Chemistry • Year 11 • Module 4 • Lesson 10

Hess’s Law Applied — Heat of Combustion & Consolidation

Extended synthesis and evaluation: real data, method comparison, and evidence-based judgement at Band 5–6 level.

Master · Extended Response

1. Data + scenario — Syngas from coal gasification (Shell Prelude context)

8 marks

Scenario. Shell’s Prelude FLNG facility, moored 475 km north-northeast of Broome, Western Australia, processes natural gas extracted from the deep-sea Browse Basin. A related industrial application is coal gasification — the conversion of coal (modelled here as carbon, C(graphite)) to synthesis gas (syngas), a mixture of CO(g) and H2(g), used as a fuel and chemical feedstock. The target reaction for syngas production from coal and steam is:

Target:   C(graphite) + H2O(g) → CO(g) + H2(g)     ΔHtarget = ?

Data — three known thermochemical equations:

(1)   C(graphite) + O2(g) → CO2(g)                  ΔH1 = −393.5 kJ mol−1

(2)   CO(g) + ½O2(g) → CO2(g)                   ΔH2 = −283.0 kJ mol−1

(3)   H2(g) + ½O2(g) → H2O(g)                   ΔH3 = −241.8 kJ mol−1

Note: H2O(g) is used in both the target equation and equation (3) — state symbol must be consistent throughout.

In your response you must:

  1. Identify which calculation method is appropriate here and justify your choice based on the data provided.
  2. Show the full Hess’s Law manipulation: for each of equations (1), (2), (3), state whether you reverse it, scale it, or use it as-is, and give the modified ΔH value.
  3. Verify the target equation by showing intermediate cancellation.
  4. State the final value of ΔHtarget with correct sign, magnitude, and units.
  5. Classify the syngas-forming reaction as exothermic or endothermic and explain its industrial significance — specifically why an endothermic or exothermic reaction changes how an industrial engineer would design a reactor.
  6. Calculate the energy per gram of CO(g) released when syngas burns in air as CO(g) + ½O2(g) → CO2(g). Use ΔH2 and M(CO) = 28.01 g mol−1. Compare this to methane (55.5 kJ g−1) and comment on syngas value as a fuel.

Write your full response here. Show all working steps.

2. Data + scenario — Comparing methods for biodiesel ΔHc determination

7 marks

Scenario. A chemical engineering team is evaluating three methods to determine the enthalpy of combustion (ΔHc) of ethyl oleate (C20H38O2), a biodiesel compound produced from the transesterification of olive oil. The three candidate methods are:

  • Method A — Bond energy method using tabulated average bond enthalpies.
  • Method B — ΔH°f method using standard enthalpies of formation from a thermochemical database.
  • Method C — Hess’s Law combining accurate calorimetric combustion data for three related ester compounds whose equations can be manipulated to give the target reaction.

Results obtained (kJ mol−1): Method A: −11 450 • Method B: −12 660 • Method C: −12 580.

Literature value for ΔHc[ethyl oleate]: −12 610 kJ mol−1.

In your response you must:

  1. Define the term standard enthalpy of combustion (ΔH°c) with reference to the conditions under which it is measured.
  2. Calculate the percentage error for each method relative to the literature value. Show the formula and working.
  3. Identify which method is most accurate and which is least accurate. Justify each in terms of the source and nature of the data used.
  4. Explain specifically why Method A gives the largest discrepancy for this large molecule. Your explanation must refer to: (a) the assumption about physical states; (b) the cumulative effect of errors across many bonds.
  5. The engineering team intends to scale up ethyl oleate combustion to generate 500 MJ of heat energy in a reactor. Using Method B’s value, calculate how many moles — and how many kilograms (M = 310.51 g mol−1) — of ethyl oleate must be burned to deliver 500 MJ.

Write your full response here. Show all working steps.

Answers — Do not peek before attempting

Q1 — Syngas from coal gasification (8 marks)

Criterion 1 — Method identification (1 mark): Hess’s Law (multi-step equation method), because a set of three thermochemical equations is provided. The ΔH°f method cannot be used because ΔH°f data for all species is not given; the bond energy method is not applicable because thermochemical equations, not bond enthalpies, are provided.

