Topic 6 · page 10 of 15

Hess cycles from formation and combustion data

With formation data, ΣΔfH(products) − ΣΔfH(reactants). With combustion data, ΣΔcH(reactants) − ΣΔcH(products). Draw the cycle first, check the arrows, then do the sum.

15 min read6.66.6(ii)

Read first: Hess's Law and enthalpy cycles

Formation data 6.66.6(ii)

The link is the elements. Every compound can be made from them, and elements have ΔfH⊖\Delta_\text{f}H^{\ominus} = 0. Both the reactants and the products are joined to the elements by arrows that point up.

Worked exampleDecomposition of calcium carbonate

Calculate ΔrH\Delta_\text{r}H for CaCOX3(s)→CaO(s)+COX2(g)\ce{CaCO3(s) -> CaO(s) + CO2(g)} using the data below.

Substance ΔfH⊖\Delta_\text{f}H^{\ominus} / kJ mol⁻¹
CaCO₃(s) −1206.9
CaO(s) −635.1
CO₂(g) −393.5
Enthalpy cycle using formation data. The elements calcium, carbon and one and a half oxygen are at the bottom. An arrow goes up from the elements to calcium carbonate, labelled delta f H of calcium carbonate, minus 1206.9. A second arrow goes up from the elements to calcium oxide plus carbon dioxide, labelled minus 1028.6. A top arrow from calcium carbonate to calcium oxide plus carbon dioxide is labelled delta r H equals question mark.
Formation cycle: both sides are formed from the same elements.
  1. Elements to CaCO₃: ΔfH = −1206.9.
  2. Elements to CaO + CO₂: −635.1 + (−393.5) = −1028.6.
  3. The route from CaCO₃ to the products goes against the first arrow, then with the second: ΔrH\Delta_\text{r}H = −(−1206.9) + (−1028.6)
  4. = 1206.9 − 1028.6 = +178.3 kJ mol⁻¹. The reaction is endothermic, as expected for thermal decomposition.

Worked exampleMultipliers and an element

Calculate ΔrH⊖\Delta_\text{r}H^{\ominus} for FeX2OX3(s)+3 CO(g)→2 Fe(s)+3 COX2(g)\ce{Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g)}. ΔfH⊖\Delta_\text{f}H^{\ominus} (kJ mol⁻¹): Fe₂O₃ −824.2, CO −110.5, CO₂ −393.5.

  1. Products: 2Fe has ΔfH\Delta_\text{f}H = 0 (an element). 3CO₂ gives 3 × (−393.5) = −1180.5.
  2. Reactants: Fe₂O₃ = −824.2. 3CO = 3 × (−110.5) = −331.5. Total = −1155.7.
  3. ΔrH⊖\Delta_\text{r}H^{\ominus} = −1180.5 − (−1155.7) = −24.8 kJ mol⁻¹

Combustion data 6.66.6(ii)

The link is the products of burning. Anything made of carbon, hydrogen and oxygen can be burned to CO₂ and H₂O, so the reactants and the products are both joined to those products by arrows that point down.

Worked exampleFormation of propane

Calculate ΔfH\Delta_\text{f}H of propane, C₃H₈(g), from ΔcH⊖\Delta_\text{c}H^{\ominus} (kJ mol⁻¹): C(s) −393.5, H₂(g) −285.8, C₃H₈(g) −2219.

Enthalpy cycle using combustion data. The elements, three carbon, four hydrogen and five oxygen, are top left. Propane plus five oxygen is top right. Three carbon dioxide plus four liquid water is at the bottom. A top arrow from the elements to propane plus oxygen is labelled delta f H of propane equals question mark. Two arrows run down to the combustion products: from the elements, labelled minus 2323.7, and from propane plus oxygen, labelled minus 2219.
Combustion cycle: both sides burn to the same products.
  1. Burn the elements: 3 × (−393.5) + 4 × (−285.8) = −1180.5 − 1143.2 = −2323.7.
  2. Burn propane: −2219.
  3. The route from the elements to propane goes with the first arrow, then against the second: ΔfH\Delta_\text{f}H = −2323.7 − (−2219)
  4. = −104.7 kJ mol⁻¹

Worked exampleHydrogenation of ethene from combustion data

Calculate ΔrH⊖\Delta_\text{r}H^{\ominus} for CX2HX4(g)+HX2(g)→CX2HX6(g)\ce{C2H4(g) + H2(g) -> C2H6(g)}. ΔcH⊖\Delta_\text{c}H^{\ominus} (kJ mol⁻¹): C₂H₄ −1411, H₂ −285.8, C₂H₆ −1560.

  1. Reactants: −1411 + (−285.8) = −1696.8.
  2. Products: −1560.
  3. ΔrH⊖\Delta_\text{r}H^{\ominus} = reactants − products = −1696.8 − (−1560) = −136.8 kJ mol⁻¹

Choosing data from a table 6.66.6(ii)

Questions often give a long table and expect you to select what the cycle needs.

Substance ΔfH⊖\Delta_\text{f}H^{\ominus} / kJ mol⁻¹
CO₂(g) −393.5
H₂O(l) −285.8
H₂O(g) −241.8
C₂H₅OH(l) −277.7
O₂(g) 0

Worked exampleCombustion of ethanol from formation data

Use the table to calculate ΔcH⊖\Delta_\text{c}H^{\ominus} of ethanol.

