1. STANDARD HEAT OF FORMATION ΔH0f or ΔHθf
It is define as the change in enthalpy that accompanies the formation of one mole of a compound from its elements at standard states of 250C and 101.3kPa.
What are the standard conditions?

2. ΔH0f or ΔHθf ΔH means enthalpy changes or heat of content
0 or θ means standard conditions:
250C and 101.3kPa
f means formation
Please open to page 530, Table 17.4

3. Enthalpy changes of formation are useful for working out enthalpy changes you can’t find directly. You need to know the ΔH0f for all the reactants and products that are compounds. The ΔH0f for elements or diatomic molecules is zero – the element is being formed from the element so there’s no change.

4. Sum all ΔH0f(reactants)
PRODUCTS
ΔH0r
SO2(g) + 2H2S(g)
3S(s) + 2H2O(l)
Sum all ΔH0f(products)
Sum all ΔH0f(reactants)
ΔH0r = ΔH0f(reactants)
ΔH0f(products)

5. EXAMPLE SO2(g) + 2H2S(g) 3S(s) + 2H2O(l) Find the ΔH0r
ΔH0f[SO2(g)] = -297 kJmol-1
ΔH0f[H2S(g)] = kJmol-1
ΔH0f[H2O(l)] = -286kJmol-1

6. ΔH0r = [0 + (-286 x 2)] – [-297 + (-20.2 x 2)]
ΔH0r = ΔH0f(reactants)
ΔH0f(products)
To find ΔH0r of this reaction;
SO2(g) + 2H2S(g) S(s) + 2H2O(l)
REACTANTS
PRODUCTS
ΔH0r = [0 + (-286 x 2)] – [ (-20.2 x 2)]
ΔH0f of sulphur is 0 because it’s an element
There’s 2 moles of H2O and 2 moles of H2S

7. ΔH0r = [0 + (-286 x 2)] – [-297 + (-20.2 x 2)] =
kJmol-1