6.4 Hess's Law

This topic explains Hess's Law, which states that the enthalpy change in a chemical reaction is independent of the path followed. It covers the statement of the law, its mathematical form, and examples of its application.

Definition

Hess's Law: The enthalpy change in a chemical reaction is the same whether the reaction takes place in a single step or in several steps.

Main Idea

Enthalpy is a state function; therefore, enthalpy change in a chemical reaction in going from some initial state to some final state is independent of the path followed by the reaction. Thus, in going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in many steps.

Core Concept

• Hess's Law: ΔH is independent of the path.
• Mathematical form: Σ Δ H_(Cycle) = 0
• State function: Enthalpy depends only on the initial and final states, not on the path.

Detailed Explanation

Statement of Hess's Law

• G.H. Hess stated this law in 1840.
• It states that the enthalpy change in a chemical reaction is the same whether the reaction takes place in a single step or in several steps.

Intermediate Level

Statement

The enthalpy change in a chemical reaction is the same whether the reaction takes place in a single step or in several steps.

Mathematical Form

Σ Δ H_(Cycle) = 0

Δ H = Δ H₁ + Δ H₂ + Δ H₃ + …

Example 6.6: Combustion of Carbon

One step: C + O₂ → CO₂ Δ H^° = -393.5 kJ

Two steps: C + 12O₂ → CO Δ H₁^° = -110.52 kJ CO + 12O₂ → CO₂ Δ H₂^° = -282.98 kJ Δ H₁^° + Δ H₂^° = -393.5 kJ

Applications

• Determine enthalpies of formation of compounds that cannot be measured directly.
• Example: CH₄, CCl₄, Al₂O₃, B₂O₃.

Example 6.7: Enthalpy of Formation of Methane

• CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH° = −890.4 kJ
• H₂ + ½O₂ → H₂O, ΔH° = −285.8 kJ
• C + O₂ → CO₂, ΔH° = −393.5 kJ

Required: - C + 2H₂ → CH₄

• Use equation (iii) as such.
• Multiply equation (ii) by 2.
• Reverse equation (i) and change sign.
• Add equations and ΔH values.

Result: C + 2H₂ → CH₄ Δ H^° = -74.6 kJ/mol

Important Points

• 1. Hess's Law: ΔH is independent of path.
• 2. Mathematical form: Σ Δ H_(Cycle) = 0.
• 3. ΔH = ΔH₁ + ΔH₂ + ΔH₃ + …
• 4. Example: C + O₂ → CO₂ (one step) = C + ½O₂ → CO + CO + ½O₂ → CO₂ (two steps).
• 5. Applications: Determining enthalpies of formation of compounds that cannot be measured directly.
• 6. Example: CH₄ formation from C and H₂, ΔH° = −74.6 kJ/mol.

Remember

• - Hess's Law: ΔH independent of path
• - Mathematical form: Σ Δ H_(Cycle) = 0
• - **ΔH = ΔH₁ + ΔH₂ + ΔH₃ + …**
• - **Example:** C + O₂ → CO₂ = −393.5 kJ
• - **Application:** CH₄ formation, ΔH° = −74.6 kJ/mol

Common Mistakes

• 1. Forgetting to reverse the sign when reversing a reaction: ΔH changes sign.
• 2. Forgetting to multiply ΔH when multiplying the equation: ΔH must be multiplied by the same factor.
• 3. Confusing one-step and multi-step processes: ΔH is the same for both.
• 4. Ignoring the state function property: Enthalpy depends only on initial and final states.

Quick Revision

• Hess's Law: ΔH is independent of path.
• Mathematical form: Σ Δ H_(Cycle) = 0.
• ΔH = ΔH₁ + ΔH₂ + ΔH₃ + …
• Example: C + O₂ → CO₂ (one step) = C + ½O₂ → CO + CO + ½O₂ → CO₂ (two steps).
• Applications: Determining enthalpies of formation of compounds that cannot be measured directly.
• Example: CH₄ formation from C and H₂, ΔH° = −74.6 kJ/mol.

60-Second Revision

• Hess's Law: ΔH independent of path
• Mathematical form: Σ Δ H_(Cycle) = 0
• **ΔH = ΔH₁ + ΔH₂ + ΔH₃ + …**
• **Example:** C + O₂ → CO₂ = −393.5 kJ
• **Application:** CH₄ formation, ΔH° = −74.6 kJ/mol

QUALITY VALIDATION

• Topic: 6.4 Hess's Law
• Chapter: 6 (Energetics)

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