Thermochemistry
Subject: Chemistry
Topic: 6
Cambridge Code: 0620 / 0971 / 5070
Energy in Reactions
Exothermic Reactions
Definition: Release energy (heat) to surroundings
- Temperature increases
- Combustion, neutralization
- ΔH = negative
- Products lower energy than reactants
Examples:
- Burning fuel: CH₄ + 2O₂ → CO₂ + 2H₂O + heat
- Neutralization: HCl + NaOH → NaCl + H₂O + heat
Endothermic Reactions
Definition: Absorb energy from surroundings
- Temperature decreases
- Photosynthesis, melting, evaporation
- ΔH = positive
- Products higher energy than reactants
Examples:
- Photosynthesis: 6CO₂ + 6H₂O + light → C₆H₁₂O₆ + 6O₂
- Melting ice: Ice + heat → Water
Enthalpy Change (ΔH)
Enthalpy (H) - Total heat content
Enthalpy change (ΔH):
Sign Convention
- ΔH < 0: Exothermic (heat released)
- ΔH > 0: Endothermic (heat absorbed)
Units
- J/mol (joules per mole)
- kJ/mol (kilojoules per mole)
- Often written: ΔH = -393 kJ/mol
Calorimetry
Calorimetry - Measure heat change in reactions
Heat Equation
where:
- q = heat energy (joules)
- m = mass (grams)
- c = specific heat capacity (J/g°C)
- ΔT = temperature change (°C)
Common Specific Heat Capacities
| Substance | c (J/g°C) |
|---|---|
| Water | 4.18 |
| Aluminum | 0.897 |
| Iron | 0.449 |
| Copper | 0.385 |
Example Calculation
Problem: 100 g water heated from 20°C to 30°C (c = 4.18)
Bomb Calorimeter
- Insulated chamber
- Combustion occurs in sealed container
- Temperature rise measured
- Accurate for combustion reactions
Simple Calorimeter
- Beaker with thermometer
- Less accurate (heat loss to surroundings)
- Laboratory use
Hess's Law
Hess's Law - Enthalpy change independent of reaction pathway
Application
Find ΔH of difficult reaction using known reactions
Example: Find ΔH for: C(s) + O₂(g) → CO₂(g)
Given:
- C(s) + ½O₂(g) → CO(g), ΔH₁ = -110.5 kJ
- CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = -283.0 kJ
Solution:
- Add equations 1 + 2:
- C(s) + O₂(g) → CO₂(g), ΔH = -110.5 + (-283.0) = -393.5 kJ
Reversing Equations
If reverse equation: ΔH changes sign
Example:
- Forward: H₂ + ½O₂ → H₂O, ΔH = -286 kJ
- Reverse: H₂O → H₂ + ½O₂, ΔH = +286 kJ
Scaling Equations
If multiply by factor n: ΔH also multiplies by n
Example:
- Given: N₂ + 3H₂ → 2NH₃, ΔH = -92 kJ
- Multiply by 2: 2N₂ + 6H₂ → 4NH₃, ΔH = -184 kJ
Standard Enthalpy Change of Formation (ΔHf°)
ΔHf° - Enthalpy change when 1 mole of compound forms from elements in standard state
Standard state: Most stable form at 25°C, 1 atm
Values
ΔHf° = 0 for elements in standard state
- O₂(g) has ΔHf° = 0
- C(s) has ΔHf° = 0
ΔHf° for compounds: Tabulated values available
Calculating ΔH°reaction
Example: CH₄ + 2O₂ → CO₂ + 2H₂O
Using standard enthalpies of formation
Breaking and Forming Bonds
Bond Energy
Bond energy - Energy needed to break bond or released when formed
Energy Calculation
Example: H₂ + Cl₂ → 2HCl
- Bonds broken: H-H (436 kJ) + Cl-Cl (244 kJ) = 680 kJ
- Bonds formed: 2 × H-Cl (432 kJ) = 864 kJ
- ΔH = 680 - 864 = -184 kJ (exothermic)
Fuel Value
Fuel value - Energy released per mass of fuel
Examples
| Fuel | Value (kJ/g) |
|---|---|
| Hydrogen | 141 |
| Methane | 55.5 |
| Ethanol | 29.6 |
| Coal | 30 |
Higher value = Better fuel
Key Points
- Exothermic: ΔH < 0 (heat released)
- Endothermic: ΔH > 0 (heat absorbed)
- Heat equation: q = mcΔT
- Hess's Law: ΔH independent of pathway
- ΔHf° for elements = 0
- Bond energy: Breaking requires energy, forming releases
- Fuel value: Energy per unit mass
Practice Questions
- Classify reactions as exothermic/endothermic
- Calculate heat using q = mcΔT
- Apply Hess's Law to calculate ΔH
- Use ΔHf° values to find ΔH°reaction
- Calculate from bond energies
- Compare fuel values
Revision Tips
- Know exothermic/endothermic clearly
- Practice calorimetry calculations
- Understand Hess's Law
- Learn ΔHf° concept
- Know bond energy calculations
- Know equation manipulation rules
- Practice multiple approaches