Edexcel International A Level Chemistry: Energetics (YCH11)

This guide teaches Topic 6: Energetics from the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It works through learning outcomes 6.1 to 6.11, including Core Practical 2. Topic 6 is International AS (IAS) content and sits in the unit the specification calls Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols.

Links: the course hub, the printable checklist, the energetics revision notes, the energetics practice questions and the free diagnostics. The calculations rely on moles, so revise amount of substance first if needed.

Topic 6 at a glance

Outcome	What you must be able to do
6.1	ΔH at constant pressure; standard conditions
6.2	Sign convention for exothermic and endothermic
6.3	Enthalpy level diagrams
6.4	Five standard enthalpy change definitions
6.5	Energy = m × c × ΔT, then ΔH in kJ mol⁻¹
6.6	Hess’s Law and enthalpy cycles
6.7	Core Practical 2
6.8	Errors, uncertainty, assumptions, cooling corrections
6.9	Bond enthalpy and mean bond enthalpy; limitations
6.10	Mean bond enthalpy from ΔH data
6.11	Bond enthalpy, which bond breaks first, and rate

6.1 and 6.2 What ΔH means

The enthalpy change, ΔH, is the heat energy change measured at constant pressure.

Standard conditions are a pressure of 100 kPa and a specified temperature, usually 298 K. Each substance is in its standard state, the physical state it takes under those conditions. A standard value carries the symbol ⦵, as in ΔH⦵.

• Exothermic: heat passes from the system to the surroundings, which warm up. ΔH is negative.
• Endothermic: heat passes into the system, so the surroundings cool. ΔH is positive.

Always write the sign: +94 kJ mol⁻¹, not 94 kJ mol⁻¹.

6.3 Enthalpy level diagrams

An enthalpy level diagram has enthalpy, H, on the vertical axis. Reactants and products are drawn as horizontal lines. A vertical arrow runs from reactants to products and is labelled with ΔH.

Exothermic                          Endothermic
H |  reactants ____                 H |              ____ products
  |                |  ΔH < 0          |             ↑
  |                ↓                  |             |  ΔH > 0
  |          ____ products            |  reactants ____

Products below reactants means exothermic. Label both levels with formulae and state symbols, and put ΔH, with its sign, beside the arrow.

Worked example 1. NH₄Cl(s) + aq → NH₄⁺(aq) + Cl⁻(aq), ΔH = +15 kJ mol⁻¹. ΔH is positive, so draw NH₄⁺(aq) + Cl⁻(aq) above NH₄Cl(s) + aq, with an upward arrow labelled +15 kJ mol⁻¹. The solution feels cold.

6.4 Standard enthalpy change definitions

Each ends “under standard conditions, with all substances in their standard states”.

• ΔrH⦵ (reaction): the enthalpy change for the molar amounts shown in the equation reacting.
• ΔfH⦵ (formation): the enthalpy change for forming one mole of a compound from its elements. For an element in its standard state, ΔfH⦵ = 0.
• ΔcH⦵ (combustion): the enthalpy change for burning one mole of a substance completely in oxygen.
• ΔneutH⦵ (neutralisation): the enthalpy change for an acid and a base reacting to give one mole of water.
• ΔatH⦵ (atomisation): the enthalpy change for the element producing one mole of gaseous atoms.

Equations must match the definition:

Formation of ethanoic acid:   2C(s) + 2H₂(g) + O₂(g) → CH₃COOH(l)
Combustion of propan-2-ol:    CH₃CH(OH)CH₃(l) + 4½O₂(g) → 3CO₂(g) + 4H₂O(l)
Neutralisation:               H⁺(aq) + OH⁻(aq) → H₂O(l)
Atomisation of bromine:       ½Br₂(l) → Br(g)

Atomisation makes one mole of Br atoms, starting from liquid bromine (its standard state).

6.5 Calculating ΔH from experiments

This covers mixing in an insulated container and combustion calorimetry. Two steps:

Step 1  energy transferred (J) = m (g) × c (J g⁻¹ °C⁻¹) × ΔT (°C)
Step 2  ΔH (kJ mol⁻¹) = − energy transferred (kJ) ÷ moles of the substance in the definition

m is the mass of solution or water that changed temperature, not the solid or fuel. A temperature rise means a negative ΔH.

