Hess's Law Applied — Heat of Combustion & Consolidation
In 2003, Brazil's National Agency of Petroleum, Natural Gas and Biofuels (ANP) calculated that ethanol fuel delivers 26.8 MJ L⁻¹ versus petrol's 34.2 MJ L⁻¹ — a 22% shortfall per litre, meaning Brazil's flex-fuel vehicles needed 22% more fuel to cover the same distance. ANP engineers used all three enthalpy calculation methods to arrive at this figure: bond energies for rapid screening, ΔH°f values for accuracy, and Hess's Law for the multi-step oxidation pathways. This lesson is your version of that consolidation.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
A fuel with a larger ΔHc (enthalpy of combustion in kJ mol⁻¹) always gives you more energy for a given mass of fuel. True or false?
Think about ethanol (C₂H₅OH, ΔHc = −1367 kJ mol⁻¹, M = 46.07 g mol⁻¹) vs methanol (CH₃OH, ΔHc = −726 kJ mol⁻¹, M = 32.04 g mol⁻¹). Ethanol releases almost twice as much energy per mole. But if you fill a fuel tank by mass — which fuel gives you more energy per kilogram?
Before this lesson: Calculate (or estimate) energy per gram for each fuel. Does the fuel with higher ΔHc per mole also have higher energy per gram? Write your prediction with any working.
Key Facts
- Energy per gram = |ΔHc| ÷ M (kJ g⁻¹) — the mass-based fuel comparison
- Bond energy method → approximate; ΔH°f method → more accurate; Hess's Law → depends on data quality
- ΔH°f method and Hess's Law combustion cycle are mathematically equivalent
Concepts
- Why the three ΔH calculation methods are equivalent when applied to the same reaction
- Why a higher ΔHc per mole does not guarantee higher energy per gram
- How to select the correct method from the data provided in an HSC question
Skills
- Calculate ΔHc using ΔH°f data (products − reactants, scaled coefficients)
- Calculate and compare energy per gram for two fuels
- Identify which ΔH method to use from question context and data type provided
In HSC, the method is almost always signalled by the data provided. The question tells you which method to use — your job is to recognise the signal and apply the correct procedure.
Table of bond enthalpies only
Table of ΔH°f values
Set of thermochemical equations
Mix of ΔH°f and experimental ΔH
Why do the methods give different numerical results?
- Bond energy: uses average values across many molecular contexts; assumes all species are gaseous (even liquids); introduces cumulative error across multiple bonds
- ΔH°f: uses experimentally measured values for each specific substance in its actual state at 25°C; no averaging; matches real conditions → most accurate for standard conditions
- Hess's Law: inherits the accuracy of whatever ΔH values are used as input; if using precise experimental values → highly accurate; if using estimated values → less so
Three ΔH calculation methods: (1) calorimetry — direct measurement via q = mcΔT; (2) bond energies — structural formula analysis, gives estimates; (3) Hess’s Law / formation enthalpies — most accurate, uses measured ΔH°f data. The data provided in the question signals which method to use.
Pause — copy the highlighted definition into your book before moving on.
Quick check: A student is given a table of ΔH°f values and is asked to calculate ΔH for the combustion of ethanol. Which method should they use?
We just saw how to choose the right ΔH calculation method based on the data provided. That raises a question: when comparing fuels, should we use molar enthalpy or energy per gram — and does it matter which we choose? This card answers it → because the two measures can give different rankings, each valid for a different purpose.
Molar enthalpy of combustion (kJ mol⁻¹) is the chemist's comparison. Energy per gram (kJ g⁻¹) is the engineer's comparison. Neither is universally "better" — you must specify which comparison you are making.
