Section 5.5 - Using Standard Enthalpies of Formation (Pg. 331)
 We are not always able to perform experiments to determine the enthalpy
change for a particular reaction.
 We can use formation equations as our working equations to solve for ΔH° of
a target reaction, using Hess’s Law.
Formation Reactions
 Are synthesis reactions that result in the formation of a single product.
 The product is formed from elements in their standard states
 The enthalpy change for this type of reaction is called the standard molar
enthalpy of formation, ΔH°f
 All ΔH°f values represent the quantity of energy that is released or absorbed
when exactly 1 mol of that substance is formed.
Ex. For the formation of carbon monoxide
C(s) + ½O2(g)  CO(g)
ΔH°f = -110.5 kJ (unit is always kJ)
 Appendix C6 (Pg. 799) lists some Standard Enthalpy of Formation Values
(ΔH°f )
 Note: Most are negative, that is because compounds tend to be more stable
than the elements are by themselves…so the formation releases energy to
the surroundings.
 By definition – the enthalpy of formation of an element in its standard state is
zero.
o Ex. Standard state of nitrogen is N2(g)
o Standard state of potassium is K(s)
 If we want an element in a state other than it’s standard state, we
must find a Formation Reaction that gives us ΔH°f
 Ex. Standard enthalpy of formation for oxygen gas, O2(g) = 0 kJ
Standard enthalpy of formation for ozone, O3(g) = 143 kJ
 This also applies to changes of state…ie. O2(g) vs. O2(l)
Writing Formation Equations
1. Write one mole of product in the state that is specified in the question.
2. Write the reactant elements in their standard states.
3. Add coefficients for the reactants to give a balanced equation that results
in 1 mole of product (You may need to use fractions for some elements)
Ex. Write the formation equation for liquid 2-propanol
3 C(S) +
4 H2(g)
+
½ O2(g)

C3H7OH
Practice for writing formation equations: Pg. 332 # 1
Using Standard Molar Enthalpies of Formation
ΔH° = Σ(nΔH°f products) - Σ(nΔH°f reactants), where Σ = sum of
n = coefficient for each compound
For the equation:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Use the table on pg. 799 to calculate the ΔH° value
ΔH° = Σ(nΔH°f products) - Σ(nΔH°f reactants)
= (nΔH°CO2(g) + nΔH°H2O(l) ) - (nΔH°CH4(g) + nΔH°O2(g))
= ( 1 mol CO2 x -393.5 KJ/mol + 2 mol H2O x -285.8 KJ/mol)
– (1 mol CH4 x -74.4 KJ/mol + 2 mol O2 x 0 KJ/mol )
= (-393.5 KJ/mol + 2(-285.8 KJ/mol) - (-74.4 KJ/mol)
= - 890.7 KJ/mol
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
ΔH°comb = -890.7 KJ/mol
Let’s do the same calculation by manipulating the equations using Hess’s Law.
Step 1: Write the formation reactions for products and reactants and their ΔH°f values.
Step 2: Manipulate the equations so that you get the target reaction.
Compare your result to the result obtained above.
The same method can be used to determine the standard enthalpy of formation for a
compound if we know the enthalpy of reaction for the equation, see Pg. 334 Example 2.
Practice Problems: Pg. 335# 2, 3, 4, 5