UNIT 2 CHEMISTRY · PEARSON AAQ · D1.3.5
Calculations Involving Energy Cycles and Standard Enthalpy Changes
FREE CHEMISTRY REVISION · TOPIC 42
Calculations Involving Energy Cycles and Standard Enthalpy Changes
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · D1.3.5
What you will learn
Calculate enthalpy changes from internal-energy data, formation and combustion enthalpies, and calorimetry. Choose the correct route, balance the equation, convert units and explain the sign of the answer. All numerical data in worked examples are supplied or rounded teaching values.
1. Quantities, signs and units
Enthalpy H and internal energy U are state functions. Heat q is energy transferred because of a temperature difference; it is not a substance stored inside a sample. A change symbol means final minus initial. For the reacting system, an exothermic constant-pressure process has ΔH < 0; an endothermic process has ΔH > 0.
| Quantity | Meaning | Compatible units |
|---|---|---|
| ΔH, ΔU | Changes for the stated amount of system | J or kJ; molar values J mol⁻¹ or kJ mol⁻¹ |
| p | Pressure, held constant in ΔH = ΔU + pΔV | Pa = N m⁻² |
| ΔV | Vfinal − Vinitial | m³; 1 dm³ = 10⁻³ m³ |
| pΔV | Pressure × volume change | Pa m³ = J |
| ΔQ or q | Heat transferred into the named body | J when m and c use matching units |
| m | Mass of the body whose temperature changes | g with J g⁻¹ K⁻¹, or kg with J kg⁻¹ K⁻¹ |
| c | Specific heat capacity | Energy per unit mass per kelvin |
| ΔT | Tfinal − Tinitial | K or °C interval; an interval of 1°C equals 1 K |
A value in kJ applies to an actual quantity of material. A molar reaction value in kJ mol⁻¹ applies per mole of reaction as defined by the balanced equation. State the equation and basis: for 2H₂ + O₂ → 2H₂O, one mole of reaction forms two moles of water. Standard data apply to standard states at a specified temperature, commonly 298 K, and standard pressure 100 kPa. The ° symbol is also written θ.
2. Internal energy and pressure–volume change
H = U + pV. At constant pressure, ΔH = ΔU + pΔV. More generally the last term is Δ(pV); pΔV requires unchanged pressure. Expansion gives positive ΔV, contraction negative ΔV. Convert all terms to one energy unit before adding. If ΔH and ΔU are molar, ΔV must refer to the same mole-of-reaction basis.
With only pressure–volume work, heat absorbed at constant pressure equals ΔH; at constant volume it equals ΔU. This explains why a constant-volume measurement need not equal a constant-pressure enthalpy change. For negligible volume change the correction is small, but do not assume it is zero when data are supplied.
Example 1 — expansion with a joule conversion
For a stated sample, ΔU = +1.50 kJ, p = 100 kPa and ΔV = +2.00 dm³ at constant pressure. Convert: p = 100000 Pa; ΔV = 0.00200 m³. Then pΔV = +200 J = +0.200 kJ. Thus ΔH = 1.50 + 0.200 = +1.70 kJ for this sample. Expansion makes ΔH more positive than ΔU.
Example 2 — contraction and a molar basis
For one mole of a specified reaction, ΔU = −50.0 kJ mol⁻¹, p = 200 kPa and ΔV = −5.00 dm³ mol⁻¹. pΔV = 200000 × (−0.00500) = −1000 J mol⁻¹ = −1.00 kJ mol⁻¹. ΔH = −50.0 −1.00 = −51.0 kJ mol⁻¹. Recheck: ΔH − ΔU = −1.00 kJ mol⁻¹, consistent with contraction.
3. Formation data: products minus reactants
ΔH° = ΣνΔfH°(products) − ΣνΔfH°(reactants). ν is each balanced coefficient; Pearson’s Σ notation includes these multipliers. Formation data refer to one mole of each substance made from its reference elements. Reference elements in their standard states have ΔfH° = 0. This rule does not assign zero to all elements in every form, or to all reactants.
