Year 12

AQA A level Chemistry Chemical Energetics Notes

Chemical Energetics

— Chemistry · AS Level (7405) · Enthalpy Changes —
★ Exam Tip: The examiner usually gives you only "ΔH is positive / negative" — he will not say endo or exo. Know the sign ⇄ type link cold. The definitions of each standard enthalpy change (and "per mole of WHAT?") decide most of the marks, and everything later (Hess's law, cycles) depends on them.

1. What is Enthalpy?

Enthalpy (H): the total heat energy content of a system (in reactants, products — wherever).

Heat energy = the kinetic energy of particles (how fast they move); temperature = average heat energy — that is how we measure it.

Enthalpy change (ΔH): the change in heat energy content of a system, measured at constant pressure.

Why ΔH, not H? Total H needs every molecule's KE — impossible. A thermometer gives the change easily.

We only ever measure the change: thermometer → ΔT → heat energy gained/lost.

Constant pressure is needed so that values are comparable between reactions (volume may change, pressure must not).

Δ anything = FINAL − INITIAL
ΔH = Hproducts − Hreactants
Note: ΔH is quoted in kJ mol⁻¹ (kilojoules, not joules — always convert J → kJ).

2. Endothermic vs Exothermic Reactions

① Endothermic — absorbs heat from the surroundings

  • Products end up with more energy than reactants (they gained it).
  • ΔH = Eproducts − Ereactants = +ve
  • All bond breaking is endothermic.
  • Examples: thermal decomposition (e.g. CaCO₃ → CaO + CO₂), photosynthesis, cracking.
Not a reaction! Melting / boiling are endothermic physical state changes, not chemical reactions — don't give them as examples of endothermic reactions. Examiner expects you to know thermal decomposition = endothermic.

② Exothermic — releases heat to the surroundings

  • Reactants had more energy; they gave it away → products have less.
  • ΔH = Eproducts − Ereactants = −ve (small − big)
  • All bond making is exothermic.
  • Examples: combustion (incl. respiration — combustion inside cells), neutralisation. Melting/cooling are state changes, so avoid them.
Magnet analogy: pull two stuck magnets apart → you must put energy IN (bond breaking, endo). Let them snap together → they release sound/KE by themselves (bond making, exo).

Bond energies (one mole of bonds broken in a molecule → atoms) are therefore always positive.

Heat loss during breaking? Small losses always exist in real experiments, but energy put in to break bonds is far larger — the reaction is still endothermic.
Endothermic Exothermic
Heat absorbed from surroundings released to surroundings
Energy of products higher than reactants lower than reactants
ΔH +ve −ve
Bonds breaking making
Temp. of surroundings falls rises
Examples thermal decomp., photosynthesis, cracking combustion, neutralisation, respiration

③ Energy profile (reaction pathway) diagrams

Eprogress of rxnreactantsproductsEaΔH = +vetransition state
ENDO: products ABOVE reactants; ΔH arrow points UP (+ve)
Eprogress of rxnreactantsproductsEaΔH = −vetransition state
EXO: products BELOW reactants; ΔH arrow points DOWN (−ve)

Axes: y = energy; x = progress of reaction or time (both accepted). Draw reactants line low/high as the type demands, join with a curve.

Rules for the arrows (examiners punish these!)① ΔH arrow: ALWAYS from the reactants' base line → products level (never reverse), for endo and exo. Up = +ve, down = −ve.
② Ea arrow: ALWAYS from the reactants' base line → top of the peak (transition state), ALWAYS pointing up — Ea can never be negative.
③ Transition state = top of the peak, where reactants have started turning into products.
Never, ever: mark ΔH from products to reactants, even in a hurry. In an endo graph a downward ΔH arrow would show −ve ΔH = wrong → cross from the examiner.
Why a hump, not a straight line? Every reaction needs a kick to start: a gas stove will not light by itself, you need a match (activation energy) — even though burning then releases lots of heat.

Activation energy (Ea): the minimum energy needed for the reaction to start (for bonds in reactants to begin breaking). Note it is transition state, not "transition energy".

