| Substance | Coeff. | ΔHf° (kJ/mol) |
|---|---|---|
| CH4(g) reactant | 1 | |
| O2(g) reactant | 2 | |
| CO2(g) product | 1 | |
| H2O(l) product | 2 |
ΔH°rxn = -890.3 kJ for the equation as balanced
- Σ n ΔHf°(products) = 1(-393.5) + 2(-285.8) =
-965.1 kJ - Σ n ΔHf°(reactants) = 1(-74.8) + 2(0) =
-74.8 kJ - ΔH°rxn = products − reactants =
-890.3 kJ
Elements in their standard states (O₂(g), H₂(g), Fe(s)…) have ΔHf° = 0. Values are standard data-table values at 298 K; sources differ by a few tenths of a kJ/mol. Any value can be edited.
How it works
Standard enthalpies of formation are the enthalpy change when one mole of a compound forms from its elements in their standard states at 298 K and 1 bar. Elements in their standard states — O₂(g), H₂(g), C(s), Fe(s) — are zero by definition. Because enthalpy is a state function, the reaction enthalpy is simply products minus reactants.
The built-in values are standard data-table values; sources differ by a few tenths of a kJ/mol. Any substance not in the table can be typed in, and every value can be edited. State symbols matter: liquid water (−285.8 kJ/mol) and steam (−241.8 kJ/mol) differ by the enthalpy of vaporisation.
Frequently asked questions
- What is Hess's law?
- The enthalpy change of a reaction is the same whatever route is taken, so it can be found by combining other known enthalpy changes — such as enthalpies of formation.
- What is the enthalpy of combustion of methane?
- CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l): [−393.5 + 2(−285.8)] − [−74.8 + 0] = −890.3 kJ/mol.
- Why is ΔHf° of O₂ zero?
- By definition, elements in their most stable form at standard conditions have an enthalpy of formation of zero; they are the reference point.
Related
Advertisement