Why this order? The route from reactants to products first reverses formation of reactants back to reference elements, then forms products. Its sum is −Σ formation(reactants) + Σ formation(products). States matter: use liquid-water data for H₂O(l), not steam data.
Example 3 — methane combustion from formation data
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). Supplied ΔfH° in kJ mol⁻¹: CH₄ −74.8; O₂ 0; CO₂ −393.5; H₂O(l) −285.8. Product sum = −393.5 + 2(−285.8) = −965.1. Reactant sum = −74.8 + 2(0) = −74.8. ΔH° = −965.1 − (−74.8) = −890.3 kJ mol⁻¹ of reaction. The negative sign means heat release at constant pressure.
Example 4 — coefficients and an endothermic decomposition
CaCO₃(s) → CaO(s) + CO₂(g). Supplied formation values: −1206.9, −635.1 and −393.5 kJ mol⁻¹ respectively. ΔH° = [−635.1 −393.5] − (−1206.9) = +178.3 kJ mol⁻¹. For 2CaCO₃(s) → 2CaO(s) + 2CO₂(g), double every term: +356.6 kJ mol⁻¹ of the doubled reaction, still +178.3 kJ per mole of CaCO₃.
Example 5 — find an unknown formation enthalpy
N₂(g) + 3H₂(g) → 2NH₃(g), ΔH° = −92.4 kJ mol⁻¹ of reaction. Both reference elements have zero formation enthalpy, so −92.4 = 2ΔfH°[NH₃(g)] − 0. Divide by 2: ΔfH°[NH₃(g)] = −46.2 kJ mol⁻¹. Do not assign the two-mole reaction value to one mole of ammonia.
4. Combustion data: reactants minus products
ΔH° = ΣνΔcH°(reactants) − ΣνΔcH°(products). Use this when the two sides can be completely burned to the same amounts and physical states of common products. Include every substance that contributes to the route. Already fully oxidised products such as CO₂(g) and H₂O(l) contribute no further combustion step in these cycles; their formation enthalpies are not zero.
Why reverse the subtraction order? Burn reactants to the common products, then reverse combustion of the target products. Therefore the route is Σ combustion(reactants) − Σ combustion(products). The formula changes because combustion arrows point towards the common state, while formation arrows point away from the common reference state.
Example 6 — hydrogenation from combustion values
C₂H₄(g) + H₂(g) → C₂H₆(g). Supplied ΔcH° values: ethene −1411, hydrogen −286, ethane −1560 kJ mol⁻¹. ΔH° = [−1411 +(−286)] − (−1560) = −137 kJ mol⁻¹. Both routes finish at 2CO₂(g) + 3H₂O(l). Add 3½O₂ to both complete upper states; it is a spectator in the target equation.
Example 7 — recover formation of ethanol using a cycle
Target: 2C(s, graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l). Supplied combustion values: C −393.5, H₂ −285.8, ethanol −1366.8 kJ mol⁻¹. Reactant combustion sum = 2(−393.5) + 3(−285.8) = −1644.4. Ethanol combustion sum = −1366.8. Target ΔH° = −1644.4 − (−1366.8) = −277.6 kJ mol⁻¹. The oxygen term has no additional combustion change. The target forms exactly one mole of ethanol; both full routes end at 2CO₂(g) + 3H₂O(l).
5. Calorimetry: whose heat are you calculating?
ΔQ = mcΔT calculates the sensible heat gained by the named body, assuming c is approximately constant and no phase change occurs. In solution calorimetry, it normally gives qsolution, not the molar reaction enthalpy. For a warming solution ΔT is positive, so qsolution is positive; the exothermic reaction has negative heat transfer.
Neglecting heat leakage, qreaction = −(qsolution + qcalorimeter). If the container’s heat capacity is negligible, qreaction ≈ −mcΔT. Divide by the moles corresponding to the reaction basis, and convert J to kJ, to estimate molar ΔH at constant pressure. For mixed solutions, use the total heated solution mass, not just the mass of the limiting solute. If density is assumed 1.00 g cm⁻³, say so.
