Topic 6 · page 4 of 15

Enthalpy changes of combustion, neutralisation and atomisation

Combustion counts per mole of fuel burnt, neutralisation per mole of water formed, atomisation per mole of gaseous atoms made. Know each definition and its equation, with state symbols.

14 min read6.46.4(iii)6.4(iv)6.4(v)

Read first: Enthalpy changes of reaction and formation

Enthalpy change of combustion 6.46.4(iii)

The equation has one mole of the fuel on the left and excess oxygen. For a compound of carbon, hydrogen and oxygen, the products are carbon dioxide gas and liquid water. Fractions of O₂ are fine.

CHX4(g)+2 OX2(g)→COX2(g)+2 HX2O(l)ΔcH⊖=−890 kJ mol−1 \ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)} \qquad \Delta_\text{c}H^{\ominus} = -890\ \text{kJ mol}^{-1}
CX2HX5OH(l)+3 OX2(g)→2 COX2(g)+3 HX2O(l)ΔcH⊖=−1367 kJ mol−1 \ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)} \qquad \Delta_\text{c}H^{\ominus} = -1367\ \text{kJ mol}^{-1}
Enthalpy level diagram for the combustion of ethanol. Liquid ethanol plus three oxygen on the upper level, two carbon dioxide plus three liquid water on the lower level, a downward arrow labelled delta c H equals minus 1367 kilojoules per mole.
Combustion is always exothermic, so the arrow always points down.

Combustion is always exothermic, so ΔcH⊖\Delta_\text{c}H^{\ominus} is always negative.

Worked exampleA combustion equation with fractions

Write the equation for the standard enthalpy change of combustion of propan-1-ol, C₃H₇OH(l).

  1. One mole of fuel on the left: C₃H₇OH(l) + O₂(g).
  2. Products: 3 carbons give 3CO₂(g); 8 hydrogens give 4H₂O(l).
  3. Oxygen needed on the right: 6 + 4 = 10 atoms. One is already in the alcohol, so 9 atoms come from O₂, which is 4½ O₂.

CX3HX7OH(l)+92 OX2(g)→3 COX2(g)+4 HX2O(l)\ce{C3H7OH(l) + 9/2 O2(g) -> 3CO2(g) + 4H2O(l)}

Some equations carry two names. C(s)+OX2(g)→COX2(g)\ce{C(s) + O2(g) -> CO2(g)} is both the formation of CO₂ and the combustion of carbon, so ΔfH⊖(CO2)=ΔcH⊖(C)=−393.5 kJ mol−1\Delta_\text{f}H^{\ominus}(\text{CO}_2) = \Delta_\text{c}H^{\ominus}(\text{C}) = -393.5\ \text{kJ mol}^{-1}. In the same way ΔcH⊖(H2)=ΔfH⊖(H2O(l))=−285.8 kJ mol−1\Delta_\text{c}H^{\ominus}(\text{H}_2) = \Delta_\text{f}H^{\ominus}(\text{H}_2\text{O(l)}) = -285.8\ \text{kJ mol}^{-1}.

Enthalpy change of neutralisation 6.46.4(iv)

For a strong acid with a strong alkali the reaction is the same ionic equation whatever the pair:

HX+(aq)+OHX−(aq)→HX2O(l)ΔneutH⊖≈−57 kJ mol−1 \ce{H+(aq) + OH-(aq) -> H2O(l)} \qquad \Delta_\text{neut}H^{\ominus} \approx -57\ \text{kJ mol}^{-1}
Two enthalpy level diagrams. Left, neutralisation: hydrogen ions plus hydroxide ions on the upper level, one mole of liquid water on the lower level, downward arrow of about minus 57 kilojoules per mole. Right, atomisation of chlorine: half a mole of chlorine gas on the lower level, one mole of gaseous chlorine atoms on the upper level, upward arrow of plus 122 kilojoules per mole.
Neutralisation (exothermic, down) and atomisation (endothermic, up), both per mole of the product.

