# Hess's Law

> IB Chemistry HL · R1: What drives chemical reactions?
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u4-hess-s-law/

Hess's law lets us calculate unknown enthalpy changes for reactions that cannot be measured directly experimentally. We cover constructing Hess cycles, core calculation methods, and HL-specific applications including Born-Haber cycles.

**Prerequisites:** [Enthalpy changes and thermochemical equations](https://www.owlsprep.com/study/ib-chemistry-hl-u4-enthalpy-introduction/); [Standard enthalpy values (formation, combustion)](https://www.owlsprep.com/study/ib-chemistry-hl-u4-standard-enthalpy-changes/)

## Learning objectives

- State Hess's law and explain its relation to the first law of thermodynamics
- Calculate unknown enthalpy changes using Hess cycles and given enthalpy values
- Apply Hess's law to construct Born-Haber cycles and calculate lattice enthalpy (HL)
- Identify and avoid common sign and scaling errors in Hess cycle calculations

## 1. The Fundamental Principle of Hess's Law

**Hess's Law** — The total enthalpy change for a chemical reaction is the same regardless of the path taken between reactants and products, as long as initial and final conditions are identical.

*Notation:* $\Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 + ... + \Delta H_n$

*Example:* If A forms C via intermediate B, $\Delta H_{A \to C} = \Delta H_{A \to B} + \Delta H_{B \to C}$

Hess's law follows directly from the first law of thermodynamics (energy cannot be created or destroyed) and the fact that enthalpy is a **state function**: its value only depends on the current state of the system, not the path taken to reach that state.

> **Key reminder: State vs path functions**
>
> State functions (enthalpy, entropy, Gibbs free energy) are path-independent. Path functions (heat, work) depend on the route taken.

**Worked example:** Given the following thermochemical equations, calculate $\Delta H$ for the target reaction $2C(s) + 2H_2(g) \to C_2H_4(g)$:
(1) $C(s) + O_2(g) \to CO_2(g) \quad \Delta H_1 = -393 \, \text{kJ mol}^{-1}$
(2) $H_2(g) + \frac{1}{2}O_2(g) \to H_2O(l) \quad \Delta H_2 = -286 \, \text{kJ mol}^{-1}$
(3) $C_2H_4(g) + 3O_2(g) \to 2CO_2(g) + 2H_2O(l) \quad \Delta H_3 = -1411 \, \text{kJ mol}^{-1}$

1. Scale equations (1) and (2) by 2 to match the moles of C and $H_2$ in the target reaction:
2. $$2 \times (1): 2C(s) + 2O_2(g) \to 2CO_2(g) \quad \Delta H = -786 \, \text{kJ mol}^{-1} \\ 2 \times (2): 2H_2(g) + O_2(g) \to 2H_2O(l) \quad \Delta H = -572 \, \text{kJ mol}^{-1}$$
3. Reverse equation (3) to get $C_2H_4$ as a product, and flip the sign of $\Delta H$:
4. $$2CO_2(g) + 2H_2O(l) \to C_2H_4(g) + 3O_2(g) \quad \Delta H = +1411 \, \text{kJ mol}^{-1}$$
5. Add all adjusted $\Delta H$ values, cancel common species, and calculate the result:
6. $$\Delta H_{\text{total}} = -786 - 572 + 1411 = +53 \, \text{kJ mol}^{-1}$$

## 2. Calculations from Standard Enthalpy Values

Hess's law gives us general formulas to calculate reaction enthalpy from tabulated standard enthalpy values. Two of the most common approaches use enthalpies of formation and enthalpies of combustion.

**Comparing methods**

The two main methods for standard enthalpy calculations are summarized below:

- **From Standard Enthalpies of Formation** — $\Delta H^\circ_r = \sum n \Delta H^\circ_f (\text{products}) - \sum m \Delta H^\circ_f (\text{reactants})$, where $n, m$ are stoichiometric coefficients.
  - Pros: Fast for any general reaction; Works with all compound types
  - Cons: Requires all formation values to be available

- **From Standard Enthalpies of Combustion** — $\Delta H^\circ_r = \sum m \Delta H^\circ_c (\text{reactants}) - \sum n \Delta H^\circ_c (\text{products})$, where $n, m$ are stoichiometric coefficients.
  - Pros: Convenient for organic reactions
  - Cons: Only works for combustible compounds

**Worked example:** Calculate the standard enthalpy of fermentation of glucose: $C_6H_{12}O_6(s) \to 2C_2H_5OH(l) + 2CO_2(g)$ given:
$\Delta H^\circ_f (C_6H_{12}O_6) = -1273 \, \text{kJ mol}^{-1}$
$\Delta H^\circ_f (C_2H_5OH) = -278 \, \text{kJ mol}^{-1}$
$\Delta H^\circ_f (CO_2) = -394 \, \text{kJ mol}^{-1}$

1. Write the formula for reaction enthalpy from standard enthalpies of formation:
2. $$\Delta H^\circ_r = \sum n \Delta H^\circ_f (\text{products}) - \sum m \Delta H^\circ_f (\text{reactants})$$
3. Substitute the values, multiplying each enthalpy by its stoichiometric coefficient:
4. $$\Delta H^\circ_r = [2(-278) + 2(-394)] - [1(-1273)]$$
5. Calculate the final result:
6. $$\Delta H^\circ_r = (-556 - 788) + 1273 = -71 \, \text{kJ mol}^{-1}$$

## 3. Born-Haber Cycles for Ionic Compounds (HL Only)

Lattice enthalpy cannot be measured directly, so we use Hess's law in the form of a Born-Haber cycle to calculate it from other experimentally measurable enthalpy values.

