Understand and apply the enthalpy change (ΔrH°) for chemical reactions under standard conditions.
Enthalpy change (ΔH): The heat transferred at constant pressure during a chemical reaction.
Standard enthalpy change of reaction (ΔrH°): The enthalpy change when reactants in their standard states form products in their standard states at 1 bar pressure and 25°C (298 K).
Standard state: The most stable form of a substance at 1 bar and a specified temperature (usually 25°C).
Positive ΔrH° → Endothermic (absorbs heat from surroundings)
Negative ΔrH° → Exothermic (releases heat to surroundings)
Reversing a chemical equation reverses the sign of ΔrH°.
Example:
H₂O(g) → H₂(g) + ½O₂(g) → ΔrH° = +241.8 kJ/mol
H₂(g) + ½O₂(g) → H₂O(g) → ΔrH° = −241.8 kJ/mol
If you multiply the coefficients in a reaction, multiply ΔrH° by the same factor.
Example:
H₂(g) + ½O₂(g) → H₂O(g), ΔrH° = −241.8 kJ/mol
2H₂(g) + O₂(g) → 2H₂O(g), ΔrH° = −483.6 kJ/mol
ΔrH° values differ for different physical states of products/reactants.
Example:
H₂(g) + ½O₂(g) → H₂O(g), ΔrH° = −241.8 kJ/mol
H₂(g) + ½O₂(g) → H₂O(l), ΔrH° = −285.8 kJ/mol
Condensation of water vapor: −44.0 kJ/mol
Use this two-step method:
mol=mass (g)molar mass (g/mol)\text{mol} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}mol=molar mass (g/mol)mass (g)
ΔH=(mol of substance)×(ΔrH∘mol-rxn)\Delta H = \left(\text{mol of substance}\right) \times \left(\frac{\Delta_rH^\circ}{\text{mol-rxn}}\right)ΔH=(mol of substance)×(mol-rxnΔrH∘)
Given Reaction:
2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l)ΔrH∘=−5755 kJ/mol-rxn2 \text{C}_4\text{H}_{10}(g) + 13 \text{O}_2(g) \rightarrow 8 \text{CO}_2(g) + 10 \text{H}_2\text{O}(l) \quad \Delta_rH^\circ = -5755 \, \text{kJ/mol-rxn}2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l)ΔrH∘=−5755kJ/mol-rxn
Given: 454 g of butane (C₄H₁₀)
454 g58.12 g/mol=7.811 mol C4H10\frac{454 \, \text{g}}{58.12 \, \text{g/mol}} = 7.811 \, \text{mol C}_4\text{H}_{10}58.12g/mol454g=7.811mol C4H10
ΔH=7.811 mol×(−5755 kJ2 mol C4H10)=−22,500 kJ\Delta H = 7.811 \, \text{mol} \times \left(\frac{-5755 \, \text{kJ}}{2 \, \text{mol C}_4\text{H}_{10}}\right) = -22,500 \, \text{kJ}ΔH=7.811mol×(2mol C4H10−5755kJ)=−22,500kJ
Concept | Description |
---|---|
ΔrH° | Heat change at constant pressure for 1 mol-rxn |
State of reactants/products | Must be specified: (s), (l), (g), or (aq) |
Reversing reaction | Flips sign of ΔrH° |
Scaling reaction | Scale ΔrH° accordingly |
Exothermic reaction | ΔrH° < 0 (releases heat) |
Endothermic reaction | ΔrH° > 0 (absorbs heat) |
Always check physical states in equations.
Pay attention to coefficients; they determine how ΔrH° scales.
Use unit conversions carefully (g → mol → kJ).
Remember: ΔrH° is per mole of reaction, not per mole of a specific substance.