Reverse reaction: ΔHrev = −ΔHfwd (sign flips) and Ea,rev = Ea,fwd + |ΔHfwd| for an exothermic forward reaction. Examiners like to give one direction and ask about the other!
Q2MCQ · 2-stage exothermic profile

Exothermic, 2 stages: reactants → intermediate → products. Ea of stage 1 = 50 kJ mol⁻¹; overall ΔH = −100 kJ mol⁻¹. Which profile?

Two humps (two stages, valley = intermediate). 1st hump is 50 above the reactants. Products end 100 below reactants (exo, ΔH = −100).

Answer: C ✓  (A/B/D fail: wrong first-hump height, wrong end level, or only one hump)

Eprogress of rxnreactantsproductsE a1 = 50−100intermediate
Q2 — profile C: 1st hump 50 above reactants; products 100 below
Eprogress of rxnXYZ
Q3 — profile A: X→Y ΔH +ve (Y above X); Y→Z ΔH −ve; Y in a valley
Q3MCQ · isolable intermediate

X → Y; ΔH positive. Y → Z; ΔH negative. Y could be isolated. Which profile?

Y is isolable ⇒ it sits in a valley between two humps (A has two humps). X→Y +ve ⇒ Y higher than X. Y→Z −ve ⇒ Z lower than Y (and here below X too).

Answer: A ✓

E₁E₂E₃transition statereactantsproductsEa = E₁ − E₂
Q4: Ea is measured from the REACTANTS level (E₂) to the top (E₁) → A (E₁ − E₂)
Q4MCQ · activation energy from diagram

Profile labelled E₁ (transition state), E₂ (reactants), E₃ (products). Which expression = Ea of the forward reaction?

Ea = top of hump − reactants.

A ✓ E₁ − E₂B ✗ E₁ − E₃ (from products = reverse Ea)C ✗ E₂ − E₃ (that is −ΔH)D ✗ mixes both

Answer: A

Q5MCQ · reverse reaction

Profile for W(g) + X(g) → Y(g) + Z(g): W+X sits higher than Y+Z (exothermic forward). What about the reverse, Y + Z → W + X?

Forward ΔH −ve ⇒ reverse ΔH +ve. Reverse must climb from the low Y+Z level to the same peak ⇒ larger Ea.

A ✓ larger Ea and positive ΔHB ✗ larger Ea but negative ΔHC ✗ smaller Ea, positive ΔHD ✗ smaller Ea, negative ΔH

Answer: A

3050reverse Ea = 80E/kJ
Q48: forward Ea = 30, ΔHfwd = −50; reverse Ea = 30 + 50 = 80, ΔHrev = +50
Q48MCQ · reversible reaction

Reversible reaction: forward Ea = 30 kJ mol⁻¹, products 50 kJ mol⁻¹ below reactants. Which statement is correct?

A ✓ Ea of reverse = +80 kJ mol⁻¹B ✗ ΔH forward = +30 (it is −50)C ✗ ΔH forward = +50 (wrong sign)D ✗ ΔH reverse = +30 (it is +50)

Answer: A

3. Standard Conditions (ΔH⊖)

To compare ΔH of different reactions we measure/calculate under standard conditions, written ΔH⊖. Memorise all four:

  1. Temperature of 25 °C (298 K)
  2. Pressure of 1 atm (1.013 × 10⁵ Pa)
  3. Concentration of any solution = 1 mol dm⁻³
  4. All chemicals in their preferred (most stable) states at these conditions
Note: a value calculated from ΔH⊖ data can be trusted as "standard"; you will meet these conditions again in ideal gases.

4. The Four Standard Enthalpy Changes

Golden rule: in each definition, ask "one mole of WHAT?" — formation: 1 mol of the compound formed; combustion: 1 mol of the substance burned; neutralisation: 1 mol of water formed; reaction: moles exactly as in the equation.

① Standard enthalpy change of REACTION, ΔH⊖r

Definition: amount of energy released or absorbed when the molar quantities of reactants as given in the chemical equation react completely to form products under standard conditions.

e.g. — double the equation, double ΔH
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)   ΔH⊖r = −92 kJ
2N₂(g) + 6H₂(g) ⇌ 4NH₃(g)   ΔH⊖r = −184 kJ

② Standard enthalpy change of FORMATION, ΔH⊖f

Definition: heat energy absorbed or released when 1 mole of a compound is formed from its constituent elements, under standard conditions.

