Hess's Law

IB Chemistry HL· 45 min read

1. 1. The Fundamental Principle of Hess's Law★★☆☆☆⏱ 15 min

📘 Definition

Hess's Law

$Δ H_{total} = Δ H_{1} + Δ H_{2} + ... + Δ H_{n}$

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.

Example:

If A forms C via intermediate B, $Δ H_{A \to C} = Δ H_{A \to B} + Δ H_{B \to C}$

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.

📐 Worked Example

Given the following thermochemical equations, calculate $Δ H$ for the target reaction $2 C (s) + 2 H_{2} (g) \to C_{2} H_{4} (g)$ : (1) $C (s) + O_{2} (g) \to C O_{2} (g) Δ H_{1} = - 393 kJ mol^{- 1}$ (2) $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) Δ H_{2} = - 286 kJ mol^{- 1}$ (3) $C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l) Δ H_{3} = - 1411 kJ mol^{- 1}$

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 $C_{2} H_{4}$ as a product, and flip the sign of $Δ H$ :
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 $Δ H$ values, cancel common species, and calculate the result:
6
$\Delta H_{\text{total}} = -786 - 572 + 1411 = +53 \, \text{kJ mol}^{-1}$

2. 2. Calculations from Standard Enthalpy Values★★★☆☆⏱ 20 min

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.

Methods compared

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

From Standard Enthalpies of Formation

$Δ H_{r}^{\circ} = \sum n Δ H_{f}^{\circ} (products) - \sum m Δ H_{f}^{\circ} (reactants)$ , 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

$Δ H_{r}^{\circ} = \sum m Δ H_{c}^{\circ} (reactants) - \sum n Δ H_{c}^{\circ} (products)$ , 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: $C_{6} H_{12} O_{6} (s) \to 2 C_{2} H_{5} O H (l) + 2 C O_{2} (g)$ given: $Δ H_{f}^{\circ} (C_{6} H_{12} O_{6}) = - 1273 kJ mol^{- 1}$ $Δ H_{f}^{\circ} (C_{2} H_{5} O H) = - 278 kJ mol^{- 1}$ $Δ H_{f}^{\circ} (C O_{2}) = - 394 kJ mol^{- 1}$

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. 3. Born-Haber Cycles for Ionic Compounds (HL Only)★★★★☆HL only⏱ 25 min

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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.

📘 Definition

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 $N a C l (s)$ given: $Δ H_{f}^{\circ} (N a C l (s)) = - 411 kJ mol^{- 1}$ $Δ H_{a t m}^{\circ} (N a (s)) = + 107 kJ mol^{- 1}$ $I E_{1} (N a (g)) = + 496 kJ mol^{- 1}$ $Δ H_{a t m}^{\circ} (\frac{1}{2} C l_{2} (g)) = + 122 kJ mol^{- 1}$ $E A_{1} (C l (g)) = - 349 kJ mol^{- 1}$

1
The target reaction for lattice enthalpy (formation of solid from gaseous ions) is: $N a^{+} (g) + C l^{-} (g) \to N a C l (s) Δ H_{L} = ?$
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 $Δ H_{L}$ :
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
$\Delta H_L = -411 - (107 + 496 + 122 - 349) = -787 \, \text{kJ mol}^{-1}$

4. Common Pitfalls

Wrong move:

Forgetting to flip the sign of ΔH when reversing a thermochemical equation

Why:

Reversing a reaction changes the direction of enthalpy flow, so the sign must change to match

Correct move:

Always reverse the sign of ΔH any time you reverse a reaction in a Hess cycle

Wrong move:

Forgetting to scale ΔH by the stoichiometric coefficient when adjusting a reaction

Why:

Enthalpy is an extensive property, so it changes proportionally with the amount of substance

Correct move:

Check all coefficients match the target, and multiply ΔH by the same scaling factor

Wrong move:

Mixing up the order of subtraction for formation vs combustion enthalpy

Why:

Memorizing the wrong order leads to systematic sign errors

Correct move:

Derive the cycle from first principles instead of relying on memorized formulas

Wrong move:

Ignoring mismatched state symbols in thermochemical equations

Why:

Enthalpy values change with state (e.g. liquid vs gaseous water), leading to incorrect results

Correct move:

Always confirm that the state symbols of all species match the given enthalpy values

Wrong move:

Using the wrong sign for lattice enthalpy based on question definition

Why:

Lattice enthalpy can be defined as formation (negative) or dissociation (positive), so sign depends on the question

Correct move:

Always check the question's definition of lattice enthalpy and adjust the sign accordingly

5. Quick Reference Cheatsheet

Method	Key Formula/Rule	Use Case
Manipulating given equations	Reverse = flip sign, scale = multiply ΔH by factor, $Δ H_{total} = \sum Δ H_{i}$	Any problem with given intermediate reactions
Enthalpies of formation	$Δ H_{r}^{\circ} = \sum n Δ H_{f} (products) - \sum m Δ H_{f} (reactants)$	General reactions with tabulated formation values
Enthalpies of combustion	$Δ H_{r}^{\circ} = \sum m Δ H_{c} (reactants) - \sum n Δ H_{c} (products)$	Organic reactions with combustible reactants/products
Born-Haber (HL)	$Δ H_{f}^{\circ} (solid) = \sum atomization/ionization/EA + Δ H_{L}$	Calculate unknown lattice enthalpy for ionic compounds

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Calculate reaction enthalpy change
2024 · 2
Born-Haber lattice enthalpy calculation
2023 · 1
Hess cycle multiple choice problem

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.