AHL: Extended enthalpy and entropy calculations

IB Chemistry HL· R1.3 Entropy and spontaneity· 20 min read

1. Calculating enthalpy change using average bond enthalpies★★☆☆☆⏱ 5 min

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Bond enthalpy calculations rely on the principle that breaking bonds is endothermic (positive energy change) and forming bonds is exothermic (negative energy change). The general formula for enthalpy change of reaction is:

\Delta H = \sum \text{(bond enthalpies of bonds broken)} - \sum \text{(bond enthalpies of bonds formed)}

📘 Definition

Average bond enthalpy

E(\text{X-Y})

The average energy required to break 1 mole of a gaseous covalent bond, measured across a range of compounds containing the bond

Example:

E(\text{C-H}) = +413 \text{ kJ mol}^{-1}

📐 Worked Example

Calculate the enthalpy of combustion of gaseous methane ( $CH_{4}$ ) using bond enthalpies.

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1. Write the balanced equation with all species gaseous:
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$CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g)$
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1. Sum bond enthalpies of bonds broken (reactants): 4 × C-H, 2 × O=O
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$\sum E_{\text{broken}} = 4(413) + 2(498) = 2648 \text{ kJ mol}^{-1}$
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1. Sum bond enthalpies of bonds formed (products): 2 × C=O, 4 × O-H
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$\sum E_{\text{formed}} = 2(805) + 4(463) = 3462 \text{ kJ mol}^{-1}$
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1. Calculate $Δ H$ using the formula:
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$\Delta H = 2648 - 3462 = -814 \text{ kJ mol}^{-1}$

Exam tip:

Bond enthalpies are only defined for gaseous species. If water is liquid in your reaction, add the total enthalpy of vaporization of water to your final result.

2. Enthalpy from enthalpy of formation and combustion★★★☆☆⏱ 6 min

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Hess's law gives us simple general formulas to calculate reaction enthalpy from tabulated standard enthalpies of formation or combustion:

📘 Definition

Standard enthalpy of formation

\Delta H^\circ_f

Enthalpy change when 1 mole of a substance is formed from its elements in their standard states under standard conditions

For enthalpy of formation: $Δ H_{rxn}^{\circ} = \sum n Δ H_{f}^{\circ} (products) - \sum m Δ H_{f}^{\circ} (reactants)$ , where $n, m$ are stoichiometric coefficients. For enthalpy of combustion, the formula is reversed: $Δ H_{rxn}^{\circ} = \sum n Δ H_{c}^{\circ} (reactants) - \sum m Δ H_{c}^{\circ} (products)$ .

📐 Worked Example

Calculate $Δ H^{\circ}$ for $2 C_{2} H_{6} (g) + 7 O_{2} (g) \to 4 C O_{2} (g) + 6 H_{2} O (l)$ , given $Δ H_{f}^{\circ} (C_{2} H_{6} (g)) = - 85$ kJ mol⁻¹, $Δ H_{f}^{\circ} (C O_{2} (g)) = - 394$ kJ mol⁻¹, $Δ H_{f}^{\circ} (H_{2} O (l)) = - 286$ kJ mol⁻¹, $Δ H_{f}^{\circ} (O_{2} (g)) = 0$ .

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1. Substitute values into the enthalpy of reaction formula:
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$\Delta H^\circ = [4(-394) + 6(-286)] - [2(-85) + 7(0)]$
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1. Calculate product and reactant sums separately:
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$\sum \text{products} = -1576 - 1716 = -3292; \quad \sum \text{reactants} = -170$
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1. Subtract reactant sum from product sum:
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$\Delta H^\circ = -3292 - (-170) = -3122 \text{ kJ mol}^{-1}$

Exam tip:

The enthalpy of formation of any element in its standard state is always 0, so you can eliminate these terms from your calculation immediately.

3. Calculating standard entropy change of reaction★★★☆☆⏱ 5 min

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By the third law of thermodynamics, the entropy of a perfect crystal at 0 K is 0, so we can measure absolute entropy values for all pure substances. This lets us directly calculate the entropy change of any reaction from tabulated standard molar entropies.

📘 Definition

Standard molar entropy

S^\circ_m

Entropy content of 1 mole of pure substance under standard conditions, units of J K⁻¹ mol⁻¹

The formula for standard reaction entropy change is analogous to the enthalpy of reaction formula:

\Delta S^\circ_{\text{rxn}} = \sum n S^\circ_m (\text{products}) - \sum m S^\circ_m (\text{reactants})

📐 Worked Example

Calculate $Δ S^{\circ}$ for the Haber process: $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ . Given $S_{m}^{\circ} (N_{2} (g)) = 192$ J K⁻¹ mol⁻¹, $S_{m}^{\circ} (H_{2} (g)) = 131$ J K⁻¹ mol⁻¹, $S_{m}^{\circ} (N H_{3} (g)) = 193$ J K⁻¹ mol⁻¹.

