5.7

🔥 Chapter 5.7 – Enthalpy Calculations Study Guide

🎯 Goals

• Apply Hess’s Law to find enthalpy changes.

• Draw and interpret energy level diagrams.

• Use standard enthalpies of formation (∆fH°) to calculate the standard enthalpy change (∆rH°) for a reaction.

🧪 1. Hess’s Law

• Used when: the enthalpy change of a reaction is not directly measurable.

• How: Manipulate and add known reactions to match the target reaction.

✅ Key Rules

• Reverse a reaction → change sign of ∆H°.

• Multiply coefficients → multiply ∆H° by same factor.

Example:
To find the enthalpy change for:
 C(s) + ½ O₂(g) → CO(g)
Use:
 CO(g) + ½ O₂(g) → CO₂(g) ∆H° = −283.0
 C(s) + O₂(g) → CO₂(g) ∆H° = −393.5
Rearrange and add to find ∆H° = −110.5 kJ/mol.

📊 2. Energy Level Diagrams

• Visual tool for showing enthalpy changes.

• Y-axis = energy; reactants/products placed according to their ∆H°.

• Direction of arrow shows exo or endo thermic nature.

Key Takeaway:

• State function: Enthalpy change depends only on initial and final states, not on the path.

📘 3. Standard Enthalpies of Formation (∆fH°)

📌 Important Notes

• Elements in standard state (e.g., O₂(g), C(s), H₂(g)) → ∆fH° = 0

• Most ∆fH° values are negative → compound formation is usually exothermic.

• Some ∆fH° values are positive → unstable compounds (e.g., NO(g)).

🔁 Examples:

• NaCl(s):
   Na(s) + ½ Cl₂(g) → NaCl(s) ∆fH° = −411.12 kJ/mol

• C₂H₅OH(ℓ):
   2 C(s) + 3 H₂(g) + ½ O₂(g) → C₂H₅OH(ℓ) ∆fH° = −277.0 kJ/mol

📐 4. Calculating ∆rH° Using ∆fH°

Formula:

ΔrH∘=∑nΔfH∘(products)−∑nΔfH∘(reactants)\Delta_rH^\circ = \sum n \Delta_fH^\circ(\text{products}) - \sum n \Delta_fH^\circ(\text{reactants})Δr​H∘=∑nΔf​H∘(products)−∑nΔf​H∘(reactants)

Where:

• n = coefficient in balanced equation

• All values must be in kJ/mol

Example:

CaCO₃(s) → CaO(s) + CO₂(g)
Given:

• ∆fH°[CaCO₃] = −1207.6

• ∆fH°[CaO] = −635.1

• ∆fH°[CO₂] = −393.5

ΔrH∘=[(−635.1)+(−393.5)]−(−1207.6)=−1028.6+1207.6=+179.0 kJ/mol\Delta_rH^\circ = [(-635.1) + (-393.5)] - (-1207.6) = -1028.6 + 1207.6 = +179.0\ \text{kJ/mol}Δr​H∘=[(−635.1)+(−393.5)]−(−1207.6)=−1028.6+1207.6=+179.0 kJ/mol

✅ Endothermic reaction (requires heat input).

📚 Key Vocabulary

• ∆rH°: Standard enthalpy change of a reaction.

• ∆fH°: Standard enthalpy of formation.

• Standard state: Most stable form of a substance at 25°C and 1 bar.

• Exothermic: Releases heat (∆H < 0).

• Endothermic: Absorbs heat (∆H > 0).

• State function: Path-independent property.

📝 Tips for Success

• Always balance the chemical equation.

• Label units carefully (kJ/mol-rxn).

• Double-check if you reverse or scale equations in Hess’s Law.

• Use Appendix L or your provided table for ∆fH° values.