released to the surroundings).
(d) M1 breaking bonds in the reactants requires an input of energy (endothermic process) oe
(d) M1 forming new bonds in the products releases energy (exothermic process), and more energy is released forming the new bonds than is absorbed breaking the original bonds oe
(d) Answer: Breaking bonds in H2 and O2 absorbs energy; forming O-H bonds in H2O releases energy. Because more energy is released forming the new bonds than is absorbed breaking the old ones, the overall reaction is exothermic.

Question 2

(a) B1 the (average) energy required to break one mole of a specified covalent bond
(a) B1 averaged over that type of bond in a range of different gaseous molecules/compounds oe
(a) Answer: The average energy needed to break one mole of a given covalent bond, averaged over many different gaseous compounds containing that bond.
(b) M1 bonds broken = C-H + Cl-Cl = 413 + 242 (= 655 kJ/mol)
(b) M1 bonds formed = C-Cl + H-Cl = 346 + 431 (= 777 kJ/mol)
(b) A1 delta H = 655 - 777 = -122 (kJ/mol) cao
(b) Answer: delta H = -122 kJ/mol
(c) M1 mean bond enthalpies are average values, taken across many different compounds containing that bond oe
(c) M1 the actual bond enthalpy for a specific bond in this particular molecule may differ from the mean value used, so the calculated delta H is only an approximation oe
(c) Answer: Mean bond enthalpies are averaged over many compounds, so the true bond enthalpy in CH4, CH3Cl and HCl may differ slightly from the mean value used, making the calculated delta H only an approximate value.
(d) B1 mean bond enthalpies apply only to gaseous species; they cannot be used directly if any reactant or product is not in the gas phase (e.g. a liquid or solid) oe
(d) Answer: Mean bond enthalpies are only valid for substances in the gaseous state, so cannot be used directly for reactions involving liquids or solids.

Question 3

(a) M1 C(s) + 2H2(g) -> CH4(g), correctly balanced
(a) A1 all state symbols correct: (s), (g), (g) cao
(a) Answer: C(s) + 2H2(g) -> CH4(g)
(b) M1 correct Hess's law relationship: delta Hf(CH4) = delta Hc(C) + 2 x delta Hc(H2) - delta Hc(CH4) (oe, allow a correctly drawn or described cycle)
(b) M1 correct substitution: -393 + 2 x (-286) - (-890)
(b) A1 delta Hf(CH4) = -75 (kJ/mol) cao
(b) Answer: delta Hf(CH4) = -75 kJ/mol
(c) B1 very good/close agreement (within about 1 kJ/mol, less than 1-2% difference) oe; any sensible comment referencing the closeness of the two values, ecf from (b)
(c) Answer: Very good agreement - the calculated value (-75 kJ/mol) is within about 1 kJ/mol of the literature value (-74.8 kJ/mol).
(d) B1 the total enthalpy change of a reaction is independent of the route taken
(d) B1 provided the initial and final conditions/states are the same in each route oe
(d) Answer: The total enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same.

Question 4

(a) B1 a measure of the number of ways particles (and their energy) can be arranged/distributed (a measure of disorder) oe
(a) Answer: A measure of the number of ways that particles (and their energy) can be arranged - i.e. a measure of disorder.
(b) B1 negative
(b) B1 because the number of moles of gas decreases (4 mol of gas reactants forms 2 mol of gas product), so there are fewer ways to arrange the particles (disorder decreases) oe
(b) Answer: Negative, because 4 mol of gas reactants form only 2 mol of gas product, decreasing the number of ways the particles can be arranged.
(c) M1 delta S = sum S(products) - sum S(reactants), set up correctly
(c) M1 correct substitution: 2 x 193 - [192 + 3 x 131]
(c) A1 delta S = -199 (J/(K mol)) cao
(c) Answer: delta S = -199 J/(K mol)

