# AHL: Extended enthalpy and entropy calculations

> IB Chemistry HL · IB Chemistry HL 2025
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u4-ahl-extended-enthalpy-and-entropy/

This sub-topic covers extended calculations of reaction enthalpy (bond enthalpies, formation/combustion, Hess cycles) and reaction entropy from absolute molar entropies for IB Chemistry HL.

**Prerequisites:** [Basic enthalpy changes and Hess's law](https://www.owlsprep.com/study/ib-chemistry-hl-u4-enthalpy-basics/); [Introduction to entropy](https://www.owlsprep.com/study/ib-chemistry-hl-u4-entropy-introduction/)

## Learning objectives

- Calculate enthalpy change of reaction using average bond enthalpies and tabulated formation/combustion values
- Calculate standard entropy change of reaction from absolute molar entropies
- Apply Hess's law to multi-step enthalpy and entropy changes
- Avoid common unit and sign errors in thermodynamics calculations

## Calculating enthalpy change using average bond enthalpies

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:

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

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

*Notation:* E(\text{X-Y})

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

**Worked example:** Calculate the enthalpy of combustion of gaseous methane ($\text{CH}_4$) using bond enthalpies.

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

> **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.

*Calculator:* allowed

## Enthalpy from enthalpy of formation and combustion

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

**Standard enthalpy of formation** — Enthalpy change when 1 mole of a substance is formed from its elements in their standard states under standard conditions

*Notation:* \Delta H^\circ_f

For enthalpy of formation: $\Delta H^\circ_{\text{rxn}} = \sum n \Delta H^\circ_f (\text{products}) - \sum m \Delta H^\circ_f (\text{reactants})$, where $n,m$ are stoichiometric coefficients. For enthalpy of combustion, the formula is reversed: $\Delta H^\circ_{\text{rxn}} = \sum n \Delta H^\circ_c (\text{reactants}) - \sum m \Delta H^\circ_c (\text{products})$.

**Worked example:** Calculate $\Delta H^\circ$ for $2 C_2H_6(g) + 7 O_2(g) \rightarrow 4 CO_2(g) + 6 H_2O(l)$, given $\Delta H^\circ_f (C_2H_6(g)) = -85$ kJ mol⁻¹, $\Delta H^\circ_f (CO_2(g)) = -394$ kJ mol⁻¹, $\Delta H^\circ_f (H_2O(l)) = -286$ kJ mol⁻¹, $\Delta H^\circ_f (O_2(g)) = 0$.

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

> **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.

*Calculator:* allowed

## Calculating standard entropy change of reaction

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.

**Standard molar entropy** — Entropy content of 1 mole of pure substance under standard conditions, units of J K⁻¹ mol⁻¹

*Notation:* S^\circ_m

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 $\Delta S^\circ$ for the Haber process: $N_2(g) + 3 H_2(g) \rightleftharpoons 2 NH_3(g)$. Given $S^\circ_m (N_2(g)) = 192$ J K⁻¹ mol⁻¹, $S^\circ_m (H_2(g)) = 131$ J K⁻¹ mol⁻¹, $S^\circ_m (NH_3(g)) = 193$ J K⁻¹ mol⁻¹.

1. 1. Substitute stoichiometric coefficients into the formula:
2. $$\Delta S^\circ = [2 \times 193] - [(1 \times 192) + (3 \times 131)]$$
3. 2. Evaluate the expression:
4. $$\Delta S^\circ = 386 - 585 = -199 \text{ J K}^{-1} \text{mol}^{-1}$$
5. 3. 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.

*Calculator:* allowed

## Hess's law for combined enthalpy and entropy changes

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

> **tip**
>
> Reverse the reaction = reverse the sign of the change; scale the reaction by a coefficient = scale the change by the same coefficient; add reactions = add the total changes.

**Worked example:** Find $\Delta S$ for $C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g)$, given: 1. $C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta S_1 = +3$ J K⁻¹ mol⁻¹ 2. $CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \quad \Delta S_2 = -86$ J K⁻¹ mol⁻¹.

1. 1. Reverse reaction 2 to get $CO_2$ as a reactant, and reverse the sign of $\Delta S_2$:
2. $$-\Delta S_2 = +86 \text{ J K}^{-1} \text{mol}^{-1}$$
3. 2. Add reaction 1 and modified reaction 2, add the entropy changes:
4. $$\Delta S_{\text{total}} = \Delta S_1 + (-\Delta S_2) = 3 + 86 = +89 \text{ J K}^{-1} \text{mol}^{-1}$$
5. 3. Cancel common species on both sides to confirm you get the target reaction, which matches.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to multiply enthalpy/entropy values by stoichiometric coefficients
  - Why it fails: Tabulated values are per mole, so they must be scaled to match the balanced equation
  - Correct: Always multiply each value by its stoichiometric coefficient from the balanced equation before summing
- **Wrong:** Mixing up product/reactant order for enthalpy of combustion vs formation
  - Why it fails: The formula for combustion is reversed from the formula for formation, leading to wrong sign
  - Correct: Remember: $\Delta H$ from formation = products minus reactants; $\Delta H$ from combustion = reactants minus products
- **Wrong:** Using liquid water in bond enthalpy calculations with no adjustment
  - Why it fails: Bond enthalpies are only defined for gaseous species
  - Correct: Add the total enthalpy of vaporization of product water to your final $\Delta H$ if water is liquid
- **Wrong:** Forgetting to convert entropy units from J to kJ for Gibbs free energy
  - Why it fails: Mismatched units lead to a 1000× error in the final Gibbs free energy value
  - Correct: Always divide $\Delta S$ by 1000 to convert to kJ K⁻¹ mol⁻¹ before combining with $\Delta H$
- **Wrong:** Assuming $\Delta S$ must be positive because all absolute $S^\circ$ are positive
  - Why it fails: The sum of product entropies can be smaller than the sum of reactant entropies
  - Correct: Always calculate $\Delta S$ using the formula, then check if the sign matches the change in moles of gas

## Cheatsheet

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

## 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.

- [AHL: Gibbs free energy and reaction spontaneity](https://www.owlsprep.com/study/ib-chemistry-hl-u4-ahl-gibbs-free-energy-and/)
- [R2: How much / how fast / how far?](https://www.owlsprep.com/study/ib-chemistry-hl-u5-overview/)
- [Stoichiometric relationships](https://www.owlsprep.com/study/ib-chemistry-hl-u5-stoichiometric-relationships/)

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