Edexcel International A Level Chemistry: Energetics (YCH11) – Revision Notes

These notes condense Topic 6: Energetics of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 6.1 to 6.11 with Core Practical 2. It is IAS material, in Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols. The energetics study guide has the full teaching and longer worked examples; start there if anything here feels unfamiliar.

Also useful: the practice questions, the course hub, the printable checklist and the free diagnostics.

6.1 to 6.3 Core ideas

• ΔH = heat energy change at constant pressure.
• Standard conditions: 100 kPa and a specified temperature, usually 298 K. Substances in their standard states. Symbol ⦵.
• Exothermic: ΔH negative; surroundings get hotter.
• Endothermic: ΔH positive; surroundings get colder.
• Enthalpy level diagram: H on the vertical axis; arrow from reactants to products. Arrow down = exothermic; arrow up = endothermic. Label both levels with formulae and state symbols.

Exothermic                    Endothermic
H  reactants ─────            H  products ─────
        │ ΔH negative                 ↑ ΔH positive
        ↓                             │
   products ─────                reactants ─────

Read a diagram by asking where the products sit: lower than the reactants means heat was released.

6.4 Definitions

All five end: “under standard conditions, with all substances in their standard states”.

Term	Symbol	Key phrase	Example equation
Reaction	ΔrH⦵	molar amounts in the equation react	N₂(g) + 3H₂(g) → 2NH₃(g)
Formation	ΔfH⦵	1 mol of compound from its elements	Na(s) + ½Cl₂(g) → NaCl(s)
Combustion	ΔcH⦵	1 mol of substance burned completely in oxygen	C₄H₁₀(g) + 6½O₂(g) → 4CO₂(g) + 5H₂O(l)
Neutralisation	ΔneutH⦵	acid and base react to form 1 mol of water	½H₂SO₄(aq) + NaOH(aq) → ½Na₂SO₄(aq) + H₂O(l)
Atomisation	ΔatH⦵	1 mol of gaseous atoms from the element	½I₂(s) → I(g)

Two quick checks: ΔfH⦵ of any element in its standard state is zero, and ΔatH⦵ is always positive.

6.5 Calorimetry: method in steps

1. Energy transferred (J) = m × c × ΔT, with m in g of the water or solution, c in J g⁻¹ °C⁻¹, ΔT in °C.
2. Divide by 1000 to get kJ.
3. Find moles of the substance named in the definition (the limiting reagent, the fuel, or water formed).
4. ΔH = energy (kJ) ÷ moles. Add − if the temperature rose, + if it fell.
5. Round to match the data (usually 3 significant figures) and give units, kJ mol⁻¹.

Small worked reminder: 25.0 cm³ of solution warms by 4.4 °C when 0.00500 mol of a metal reacts.

energy = 25.0 × 4.18 × 4.4 = 459.8 J = 0.4598 kJ
ΔH = −0.4598 ÷ 0.00500 = −92.0 kJ mol⁻¹ → −92 kJ mol⁻¹ (ΔT has 2 s.f.)

6.6 Hess’s Law: method in steps

Statement: whichever route a reaction takes, the overall enthalpy change is the same, as long as the start and end conditions match.

1. Write the target equation across the top.
2. Choose the bottom box: elements (formation data) or combustion products (combustion data).
3. Draw arrows: formation arrows point up from elements; combustion arrows point down to oxides.
4. Multiply each value by its coefficient.
5. Add round the indirect route, reversing the sign of any arrow you go against.

Data given	Rule
ΔfH	ΔrH = ΣΔfH(products) − ΣΔfH(reactants)
ΔcH	ΔrH = ΣΔcH(reactants) − ΣΔcH(products)
Two measured reactions with a common product	ΔH(target) = ΔH(route A) − ΔH(route B), scaled by coefficients

Small worked reminder: find ΔrH for 2H₂O₂(l) → 2H₂O(l) + O₂(g), using ΔfH (kJ mol⁻¹) H₂O₂(l) −188 and H₂O(l) −286.

ΣΔfH(products) = 2(−286) + 0 = −572     (O₂ is an element: zero)
ΣΔfH(reactants) = 2(−188) = −376
ΔrH = −572 − (−376) = −196 kJ for the equation as written

6.7 Core Practical 2 in brief

• Aim: find ΔH for a reaction you cannot measure directly, such as a thermal decomposition.
• Idea: react each solid separately with excess acid in a polystyrene cup. Both reactions finish at the same solution, so a Hess cycle links them.
• Readings: mass of solid added (weigh by difference), volume of acid, temperatures before and after mixing.
• Calculation: mcΔT for each run, then ΔH per mole of each solid, then combine with the cycle.
• Check: the acid must be in excess in both runs, so the solid is the limiting reagent.

6.8 Evaluation toolkit

Cooling correction: plot temperature against time; extrapolate the cooling line back to the moment of mixing; ΔT = extrapolated value − initial temperature. This adds back heat lost during the reaction.

Percentage uncertainty: uncertainty ÷ measured value × 100. For ΔT, double the reading uncertainty, because two readings are subtracted.