Question 5

(a) B1 22.0 (degC) cao
(a) Answer: 22.0 degC
(b) B1 0.84 (g) cao
(b) Answer: 0.84 g
(c) M1 correct substitution q = 150 x 4.18 x 22.0 (= 13794 J), ecf from (a)
(c) A1 13.8 (kJ) awrt (accept 13794 J), cao
(c) Answer: q = 13.8 kJ (13794 J)
(d) B1 0.0183 (mol) cao, ecf from (b)
(d) Answer: 0.0183 mol
(e) M1 delta Hc = -q / n, ecf from (c) and (d)
(e) A1 -755 (kJ/mol) awrt, cao, ecf
(e) Answer: delta Hc = -755 kJ/mol
(f) M1 correct substitution: (1367-755)/1367 x 100, ecf from (e)
(f) A1 44.8 (%) awrt (accept 44.7-44.8%), ecf
(f) Answer: 44.8% (the experimental value is about 44.8% less exothermic than the literature value)
(g) B1 heat loss to the surroundings (not all the energy released is transferred to the water) oe
(g) B1 incomplete combustion of the ethanol (some carbon forms soot/CO rather than CO2, releasing less energy) oe; accept heat capacity of the calorimeter/apparatus not accounted for, or evaporation of ethanol before combustion - max 2 sensible reasons
(g) Answer: Any two of: heat loss to the surroundings; incomplete combustion (soot/CO formed); heat capacity of the calorimeter not included; evaporation of ethanol.

Question 6

(a) B1 the enthalpy change when one mole of gaseous ions is dissolved in (sufficient) water
(a) B1 to form one mole of aqueous ions, under standard conditions (at infinite dilution) oe
(a) Answer: The enthalpy change when one mole of gaseous ions dissolves in water to form one mole of aqueous ions, under standard conditions.
(b) M1 correct cycle relationship: delta Hsol = delta H(lattice dissociation) + delta Hhyd(Na+) + delta Hhyd(Cl-) (oe, allow cycle described or drawn)
(b) M1 correct substitution: 3.9 = 771 + (-406) + delta Hhyd(Cl-)
(b) A1 correct rearrangement: delta Hhyd(Cl-) = 3.9 - 365 (= -361.1)
(b) A1 -361 (kJ/mol) awrt, cao
(b) Answer: delta Hhyd(Cl-) = -361 kJ/mol
(c) B1 the lattice dissociation enthalpy and the total (sum of the) hydration enthalpies of the ions are very similar in magnitude (almost equal and opposite) oe
(c) B1 because the small overall enthalpy of solution is the (small) difference between these two large, opposing enthalpy values oe
(c) Answer: The lattice dissociation enthalpy (+771 kJ/mol) and the total hydration enthalpy (-406 + -361 = -767 kJ/mol) are almost equal and opposite; the small enthalpy of solution is the small difference between these two large values.

Question 7

(a) B1 50.0 (g) cao
(a) Answer: 50.0 g
(b) M1 correct substitution: q = 50.0 x 4.18 x (26.0-19.2), ecf from (a)
(b) A1 1.42 (kJ) awrt (accept 1421 J), cao
(b) Answer: q = 1.42 kJ (1421 J)
(c) B1 0.0250 (mol) cao
(c) Answer: 0.0250 mol
(d) M1 correct method: delta Hneut = -q/n, ecf from (b) and (c)
(d) A1 -56.8 (kJ/mol) awrt, cao, ecf
(d) Answer: delta Hneut = -56.8 kJ/mol

Question 8

Level 3 (5-6): A full, coherent explanation that correctly links the partial (incomplete) dissociation of ethanoic acid to an additional endothermic process during the reaction, and explains that this reduces the overall exothermic enthalpy change, using correct chemical terms throughout.
Level 2 (3-4): The answer identifies that ethanoic acid is a weak acid that is only partially dissociated/ionised and links this to a reduced exothermic enthalpy change, but the explanation of the underlying energy changes is incomplete or contains minor inaccuracies.
Level 1 (1-2): A basic statement that ethanoic acid is a weak acid or is only partially dissociated, or that the reaction is still exothermic overall, with little or no linked reasoning about energy changes.
Level 0 (0): No relevant content, or the answer does not relate to the question.
Indicative content:
- Hydrochloric acid is a strong acid and is fully (completely) dissociated/ionised in aqueous solution before the reaction takes place.
- Ethanoic acid is a weak acid and is only partially dissociated/ionised in solution; an equilibrium exists between CH3COOH and CH3COO- + H+.
- The core neutralisation reaction, H+(aq) + OH-(aq) -> H2O(l), releases the same fixed amount of energy in both cases.
- As the reaction with ethanoic acid proceeds, H+ ions are removed from the equilibrium, so more CH3COOH molecules must ionise to replace them; this ionisation step requires an input of energy (it is endothermic).
- This additional endothermic ionisation energy partly cancels out the exothermic enthalpy of neutralisation, so the overall (measured) enthalpy change is less exothermic (smaller in magnitude, less negative) than for a strong acid.
- Correct reference to typical values supports the explanation, e.g. HCl with NaOH is about -57 kJ/mol, while ethanoic acid with NaOH is numerically less negative (a smaller magnitude, e.g. around -50 to -55 kJ/mol), consistent with the smaller temperature rise observed.