Fuel efficiency data — selected fuels:
| Fuel | Formula | M (g mol⁻¹) | ΔHc (kJ mol⁻¹) | Energy per gram (kJ g⁻¹) |
|---|---|---|---|---|
| Hydrogen | H₂(g) | 2.02 | −286 | 141.6 |
| Methane (natural gas) | CH₄(g) | 16.04 | −890 | 55.5 |
| Octane (petrol) | C₈H₁₈(l) | 114.23 | −5471 | 47.9 |
| Ethanol | C₂H₅OH(l) | 46.07 | −1367 | 29.7 |
| Methanol | CH₃OH(l) | 32.04 | −726 | 22.7 |
Key observations from the data:
- Hydrogen has by far the highest energy per gram (141.6 kJ g⁻¹) — this is why it is used in space rockets where mass matters critically
- Octane has higher energy per gram than ethanol (47.9 vs 29.7 kJ g⁻¹) — so an ethanol-fuelled car needs more fuel by mass for the same range
- Methanol has lower energy per gram than ethanol, despite being a simpler molecule — because its lower molar mass does not compensate enough for its lower ΔHc
Molar enthalpy of combustion (kJ mol¹) compares equal moles; energy per gram (kJ g¹) compares equal mass. Energy per gram = |ΔHc| ÷ M. These measures can give different fuel rankings — always specify which comparison is being made.
Add the highlighted point to your notes before the check below.
Explain it: A student claims: "Octane must give more energy per gram than ethanol because it releases 5471 kJ mol⁻¹ compared to ethanol's 1367 kJ mol⁻¹." Explain why this reasoning is flawed and describe the correct approach.
We just saw that molar vs per-gram comparisons give different fuel rankings, each valid for different purposes. That raises a question: are the formation method and Hess’s Law combustion cycle two separate methods, or expressions of the same principle? This card answers it → because ΔH is a state function, both approaches always give identical results.
The enthalpy of combustion of any fuel can be calculated from standard enthalpies of formation — and this is mathematically identical to using the Hess's Law combustion cycle via elements. Understanding why they are equivalent is the key consolidation insight of this lesson.
The ΔH°f method — direct application:
For the combustion of propan-1-ol: C₃H₇OH(l) + 9/2 O₂(g) → 3CO₂(g) + 4H₂O(l)
- Step 1: Sum ΔH°f(products) × stoichiometric coefficients
- Step 2: Sum ΔH°f(reactants) × stoichiometric coefficients
- Step 3: ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants)
- O₂(g) always has ΔH°f = 0 — include the step to show working, even though it adds zero
Why ΔH°f and Hess's Law combustion cycle give the same answer:
The Hess's Law combustion cycle uses three equations:
(Reverse) Fuel → Products: ΔH = ΔHc[fuel]
(Forward) Elements → Products: ΔH = ΣΔH°f[products]
∴ ΔH°f[fuel] + ΔHc[fuel] = ΣΔH°f[products]
∴ ΔHc[fuel] = ΣΔH°f[products] − ΔH°f[fuel]
— identical to the ΔH°f formula
ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants), each multiplied by stoichiometric coefficients. This is mathematically identical to a Hess’s Law combustion cycle because ΔH is a state function — path independent. ΔH°f[O&sub2;(g)] = 0.
Pause — write the highlighted equation into your book.
Fill in the blanks: Complete the ΔH°f method procedure. ΔH°rxn = Σ_____(products) − Σ_____(reactants). Multiply each value by its _____ coefficient. O₂(g) always has ΔH°f = _____. Products come _____ in the formula (first/second).
We just saw that the formation method and Hess’s Law are equivalent via the state function principle. That raises a question: what misconceptions trip students up when choosing and applying ΔH methods in exams? This card answers it → with three specific traps that HSC marking guidelines repeatedly penalise.
Three persistent misconceptions about ΔH calculation methods appear repeatedly in HSC marking guidelines. Identifying and fixing these earns you marks that weaker students reliably lose.
Misconception 1: Higher ΔHc per mole always means more energy per gram.
Why students think this: ΔHc is the most visible number and larger magnitude seems "better."
What is actually true: Energy per gram = |ΔHc| ÷ M. A fuel can have large ΔHc per mole but low energy per gram if its molar mass is large. Always calculate both before comparing fuels.
Misconception 2: The bond energy method and ΔH°f method always give the same ΔH for a reaction.
