Enthalpy of Reaction Formula
Calculate heat change using standard enthalpies of formation
Main Formula
ΔH°rxn = Σn·ΔH°f(products) - Σn·ΔH°f(reactants)
ΔH°rxn = standard enthalpy of reaction (kJ/mol)
ΔH°f = standard enthalpy of formation (kJ/mol)
n = stoichiometric coefficient
Σ = sum of all species
Standard conditions: 25°C (298 K), 1 atm
Products (Subtract From)
- • Use ΔH°f values from tables
- • Multiply by coefficients
- • Sum all products
- • Positive contribution
Reactants (Subtract)
- • Use ΔH°f values from tables
- • Multiply by coefficients
- • Sum all reactants
- • Negative contribution
Common ΔH°f Values (kJ/mol)
Important Elements
| H₂(g) | 0 |
| O₂(g) | 0 |
| N₂(g) | 0 |
| C (graphite) | 0 |
| Br₂(l) | 0 |
| Elements in standard state = 0 | |
Common Compounds
| H₂O(l) | -285.8 |
| H₂O(g) | -241.8 |
| CO₂(g) | -393.5 |
| CO(g) | -110.5 |
| NH₃(g) | -45.9 |
| CH₄(g) | -74.6 |
| C₂H₅OH(l) | -277.6 |
Ionic Compounds
| NaCl(s) | -411.2 |
| CaO(s) | -635.1 |
| Fe₂O₃(s) | -824.2 |
| Al₂O₃(s) | -1675.7 |
| MgO(s) | -601.6 |
Acids & Bases
| HCl(g) | -92.3 |
| H₂SO₄(l) | -814.0 |
| HNO₃(l) | -174.1 |
| NaOH(s) | -425.6 |
Interpreting ΔH°rxn
ΔH° < 0 (Negative)
Exothermic Reaction
- Heat released to surroundings
- Products lower enthalpy than reactants
- Thermodynamically favorable
- Temperature increases
- Example: Combustion reactions
ΔH° > 0 (Positive)
Endothermic Reaction
- Heat absorbed from surroundings
- Products higher enthalpy than reactants
- Requires energy input
- Temperature decreases
- Example: Photosynthesis, decomposition
Worked Examples
Example 1: Combustion of Methane
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Given ΔH°f values:
CH₄(g) = -74.6 kJ/mol
O₂(g) = 0 kJ/mol (element)
CO₂(g) = -393.5 kJ/mol
H₂O(l) = -285.8 kJ/mol
Solution:
Products:
1×(-393.5) + 2×(-285.8)
= -393.5 - 571.6 = -965.1 kJ
Reactants:
1×(-74.6) + 2×(0)
= -74.6 kJ
Calculate ΔH°rxn:
ΔH° = -965.1 - (-74.6)
ΔH° = -965.1 + 74.6
ΔH°rxn = -890.5 kJ/mol (Highly Exothermic!)
Example 2: Formation of Ammonia
N₂(g) + 3H₂(g) → 2NH₃(g)
Given:
N₂(g) = 0 kJ/mol
H₂(g) = 0 kJ/mol
NH₃(g) = -45.9 kJ/mol
Solution:
Products: 2×(-45.9) = -91.8 kJ
Reactants: 0 + 0 = 0 kJ
ΔH° = -91.8 - 0 = -91.8 kJ/mol
ΔH°rxn = -91.8 kJ/mol (Exothermic)
Example 3: Decomposition of Calcium Carbonate
CaCO₃(s) → CaO(s) + CO₂(g)
Given:
CaCO₃(s) = -1206.9 kJ/mol
CaO(s) = -635.1 kJ/mol
CO₂(g) = -393.5 kJ/mol
Solution:
Products: -635.1 + (-393.5) = -1028.6 kJ
Reactants: -1206.9 kJ
ΔH° = -1028.6 - (-1206.9)
ΔH° = -1028.6 + 1206.9
ΔH°rxn = +178.3 kJ/mol (Endothermic)
Requires heat to decompose limestone!
Common Mistakes
Reversing Products and Reactants
It's PRODUCTS - REACTANTS, not REACTANTS - PRODUCTS! Order matters.
Forgetting Stoichiometric Coefficients
For 2H₂O, multiply ΔH°f by 2! Coefficients are critical.
Assuming Elements Have ΔH°f ≠ 0
Elements in standard state ALWAYS have ΔH°f = 0 by definition!
Using Wrong State
H₂O(l) ≠ H₂O(g). Phase matters! Check (s), (l), (g), or (aq).
Standard State Reference
O₂(g), N₂(g), H₂(g), C(graphite), Br₂(l), I₂(s), Hg(l) at 25°C, 1 atm all have ΔH°f = 0