Examples (products always ×1 mol):
Na(s) + ½Cl₂(g) → NaCl(s)   ΔH⊖f = −41 kJ mol⁻¹  (board value; data-book value is ≈ −411)
C(graphite) + O₂(g) → CO₂(g)   ΔH⊖f = −394 kJ mol⁻¹
½H₂(g) + ½Br₂(g) → HBr(g)   ΔH⊖f = −36 kJ mol⁻¹
Per mole of… the compound formed (right-hand side). Use fractions (½) freely to keep it at 1 mol.

Carbon: write C(graphite) rather than C(s) — graphite is the most stable form. (C(s) is not wrong, graphite is just better.)

ΔH⊖f of an ELEMENT (in its standard state) = 0
e.g. O₂(g) → O₂(g), ΔH⊖f = 0 — nothing changed

③ Standard enthalpy change of COMBUSTION, ΔH⊖c

Definition: amount of heat energy released when 1 mole of a substance is completely burned in oxygen under standard conditions.

Examples (per mole of substance BURNED = left side):
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)   ΔH⊖c = −890 kJ mol⁻¹
C(graphite) + O₂(g) → CO₂(g)   ΔH⊖c = −394 kJ mol⁻¹
Key idea: combustion is always exothermic (ΔH⊖c −ve). Keep 1 mol of the fuel — put fractions like 7/2 O₂ in front of oxygen, never change the 1 in front of the fuel. Water must be liquid, and burning must be complete.
Q. What if incomplete combustion occurred? → ΔH⊖c would be less negative (less exothermic) — less energy released.
Q. What if steam formed instead of liquid water? → ΔH⊖c less negative (less exothermic) since hydrogen bonds would not be made — and bond formation is exothermic.
Why C(graphite) + O₂ → CO₂ is both: the same equation is ΔHf of CO₂ and ΔHc of carbon = −394. Same for H₂ + ½O₂ → H₂O(l): formation of water and combustion of hydrogen.

④ Standard enthalpy change of NEUTRALISATION, ΔH⊖n

Definition: amount of heat energy released when solutions of acid and alkali react together to form 1 mole of water under standard conditions.

Amount of acid/alkali does not matter — only 1 mol H₂O (so halve equations where needed!). Products are always salt + water.

e.g. NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)   ΔH⊖n = −57 kJ mol⁻¹
(board wrote NaCl(s); the salt is dissolved so aq is more accurate — same idea)
Ionic: Na⁺ + OH⁻ + H⁺ + Cl⁻ → Na⁺ + Cl⁻ + H₂O  (cross out spectator ions)
⇒ OH⁻(aq) + H⁺(aq) → H₂O(l)   ΔH⊖n = −57 kJ mol⁻¹
e.g. ½H₂SO₄(aq) + KOH(aq) → H₂O(l) + ½K₂SO₄(aq)   ΔH⊖n = −57 kJ mol⁻¹
Ionic: H⁺ + ½SO₄²⁻ + K⁺ + OH⁻ → H₂O + K⁺ + ½SO₄²⁻ ⇒ H⁺(aq) + OH⁻(aq) → H₂O(l)   −57 kJ mol⁻¹
Watch out: H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O makes 2 mol water — not a neutralisation equation. Halve it so H₂O has coefficient 1. Salt formula: K⁺ and SO₄²⁻ swap charges → K₂SO₄ (learn polyatomic ions).
  • Strong acid + strong alkali: always ΔH⊖n = −57 kJ mol⁻¹ (same ionic equation every time).
  • Weak acid and/or weak alkali: slightly less negative than −57 — some energy is needed to complete the ionisation.
  • Always exothermic since H⁺ and OH⁻ form a bond (bond making). It is never +57.

Summary — "one mole of what?"

ΔH⊖ 1 mol of… Sign
reaction (r) as in equation ±
formation (f) compound formed (RHS) ± (elements = 0)
combustion (c) substance burned (LHS) always −ve
neutralisation (n) water formed always −ve (≈ −57)
Q1MCQ · spot the formation equation

Which equation is an enthalpy change of formation?

