Enthalpy of Formation (College Board AP® Chemistry)
Study Guide
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
Step 3: Calculate the enthalpy change:
ΔH° = ΔH°f products - ΔH°f reactants = (-1996) - (+ 31.4) = -2027.4 kJ
Examiner Tip
Enthalpy of formation data are given to you in the data booklet.
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