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5.5

📘 Chapter 5.5 Study Guide: Enthalpy Changes for Chemical Reactions

🎯 Goal of Section 5.5

Understand and apply the enthalpy change (ΔrH°) for chemical reactions under standard conditions.

🔑 Key Definitions

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).

📐 Important Concepts

1. Signs of ΔrH°

Positive ΔrH° → Endothermic (absorbs heat from surroundings)
Negative ΔrH° → Exothermic (releases heat to surroundings)

2. Reversing Reactions

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

3. Scaling the Reaction

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

4. State of Matter Matters

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

🧮 Calculating Energy Changes for Given Masses

Use this two-step method:

Step 1: Convert grams to moles

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)

Step 2: Use ΔrH° to calculate ΔH for actual amount

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∘)

🧪 Example Problem

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)

Step 1: Convert to moles

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

Step 2: Use stoichiometry (2 mol C₄H₁₀ per 1 mol-rxn)

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 Summary

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)

📝 Tips for Exam

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.

Note