Hess's Law

Overview

Hess's Law states that the enthalpy change for a reaction is the same whether it occurs in one step or multiple steps. This allows us to calculate ΔH for reactions that are difficult to measure directly.

Statement of Hess's Law

$$\Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \cdots$$

Enthalpy is a state function - it depends only on initial and final states, not the path taken.

Using Hess's Law

Method: Algebraic Manipulation

1. Write the target equation
2. Manipulate given equations to match
3. Apply the same operations to ΔH values
4. Add the adjusted ΔH values

Rules for Manipulating Equations

Operation	Effect on ΔH
Reverse reaction	Change sign of ΔH
Multiply by coefficient	Multiply ΔH by same factor
Divide by coefficient	Divide ΔH by same factor

Example 1: Basic Application

Target: $\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)$ , $\Delta H = ?$

Given:

$$(1) \quad \text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_1 = -393.5 \text{ kJ}$$ $$(2) \quad \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_2 = -283.0 \text{ kJ}$$

Solution:

Keep equation (1):

$$\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -393.5 \text{ kJ}$$

Reverse equation (2):

$$\text{CO}_2(g) \rightarrow \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \quad \Delta H = +283.0 \text{ kJ}$$

Add and cancel CO₂ and simplify:

$$\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g) \quad \Delta H = -110.5 \text{ kJ}$$

Example 2: Three Equations

Target: $2\text{C}(s) + \text{H}_2(g) \rightarrow \text{C}_2\text{H}_2(g)$ , $\Delta H = ?$

Given:

$$(1) \quad \text{C}_2\text{H}_2(g) + \frac{5}{2}\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + \text{H}_2\text{O}(l) \quad \Delta H_1 = -1299.6 \text{ kJ}$$ $$(2) \quad \text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_2 = -393.5 \text{ kJ}$$ $$(3) \quad \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H_3 = -285.8 \text{ kJ}$$

Solution:

Reverse equation (1):

$$2\text{CO}_2(g) + \text{H}_2\text{O}(l) \rightarrow \text{C}_2\text{H}_2(g) + \frac{5}{2}\text{O}_2(g) \quad \Delta H = +1299.6 \text{ kJ}$$

Multiply equation (2) by 2:

$$2\text{C}(s) + 2\text{O}_2(g) \rightarrow 2\text{CO}_2(g) \quad \Delta H = -787.0 \text{ kJ}$$

Keep equation (3):

$$\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H = -285.8 \text{ kJ}$$

Add and simplify:

$$2\text{C}(s) + \text{H}_2(g) \rightarrow \text{C}_2\text{H}_2(g) \quad \Delta H = +226.8 \text{ kJ}$$

Using Standard Enthalpies of Formation

A direct application of Hess's Law:

$$\Delta H_{rxn}° = \sum n \Delta H_f°(\text{products}) - \sum n \Delta H_f°(\text{reactants})$$

Example

Calculate $\Delta H°$ for: $2\text{NH}_3(g) + 3\text{Cl}_2(g) \rightarrow \text{N}_2(g) + 6\text{HCl}(g)$

$\Delta H_f°$ values (kJ/mol):

• $\text{NH}_3(g) = -46.1$
• $\text{Cl}_2(g) = 0$
• $\text{N}_2(g) = 0$
• $\text{HCl}(g) = -92.3$

$$\Delta H° = [0 + 6(-92.3)] - [2(-46.1) + 3(0)]$$ $$\Delta H° = [-553.8] - [-92.2] = -461.6 \text{ kJ}$$

Born-Haber Cycle

A special application of Hess's Law for calculating lattice energy of ionic compounds.

Steps for NaCl

1. Sublimation: $\text{Na}(s) \rightarrow \text{Na}(g)$ → $\Delta H_{sub}$
2. Ionization: $\text{Na}(g) \rightarrow \text{Na}^+(g) + e^-$ → IE
3. Dissociation: $\frac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl}(g)$ → $\frac{1}{2}D$
4. Electron affinity: $\text{Cl}(g) + e^- \rightarrow \text{Cl}^-(g)$ → EA
5. Lattice formation: $\text{Na}^+(g) + \text{Cl}^-(g) \rightarrow \text{NaCl}(s)$ → $-U$

Overall: $\text{Na}(s) + \frac{1}{2}\text{Cl}_2(g) \rightarrow \text{NaCl}(s)$ → $\Delta H_f°$

$$\Delta H_f° = \Delta H_{sub} + IE + \frac{1}{2}D + EA + (-U)$$

Solving for lattice energy (U):

$$U = \Delta H_{sub} + IE + \frac{1}{2}D + EA - \Delta H_f°$$

Energy Cycle Diagrams

Visual Representation

                    Elements
                   /         \
          Direct  /           \ Step-wise
         Path    /             \ Path
                /               \
               ↓                 ↓
           Products ←———————→ (Intermediates)
                    ΔHtotal

Key Points

1. Enthalpy is a state function - path independent
2. Energy is conserved - total ΔH is same regardless of route
3. Reactions can be added - like algebraic equations
4. Signs and coefficients matter - be careful with manipulations
5. All ΔH values must have consistent units - usually kJ or kJ/mol

Problem-Solving Strategy

1. Write the target equation clearly
2. Identify what species need to be products vs reactants
3. Manipulate given equations (reverse/multiply as needed)
4. Check that intermediate species cancel
5. Sum the adjusted ΔH values
6. Verify final equation matches target