Why students think this: Both methods calculate ΔH for the same reaction, so the answer should be the same.
What is actually true: Bond energies are averages for bond types across many molecules; ΔH°f values are experimentally measured for specific substances in actual states. They give different numerical results, with ΔH°f being more accurate.
Misconception 3: Hess's Law only works when you have exactly two equations to add together.
Why students think this: The NESA prototype (L08) uses two equations, and students generalise from this.
What is actually true: Hess's Law works with any number of equations. The ΔH°f method is implicitly a Hess's Law cycle using as many formation equations as there are species in the balanced equation. The principle (path independence of enthalpy) is not limited by the number of steps.
Method Selection Guide — copy into your book
Method Selection
- Bond energies given → bond energy method (approximate)
- ΔHf° table given → enthalpy of formation method (accurate)
- Thermochemical equations given → Hess's Law
- Never use bond energy method when ΔHf° data is available
Energy per Gram
- E (kJ g⁻¹) = |ΔHc| ÷ M
- Higher ΔHc per mole ≠ higher energy per gram
- Hydrogen: 141.6 · Octane: 47.9 · Ethanol: 29.7
Hess's Law Combustion Cycle
- Triangle: elements → fuel, elements → products, fuel → products
- ΔHf°(fuel) + ΔHc = ΣΔHf°(products)
- ∴ ΔHc = ΣΔHf°(products) − ΔHf°(fuel) — identical to ΔHf° formula
Three exam-critical misconceptions: (1) higher ΔHc per mole always means more energy per gram — wrong, divide by molar mass; (2) ΔH°f of all substances is zero — wrong, only elements in standard state; (3) any method suits any data — wrong, data provided dictates the method.
Add the highlighted point to your notes before the check below.
True or False: Hess's Law only applies when exactly two thermochemical equations are added together.
Worked examples · reveal as you go
Enthalpy of combustion of propan-1-ol. Calculate the standard enthalpy of combustion of propan-1-ol (C₃H₇OH) using ΔH°f data:
C₃H₇OH(l) + 9/2 O₂(g) → 3CO₂(g) + 4H₂O(l)
ΔH°f values: C₃H₇OH(l) = −303 kJ mol⁻¹ · CO₂(g) = −393.5 kJ mol⁻¹ · H₂O(l) = −285.8 kJ mol⁻¹ · O₂(g) = 0
Then calculate the energy per gram. (M = 60.10 g mol⁻¹)
ΣΔH°f(products) = 3(−393.5) + 4(−285.8) = −1180.5 + (−1143.2) = −2323.7 kJ mol⁻¹
ΣΔH°f(reactants) = 1(−303) + (9/2)(0) = −303 + 0 = −303 kJ mol⁻¹
ΔH°rxn = −2323.7 − (−303) = −2323.7 + 303 = −2020.7 kJ mol⁻¹
Eg = |ΔHc| ÷ M = 2020.7 ÷ 60.10 = 33.6 kJ g⁻¹
Comparison: propan-1-ol (33.6) > ethanol (29.7 kJ g⁻¹) per gram, despite ethanol having a smaller absolute ΔHc.
Fuel efficiency comparison: Methanol vs Ethanol. Methanol (CH₃OH, M = 32.04 g mol⁻¹, ΔHc = −726 kJ mol⁻¹) and ethanol (C₂H₅OH, M = 46.07 g mol⁻¹, ΔHc = −1367 kJ mol⁻¹) are compared as alternative automotive fuels.
(a) Which releases more energy per mole? (b) Which releases more energy per gram? (c) Suggest two reasons other than energy content why ethanol is often preferred as a fuel additive over methanol.
Methanol: |ΔHc| = 726 kJ mol⁻¹ · Ethanol: |ΔHc| = 1367 kJ mol⁻¹
Ethanol releases more energy per mole — almost twice as much.
Methanol: 726 ÷ 32.04 = 22.7 kJ g⁻¹
Ethanol: 1367 ÷ 46.07 = 29.7 kJ g⁻¹
Ethanol also releases more energy per gram — 29.7 vs 22.7 kJ g⁻¹.