  1. The equation is CX2HX5OH(l)+3 OX2(g)→2 COX2(g)+3 HX2O(l)\ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)}. Combustion is defined with liquid water, so choose H₂O(l), not H₂O(g).
  2. Products: 2(−393.5) + 3(−285.8) = −787.0 − 857.4 = −1644.4.
  3. Reactants: −277.7 + 3(0) = −277.7.
  4. ΔcH⊖\Delta_\text{c}H^{\ominus} = −1644.4 − (−277.7) = −1366.7 kJ mol⁻¹, the −1367 quoted earlier.

MethodChoosing a method

  1. List the data you have. Formation values: use ΣΔfH(products) − ΣΔfH(reactants). Combustion values: use ΣΔcH(reactants) − ΣΔcH(products).
  2. Write the equation with state symbols and decide which table entries match, in the right state.
  3. Multiply by the coefficients. Elements in their standard states are zero.
  4. Sketch the cycle and check that each route follows the arrows and gives the same sign as your sum.

Check yourself

  1. Calculate ΔrH⊖\Delta_\text{r}H^{\ominus} for CHX4(g)+2 OX2(g)→COX2(g)+2 HX2O(l)\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)} using ΔfH⊖\Delta_\text{f}H^{\ominus} (kJ mol⁻¹): CH₄(g) −74.8, CO₂(g) −393.5, H₂O(l) −285.8.
  2. Calculate ΔfH⊖\Delta_\text{f}H^{\ominus} of ethanol, C₂H₅OH(l), using ΔcH⊖\Delta_\text{c}H^{\ominus} (kJ mol⁻¹): C(s) −393.5, H₂(g) −285.8, C₂H₅OH(l) −1367.
  3. Calculate ΔfH⊖\Delta_\text{f}H^{\ominus} of ethene, C₂H₄(g), using ΔcH⊖\Delta_\text{c}H^{\ominus} (kJ mol⁻¹): C(s) −393.5, H₂(g) −285.8, C₂H₄(g) −1411.
  4. For CaCOX3(s)→CaO(s)+COX2(g)\ce{CaCO3(s) -> CaO(s) + CO2(g)}, ΔrH\Delta_\text{r}H = +178.3 kJ mol⁻¹. State ΔrH\Delta_\text{r}H for (a) CaO(s)+COX2(g)→CaCOX3(s)\ce{CaO(s) + CO2(g) -> CaCO3(s)}; (b) 2 CaO(s)+2 COX2(g)→2 CaCOX3(s)\ce{2CaO(s) + 2CO2(g) -> 2CaCO3(s)}.
  5. Calculate ΔrH⊖\Delta_\text{r}H^{\ominus} for 2 Al(s)+FeX2OX3(s)→AlX2OX3(s)+2 Fe(s)\ce{2Al(s) + Fe2O3(s) -> Al2O3(s) + 2Fe(s)} using ΔfH⊖\Delta_\text{f}H^{\ominus} (kJ mol⁻¹): Al₂O₃ −1675.7, Fe₂O₃ −824.2.
  6. Explain why ΔfH\Delta_\text{f}H of propane is calculated from combustion data and not measured directly.

Answers cover these until you have tried the questions

  1. Products: −393.5 + 2(−285.8) = −965.1. Reactants: −74.8 + 0 = −74.8. ΔrH⊖\Delta_\text{r}H^{\ominus} = −965.1 − (−74.8) = −890.3 kJ mol⁻¹.
  2. Equation: 2 C(s)+3 HX2(g)+12 OX2(g)→CX2HX5OH(l)\ce{2C(s) + 3H2(g) + 1/2 O2(g) -> C2H5OH(l)}. Burn the elements: 2(−393.5) + 3(−285.8) = −787.0 − 857.4 = −1644.4. Burn ethanol: −1367. ΔfH⊖\Delta_\text{f}H^{\ominus} = −1644.4 − (−1367) = −277.4 kJ mol⁻¹.
  3. Equation: 2 C(s)+2 HX2(g)→CX2HX4(g)\ce{2C(s) + 2H2(g) -> C2H4(g)}. Burn the elements: 2(−393.5) + 2(−285.8) = −787.0 − 571.6 = −1358.6. Burn ethene: −1411. ΔfH⊖\Delta_\text{f}H^{\ominus} = −1358.6 − (−1411) = +52.4 kJ mol⁻¹ (endothermic: ethene has more enthalpy than its elements).
  4. Reversing the equation changes the sign: (a) −178.3 kJ mol⁻¹. Doubling it doubles the value: (b) −356.6 kJ mol⁻¹.
  5. Products: −1675.7 + 2(0) = −1675.7. Reactants: 2(0) + (−824.2) = −824.2. ΔrH⊖\Delta_\text{r}H^{\ominus} = −1675.7 − (−824.2) = −851.5 kJ mol⁻¹.
  6. Carbon and hydrogen do not combine to make only propane: a mixture of hydrocarbons forms, so the reaction cannot be run cleanly in a calorimeter. Burning each substance can be measured, so a cycle through the combustion products gives the value.