Formation: from elements. Combustion: burning in O₂. Standard conditions.

Enthalpy of Formation vs Enthalpy of Combustion

Question

State the difference between enthalpy of formation and enthalpy of combustion. A compound has $\Delta H_f° = -286$ kJ/mol. Is this the same as its enthalpy of combustion? Explain with an example.

Solution — Step by Step

Enthalpy of formation ( $\Delta H_f°$ ) is the heat change when one mole of a compound is formed from its constituent elements in their standard states. The key phrase is “from elements” — we start with free elements like $\text{C}_{(s)}$ , $\text{H}_{2(g)}$ , $\text{O}_{2(g)}$ .

\text{C}_{(s)} + \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} \quad \Delta H_f° = -393.5 \text{ kJ/mol}

Enthalpy of combustion ( $\Delta H_c°$ ) is the heat released when one mole of a substance completely burns in excess oxygen. We always start with the compound itself, not elements.

\text{CH}_{4(g)} + 2\text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + 2\text{H}_2\text{O}_{(l)} \quad \Delta H_c° = -890 \text{ kJ/mol}

Yes — only when the element itself burns to form exactly one mole of product. Carbon is the classic case. The formation of $\text{CO}_2$ from $\text{C}_{(s)}$ and $\text{O}_{2(g)}$ is also the combustion of carbon.

So for carbon: $\Delta H_f°(\text{CO}_2) = \Delta H_c°(\text{C}) = -393.5$ kJ/mol. Same reaction, two different names.

$-286$ kJ/mol is $\Delta H_f°$ of liquid water:

\text{H}_{2(g)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{H}_2\text{O}_{(l)} \quad \Delta H_f° = -286 \text{ kJ/mol}

This is also the enthalpy of combustion of hydrogen gas! Here again, forming water from elements is the same as burning $\text{H}_2$ . So yes, for $\text{H}_2$ , these values coincide.

For $\text{CH}_4$ : $\Delta H_f° = -74.8$ kJ/mol (forming methane from C and $\text{H}_2$ ). But $\Delta H_c°(\text{CH}_4) = -890$ kJ/mol (burning methane in oxygen). Completely different values, completely different reactions.

Why This Works

The distinction comes down to reactants. Formation always starts from elements in standard states. Combustion always starts from the compound itself. Same product, different starting points — so different energy changes.

The special overlap case (carbon, hydrogen, sulfur) happens when an element burns to give a single, well-defined product. There’s no ambiguity — formation and combustion describe the same chemical event.

T = 298 \text{ K}, \quad P = 1 \text{ bar}

All standard enthalpies ( $\Delta H°$ ) are defined at these conditions. The ° symbol is your signal that standard states are being used.

Alternative Method — Using Hess’s Law Connection

You can calculate $\Delta H_c°$ of a compound using $\Delta H_f°$ values of reactants and products:

\Delta H_c° = \sum \Delta H_f°(\text{products}) - \sum \Delta H_f°(\text{reactants})

For methane combustion:

\Delta H_c°(\text{CH}_4) = [\Delta H_f°(\text{CO}_2) + 2\Delta H_f°(\text{H}_2\text{O})] - [\Delta H_f°(\text{CH}_4)]

= [(-393.5) + 2(-286)] - (-74.8) = -890.7 \text{ kJ/mol}

This confirms the experimentally measured value. In board exams and JEE, this calculation route appears more often than the direct definition questions.

By definition, $\Delta H_f°$ of any element in its standard state is zero. So $\Delta H_f°(\text{O}_{2(g)}) = 0$ , $\Delta H_f°(\text{C}_\text{graphite}) = 0$ . This is why elements don’t appear in the products’ formation sum when you write out Hess’s law.

Common Mistake

Students write $\Delta H_f°(\text{H}_2\text{O}) = -286$ kJ/mol and then claim “water’s enthalpy of combustion is $-286$ kJ/mol.” Water doesn’t combust — it’s already oxidised completely. The $-286$ kJ/mol is the enthalpy of formation of water, which equals the combustion of hydrogen. The subject of the combustion reaction is $\text{H}_2$ , not $\text{H}_2\text{O}$ .

Always ask: what substance is burning? That’s whose $\Delta H_c°$ you’re calculating.

Quick Reference

Property	$\Delta H_f°$	$\Delta H_c°$
Reactants	Elements in standard states	Compound + $\text{O}_2$
Product	1 mol of compound	$\text{CO}_2$ , $\text{H}_2\text{O}$ , etc.
Sign (usually)	Negative (exothermic)	Always negative
Equal when?	Element that burns to one product (C, H₂, S)	—

For CBSE Class 11 boards, the definition question carries 2 marks. The Hess’s law calculation using $\Delta H_f°$ values is typically a 3-mark numerical in JEE Main.