A ✗ 2NO → N₂ + O₂  (decomposition; reactant is a compound)B ✗ 2CO + O₂ → 2CO₂  (reactant is a compound; 2 mol formed)C ✗ H₂O(l) + NaCl(s) → NaCl(aq)  (solution)D ✓ K(s) + Mn(s) + 2O₂(g) → KMnO₄(s)  (elements → 1 mol compound)

Answer: D

Q16MCQ · why no spontaneous burning?

N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(g), ΔH = −534 kJ mol⁻¹, yet hydrazine does not burn spontaneously. Why?

Highly exothermic but still needs a start: the activation energy is too high (match/spark needed).

A ✗ being a liquid isn't the reason   C ✗ strong N≡N bond is in the products   D ✗ exothermic reactions still need Ea

Answer: B

Q17MCQ · ΔHf from a given equation

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s); ΔH⊖ = −1648 kJ. Find ΔH⊖f of iron(III) oxide.

Formation needs 1 mol of Fe₂O₃ ⇒ halve the equation and halve the energy.

2Fe(s) + 1½O₂(g) → Fe₂O₃(s)ΔH⊖f = −1648 ÷ 2−824 kJ mol⁻¹

C ✗ −1648 is for 2 mol   D ✗ doubled instead of halved   A ✗ only elements are 0

Answer: B

Q40MCQ · H₂ + ½O₂ → H₂O(l)

Which pair of descriptions is fully correct for this reaction (ΔH = x kJ mol⁻¹)?

1 mol H₂ burned and 1 mol H₂O formed from elements ⇒ both combustion and formation. Combustion is exothermic ⇒ x is negative.

A, B ✗ "formation only" misses combustion   C ✗ positive x is impossible for combustion

Answer: D — combustion, formation; x negative

Q61MCQ · formation of propane

Which reaction has ΔH = ΔH⊖f of propane, C₃H₈?

A ✗ 3C(g) + 4H₂(g) → C₃H₈(g) — carbon not in standard stateB ✗ 3C + 8H(g) → … — atomic H is not the element's standard state (H₂)C ✓ 3C(s) + 4H₂(g) → C₃H₈(g)D ✗ product is a liquid, but propane is a gas at standard conditions

Answer: C

5. Calorimetry — Measuring ΔH Experimentally

MethodQ = m c Δθ  (J)   — heat absorbed by the water/solution
ΔH = Q × 10⁻³no. of moles  (kJ mol⁻¹)
Where: m = mass of solution/water (g) (1 cm³ of solution ≈ 1 g) | c = 4.18 J g⁻¹ K⁻¹ (it takes 4.18 J to warm 1 g of water by 1 °C or 1 K; use 4.2 if the examiner says so) | Δθ = change in temp. | ×10⁻³ converts J → kJ | moles: water for ΔHn, substance burned for ΔHc
Signs: put − / + as required: temperature rises ⇒ exothermic ⇒ ΔH −ve (temperature falls ⇒ +ve). Give ΔH in kJ mol⁻¹, never J. Use exponent notation (×10⁻³) to convert.

① Finding ΔH of neutralisation

thermometerpolystyrenecupsolution
thermometer suspended in the solution — must NOT touch the walls/bottom

50.0 cm³ of 1.00 mol dm⁻³ HCl (21.5 °C) in a polystyrene cup; 50.0 cm³ of 1.00 mol dm⁻³ NaOH (same temp.) added; stirred with the thermometer; highest temperature 28.0 °C.