1. Toxicity: Methanol is highly toxic — ingestion or inhalation of small quantities causes blindness and death; ethanol far less acutely toxic.
2. Renewability: Ethanol is efficiently produced by fermentation of plant sugars (sugarcane, corn) — renewable; methanol primarily from natural gas (non-renewable).
(a) Ethanol releases more per mole (1367 vs 726 kJ mol⁻¹)
(b) Ethanol also releases more per gram (29.7 vs 22.7 kJ g⁻¹)
(c) Toxicity; renewability; miscibility with petrol; lower net CO₂ — any two valid reasons
Calculate ΔH for the combustion of methane using standard enthalpies of formation. ΔH°f: CO&sub2; = −393, H&sub2;O(l) = −286, CH&sub4; = −75 kJ mol¹.
ΔH° = [ΔH°f(CO&sub2;) + 2×ΔH°f(H&sub2;O)] − ΔH°f(CH&sub4;)
= (−393) + 2×() − ()
= −393 + () + 75
= kJ mol¹
Formula Reference — This Lesson
Common errors · the 3 traps that cost marks
"Reactants minus products" in the ΔH°f formula
Students write ΔH°rxn = ΣΔH°f(reactants) − ΣΔH°f(products) — reversing the formula.
Fix: The ΔH°f formula is products minus reactants: ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants). This is the opposite of the bond energy formula (reactants minus products). Remember: "P − R" for ΔH°f; "R − P" for bond energy. Confusing these gives the wrong sign for ΔH every time.
"Higher ΔHc per mole = more energy per gram"
Students compare fuels using ΔHc per mole values alone and conclude the larger value means "more efficient fuel."
Fix: You must divide by molar mass: Eg = |ΔHc| ÷ M. A fuel can have a massive ΔHc per mole but low energy per gram if its molar mass is high. Always calculate both comparisons and specify which one you are using.
Forgetting to include ΔH°f[O₂(g)] = 0 step
Students skip the step of writing 1(−303) + (9/2)(0) and just write ΣΔH°f(reactants) = −303, without showing O₂ explicitly.
Fix: Always write out O₂ in the ΣΔH°f(reactants) sum explicitly — even though ΔH°f[O₂(g)] = 0, the step shows you understand that all species must be included in the sum. Marking guidelines commonly award a method mark for this step.
Quick-fire practice · 5 reps +2 XP per reveal
Hess's Law — ΔH°f of ethane: Given: (1) C(graphite) + O₂(g) → CO₂(g), ΔH = −393.5 kJ mol⁻¹; (2) H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8 kJ mol⁻¹; (3) C₂H₆(g) + 7/2 O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH = −1560.7 kJ mol⁻¹.
Calculate ΔH°f[C₂H₆(g)] using Hess's Law.
Scale (1) ×2: 2C + 2O₂ → 2CO₂, ΔH = 2(−393.5) = −787.0 kJ mol⁻¹
Scale (2) ×3: 3H₂ + 3/2 O₂ → 3H₂O, ΔH = 3(−285.8) = −857.4 kJ mol⁻¹
Reverse (3): 2CO₂ + 3H₂O → C₂H₆ + 7/2 O₂, ΔH = +1560.7 kJ mol⁻¹
Cancel: 2CO₂ both sides ✓ | 3H₂O both sides ✓ | O₂ terms cancel ✓
ΔH°f = −787.0 + (−857.4) + 1560.7 = −83.7 kJ mol⁻¹
ΔH°f method — butan-1-ol combustion: Write the balanced combustion equation for C₄H₉OH and calculate ΔHc.
ΔH°f (kJ mol⁻¹): C₄H₉OH(l) = −327; CO₂(g) = −393.5; H₂O(l) = −285.8; O₂(g) = 0.
ΣΔH°f(products) = 4(−393.5) + 5(−285.8) = −1574.0 + (−1429.0) = −3003.0 kJ mol⁻¹
ΣΔH°f(reactants) = 1(−327) + 6(0) = −327 kJ mol⁻¹
ΔHc = −3003.0 − (−327) = −2676 kJ mol⁻¹
Energy per gram trend: Calculate energy per gram for the four C1–C4 alcohols and describe the trend.