**Born-Haber Cycle** — A Hess cycle that connects the enthalpy of formation of an ionic solid to its lattice enthalpy, via intermediate steps of atomization, ionization, and electron affinity.

**Worked example:** Calculate the lattice enthalpy of $NaCl(s)$ given:
$\Delta H^\circ_f (NaCl(s)) = -411 \, \text{kJ mol}^{-1}$
$\Delta H^\circ_{atm} (Na(s)) = +107 \, \text{kJ mol}^{-1}$
$IE_1 (Na(g)) = +496 \, \text{kJ mol}^{-1}$
$\Delta H^\circ_{atm} (\frac{1}{2}Cl_2(g)) = +122 \, \text{kJ mol}^{-1}$
$EA_1 (Cl(g)) = -349 \, \text{kJ mol}^{-1}$

1. The target reaction for lattice enthalpy (formation of solid from gaseous ions) is: $Na^+(g) + Cl^-(g) \to NaCl(s) \quad \Delta H_L = ?$
2. Equate the two routes from elements to solid NaCl via Hess's law:
3. $$\Delta H^\circ_f (NaCl(s)) = \Delta H^\circ_{atm}(Na) + IE_1(Na) + \Delta H^\circ_{atm}(Cl) + EA_1(Cl) + \Delta H_L$$
4. Rearrange to solve for $\Delta H_L$:
5. $$\Delta H_L = \Delta H^\circ_f (NaCl) - [\Delta H^\circ_{atm}(Na) + IE_1(Na) + \Delta H^\circ_{atm}(Cl) + EA_1(Cl)]$$
6. Substitute values and calculate:
7. $$\Delta H_L = -411 - (107 + 496 + 122 - 349) = -787 \, \text{kJ mol}^{-1}$$

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to flip the sign of ΔH when reversing a thermochemical equation
  - Why it fails: Reversing a reaction changes the direction of enthalpy flow, so the sign must change to match
  - Correct: Always reverse the sign of ΔH any time you reverse a reaction in a Hess cycle
- **Wrong:** Forgetting to scale ΔH by the stoichiometric coefficient when adjusting a reaction
  - Why it fails: Enthalpy is an extensive property, so it changes proportionally with the amount of substance
  - Correct: Check all coefficients match the target, and multiply ΔH by the same scaling factor
- **Wrong:** Mixing up the order of subtraction for formation vs combustion enthalpy
  - Why it fails: Memorizing the wrong order leads to systematic sign errors
  - Correct: Derive the cycle from first principles instead of relying on memorized formulas
- **Wrong:** Ignoring mismatched state symbols in thermochemical equations
  - Why it fails: Enthalpy values change with state (e.g. liquid vs gaseous water), leading to incorrect results
  - Correct: Always confirm that the state symbols of all species match the given enthalpy values
- **Wrong:** Using the wrong sign for lattice enthalpy based on question definition
  - Why it fails: Lattice enthalpy can be defined as formation (negative) or dissociation (positive), so sign depends on the question
  - Correct: Always check the question's definition of lattice enthalpy and adjust the sign accordingly

## Cheatsheet

| Method | Key Formula/Rule | Use Case |
| --- | --- | --- |
| Manipulating given equations | Reverse = flip sign, scale = multiply ΔH by factor, $\Delta H_{\text{total}} = \sum \Delta H_i$ | Any problem with given intermediate reactions |
| Enthalpies of formation | $\Delta H^\circ_r = \sum n\Delta H_f(\text{products}) - \sum m\Delta H_f(\text{reactants})$ | General reactions with tabulated formation values |
| Enthalpies of combustion | $\Delta H^\circ_r = \sum m\Delta H_c(\text{reactants}) - \sum n\Delta H_c(\text{products})$ | Organic reactions with combustible reactants/products |
| Born-Haber (HL) | $\Delta H^\circ_f (\text{solid}) = \sum \text{atomization/ionization/EA} + \Delta H_L$ | Calculate unknown lattice enthalpy for ionic compounds |

## What's next

Hess's law is a foundational concept for all thermodynamics and energetics topics in IB Chemistry HL. It underpins all subsequent calculations of lattice enthalpy, entropy changes, and Gibbs free energy of reaction. Mastery of Hess cycle construction and error avoidance is critical for both paper 1 multiple choice and paper 2 extended response questions, and it frequently appears in combined questions with ionic bonding and spontaneity.

- [Entropy and reaction spontaneity](https://www.owlsprep.com/study/ib-chemistry-hl-u4-entropy-and-reaction-spontaneity/)
- [Bond enthalpies](https://www.owlsprep.com/study/ib-chemistry-hl-u4-bond-enthalpies/)
- [Energy cycles](https://www.owlsprep.com/study/ib-chemistry-hl-u4-energy-cycles/)

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