MolesHCl + NaOH → H₂O + NaCln(HCl) = 1 × 50 × 10⁻³ = 0.05 moln(NaOH) = 0.05 mol ⇒ n(H₂O) = 0.05 mol
HeatQ = mcΔθQ = 100 × 4.18 × (28 − 21.5)Q = 2717 J
ΔHΔHn = (2717 × 10⁻³) / 0.05= 54.34 → −54.3 kJ mol⁻¹
Why polystyrene, not glass? It is a good insulator — we want the heat to stay in the solution, not escape. Thermometer must touch only the solution. 1 cm³ of solution has 1 g of mass; total volume 50 + 50 = 100 cm³ ⇒ m = 100 g.
Q. How can heat losses be reduced? ① use a lid (most heat escapes from the open top) ② insulate the cup further (use 2 or 3 polystyrene cups).
Q. Why may the value not be accurate? ① heat loss to surroundings (the most significant — always your first point) ② precision of thermometer is limiting (27.888 shown as 28) ③ initial temperature is non-standard ④ assumed c = 4.18 J g⁻¹ K⁻¹ (the mixture is not pure water: it contains acid, alkali, NaCl).
Result −54.3 vs data −57: a decent experiment, but not exact.

② Finding ΔH of combustion

clampdraughtshieldthermometerstirrerwaterspiritburnerpropan-1-ol
weigh burner before and after; heat water in a clamped beaker; draught shield cuts losses

Q. Spirit burner with propan-1-ol: mass 86.27 g. Heats 100 g of water in a beaker from 20.2 °C to 30.9 °C. Burner re-weighed: 86.06 g. Find ΔHc of propan-1-ol (CH₃CH₂CH₂OH, Mr = 60).

Q = mcΔθ = 100 (4.18)(30.9 − 20.2)Q = 4472.6 Jmass burnt = 86.27 − 86.06 = 0.21 gn = 0.21 / 60 = 0.0035 molΔH = (4472.6 × 10⁻³) / (0.21/60) = 1277.88ΔHc = −1280 kJ mol⁻¹
Q. Data book value is −2021 kJ mol⁻¹. Why is yours so different? ① heat losses ② incomplete combustion of the fuel ③ small mass of fuel ⇒ high percentage error ④ precision of thermometer is limiting ⑤ non-standard conditions ⑥ assumed c = 4.18
Extra (teacher): when you subtract two readings (e.g. 86.27 − 86.06), their uncertainties add — ±0.01 each gives ±0.02 in the difference, a big % error on 0.21 g.
Q8MCQ · expression for ΔHn

50 cm³ of 2.50 mol dm⁻³ HCl + 50 cm³ of 2.50 mol dm⁻³ NaOH; temp. rose by 17 °C; c = 4.2 J g⁻¹ K⁻¹. Which expression gives ΔHn?

Q = mcΔθ = (100)(4.2)(17) Jn(H₂O) = 2.5 × 50 × 10⁻³ molΔH = −100 × 4.2 × 1750 × 2.5 J mol⁻¹ (the 10⁻³ cancels: J and cm³)

A, B ✗ use 50 g not 100 g   C ✗ "0.050 × 2.5" mixes units (cm³ → dm³) with J

Answer: D

Q9MCQ · % of energy absorbed

1.60 g of fuel burnt; only 45 % of the energy is absorbed by 200 g water (18 → 66 °C, c = 4.2). Total energy per gram?

Q = (200)(4.2)(48) = 40 320 J (the 45 %)45 % = 40 320 ⇒ 100 % = 40 320 ÷ 0.45 = 89 600 Jper gram: 89 600 / 1.6 = 56 000 J

Answer: B

Q50MCQ · per mole of NaOH

25 cm³ 0.10 mol dm⁻³ NaOH + 25 cm³ 0.10 mol dm⁻³ HCl; temp. rise 2.5 °C. ΔH per mole of NaOH?

Q = 50 (4.18)(2.5) = 522.5 Jn(NaOH) = 0.1 × 25 × 10⁻³ = 2.5 × 10⁻³ molΔH = −522.5 × 10⁻³ / 2.5 × 10⁻³ = −209 kJ mol⁻¹

B ✗ half (used wrong moles)   C, D ✗ J not kJ

Answer: A

Q55MCQ · methanol burner — heat into water

Beaker+water 150.00 g, empty beaker 50.00 g ⇒ 100 g water (20 → 53 °C). Alcohol burner 259.65 → 259.15 g. How much heat went into the water?

Q = mcΔθ= 100 × 4.18 × (53 − 20)= 13 794 J

Only asks for heat into water, so no division by moles needed. Others use wrong mass (e.g. the 0.50 g of methanol).