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1. Substitute stoichiometric coefficients into the formula:
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$\Delta S^\circ = [2 \times 193] - [(1 \times 192) + (3 \times 131)]$
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1. Evaluate the expression:
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$\Delta S^\circ = 386 - 585 = -199 \text{ J K}^{-1} \text{mol}^{-1}$
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1. The negative sign makes sense: 4 moles of gas are converted to 2 moles of gas, decreasing disorder.

Exam tip:

Always check units: entropy is J K⁻¹ mol⁻¹, enthalpy is kJ mol⁻¹. You must convert entropy to kJ when calculating Gibbs free energy later.

4. Hess's law for combined enthalpy and entropy changes★★★★☆⏱ 4 min

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Hess's law applies equally to entropy changes as it does to enthalpy changes. The rules for manipulating reactions are identical:

📐 Worked Example

Find $Δ S$ for $C (s) + \frac{1}{2} O_{2} (g) \to C O (g)$ , given: 1. $C (s) + O_{2} (g) \to C O_{2} (g) Δ S_{1} = + 3$ J K⁻¹ mol⁻¹ 2. $C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g) Δ S_{2} = - 86$ J K⁻¹ mol⁻¹.

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1. Reverse reaction 2 to get $C O_{2}$ as a reactant, and reverse the sign of $Δ S_{2}$ :
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$-\Delta S_2 = +86 \text{ J K}^{-1} \text{mol}^{-1}$
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1. Add reaction 1 and modified reaction 2, add the entropy changes:
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$\Delta S_{\text{total}} = \Delta S_1 + (-\Delta S_2) = 3 + 86 = +89 \text{ J K}^{-1} \text{mol}^{-1}$
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1. Cancel common species on both sides to confirm you get the target reaction, which matches.

5. Common Pitfalls

Wrong move:

Forgetting to multiply enthalpy/entropy values by stoichiometric coefficients

Why:

Tabulated values are per mole, so they must be scaled to match the balanced equation

Correct move:

Always multiply each value by its stoichiometric coefficient from the balanced equation before summing

Wrong move:

Mixing up product/reactant order for enthalpy of combustion vs formation

Why:

The formula for combustion is reversed from the formula for formation, leading to wrong sign

Correct move:

Remember: $Δ H$ from formation = products minus reactants; $Δ H$ from combustion = reactants minus products

Wrong move:

Using liquid water in bond enthalpy calculations with no adjustment

Why:

Bond enthalpies are only defined for gaseous species

Correct move:

Add the total enthalpy of vaporization of product water to your final $Δ H$ if water is liquid

Wrong move:

Forgetting to convert entropy units from J to kJ for Gibbs free energy

Why:

Mismatched units lead to a 1000× error in the final Gibbs free energy value

Correct move:

Always divide $Δ S$ by 1000 to convert to kJ K⁻¹ mol⁻¹ before combining with $Δ H$

Wrong move:

Assuming $Δ S$ must be positive because all absolute $S^{\circ}$ are positive

Why:

The sum of product entropies can be smaller than the sum of reactant entropies

Correct move:

Always calculate $Δ S$ using the formula, then check if the sign matches the change in moles of gas

6. Quick Reference Cheatsheet

Calculation Type	Formula
Bond enthalpy	$Δ H = \sum E_{bonds broken} - \sum E_{bonds formed}$
Enthalpy from $Δ H_{f}$	$Δ H^{\circ} = \sum n Δ H_{f}^{\circ} (products) - \sum m Δ H_{f}^{\circ} (reactants)$
Enthalpy from $Δ H_{c}$	$Δ H^{\circ} = \sum n Δ H_{c}^{\circ} (reactants) - \sum m Δ H_{c}^{\circ} (products)$
Entropy of reaction	$Δ S^{\circ} = \sum n S_{m}^{\circ} (products) - \sum m S_{m}^{\circ} (reactants)$
Hess's law rule	Reverse reaction = reverse sign; scale = scale change; add reactions = add changes

7. Frequently Asked

Why are bond enthalpy calculations always approximate?

Average bond enthalpies are averaged across many different compounds containing the same bond, and do not account for specific bonding environments in individual reactant/product molecules.

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 · Paper 1
Calculate entropy change from S° values
2024 · Paper 2
Bond enthalpy enthalpy calculation
2023 · Paper 2
Hess cycle for entropy change

Going deeper

syllabusOfficial IB Chemistry 2025 SyllabusFull topic reference

What's Next

Mastering these extended enthalpy and entropy calculations is the critical foundation for the next core topic in IB Chemistry HL: calculating Gibbs free energy change and predicting the spontaneity of chemical reactions. These calculations also underpin understanding of how temperature affects spontaneity, equilibrium constants, and electrochemical cell potentials, all heavily assessed topics in the final exam. Once you are confident with the methods and common errors covered here, you are ready to move on to connecting enthalpy and entropy to reaction spontaneity.