Criterion 2 — Equation manipulations (3 marks):

  • Equation (1): C(graphite) needs to be a reactant → use as-is. ΔH = −393.5 kJ mol−1.
  • Equation (2): CO(g) needs to be a product, but in (2) it is a reactant → reverse equation (2). CO2(g) → CO(g) + ½O2(g). ΔH = +283.0 kJ mol−1.
  • Equation (3): H2O(g) needs to be a reactant, but in (3) it is a product → reverse equation (3). H2O(g) → H2(g) + ½O2(g). ΔH = +241.8 kJ mol−1.

Criterion 3 — Cancellation and verification (1 mark): Add: C(g) + O2 + CO2(g) + H2O(g) → CO2(g) + CO(g) + ½O2(g) + H2(g) + ½O2(g). Cancel CO2(g) both sides; cancel O2 + ½O2 = &frac32;O2 − O2 = ½O2 from each side. Remaining: C(graphite) + H2O(g) → CO(g) + H2(g) — this matches the target exactly.

Criterion 4 — Final ΔH value (1 mark): ΔHtarget = −393.5 + 283.0 + 241.8 = +131.3 kJ mol−1.

Criterion 5 — Classification and industrial significance (1 mark): The reaction is endothermic (+131.3 kJ mol−1). An industrial engineer must supply heat energy to drive the reaction forward — typically by partial combustion of some coal to provide this heat (auto-thermal gasification), or by external heating. If the reactor heat supply falls, the reaction slows or reverses. This requires careful thermal management, reactor insulation, and continuous heat input, unlike an exothermic reaction which must be cooled to prevent runaway.

Criterion 6 — Energy per gram of CO and comparison (1 mark): ΔHc[CO] = −283.0 kJ mol−1. Eg(CO) = 283.0 ÷ 28.01 = 10.1 kJ g−1. This is far below methane (55.5 kJ g−1). Syngas’s value as a fuel lies not in its energy density by mass but in (a) the inclusion of H2(g) which is much more energy-dense per gram (H2(g) energy per gram = 286 ÷ 2.02 = 141.6 kJ g−1), so the combined CO + H2 mixture has a significantly higher effective energy density than CO alone, and (b) its ability to be burned cleanly in industrial burners and gas turbines, producing CO2 and H2O as the only combustion products.

Q2 — Biodiesel method comparison (7 marks)

Criterion 1 — Definition (1 mark): The standard enthalpy of combustion (ΔH°c) is the enthalpy change when exactly one mole of a substance burns completely in excess O2 under standard conditions (25°C, 100 kPa), with all reactants and products in their standard states (e.g. H2O(l), CO2(g)). The value is always negative for combustion (exothermic).

Criterion 2 — Percentage errors (2 marks; 0.5 each with working):
Formula: % error = |(experimental − literature)| ÷ |literature| × 100.
Method A: |−11 450 − (−12 610)| ÷ 12 610 × 100 = 1160 ÷ 12 610 × 100 = 9.2%.
Method B: |−12 660 − (−12 610)| ÷ 12 610 × 100 = 50 ÷ 12 610 × 100 = 0.40%.
Method C: |−12 580 − (−12 610)| ÷ 12 610 × 100 = 30 ÷ 12 610 × 100 = 0.24%.

Criterion 3 — Ranking and justification (2 marks): Most accurate: Method C (0.24% error) — uses precise experimental calorimetric data for real ester compounds; no averaging involved; the Hess’s Law combination uses compound-specific ΔH values. Second: Method B (0.40%) — uses ΔH°f values measured experimentally for each specific substance in its actual state at 25°C; very accurate when high-quality database values are available for the molecule. Least accurate: Method A (9.2%) — uses average bond enthalpies and assumes all species are gaseous.

Criterion 4 — Why Method A diverges most for a large molecule (1 mark): (a) The bond energy method assumes gaseous products including H2O(g), but the correct product is H2O(l) for a combustion at standard conditions; the latent heat of condensation is not counted. For ethyl oleate, 19 moles of H2O are produced — missing this contributes ∼19 × 44 = 836 kJ mol−1 of error. (b) Each of the many bond types in C20H38O2 (C–H, C–C, C=C, C–O, C=O, O–H, and O=O) has a slightly different actual value from the tabulated average; over the large number of bonds in this molecule, these small errors accumulate into a 9.2% total discrepancy.

Criterion 5 — Scale-up calculation (1 mark): Energy needed = 500 MJ = 500 000 kJ. n(ethyl oleate) = 500 000 ÷ 12 660 = 39.5 mol. Mass = 39.5 × 310.51 = 12 265 g ≈ 12.3 kg.

HSCScience • Chemistry • Year 11 • Module 4 • Lesson 10 • Worksheet 3 (Master)