ΔHc: methanol = −726; ethanol = −1367; propan-1-ol = −2021; butan-1-ol = −2676 kJ mol⁻¹
M (g mol⁻¹): 32.04; 46.07; 60.10; 74.12
Ethanol: 1367 ÷ 46.07 = 29.7 kJ g⁻¹
Propan-1-ol: 2021 ÷ 60.10 = 33.6 kJ g⁻¹
Butan-1-ol: 2676 ÷ 74.12 = 36.1 kJ g⁻¹
Trend: Energy per gram consistently increases with chain length — each CH₂ group increases ΔHc faster than it increases molar mass. Longer-chain alcohols approach (but do not reach) octane's 47.9 kJ g⁻¹.
Method identification: For each scenario, identify which ΔH calculation method should be used:
(a) A question provides average C–H and O=O bond enthalpies only.
(b) A question provides ΔH°f values for all reactants and products.
(c) A question provides three balanced thermochemical equations and asks you to find ΔH for a fourth reaction.
(d) A question provides both ΔH°f and bond data and asks you to compare two methods.
(b) Enthalpy of formation method — ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants); most accurate
(c) Hess's Law — reverse/scale the three given equations to cancel intermediates and obtain the target
(d) Use ΔH°f method for the accurate value; use bond energy separately for comparison; explain why they differ (averages vs experimental values, gaseous state assumption)
Extension — E10 fuel calculation: A car's fuel tank holds 50 L of E10 (10% ethanol by volume). The ethanol component has density = 0.789 g mL⁻¹, M = 46.07 g mol⁻¹, ΔHc = −1367 kJ mol⁻¹. Estimate the energy (in MJ) released by the ethanol component in a full tank.
Mass = 5000 mL × 0.789 g mL⁻¹ = 3945 g
n(ethanol) = 3945 ÷ 46.07 = 85.6 mol
Energy = 85.6 × 1367 = 117,000 kJ = 117 MJ
Go back to your Think First prediction. The statement was: "A fuel with a larger ΔHc per mole always gives more energy per gram." Let's resolve it precisely — connecting to the 2003 Brazil ANP ethanol vs petrol comparison at 26.8 vs 34.2 MJ L⁻¹:
- The statement is FALSE — but context-dependent. Brazil's ANP found ethanol delivers 22% less energy per litre than petrol — but ethanol has a higher energy-to-carbon ratio and lower CO₂ per MJ. Per mole, per gram, and per litre can give different rankings. For arbitrary fuel pairs, the comparison can go either way — always calculate all three.
- The calculation method must match the data given. Bond energies → bond energy method; ΔH°f table → ΔH°f method; set of equations → Hess's Law. Read the question, identify the data type, select accordingly — exactly as ANP engineers did in 2003.
- All three methods are connected. They all derive from Hess's Law and conservation of energy. Their differences are in the precision of input data, not in principle. The same thermochemical reasoning that benchmarked Brazil's biofuel policy is what you've built across L06–L10.
Pick your answer, then rate your confidence — that tells the system what to drill next.
Wrong: Use the bond energy method whenever ΔH data is available.
Right: Match the method to the data type. Bond energy data → bond energy method. ΔH°f table → ΔH°f method (more accurate). Thermochemical equations → Hess's Law. Using bond energy when ΔH°f is given wastes time and loses accuracy.
Wrong: "A fuel with higher ΔHc per mole gives more energy per gram."
Right: E(per gram) = |ΔHc| ÷ M. You must divide by molar mass. The relationship between per-mole and per-gram comparisons depends on the specific ΔHc and M values — it is not automatic.
Wrong: "The ΔH°f formula puts reactants first."
Right: ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants). Products minus reactants. This is the opposite of the bond energy formula. Getting this backwards flips the sign of every answer.
Q1. A student uses two different methods to calculate ΔH for the combustion of methane (CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)).
