Standard Molar Enthalpy of Formation (∆H°f)
This is the change in heat that accompanies the formation of one mole of a compound from its elements in
their most stable states under standard state conditions. (at SATP: 25 C and 100 kPa)
Example: H2(g) + ½ O2(g) → H2O(l)
∆H°f = -286 kJ
(This means that 286 kJ of energy was released in the formation of one mole of the compound water from
the elements hydrogen and oxygen.)
∆H°f values have been measured for most compounds and collected into tables (see Table E.13: Standard
Molar Enthalpies of Formation in your handout). We can use these values to calculate the overall ∆H° value
for any chemical change. Note: All elements in their standard state have a ΔHf = 0
Example: What is the enthalpy change that occurs in this reaction?
2 SO2(g) + O2(g) → 2 SO3(g)
∆H° = ?
We can calculate this by subtracting the total enthalpy change of the reactants from the total enthalpy
change of the products. It is important to include the number of moles (coefficients) in all calculations.
2 SO2(g) + O2(g)
→
2 SO3(g)
2 mol (-297 kJ/ mol) + 1 mol (0 kJ/ mol) → 2 mol (-396 kJ/ mol)
-594 kJ → -792 kJ
∆H° = -198 kJ
∆H⁰rxn = Σ n∆H⁰f (products) - Σ n∆H⁰f (reactants)
= -792 KJ - -594 KJ = -198 kJ (the reaction is exothermic)
Therefore, the thermochemical equation can be written as: 2 SO2(g) + O2(g) → 2 SO3(g) + 198 kJ
Sample Problem on p.687: Iron(III) oxide reacts with carbon monoxide to produce elemental iron and
carbon dioxide. Determine the enthalpy change of this reaction, using known enthalpies of formation.
Fe2O3(s) + 3CO(g) → 3CO2(g) + 2Fe(s)
Step 1: From Appendix E, you can obtain the enthalpies of formation.
ΔH°f of Fe2O3(s) = −824.2 kJ/mol
ΔH° f of CO(g) = −110.5 kJ/mol
ΔH° f of CO2(g) = −393.5 kJ/mol
ΔH° f of Fe(s) = 0 kJ/mol (it is an element in its natural state: solid)
Step 2: Multiply the enthalpy value by the number of moles (coefficient). Then find the sum of the
enthalpies of all reactants, then products, then find the difference.
Fe2O3(s) + 3CO(g) → 3CO2(g) + 2Fe(s)
1 mol (−824.2 kJ/ mol) + 3 mol (−110.5 kJ/ mol) → 3 mol (−393.5 kJ/ mol)
+ 2 mol (0 kJ/ mol)
−824.2 kJ + -331.5 kJ → -1180.5 kJ + 0 kJ
-1155.7 kJ → -1180.5 kJ
∆H° = 24.8 kJ (when adding or subtracting, don’t forget to represent the final answer with the least # of
decimal places)
Sample Problem 2: A Car racing circuit has two fuels, methanol and octane. Determine which of the two
fuels will give greater energy output per mole?
Plan: Calculate ∆H for the combustion of each fuel, using standard molar heats of formation (∆H° f).
Part A: Focus on methanol
Step 1: Write a balanced chemical reaction to represent the combustion of methanol.
2CH3OH (l) + 3 O2 (g) → 2 CO2 (g)+ 4H2 O (l)
Step 2: Refer to the Standard Molar Enthalpies of Formation Table to determine the enthalpies of
formation for each individual reactant/product, and multiply them by the number of moles.Then find the
sum of the enthalpies of all reactants, then products.
2CH3OH (l) + 3 O2 (g)
→
2 CO2 (g)+ 4H2 O (l)
2 (-239.1 kJ/mol) + 3 (0 kJ/mol)
→
2(-393.5 kJ/mol) + 4 (-285.8 kJ/mol)
-478.2 + 0
→
-787.0 + -1143.2
-478.2
→
-1930.2
Step 3: Calculate the difference between the final ∆Hf of the products and the ∆Hf of the reactants
∆H°rxn = ∆H°f [final (products)] - ∆H°f [initial (reactants)] = -1930.2 - (-478.2) = -1452.0
Step 4: Determine the energy output of the fuel per mole:
The combustion of 2 moles of CH3OH (l) = -1452.0 kJ,
so -1452.0 kJ/2 mol = -726.0 kJ/mol for CH3OH
Part B: Focus on octane
Step 1: Write a balanced chemical reaction to represent the combustion of methanol.
2C8H18 (l) + 25 O2 (g)
→
16 CO2 (g)+ 18H2 O (l)
Step 2: Refer to the Standard Molar Enthalpies of Formation Table to determine the enthalpies of
formation for each individual reactant/product, and multiply them by the number of moles.Then find the
sum of the enthalpies of all reactants, then products.
2C8H18 (l) + 25 O2 (g)
→
16 CO2 (g)+ 18H2 O (l)
2 (-250 kJ/mol) + 25 (0 kJ/mol)
→
16(-393.5 kJ/mol) + 18(-285.8 kJ/mol)
-500.0 + 0
→
-6296.0 + -5144.4
-500.0
→
-11440.4
Step 3: Calculate the difference between the final ∆Hf of the products and the ∆Hf of the reactants
∆H°rxn = ∆H°f [final (products)] - ∆H°f [initial (reactants)] = -11440.4 - (-500.0) = -10940.4
Step 4: Determine the energy output of the fuel per mole:
The combustion of 2 moles of C8H18 (l) = -10940.4
so -10940.4 kJ/2 mol = -5470.0 kJ/mol for C8H18
Try this: Find the molar enthalpy of reaction for each of the following, then indicate whether the
reaction is endothermic or exothermic:
1. 2 CO2(g) → 2 CO(g) + O2(g)
2. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
3. 4 CO2(g) + 2 H2O(g) → 2 C2H2(g) + 5 O2(g)
4. P4O6(s) + 2 O2(g) → P4O10(s)
5. 2 Fe(s) + O2(g) → 2 FeO(s)
6. 2 Fe2O3(s) → 4 FeO(s) + O2(g)
7. SO2(g) + H2O(l) → H2SO3(aq) (note: ∆H°f of H2SO3(aq) is - 609.0)