The count is per mole of water, not per mole of acid. Sulfuric acid is dibasic, so HX2SOX4+2 NaOH→NaX2SOX4+2 HX2O\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O} makes two moles of water and releases about 2 × 57 = 114 kJ per mole of H₂SO₄. Neutralisation is always exothermic, so the sign is always negative.

Enthalpy change of atomisation 6.46.4(v)

Na(s)→Na(g)ΔatH⊖=+107 kJ mol−1 \ce{Na(s) -> Na(g)} \qquad \Delta_\text{at}H^{\ominus} = +107\ \text{kJ mol}^{-1}
12 ClX2(g)→Cl(g)ΔatH⊖=+122 kJ mol−1 \ce{1/2 Cl2(g) -> Cl(g)} \qquad \Delta_\text{at}H^{\ominus} = +122\ \text{kJ mol}^{-1}

The product is atoms, so chlorine, a gas of Cl₂ molecules, needs ½Cl₂. Particles must be pulled apart, which takes energy, so atomisation is always endothermic and ΔatH⊖\Delta_\text{at}H^{\ominus} is always positive.

The five at a glance

Enthalpy change Per one mole of The equation features Sign
reaction, ΔrH⊖\Delta_\text{r}H^{\ominus} the reaction as written any balanced equation + or −
formation, ΔfH⊖\Delta_\text{f}H^{\ominus} compound formed elements in standard states on the left + or −
combustion, ΔcH⊖\Delta_\text{c}H^{\ominus} substance burned O₂ in; CO₂(g) and H₂O(l) out always −
neutralisation, ΔneutH⊖\Delta_\text{neut}H^{\ominus} water formed H⁺(aq) + OH⁻(aq) → H₂O(l) always −
atomisation, ΔatH⊖\Delta_\text{at}H^{\ominus} gaseous atoms formed element → gaseous atoms always +

Check yourself

  1. Write the equation for the standard enthalpy change of combustion of butane, C₄H₁₀(g).
  2. Explain why the water in a combustion equation is written as H₂O(l) and not H₂O(g).
  3. ΔcH⊖\Delta_\text{c}H^{\ominus} of methane is −890 kJ mol⁻¹. How much heat is released when 8.00 g of methane burns completely? (MrM_\text{r} = 16.0)
  4. Sulfuric acid reacts with sodium hydroxide: HX2SOX4(aq)+2 NaOH(aq)→NaX2SOX4(aq)+2 HX2O(l)\ce{H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l)}. Estimate the enthalpy change for this equation as written, and explain why it is not −57 kJ mol⁻¹.
  5. Write the equation for the standard enthalpy change of atomisation of (a) bromine, Br₂(l); (b) carbon, C(s). State the sign of each.
  6. State two different names for the enthalpy change of HX2(g)+12 OX2(g)→HX2O(l)\ce{H2(g) + 1/2 O2(g) -> H2O(l)}, and give its value.

Answers cover these until you have tried the questions

  1. CX4HX10(g)+132 OX2(g)→4 COX2(g)+5 HX2O(l)\ce{C4H10(g) + 13/2 O2(g) -> 4CO2(g) + 5H2O(l)}. Check oxygen: 8 + 5 = 13 atoms, so 13/2 O₂.
  2. The definition uses standard states, and water is a liquid at 298 K and 100 kPa. Water vapour would give a less negative value, because condensing to a liquid gives out extra heat.
  3. n(CH₄) = 8.00 ÷ 16.0 = 0.500 mol. Heat released = 0.500 × 890 = 445 kJ.
  4. Two moles of water form, so ΔH ≈ 2 × (−57) = −114 kJ mol⁻¹. The value −57 is per mole of water, and this equation as written makes two.
  5. (a) 12 BrX2(l)→Br(g)\ce{1/2 Br2(l) -> Br(g)} (b) C(s)→C(g)\ce{C(s) -> C(g)}. Both are positive: atomisation is always endothermic.
  6. It is the standard enthalpy change of formation of water, and also the standard enthalpy change of combustion of hydrogen. Both equal −285.8 kJ mol⁻¹.