Method 1 (Bond energy): Bond energies: C–H = 413; O=O = 498; C=O = 743 (in CO₂); O–H = 463 kJ mol⁻¹. ΔH = −674 kJ mol⁻¹
Method 2 (ΔH°f): ΔH°f: CH₄(g) = −74.8; CO₂(g) = −393.5; H₂O(l) = −285.8 kJ mol⁻¹. ΔH = −890.3 kJ mol⁻¹
(a) Which result is more accurate? Justify in terms of the data used. (2 marks)
(b) The bond energy method gave −674 kJ mol⁻¹ while ΔH°f gave −890.3 kJ mol⁻¹ — a difference of 216 kJ mol⁻¹. Identify and explain TWO specific reasons for this large discrepancy. (4 marks)
6 MARKS
Q2. Propan-1-ol (C₃H₇OH, M = 60.10 g mol⁻¹) and butan-1-ol (C₄H₉OH, M = 74.12 g mol⁻¹) are being evaluated as biofuel alternatives to octane.
(a) Using ΔH°f data, confirm ΔHc[propan-1-ol] = −2021 kJ mol⁻¹. (ΔH°f: C₃H₇OH(l) = −303; CO₂(g) = −393.5; H₂O(l) = −285.8 kJ mol⁻¹) (3 marks)
(b) Calculate the energy per gram for propan-1-ol and butan-1-ol (ΔHc[butan-1-ol] = −2676 kJ mol⁻¹). Which has higher energy density per gram? (2 marks)
(c) Both values are lower than octane (47.9 kJ g⁻¹). Suggest one engineering advantage of alcohol fuels over octane that offsets their lower energy density. (1 mark)
6 MARKS
Q3. Australia's transport sector is exploring ethanol (E10 blend, 10% ethanol by volume) as a fuel additive to petrol.
(a) Calculate ΔHc[ethanol] from ΔH°f data: C₂H₅OH(l) = −277.7; CO₂(g) = −393.5; H₂O(l) = −285.8 kJ mol⁻¹. Balanced equation: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). (3 marks)
(b) A car fuel tank holds 50 L of fuel. The ethanol component is 10% of volume, density 0.789 g mL⁻¹. Estimate the energy (in MJ) from the ethanol component in a full 50 L tank. (M[ethanol] = 46.07 g mol⁻¹) (3 marks)
(c) The Australian government promotes E10 despite its lower energy density. Evaluate whether energy density alone is a valid basis for fuel policy decisions. (2 marks)
8 MARKS
Show comprehensive answers ▼
Drill 1 — ΔH°f of Ethane (Hess's Law)
Scale (1) ×2: ΔH = −787.0; Scale (2) ×3: ΔH = −857.4; Reverse (3): ΔH = +1560.7. After cancellation: 2C + 3H₂ → C₂H₆ ✓. ΔH°f = −787.0 + (−857.4) + 1560.7 = −83.7 kJ mol⁻¹. Accepted value = −84.7 kJ mol⁻¹ — very close match.
Drill 2 — ΔHc for Butan-1-ol
Balanced: C₄H₉OH(l) + 6O₂(g) → 4CO₂(g) + 5H₂O(l). ΣΔH°f(p) = 4(−393.5) + 5(−285.8) = −3003.0; ΣΔH°f(r) = −327. ΔHc = −3003.0 − (−327) = −2676 kJ mol⁻¹.
Drill 3 — Alcohol Energy per Gram Trend
Methanol: 22.7 | Ethanol: 29.7 | Propan-1-ol: 33.6 | Butan-1-ol: 36.1 kJ g⁻¹. Trend: consistently increases with chain length — each additional CH₂ group increases ΔHc faster than it increases molar mass, approaching (but not reaching) alkane energy density.
Multiple Choice (from question bank)
1. B — When ΔH°f data is provided, use the ΔH°f method — more accurate (experimental data, actual states). Bond energy method is only used when ΔH°f data is unavailable.