Worked example 2 (solution). 40.0 cm³ of 0.750 mol dm⁻³ sulfuric acid is mixed with 40.0 cm³ of 1.60 mol dm⁻³ potassium hydroxide in a polystyrene cup. The temperature rises by 10.0 °C. Take the density of the mixture as 1.00 g cm⁻³ and c = 4.18 J g⁻¹ °C⁻¹. Calculate the enthalpy change of neutralisation.

m = 80.0 cm³ × 1.00 g cm⁻³ = 80.0 g
energy = 80.0 × 4.18 × 10.0 = 3344 J = 3.344 kJ
n(H₂SO₄) = 0.750 × 40.0 ÷ 1000 = 0.0300 mol, so n(H⁺) = 0.0600 mol
n(KOH) = 0.0640 mol (excess), so n(H₂O formed) = 0.0600 mol
ΔneutH = −3.344 ÷ 0.0600 = −55.7 kJ mol⁻¹

Worked example 3 (combustion). A spirit burner of propan-2-ol (Mr = 60.0) loses 0.450 g while heating 200 g of water in a metal can. The water warms by 15.2 °C. Calculate ΔcH.

energy = 200 × 4.18 × 15.2 = 12 707 J = 12.707 kJ
n(propan-2-ol) = 0.450 ÷ 60.0 = 0.00750 mol
ΔcH = −12.707 ÷ 0.00750 = −1694 kJ mol⁻¹ ≈ −1690 kJ mol⁻¹

Against an accepted value of −2006 kJ mol⁻¹, this captured only about 84% of the heat (see 6.8).

6.6 Hess’s Law and enthalpy cycles

Hess’s Law: when a reaction can take more than one route, the overall enthalpy change is the same for every route, as long as the starting and finishing conditions match.

Build a cycle: the direct arrow on top, an indirect route underneath. Both routes give the same total. Travelling against an arrow reverses its sign.

Cycles from formation data

Elements go at the bottom, arrows pointing up:

ΔrH = ΣΔfH(products) − ΣΔfH(reactants)

Worked example 4. Calculate ΔrH for MgO(s) + CO₂(g) → MgCO₃(s). Use these values (kJ mol⁻¹): ΔfH MgO(s) −602, CO₂(g) −394, MgCO₃(s) −1096.

products:  −1096
reactants: (−602) + (−394) = −996
ΔrH = −1096 − (−996) = −100 kJ mol⁻¹

Cycles from combustion data

Combustion products go at the bottom, arrows pointing down:

ΔrH = ΣΔcH(reactants) − ΣΔcH(products)

Worked example 5. Calculate ΔfH of propanoic acid, 3C(s) + 3H₂(g) + O₂(g) → CH₃CH₂COOH(l). Use these values (kJ mol⁻¹): ΔcH C(s) −394, H₂(g) −286, CH₃CH₂COOH(l) −1527.

reactants: 3(−394) + 3(−286) = −1182 − 858 = −2040    (O₂ has no ΔcH)
product:   −1527
ΔfH = −2040 − (−1527) = −513 kJ mol⁻¹

The two subtraction rules point opposite ways, so draw the cycle each time.

6.7 Core Practical 2: an enthalpy change by Hess’s Law

A thermal decomposition needs continuous heating, so you cannot measure its ΔH in a cup. Instead, measure two reactions that can be done in a cup and combine them with a cycle. One common version uses potassium hydrogencarbonate:

target:  2KHCO₃(s) → K₂CO₃(s) + CO₂(g) + H₂O(l)
ΔH₁:     KHCO₃(s) + HCl(aq) → KCl(aq) + CO₂(g) + H₂O(l)
ΔH₂:     K₂CO₃(s) + 2HCl(aq) → 2KCl(aq) + CO₂(g) + H₂O(l)
cycle:   ΔH(target) = 2ΔH₁ − ΔH₂

Both routes end in the same solution, so the cycle closes. Method for each solid: put excess hydrochloric acid in a polystyrene cup, record its temperature for a few minutes, add a weighed solid at a noted time, stir, and keep recording. Reweigh the empty weighing bottle.

Worked example 6. Use Ar: H = 1.0, C = 12.0, O = 16.0, K = 39.1. In each run, 30.0 cm³ of 2.00 mol dm⁻³ HCl is used (treat the solution as 30.0 g, c = 4.18 J g⁻¹ °C⁻¹).

• 3.00 g KHCO₃ (Mr 100.1): temperature falls by 7.2 °C.
• 2.76 g K₂CO₃ (Mr 138.2): temperature rises by 5.4 °C.