Question 9

(a) B1 lattice enthalpy becomes less exothermic (decreases in magnitude) from NaF to NaI oe
(a) Answer: Lattice enthalpy becomes less exothermic (decreases in magnitude) going from NaF to NaI.
(b) M1 ionic radius of the halide ion increases from F- to I- (going down the group)
(b) M1 as ionic radius increases, the distance between the centres of the oppositely charged ions (Na+ and the halide ion) increases
(b) M1 this increased distance weakens the electrostatic force of attraction between the ions, resulting in a lower (less exothermic) lattice enthalpy oe
(b) Answer: Ionic radius increases from F- to I-, increasing the distance between ion centres; this weakens the electrostatic attraction between the ions, giving a smaller (less exothermic) lattice enthalpy.
(c) M1 the ions in MgO carry a greater ionic charge (2+ and 2-) than those in NaCl (1+ and 1-)
(c) M1 greater ionic charge results in a much stronger electrostatic force of attraction between the ions, increasing the lattice enthalpy oe
(c) Answer: MgO ions have greater charge (2+ and 2-) than NaCl ions (1+ and 1-); greater charge gives a much stronger electrostatic attraction, and so a much larger (more exothermic) lattice enthalpy.

Question 10

(a) B1 the energy required to remove one electron from each atom in one mole of gaseous atoms
(a) B1 to form one mole of gaseous 1+ ions oe (gaseous state must be specified for full credit)
(a) Answer: The energy required to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous 1+ ions.
(b) B1 atomisation is defined as forming one mole of gaseous atoms; Cl2(g) -> 2Cl(g) produces 2 mol of atoms per mole of Cl2, so the value must be expressed (halved) per mole of atoms oe
(b) Answer: Enthalpy of atomisation is defined per mole of gaseous atoms formed; since 1 mol of Cl2 gives 2 mol of Cl atoms, the value is expressed per mole of atoms.
(c) M1 correct cycle relationship set up: delta Hf = delta Hat(Mg) + IE1 + IE2 + 2 x delta Hat(Cl) + 2 x EA(Cl) + delta H(lattice) (oe, allow cycle drawn or described)
(c) M1 correct doubling of the chlorine terms: 2 x 122 and 2 x (-349), since 2 mol of Cl are needed per mole of MgCl2
(c) A1 correct sum of all non-lattice terms: 148 + 738 + 1451 + 244 - 698 = 1883 (kJ/mol)
(c) A1 delta H(lattice) = -641 - 1883 = -2524 (kJ/mol) cao
(c) Answer: Lattice enthalpy of formation of MgCl2 = -2524 kJ/mol
(d) B1 the experimental (Born-Haber) value is more exothermic/larger in magnitude than the theoretical ionic value, ecf from (c)
(d) B1 this indicates that the bonding in MgCl2 has some covalent character in addition to ionic bonding (the ionic model is not a perfect description) oe
(d) Answer: The experimental value (-2524 kJ/mol) is more exothermic than the theoretical ionic value (-2326 kJ/mol), suggesting the bonding in MgCl2 has some covalent character (polarisation of the Cl- ion), so it is not purely ionic.

