Standard enthalpy of formation
The standard enthalpy of formation (ΔHf°) is the heat change when one mole of a compound forms from its elements in their standard states. Elements in their standard state have ΔHf° = 0. Then ΔH°rxn = ΣΔHf°(products) − ΣΔHf°(reactants).
Last updated
Analogy
A price list: each compound has a 'cost' from its elements, and the reaction's total is products minus reactants.
Example
ΔHf°: CH₄ −74.8, CO₂ −393.5, H₂O(l) −285.8ΔH = (−393.5 + 2(−285.8)) − (−74.8)ΔH = −890.3 kJ
In practice
Tables of ΔHf° let you predict the heat of thousands of reactions without running them.
Practice
Flashcards
- How do you get ΔH from formation enthalpies?ΣΔHf°(products) − ΣΔHf°(reactants).
Interview questions
- Use standard enthalpies of formation to calculate ΔH for burning methane, and say if it is endothermic or exothermic.
- ΔH = ΣΔHf°(products) − ΣΔHf°(reactants)
- Use CO₂ −393.5, H₂O(l) −285.8, CH₄ −74.8
- Result about −890 kJ/mol
- Negative, so exothermic
Questions people ask
Why is ΔHf° of O₂ zero?
The enthalpy of formation is the energy change to make a compound from its elements. O₂ is already an element in its standard state, so nothing needs to be made, and the value is defined as zero. The same applies to H₂, N₂, solid carbon as graphite, and liquid bromine.
Does a more negative ΔHf° mean a more stable compound?
A more negative ΔHf° means the compound is lower in enthalpy than its elements, so energy is released on forming it. That suggests stability against breaking into its elements. But whether a reaction actually happens also depends on entropy and on how fast it goes.