Answer: B

Q56MCQ · molar ΔH (NaOH + HCl)

25.0 cm³ 0.350 mol dm⁻³ NaOH + 25.0 cm³ 0.350 mol dm⁻³ HCl; ΔT = 2.50 °C; c = 4.20; no heat lost.

n = 0.35 × 25 × 10⁻³ = 8.75 × 10⁻³ molQ = (50)(4.2)(2.5) = 525 JΔH = −525 × 10⁻³ / 8.75 × 10⁻³ = −60 kJ mol⁻¹

Answer: B

6. Hess's Law & Energy Cycles

The enthalpy change of a reaction is independent of the path taken, provided all measurements are made under the same conditions.

Comes from the 1st law of thermodynamics: energy cannot be created or destroyed, it can only be converted from one form to another. So A → B has the same ΔH via any route with the same start and end.

Useful for ΔH that cannot be measured in the laboratory; we route via reactions we do know (ΔHf or ΔHc data).

Two shortcuts (for the cycles)Using ΔHf data:   ΔHr = ΣΔHf(products) − ΣΔHf(reactants)
Using ΔHc data:   ΔHr = ΣΔHc(reactants) − ΣΔHc(products)
How to draw a cycle: ① look at what is given (ΔHf or ΔHc?). ② copy the reaction on top with ΔH1. ③ Blue arrows follow the definition (formation: elements → compound; combustion: substance + O₂ → CO₂ + H₂O). ④ Red arrow shows the direction you actually travel for ΔH1 — going with a blue arrow = +, against it = −. ⑤ Multiply by the number of moles in the equation (this is not "the same factor every time").

Worked examples from the board

Hydrazine · ΔHf routeQ6(i)

N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(g). ΔHf: N₂H₄(l) = +50.6; H₂O(g) = −241.8 kJ mol⁻¹. Find ΔH⊖.

Cycle via elements N₂ + 2H₂ + O₂. ΔH₂ = +50.6 (N₂H₄ formation, travelled backwards ⇒ −50.6); ΔH₃ = 2(−241.8).

ΔH₁ = −ΔH₂ + ΔH₃= −50.6 + 2(−241.8)ΔH⊖ = −534.2 kJ mol⁻¹

(ii) Why doesn't it burn spontaneously though very exothermic? → high activation energy. (iii) "Environmentally friendly"? → produces nitrogen and water, which are not hazardous.

CH₄ + Cl₂ → CH₃Cl + HClQ5(a)(i) · ΔHf

ΔHf: CH₄ −75; CH₃Cl −82; HCl −92 kJ mol⁻¹. (Cl₂ = element, 0.)

cycle via C + 2H₂ + Cl₂ΔH₁ = +75 + (−82 − 92)ΔH⊖reaction = −99 kJ
Urea hydrolysisQ4 · ΔHf

H₂O(l) + CO(NH₂)₂(aq) → CO₂(aq) + 2NH₃(aq). ΔHf: H₂O(l) −287.0; urea −320.5; CO₂(aq) −414.5; NH₃(aq) −81.0.

ΔH₂ = −287 − 320.5  (reactants, go backwards)ΔH₃ = −414.5 + 2(−81)ΔH₁ = −ΔH₂ + ΔH₃= −(−287 − 320.5) + (−414.5 + 2(−81))ΔH₁ = +31 kJ mol⁻¹
Methanol synthesis (CO₂ + 3H₂)Q14 · ΔHf

(a) Define ΔH⊖f of CO₂ with an equation with state symbols: C(graphite) + O₂(g) → CO₂(g). Energy released or absorbed (here released) when 1 mol of CO₂ is made from carbon and oxygen under standard conditions.

(b) CO₂(g) + 3H₂(g) ⇌ CH₃OH(g) + H₂O(g). ΔHf: CO₂ −394; CH₃OH −201; H₂O −242.

ΔH₂ = −394; ΔH₃ = −201 − 242ΔH₁ = −ΔH₂ + ΔH₃ = +394 − 201 − 242ΔH⊖reaction = −49 kJ mol⁻¹

(ii) Environmental advantage: CO₂ gas is removed, which is a greenhouse gas.

