Energy cycles

IB Chemistry Higher Level· 30 min read

1. Hess's Law and General Enthalpy Cycles★★☆☆☆⏱ 15 min

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📘 Definition

Hess's Law

The total enthalpy change for a reaction is the same regardless of the path taken between reactants and products, since enthalpy is a state function.

To construct an energy cycle, connect reactants and products to a common set of intermediate species (usually elements for enthalpy of formation, or combustion products for enthalpy of combustion). The unknown enthalpy change is found by adding and reversing known enthalpy changes to match the alternative route.

📐 Worked Example

Calculate the standard enthalpy of formation of ethanol ( $C_{2} H_{5} O H (l)$ ) given: $ \Delta H_c^\circ (C(graphite)) = -393.5\ kJ\ mol^{-1}, \Delta H_c^\circ (H_2(g)) = -285.8\ kJ\ mol^{-1}, \Delta H_c^\circ (C_2H_5OH(l)) = -1367\ kJ\ mol^{-1}$

1
Write the target equation for formation of 1 mole of ethanol:
$2C(graphite) + 3H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_2H_5OH(l) \quad \Delta H_f^\circ = ?$
2
Apply the rule for enthalpy of reaction from combustion data: sum of reactant combustion enthalpies minus sum of product combustion enthalpies:
$\Delta H_f^\circ = 2\Delta H_c^\circ(C) + 3\Delta H_c^\circ(H_2) - \Delta H_c^\circ(C_2H_5OH)$
3
Substitute values and calculate the result:
$(2 \times -393.5) + (3 \times -285.8) - (-1367) = -277.4\ kJ\ mol^{-1}$

Exam tip:

Always check the direction of your route: reverse the sign of any enthalpy change you traverse opposite to its standard definition.

2. Born-Haber Cycles for Ionic Compounds★★★☆☆HL only⏱ 20 min

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📘 Definition

Lattice Enthalpy

\Delta H_{latt}

The enthalpy change when one mole of an ionic solid is formed from its gaseous ions under standard conditions. Dissociation of the lattice has the opposite sign.

Example:

Dissociation of NaCl: $Δ H_{l a tt} = + 787 k J m o l^{- 1}$

A Born-Haber cycle is a specialized energy cycle that applies Hess's Law to calculate lattice enthalpy, which cannot be measured directly. The cycle relates lattice enthalpy to measurable terms: enthalpy of formation, atomization enthalpy, ionization energy, and electron affinity.

📐 Worked Example

Calculate the lattice enthalpy of dissociation for KCl(s) given: $Δ H_{f}^{\circ} (K C l (s)) = - 437 k J m o l^{- 1}$ , $Δ H_{a t}^{\circ} (K (s)) = + 90 k J m o l^{- 1}$ , $Δ H_{a t}^{\circ} (\frac{1}{2} C l_{2} (g)) = + 122 k J m o l^{- 1}$ , $I E_{1} (K (g)) = + 419 k J m o l^{- 1}$ , $E A_{1} (C l (g)) = - 349 k J m o l^{- 1}$

1
Apply Hess's Law to the full Born-Haber cycle, solving for lattice enthalpy of formation first:
$\Delta H_f^\circ = \Delta H_{at}(K) + \Delta H_{at}(Cl_2) + IE_1(K) + EA_1(Cl) + \Delta H_{latt(formation)}$
2
Rearrange to find $Δ H_{l a tt (f or ma t i o n)}$ :
$\Delta H_{latt(formation)} = -437 - (90 + 122 + 419 - 349) = -719\ kJ\ mol^{-1}$
3
Dissociation is the reverse process, so reverse the sign:
$\Delta H_{latt(dissociation)} = +719\ kJ\ mol^{-1}$

3. Drawing and Interpreting Exam Energy Cycles★★★☆☆⏱ 15 min

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IB examiners regularly require you to draw fully labelled energy cycles, not just complete calculations. Every arrow must be correctly directed from starting species to product species, and each step must be clearly labelled with the correct enthalpy term name.

✓ Quick check

Check your understanding of enthalpy sign conventions:

What is the sign of atomization enthalpy for solid sodium?
- Negative, because bonds are broken
- Positive, because bonds are broken
- Negative, because bonds are formed
- Positive, because bonds are formed
Reveal answer
1 —
Atomization breaks metallic bonds in solid sodium to form gaseous atoms, which is endothermic, so ΔH is positive.

4. Common Pitfalls

Wrong move:

Forgetting to reverse the sign of an enthalpy change when traversing a step opposite to its definition

Why:

Enthalpy change sign depends on direction of heat flow; ignoring this flips the sign of the final result

Correct move:

Always label arrows on your cycle, and add a negative sign to any enthalpy term you move through backwards

Wrong move:

Not scaling enthalpy values by the stoichiometric coefficients in the target equation

Why:

Standard enthalpy values are reported per mole, so they must be scaled for the number of moles in the reaction

Correct move:

Check the stoichiometry of every step before calculating, and multiply each enthalpy value by its mole count

Wrong move:

Confusing formation and dissociation lattice enthalpy in Born-Haber calculations

Why:

Formation of solid from ions is exothermic, dissociation is endothermic, so they have opposite signs

Correct move:

Always read the question carefully to confirm whether a positive (dissociation) or negative (formation) answer is required

Wrong move:

Drawing arrows pointing in the wrong direction for steps in the cycle

Why:

Examiners penalize incorrectly directed arrows even if the final numerical answer is correct

Correct move:

Always draw the arrow starting at the reactant of the step and pointing to the product of the step

Wrong move:

Missing bonds when calculating enthalpy change from average bond enthalpies

Why:

Students often miss bonds that do not change in the reaction or double count bonds in cyclic structures

Correct move:

Draw full structural formulas for all reactants and products, count every bond before calculating

5. Quick Reference Cheatsheet

Calculation Type	Formula	Key Rule
Enthalpy from formation	$Δ H_{r x n} = \sum Δ H_{f} (p r o d u c t s) - \sum Δ H_{f} (r e a c t an t s)$	Reverse sign for reactants
Enthalpy from combustion	$Δ H_{r x n} = \sum Δ H_{c} (r e a c t an t s) - \sum Δ H_{c} (p r o d u c t s)$	Reverse sign for products
Bond enthalpy reaction	$Δ H_{r x n} = \sum b o n d s b r o k e n - \sum b o n d s f or m e d$	Breaking = positive, forming = negative
Born-Haber dissociation	$Δ H_{l a tt (d i ssoc)} = - Δ H_{l a tt (f or ma t i o n)}$	Dissociation is always positive

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.

2022 · 1
Enthalpy of formation calculation
2023 · 2
Born-Haber lattice enthalpy calculation
2024 · 1
Hess's Law route identification

What's Next

Energy cycles are the foundation of all thermochemical calculations in IB Chemistry, and underpin all advanced topics related to reaction spontaneity. Mastering Hess's Law and energy cycle construction will make it much easier to tackle enthalpy problems across all papers, from multiple choice to extended response. Next, you will build on this knowledge to explore trends in lattice enthalpy and their relationship to ionic bonding, before moving on to entropy and Gibbs free energy, which explain what truly drives chemical reactions.