Example 8 — neutralisation: heat and moles
Mix 50.0 cm³ of 1.00 mol dm⁻³ HCl with 50.0 cm³ of 1.00 mol dm⁻³ NaOH. HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l). Both provide 0.0500 mol; 0.0500 mol water forms. Assume total solution mass 100.0 g, c = 4.18 J g⁻¹ K⁻¹ and ΔT = +6.80 K. qsolution = 100.0 × 4.18 × 6.80 = +2842.4 J. Neglecting losses and container heat, qreaction = −2.8424 kJ. ΔH ≈ −2.8424/0.0500 = −56.8 kJ mol⁻¹ of water formed (3 significant figures). This experiment estimates a value; it does not automatically establish a standard enthalpy.
Example 9 — an endothermic temperature fall
Dissolve 0.0250 mol of a salt in water. The total final solution mass is 100.0 g, c = 4.18 J g⁻¹ K⁻¹ and temperature falls from 22.0°C to 18.5°C. ΔT = −3.5 K. qsolution = 100.0 × 4.18 × (−3.5) = −1463 J. Hence qdissolving ≈ +1.463 kJ and ΔH ≈ 1.463/0.0250 = +58.5 kJ mol⁻¹ before rounding to the precision of the temperature change (about +59 kJ mol⁻¹). Cooling of the solution indicates heat absorption by dissolution.
Example 10 — fuel calorimetry and heat losses
Burn 0.460 g ethanol (Mr = 46.0) beneath 200.0 g water. The water warms by 12.0 K; c = 4.18 J g⁻¹ K⁻¹. n = 0.460/46.0 = 0.0100 mol. qwater = 200.0 × 4.18 × 12.0 = 10032 J = 10.032 kJ. Estimated ΔcH = −10.032/0.0100 = −1003.2 kJ mol⁻¹, or −1.00 × 10³ kJ mol⁻¹ to 3 significant figures. Heat lost to air and apparatus makes the measured magnitude too small; incomplete combustion and evaporation can also affect the result.
Example 11 — include calorimeter heat capacity
A reaction warms 100.0 g solution by 5.00 K. c = 4.18 J g⁻¹ K⁻¹; calorimeter heat capacity Ccal = 50.0 J K⁻¹. qsolution = 2090 J; qcal = CcalΔT = 250 J. Total surroundings gain = 2340 J. For 0.0400 mol reaction, ΔH ≈ −2.340/0.0400 = −58.5 kJ mol⁻¹. Omitting the container gives −52.3 kJ mol⁻¹, underestimating the exothermic magnitude.
Common misconceptions
- Use the same subtraction order for both tables: formation is products minus reactants; combustion is reactants minus products. Trace arrows to justify it.
- Ignore coefficients: multiply each tabulated molar value by its balanced coefficient.
- Mix J and kJ: divide joules by 1000 before combining with kilojoules.
- Positive mcΔT means endothermic reaction: name the body; warming solution usually accompanies an exothermic reaction.
- q equals molar ΔH: divide the reaction heat by the appropriate number of moles.
- Standard means all temperatures are 298 K: state the temperature and physical states used.
- Use pΔV at variable pressure without qualification: the stated relationship assumes constant pressure.
Exam tips
- Write the balanced target equation and state symbols first.
- List supplied data with units and multiply by coefficients before adding.
- Keep brackets around negative sums.
- Show the mass, temperature interval and moles explicitly in calorimetry.
- Check the sign using the physical heat flow.
- Keep extra digits during working, then round sensibly.
- Quote assumptions such as negligible heat loss and water-like heat capacity.
Check your understanding
Choose one answer, select Check Answer, then read the explanation. Try Again resets the question. Use Tab between controls, arrow keys within a radio group, and Enter or Space on buttons. Answers are not saved.