Ethene + water → ethanolQ8(a) · ΔHc

C₂H₄(g) + H₂O(l) → C₂H₅OH(l). ΔHc: C₂H₄ −1411; C₂H₅OH −1367 kJ mol⁻¹. Cycle down to 2CO₂ + 3H₂O.

ΔH₁ = ΔH₂ − ΔH₃= −1411 − (−1367)ΔH₁ = −44 kJ mol⁻¹
CO + 2H₂ → CH₃OHQ8(b) · ΔHc

ΔHc: CO −283; H₂ −286; CH₃OH −726. Cycle via CO₂ + 2H₂O (with +3⁄2 O₂ each side).

ΔH₂ = −283 + 2(−286); ΔH₃ = −726ΔH₁ = ΔH₂ − ΔH₃= −283 + 2(−286) + 726ΔH⊖reaction = −129 kJ mol⁻¹
ΔHf of propanol C₃H₇OH(l)from ΔHc

ΔHc: C(s) −393.5; H₂(g) −285.8; C₃H₇OH(l) −2021.0. Labelled cycle: 3C(s) + 4H₂(g) + ½O₂(g) → C₃H₇OH(l), via 3CO₂ + 4H₂O (+9⁄2 O₂).

ΔH₂ = 3(−393.5) + 4(−285.8); ΔH₃ = −2021ΔHf = ΔH₂ − ΔH₃= 3(−393.5) + 4(−285.8) + 2021ΔHf = −302.7 kJ mol⁻¹
ΔHf of etheneQ9 · ΔHc data, 298 K

ΔHc: C −393.7; H₂ −285.9; C₂H₄ −1411.0 kJ mol⁻¹. Cycle: 2C + 2H₂ → C₂H₄ via 2CO₂ + 2H₂O.

ΔH₁ = 2(−393.7) + 2(−285.9) − (−1411)= −787.4 − 571.8 + 1411ΔHf = +51.8 kJ mol⁻¹
ΔHf of CS₂Q10 · mixed data

ΔHc(CS₂) = −1110; ΔHf(CO₂) = −395; ΔHf(SO₂) = −298. C + ¼S₈ → CS₂ via CO₂ + 2SO₂ (+3O₂).

ΔH₂ = −395 + 2(−298); ΔH₃ = −1110ΔH₁ = ΔH₂ − ΔH₃= −395 + 2(−298) + 1110ΔHf = +119 kJ mol⁻¹
ΔHf of ketene C₂H₂OQ11

ΔHf(CO₂) = −395; ΔHc(H₂) = −286; ΔHc(CH₂=C=O) = −1028. Cycle: 2C + H₂ + ½O₂ → C₂H₂O via 2CO₂ + H₂O (+2O₂).

ΔH₁ = 2(−395) − 286 − (−1028)= −790 − 286 + 1028ΔHf = −48 kJ mol⁻¹
C₂H₂ (ethyne)Q7 · equations + ΔHf

ΔH⊖c(C₂H₂) = −1300; ΔHf: CO₂ −394; H₂O(l) −286.
(i) combustion: C₂H₂(g) + 5⁄2 O₂(g) → 2CO₂ + H₂O
formation: 2C(graphite) + H₂(g) → C₂H₂(g)

ΔH₁ = 2(−394) − 286 − (−1300)= −788 − 286 + 1300 = 226ΔH⊖f = +226 kJ mol⁻¹ (positive)
Cycle check: data type decides the arrow. Using ΔHf: go via elements. Using ΔHc: go via combustion products (CO₂, H₂O). Products − reactants for ΔHf; reactants − products for ΔHc.

