What are standard conditions in A Level Chemistry?

A pressure of 100 kPa and a temperature of 298 K (25 °C). Some courses give the pressure as 101 kPa (1 atm). 273 K is 0 °C, which is a common mix-up.

Why is the enthalpy of formation of an element zero?

Forming an element in its standard state from itself involves no change, so its standard enthalpy of formation is zero by definition.

Why do formation equations have fractions in them?

The definition is for one mole of the compound, so the elements are balanced to make exactly one mole. For water that means H₂(g) + ½O₂(g) → H₂O(l).

Standard enthalpy of formation definition

In short: The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states, under standard conditions of 100 kPa and 298 K.

Definitions are easy marks, but only if every word is right. This one looks simple, and that is exactly why small slips cost students the mark.

Spot the mistake

In the clip, a student defines the standard enthalpy of formation as "the enthalpy change when one mole of a compound is formed from its elements in their standard states, at 100 kPa and 273 K."

The mistake is the temperature. Standard conditions are 100 kPa and 298 K (25 °C). 273 K is 0 °C, and it is a common mix-up. (Some courses give the pressure as 101 kPa, or 1 atm; use the value in your own course's data booklet.)

Each key word earns its place

Enthalpy change: the heat energy change at constant pressure, given the symbol ΔfH⦵.
One mole of a compound: the definition is per mole of the product, which is why formation equations often need fractions.
From its elements: only elements go on the left of the equation, never compounds.
In their standard states: each element as it is at 100 kPa and 298 K, for example O₂(g), H₂(g) and C(s).
Standard conditions: 100 kPa and 298 K.

Rich's tip: The devil is in the detail with definitions. Learn them word for word, then write them from memory.

Writing formation equations

Make one mole of the compound, from elements in their standard states, with state symbols. For water:

H₂(g) + ½O₂(g) → H₂O(l)

For ethanol:

2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)

The ½O₂ is right: halving the oxygen is how you keep exactly one mole of product.

Why an element's enthalpy of formation is zero

Forming an element in its standard state from itself involves no change at all. So the standard enthalpy of formation of any element in its standard state, such as O₂(g) or C(s), is zero. Remember that when you use the values in a Hess's law calculation.

How it is used: Hess's law

Enthalpies of formation let you work out the enthalpy change of a reaction you cannot measure directly:

ΔrH = ΣΔfH(products) − ΣΔfH(reactants)

Multiply each value by the number of moles in the balanced equation before you add them up.

Formation or combustion?

The two definitions are often mixed up.

Enthalpy of formation: one mole of a compound is formed from its elements.
Enthalpy of combustion: one mole of a substance is burned completely in oxygen.

Both are under standard conditions, with everything in its standard state. In formation, the one mole is the product. In combustion, the one mole is the substance being burned.

If you want to test yourself on definitions like this one, Dexter, the A Level Chemistry tutor I built, can quiz you on the definitions in your course and point out any key words you missed.