AM

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

If a chemical reaction is the sum of two or more other reactions, then:

∆rH° (overall) = ∆rH°₁ + ∆rH°₂ + ...

  • 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°)

∆fH° = enthalpy change when 1 mol of a compound forms from its elements in their standard states at 25°C and 1 bar.

📌 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.