Question 11

(a) M1 correct conversion so that delta H and T x delta S are in the same units (e.g. delta H = -92000 J)
(a) M1 correct substitution: delta G = -92000 - (298 x -199)
(a) A1 -32.7 (kJ/mol) awrt, cao
(a) Answer: delta G = -32.7 kJ/mol
(b) B1 yes, feasible, because delta G is negative at this temperature, ecf from (a)
(b) Answer: Yes, feasible - delta G is negative (-32.7 kJ/mol) at 298 K.
(c) M1 delta G = 0, so delta H = T x delta S, giving T = delta H / delta S
(c) M1 correct substitution using consistent units: T = -92000 / -199
(c) A1 462 (K) awrt, cao
(c) Answer: T = 462 K
(d) M1 above the temperature found in (c), delta G for the forward reaction becomes positive, so the reverse reaction (decomposition of ammonia) becomes more thermodynamically favourable, and the equilibrium yield of ammonia is lower oe
(d) M1 however, a higher temperature increases the rate of reaction (more particles have energy greater than or equal to the activation energy, so more frequent successful collisions), so equilibrium is reached faster oe
(d) M1 450 degC is chosen as a compromise between a reasonable rate of reaction and an acceptable (economically viable) yield, since delta G determines feasibility/position of equilibrium but not the rate at which equilibrium is reached oe
(d) Answer: Above 462 K, delta G for the forward reaction becomes positive so the equilibrium yield of ammonia falls, but a higher temperature increases the rate of reaction; 450 degC is a compromise between reaction rate and equilibrium yield, since delta G governs feasibility/yield, not rate.

Question 12

(a) B1 ± 1.0 (degC) cao
(a) Answer: ± 1.0 degC
(b) B1 4.5 (%) awrt, ecf from (a)
(b) Answer: 4.5%
(c) M1 total absolute uncertainty in mass = 0.01 + 0.01 = ± 0.02 (g)
(c) A1 2.4 (%) awrt (accept 2.3-2.4%), cao
(c) Answer: 2.4%
(d) M1 total % uncertainty = sum of individual % uncertainties = 4.5 + 2.4 (= 6.9%), ecf from (b) and (c)
(d) M1 absolute uncertainty = 6.9% of 755 (= 52.1 kJ/mol), ecf
(d) A1 delta Hc = -755 (± 52) kJ/mol (accept sensible rounding, e.g. -760 ± 50), ecf
(d) Answer: delta Hc = -755 ± 52 kJ/mol

Question 13

(a) B1 a gas (CO2) is produced from a solid reactant only, so the number of moles of gas (and the disorder) increases oe
(a) Answer: A gas (CO2) is formed from solids only, increasing the number of gas particles and so increasing entropy/disorder.
(b) M1 delta G = 0, so T = delta H / delta S
(b) M1 correct substitution using consistent units: T = 178000 / 161
(b) A1 1110 (K) awrt (accept 1100-1110 K; exact value 1105.6 K), cao
(b) Answer: T = 1110 K (awrt; exact value 1105.6 K)

Question 14

Level 3 (5-6): A full and coherent explanation that correctly links the positive delta H and positive delta S to delta G = delta H - T x delta S, correctly explains that at low T the T x delta S term is small so delta G is positive (not feasible), and that as T increases the T x delta S term grows, eventually making delta G negative, with reference to the calculated minimum feasibility temperature.
Level 2 (3-4): The answer shows understanding that delta G depends on both delta H and delta S and on temperature, and that increasing temperature makes delta G more negative (or includes a correct comparison of the sign of delta G at different temperatures), but the reasoning is incomplete or lacks full linkage/detail.
Level 1 (1-2): Basic recognition that the reaction is endothermic and/or that entropy increases and/or that a high temperature is needed, without clear linkage to the Gibbs free energy equation or feasibility.
Level 0 (0): No relevant content, or the answer does not relate to the question.
Indicative content:
- delta H is positive (endothermic, +178 kJ/mol) and delta S is positive (+161 J/(K mol)) for this reaction.
- Feasibility (spontaneity) is determined by the sign of delta G, where delta G = delta H - T x delta S.
- At 298 K, T x delta S = 298 x 161 = 47978 J = 48.0 kJ (approximately), which is smaller in magnitude than delta H (+178 kJ), so delta G = delta H - T x delta S is positive; the reaction is not feasible at room temperature.
- As temperature increases, the magnitude of the T x delta S term increases, while delta H and delta S are assumed to remain roughly constant with temperature.
- Above the calculated minimum feasibility temperature (approximately 1106 K, i.e. 1110 K to 3 s.f.), T x delta S becomes greater in magnitude than delta H, so delta G becomes negative and the reaction becomes thermodynamically feasible.
- At 1200 K (the temperature used industrially), the temperature is above this minimum feasibility temperature, so delta G is negative and decomposition is thermodynamically favourable/feasible at this temperature.

Energetics and Thermodynamics - Worksheets, Questions and Revision