Run 1: energy = 30.0 × 4.18 × 7.2 = 902.9 J
       n(KHCO₃) = 3.00 ÷ 100.1 = 0.02997 mol
       ΔH₁ = +0.9029 ÷ 0.02997 = +30.1 kJ mol⁻¹   (temperature fell)
Run 2: energy = 30.0 × 4.18 × 5.4 = 677.2 J
       n(K₂CO₃) = 2.76 ÷ 138.2 = 0.01997 mol
       ΔH₂ = −0.6772 ÷ 0.01997 = −33.9 kJ mol⁻¹
Target: 2(+30.1) − (−33.9) = +94.1 ≈ +94 kJ mol⁻¹

The acid is in excess: 0.0600 mol HCl, against 0.0300 mol (run 1) and 0.0400 mol (run 2) needed.

6.8 Evaluating calorimetry results

Cooling curve corrections

Heat leaks out while the reaction is still going. Record temperature at regular intervals before and after mixing and plot it against time. Draw a best-fit line through the points after the peak and extrapolate it back to the time of mixing. ΔT is the gap between that value and the starting temperature. For an endothermic reaction, extrapolate the warming line after the lowest reading back in the same way.

Uncertainty

ΔT comes from two readings, so with a thermometer read to ±0.1 °C, ΔT carries ±0.2 °C. In worked example 6 that is 0.2 ÷ 7.2 × 100 = 2.8% (run 1) and 0.2 ÷ 5.4 × 100 = 3.7% (run 2). Small changes give large percentage uncertainties.

Assumptions to state

• The solution has the density and specific heat capacity of water.
• The solid added does not change m.
• The cup and thermometer absorb no heat.
• The reaction is complete.

Sources of error in combustion experiments

A spirit burner value is usually less exothermic than the accepted one because heat warms the air and can, combustion is incomplete (soot), fuel evaporates from the wick, and conditions are not standard. Improve it with a draught shield, a smaller gap between flame and can, and a cap on the burner between weighings.

6.9 Bond enthalpy and mean bond enthalpy

Bond enthalpy: the enthalpy change to break one mole of a particular covalent bond, with all species gaseous. It is always positive.

Mean bond enthalpy: the average for that bond over a range of compounds. O–H in water and O–H in methanol differ slightly; the mean blends them.

ΔH = total enthalpy of bonds broken − total enthalpy of bonds made

Worked example 7. Estimate ΔH for 2H₂(g) + O₂(g) → 2H₂O(g). Use these values (kJ mol⁻¹): H–H 436, O=O 498, O–H 464.

broken: 2(H–H) + 1(O=O) = 872 + 498 = 1370
made:   4(O–H) = 1856
ΔH = 1370 − 1856 = −486 kJ for the equation as written
per mole of H₂O(g): −243 kJ mol⁻¹

Limitations. Mean values are not the exact values in your molecules, so the answer is an estimate. The method only applies to gases, so any change of state is missed. ΔfH or ΔcH data give more reliable values.

6.10 Mean bond enthalpies from enthalpy changes

Worked example 8. Calculate the mean N–H bond enthalpy in ammonia. Use these values (kJ mol⁻¹): ΔfH NH₃(g) −46; ΔatH nitrogen +473; ΔatH hydrogen +218.

1. Breaking three N–H bonds turns NH₃(g) into N(g) + 3H(g).
2. Via the elements: reverse formation (+46), then atomise: 473 + 3(218) = 1127.
3. Total = 46 + 1127 = 1173 kJ mol⁻¹, so mean N–H = 1173 ÷ 3 = 391 kJ mol⁻¹.

6.11 Bond enthalpy and reaction rate

The bond with the lowest bond enthalpy usually breaks first. If the bond that must break is strong, the reaction is difficult to start and slow at room temperature.

Worked example 9. Using C–Cl 346, C–Br 290 and C–I 228 kJ mol⁻¹, rank the hydrolysis rates of chloro-, bromo- and iodoethane.

The C–halogen bond breaks. C–I is weakest, so iodoethane reacts fastest and chloroethane slowest. Likewise, the strong N≡N bond (945 kJ mol⁻¹) explains why nitrogen barely reacts at room temperature.

Common errors

• Using the mass of fuel or solid as m.
• Giving + for a temperature rise, or no sign at all.
• Dividing by moles of the reagent in excess.
• ΔfH equations that make two moles or use non-standard states.
• Ignoring coefficients when counting bonds.
• Taking ΔT from the highest reading, not the extrapolated value.

For exam technique, see the exam preparation guide.

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.