Assumptions: solution behaves like water (density 1.00 g cm⁻³, c = 4.18 J g⁻¹ °C⁻¹); no heat absorbed by the container; reaction complete; solid does not add to m.

Why combustion values are too small: heat lost to air and the can; incomplete combustion; fuel evaporation; non-standard conditions.

Improvements: lid and insulation; draught shield; flame close to the calorimeter; a larger temperature change to cut percentage uncertainty; a more precise thermometer.

6.9 to 6.10 Bond enthalpies: method in steps

Bond enthalpy: enthalpy change to break one mole of a stated covalent bond, all species gaseous. Mean bond enthalpy: the average for that bond over many compounds.

1. Draw displayed formulae and count every bond, using the coefficients.
2. Sum the bonds broken (endothermic, +).
3. Sum the bonds made (exothermic, −).
4. ΔH = Σ(broken) − Σ(made).
5. To find an unknown bond, put it in as x and solve the same equation.

Limitations: mean values are not exact for the molecule involved; the method assumes everything is gaseous, so it ignores enthalpy changes of state. Expect a different answer from a Hess’s Law calculation using ΔfH or ΔcH.

6.11 Bond enthalpy and rate

• The weakest bond tends to break first.
• A high bond enthalpy for the bond that must break means a slow reaction at room temperature.
• Example pattern: C–I (weakest) > C–Br > C–Cl for rate of halogenoalkane hydrolysis.

Must-know distinctions

• Enthalpy change vs energy transferred: energy transferred is in J and has no sign; ΔH is in kJ mol⁻¹ and has a sign.
• Formation vs atomisation: formation makes 1 mol of a compound; atomisation makes 1 mol of gaseous atoms.
• Bond enthalpy vs mean bond enthalpy: one specific bond in one molecule vs an average over many.
• Bond enthalpy route vs Hess route: estimate vs more reliable value.

Quick self-test

1. State the standard pressure used for standard enthalpy changes.
2. A reaction in a cup makes the temperature fall. What is the sign of ΔH?
3. 75.0 g of water warms by 6.4 °C. Calculate the energy transferred in J.
4. 0.0150 mol of a reactant releases 1.20 kJ. Calculate ΔH.
5. Write the equation for the standard enthalpy change of atomisation of chlorine.
6. Use ΔfH (kJ mol⁻¹): C₂H₄(g) +52, H₂O(g) −242, C₂H₅OH(g) −235. Calculate ΔrH for C₂H₄(g) + H₂O(g) → C₂H₅OH(g).
7. Use ΔcH (kJ mol⁻¹): C(s) −394, H₂(g) −286, HCOOH(l) −255. Calculate ΔfH of methanoic acid, HCOOH(l).
8. Use bond enthalpies (kJ mol⁻¹): H–H 436, I–I 151, H–I 299. Estimate ΔH for H₂(g) + I₂(g) → 2HI(g).
9. Use ΔatH (kJ mol⁻¹): C +717, Cl +122; ΔfH CCl₄(g) −96. Calculate the mean C–Cl bond enthalpy in CCl₄.
10. ΔT is 4.0 °C and each reading is ±0.1 °C. Calculate the percentage uncertainty in ΔT.
11. Give one reason why a bond enthalpy calculation differs from a Hess’s Law value.

Answers

1. 100 kPa.
2. Positive (endothermic).
3. 75.0 × 4.18 × 6.4 = 2006.4 J (2010 J to 3 s.f.).
4. −1.20 ÷ 0.0150 = −80.0 kJ mol⁻¹ (negative, because energy is released).
5. ½Cl₂(g) → Cl(g).
6. −235 − (52 + (−242)) = −235 + 190 = −45 kJ mol⁻¹.
7. (−394) + (−286) − (−255) = −425 kJ mol⁻¹.
8. Broken 436 + 151 = 587; made 2 × 299 = 598; ΔH = −11 kJ for the equation as written.
9. Energy to break four C–Cl bonds = 96 + 717 + 4(122) = 1301; mean C–Cl = 1301 ÷ 4 = 325 kJ mol⁻¹.
10. 0.2 ÷ 4.0 × 100 = 5%.
11. Mean values are averages over many compounds, or the method assumes all species are gaseous.

Where marks are usually lost

• Putting the solid or fuel mass into the formula where the water or solution mass belongs.
• Giving ΔH in J rather than kJ mol⁻¹, or missing the unit entirely.
• Leaving off the sign, or giving a positive value after a temperature rise.
• Dividing by the moles of the reagent in excess.
• Definitions that miss “one mole”, “completely” or “standard states”.
• Atomisation equations written for Cl₂ → 2Cl, which make two moles of atoms.
• Forgetting to multiply by coefficients in Hess cycles and bond counts.
• Using the maximum reading instead of an extrapolated temperature.
• Saying heat loss makes a combustion value “more exothermic”; it makes it less exothermic.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Topic 6: Energetics (outcomes 6.1 to 6.11, including Core Practical 2). Published by Pearson Education Limited.