2. A — Propane: 2220 ÷ 44.1 = 50.3 kJ g⁻¹; Butane: 2877 ÷ 58.1 = 49.5 kJ g⁻¹. Propane has slightly higher energy per gram. Option D demonstrates the exact misconception this lesson addresses.
3. B — The ΔH°f method uses experimentally measured values for each specific substance in its actual state — most accurate because no averaging is involved.
4. C — ΣΔH°f(p) = 4(−393.5) + 5(−285.8) = −3003.0; ΣΔH°f(r) = −126.2; ΔH = −3003.0 − (−126.2) = −2876.8 ≈ −2877 kJ mol⁻¹.
5. A — The Hess's Law combustion cycle gives ΔHc = ΣΔH°f(products) − ΔH°f(fuel), identical to the ΔH°f formula — mathematically equivalent, different representations of the same principle.
Short Answer Model Answers
Q1 (6 marks):
(a) ΔH°f method gives −890.3 kJ mol⁻¹ — more accurate [½]. It uses experimentally measured values for each substance in its actual state (CO₂(g) and H₂O(l)) — not averages, and not assuming gaseous water [1½].
(b) Reason 1: Bond energy method assumed H₂O(g) as product, but actual product is H₂O(l) [1]. Condensation of 2 mol H₂O releases 2 × 44 ≈ 88 kJ mol⁻¹ that is not counted in the bond energy result — this alone accounts for ~88 of the 216 kJ mol⁻¹ discrepancy [1]. Reason 2: Bond energy values are averages across many molecular environments [1]. The C–H, C=O and O–H bonds have slightly different actual values from tabulated averages; errors accumulate across 8 bond values in this calculation [1].
Q2 (6 marks):
(a) ΣΔH°f(p) = 3(−393.5) + 4(−285.8) = −1180.5 + (−1143.2) = −2323.7 kJ mol⁻¹ [1]; ΣΔH°f(r) = 1(−303) + 9/2(0) = −303 kJ mol⁻¹ [1]; ΔHc = −2323.7 − (−303) = −2020.7 ≈ −2021 kJ mol⁻¹ ✓ [1].
(b) Propan-1-ol: 2021 ÷ 60.10 = 33.6 kJ g⁻¹ [1]; Butan-1-ol: 2676 ÷ 74.12 = 36.1 kJ g⁻¹ [1]; Butan-1-ol has higher energy density per gram [½]. Both still below octane's 47.9 kJ g⁻¹ [½].
(c) Any valid advantage — e.g. higher octane rating (reduces engine knock, allows higher compression ratios) [1]; or renewable/carbon-neutral production from biomass; or reduced CO and particulate emissions; or improved combustion completeness due to oxygen content [1].
Q3 (8 marks):
(a) ΣΔH°f(p) = 2(−393.5) + 3(−285.8) = −787.0 + (−857.4) = −1644.4 kJ mol⁻¹ [1]; ΣΔH°f(r) = 1(−277.7) + 3(0) = −277.7 kJ mol⁻¹ [1]; ΔHc = −1644.4 − (−277.7) = −1366.7 ≈ −1367 kJ mol⁻¹ [1].
(b) Volume ethanol = 10% × 50 L = 5 L = 5000 mL [½]. Mass = 5000 × 0.789 = 3945 g [½]. n = 3945 ÷ 46.07 = 85.6 mol [1]. Energy = 85.6 × 1367 = 117,000 kJ = 117 MJ [1].
(c) Energy density alone is not a valid sole basis for fuel policy [½]. Other critical factors: (1) Renewability — ethanol from sugarcane is carbon-neutral over its lifecycle, reducing greenhouse gas emissions, a policy priority that outweighs the minor reduction in fuel economy [1]. (2) Energy security — domestic production reduces oil import dependence [½]. (3) Co-benefits — higher oxygen content reduces CO and hydrocarbon emissions, improving urban air quality [1]. Policy requires balancing multiple competing objectives.
Five timed questions on Hess's Law — heat of combustion and consolidation. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared). Pool: lessons 1–10.
Climb platforms, hit checkpoints, and answer questions on heat of combustion and method selection.
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