7. Enthalpy Changes from Bond Energies

  • Bond breaking is endothermic (ΔH = +ve); bond formation is exothermic (ΔH = −ve).
ΔH = Σ E(bonds broken, reactants) − Σ E(bonds formed, products)
= Ereactants − Eproducts (bond energies) — opposite to the ΔHf formula
Data-booklet values used: C–H 410 | C≡C 840 | O=O 496 | C=O 805 | O–H 460 (kJ mol⁻¹). Draw displayed formulae so you count every bond.
Bond-energy answers are only estimates because the values are average bond enthalpies, and bond energies are for gases: in the standard state H₂O is liquid, so hydrogen bonds are not accounted for in the calculation.
Ethyne combustionQ3(d)/(e) · calculation

C₂H₂(g) + 5⁄2 O₂(g) → 2CO₂(g) + H₂O(g). H–C≡C–H + 5⁄2 O=O → 2 O=C=O + H–O–H.

broken: 2(C–H) + 1(C≡C) + 5⁄2(O=O)formed: 4(C=O) + 2(O–H)ΔH = 2(410) + 840 + 5⁄2(496) − 4(805) − 2(460)ΔH = −1240 kJ mol⁻¹

(e)(ii) Why does −1240 differ from ΔH⊖c = −1300? ① average bond enthalpies are used ② in the standard state H₂O is liquid, so hydrogen bonds have not been accounted for (calculation gives steam).

Q35 · N–F bondMCQ

2NF₃(g) → 2N(g) + 6F(g); ΔH = +1668 kJ. Bond energy of N–F?

2 NF₃ has 6 N–F bonds, all broken6(N–F) = 1668N–F = +278 kJ mol⁻¹

A, B ✗ bond energy cannot be negative   D ✗ 556 = per 2 mol NF₃ (divided wrongly)

Answer: C

CO + 2H₂ → CH₃OHQ11 · MCQ

E(C≡O) 1077, E(C–O) 360, E(C–H) 410, E(H–H) 436, E(O–H) 460 kJ mol⁻¹. CH₃OH has 3 C–H, 1 C–O, 1 O–H.

broken: (C≡O) + 2(H–H)formed: 3(C–H) + (C–O) + (O–H)ΔH = 1077 + 2(436) − 3(410) − 360 − 460ΔH = −101 kJ mol⁻¹

Answer: B

CH₄ + I₂ → CH₃I + HI(ii)–(iii)

C–H 410, I–I 151, C–I 240, H–I 299. Break 4 C–H + I–I; make 3 C–H + C–I + H–I.

ΔH = 4(C–H) + (I–I) − 3(C–H) − (C–I) − (H–I)= 4(410) + 151 − 3(410) − 240 − 299ΔHreaction = +22 kJ mol⁻¹

(iii) Why does this reaction not occur (methane does not react with iodine)? → high activation energy (and it is slightly endothermic).

8. Quick Recap — Full Chapter Cheat-Sheet

✓ Enthalpy basics
  • ΔH = Hproducts − Hreactants (final − initial); measured at constant pressure; units kJ mol⁻¹.
  • Endo: ΔH +ve, absorbs heat, bond breaking, e.g. thermal decomposition, photosynthesis, cracking.
  • Exo: ΔH −ve, releases heat, bond making, e.g. combustion, neutralisation, respiration.
  • Profile: ΔH arrow reactants → products (never reversed); Ea arrow reactants → peak (always up); peak = transition state.
  • Standard: 298 K | 1 atm | 1 mol dm⁻³ | preferred states. Elements have ΔHf = 0.
✓ Definitions ("1 mole of…")
  • ΔHr: moles as in the equation react completely. ΔHf: 1 mol of a compound from its elements (C = graphite).
  • ΔHc: 1 mol of substance burned completely in O₂ (always −ve). ΔHn: acid + alkali → 1 mol water (strong/strong = −57).
✓ Formulas
  • Q = mcΔθ (m = mass of solution/water in g; c = 4.18 J g⁻¹ K⁻¹); ΔH = Q × 10⁻³ ÷ moles (kJ mol⁻¹); sign − if temp. rises.
  • Moles: water (n) or substance burnt (c). Heat-loss fixes: lid, more insulation, polystyrene cup; thermometer not touching walls.
  • ΔHr = ΣΔHf(products) − ΣΔHf(reactants)  |  ΔHr = ΣΔHc(reactants) − ΣΔHc(products)
  • Bond energies: ΔH = Σ(bonds broken) − Σ(bonds formed). Hess: ΔH independent of route (same conditions), from conservation of energy.
  • Reasons values differ: heat loss, incomplete combustion, thermometer precision, non-standard conditions, assumed c = 4.18, small mass of fuel, average bond enthalpies.