Enthalpy of Formation Calculations
- Standard Enthalpy of Formation is defined as
“The enthalpy change when one mole of a compound is formed from its elements under standard conditions”
- We can use enthalpy of formation of substances to find an unknown enthalpy change using a Hess's Law cycle
- In this type of cycle the elements are always placed at the bottom of the diagram:
Hess's Law Cycle
Enthalpy changes using enthalpy of formation
- In this cycle the arrows will always be pointing upwards because the definition of the enthalpy of formation must go from elements to compounds
- This means the Hess's Law calculation of ΔH will always be in the same arrangement
ΔHr = ∑ΔHf products - ∑ΔHf reactants
- Try the following worked example:
Worked example
Given the data:
| Substance | B2H6 (g) | B2O3 (g) | H2O (g) |
| ∆H°f / kJ mol-1 | +31.4 | -1270 | -242 |
Calculate the enthalpy of combustion of gaseous diborane given that it burns according to the following equation:
B2H6 (g) + 3O2 (g) → B2O3 (s) + 3H2O (g)
Answer:
- Step 1: Find the sum of the enthalpies of combustion of the products:
- ΔH°f = + (-1270) + (-242 x 3) = -1996 kJ
- Step 2: Find the sum of the enthalpies of combustion of the reactants:
- ΔH°f = + (+31.4) + 0 = + 31.4 kJ
- There is no enthalpy of formation for oxygen as ΔHf of elements by definition is zero
- ΔH°f = + (+31.4) + 0 = + 31.4 kJ
- Step 3: Calculate the enthalpy change:
- ΔH° = ΔH°f products - ΔH°f reactants = (-1996) - (+ 31.4) = -2027.4 kJ
Exam Tip
Enthalpy of formation data are given to you in the data booklet.
