Combustion cycles: burning your way to an answer

By the end of this lesson

By the end you can build a cycle in which everything burns to the same products, and use combustion enthalpies to find $\Delta H_f^{\ominus}$ or $\Delta H_r^{\ominus}$ .

Why burn things?

Enthalpy changes of combustion are the easiest to measure: put the substance in a burner, supply excess oxygen, and measure the heat released. If combustion is complete, carbon ends up as CO₂ and hydrogen as H₂O — always the same products.

That shared finishing point makes a perfect meeting place for a cycle. Reactants can be burned; the products can be burned. Both routes end at the same oxides.

Definition

Standard enthalpy change of combustion

The enthalpy change when one mole of a substance is burned completely in excess oxygen under standard conditions, with all reactants and products in their standard states.

Symbol: $\Delta H_c^{\ominus}$ , in $\text{kJ mol}^{-1}$ . It is almost always negative.

The pattern for the calculation

For a substance containing only carbon and hydrogen (a hydrocarbon), complete combustion gives CO₂ and H₂O. The elements carbon and hydrogen burn to the same two oxides.

So, per mole of the compound:

$\Delta H_c^{\ominus}(\text{compound}) = \Delta H_f^{\ominus}(\text{compound}) - \sum \Delta H_c^{\ominus}(\text{elements in it})$

which rearranges to the version you will actually use:

$\Delta H_f^{\ominus}(\text{compound}) = \Delta H_c^{\ominus}(\text{compound}) - \sum \Delta H_c^{\ominus}(\text{elements})$

Careful with what "the elements in it" means: one mole of the compound contains some number of moles of C and some of H₂, and each of those burns separately.

Worked example: Worked example: formation of ethane from combustion data

Find $\Delta H_f^{\ominus}$ of ethane using:

$\Delta H_c^{\ominus}[\text{C}_2\text{H}_6\text{(g)}] = -1560\ \text{kJ mol}^{-1}$
$\Delta H_c^{\ominus}[\text{C(s)}] = -393.5\ \text{kJ mol}^{-1}$
$\Delta H_c^{\ominus}[\text{H}_2\text{(g)}] = -285.8\ \text{kJ mol}^{-1}$

The formation reaction is $2\text{C(s)} + 3\text{H}_2\text{(g)} \rightarrow \text{C}_2\text{H}_6\text{(g)}$ .

Have a go first?

Worked example: Worked example: a reaction enthalpy from combustion data

Find $\Delta H_r^{\ominus}$ for

$\text{C}_2\text{H}_4\text{(g)} + \text{H}_2\text{(g)} \rightarrow \text{C}_2\text{H}_6\text{(g)}$

given $\Delta H_c^{\ominus}$ : C₂H₄ $= -1411$ , H₂ $= -286$ , C₂H₆ $= -1560\ \text{kJ mol}^{-1}$ .

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Quick check

Multiple choiceQuestion 1 · Recall

Multiple choiceQuestion 2 · Recall

Multiple choiceQuestion 3 · Apply

Answer the 3 questions above to continue.