# Enthalpy of reaction

> IB Chemistry SL · Reactivity 2: Energetics and Kinetics
> Source: https://www.owlsprep.com/study/ib-chemistry-sl-u5-enthalpy-of-reaction/

This sub-topic covers the definition of enthalpy of reaction, sign conventions for exothermic and endothermic processes, and core calculation methods using bond enthalpies and standard enthalpies of formation for IB SL Chemistry.

**Prerequisites:** [Enthalpy basics](https://www.owlsprep.com/study/ib-chemistry-sl-u5-enthalpy-basics/); [Standard state conditions](https://www.owlsprep.com/study/ib-chemistry-sl-u5-standard-states/)

## Learning objectives

- Define enthalpy of reaction and correct sign conventions for endothermic and exothermic processes
- Calculate enthalpy of reaction using average bond enthalpies
- Calculate standard enthalpy of reaction from standard enthalpies of formation
- Recognize common exam traps and avoid calculation errors

## Definition and Sign Conventions

**Enthalpy of Reaction** — The total change in enthalpy (heat content at constant pressure) when a reaction proceeds according to its given balanced stoichiometric equation.

*Notation:* \(\Delta H\)

*Example:* For \(2H_2(g) + O_2(g) \rightarrow 2H_2O(l)\), ΔH is the enthalpy change for 2 moles of H₂ reacting with 1 mole of O₂.

The sign of ΔH always describes the change in enthalpy of the chemical system (not the surroundings). The IB uses a consistent convention that is frequently tested in multiple choice questions.

- **Exothermic reactions**: Release heat to the surroundings, so the system's enthalpy decreases. $\Delta H < 0$ (negative).
- **Endothermic reactions**: Absorb heat from the surroundings, so the system's enthalpy increases. $\Delta H > 0$ (positive).

> **warning**
>
> Never reverse the sign convention. This is one of the most common mistakes on IB Paper 1 multiple choice.

**Worked example:** State the sign of ΔH for the melting of ice to liquid water, and classify the process as exothermic or endothermic.

1. Melting requires heat energy to be absorbed from the surroundings to break the intermolecular forces in ice.
2. The heat is absorbed by the system (the water), so the enthalpy of the system increases.
3. Conclusion: ΔH is positive, and melting is endothermic.

## Calculation from Bond Enthalpies

Breaking covalent bonds always requires energy (endothermic, positive energy contribution), and forming covalent bonds always releases energy (exothermic, negative energy contribution). We use this to calculate ΔH for reactions with all gaseous species:

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

> **mnemonic**
>
> Broken First, Formed Second: ΔH = Broken minus Formed

**Worked example:** Calculate ΔH for the reaction: $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$. Bond enthalpies (kJ mol⁻¹): H-H = 436, Cl-Cl = 242, H-Cl = 431.

1. Count all bonds broken in reactants: 1 H-H bond and 1 Cl-Cl bond.
2. Sum of bonds broken: $436 + 242 = 678$ kJ mol⁻¹
3. Count all bonds formed in products: 2 H-Cl bonds.
4. Sum of bonds formed: $2 \times 431 = 862$ kJ mol⁻¹
5. Apply formula: $\Delta H = 678 - 862 = -184$ kJ mol⁻¹
6. Negative sign confirms the reaction is exothermic, which matches experimental data.

**Check your understanding**

Check your understanding

1. Which of the following is the correct formula for ΔH from bond enthalpies?

   - ΔH = bonds formed - bonds broken
   - ΔH = bonds broken - bonds formed
   - ΔH = bonds broken + bonds formed
   - ΔH = -(bonds broken + bonds formed)

   *Answer:* ΔH = bonds broken - bonds formed

   *Why:* Correct! Breaking bonds requires energy (positive input) and forming bonds releases energy (negative output), so ΔH equals total bonds broken minus total bonds formed.

*Calculator:* allowed

## Calculation from Standard Enthalpies of Formation

The standard enthalpy of formation of any element in its standard state is zero. This method can be used for reactions with any state (solid, liquid, gas) and gives the exact standard enthalpy of reaction, unlike average bond enthalpies.

$$\Delta H^\ominus_{\text{rxn}} = \sum \Delta H_f^\ominus (\text{products}) - \sum \Delta H_f^\ominus (\text{reactants})$$

**Standard Enthalpy of Reaction** — Enthalpy change for a reaction when all reactants and products are in their standard states at 298 K and 100 kPa.

*Notation:* $\Delta H^\ominus_{rxn}$

**Worked example:** Calculate $\Delta H^\ominus_{rxn}$ for: $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$. Given $\Delta H_f^\ominus$ (kJ mol⁻¹): $C_3H_8(g) = -104$, $CO_2(g) = -394$, $H_2O(l) = -286$, $O_2(g) = 0$.

1. Calculate sum of $\Delta H_f^\ominus$ for products: $(3 \times -394) + (4 \times -286) = -1182 - 1144 = -2326$ kJ mol⁻¹
2. Calculate sum of $\Delta H_f^\ominus$ for reactants: $(1 \times -104) + (5 \times 0) = -104$ kJ mol⁻¹
3. Apply formula: $\Delta H^\ominus_{rxn} = (-2326) - (-104) = -2326 + 104 = -2222$ kJ mol⁻¹
4. Negative value confirms combustion is exothermic, which is expected.

**Exam command terms**

Common command terms for this topic in IB exams:

- **Calculate** — Full working (formula, substitution, final answer with units and sign) is required. Working is worth most marks. *(Typically 3-4 marks for a full calculation question.)*

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Claiming ΔH is positive for exothermic reactions
  - Why it fails: Confusing the perspective of the surroundings vs the chemical system. ΔH always measures change in the system.
  - Correct: Remember: Exothermic = system loses enthalpy, so ΔH is negative. Endothermic = system gains enthalpy, so ΔH is positive.
- **Wrong:** Reversing the bond enthalpy formula (ΔH = bonds formed - bonds broken)
  - Why it fails: Forgetting that bond formation releases energy, which reduces the total enthalpy change of the system.
  - Correct: Use the mnemonic: Broken minus Formed, ΔH = broken - formed.
- **Wrong:** Forgetting to multiply enthalpy values by stoichiometric coefficients
  - Why it fails: Enthalpy of formation and bond enthalpy are per mole values, so they must scale with the balanced equation.
  - Correct: Always multiply each value by the number of moles (or number of bonds) given by the reaction stoichiometry.
- **Wrong:** Using bond enthalpies for liquid or solid species
  - Why it fails: Bond enthalpies are only defined for gaseous molecules, and are average values anyway.
  - Correct: Always use the standard enthalpy of formation method for reactions that include liquid or solid reactants or products.

## Cheatsheet

| Concept | Formula | Key Rule |
| --- | --- | --- |
| Enthalpy of Reaction | $\Delta H = H_{products} - H_{reactants}$ | Exothermic: ΔH < 0; Endothermic: ΔH > 0 |
| Bond Enthalpy Method | $\Delta H = \sum (bonds\ broken) - \sum (bonds\ formed)$ | Only for all-gaseous reactions, average values |
| Enthalpy of Formation Method | $\Delta H^\ominus_{rxn} = \sum \Delta H_f^\ominus(products) - \sum \Delta H_f^\ominus(reactants)$ | $\Delta H_f^\ominus = 0$ for elements in standard state |

## What's next

Enthalpy of reaction forms the foundation of all thermochemistry topics in IB SL Chemistry. The sign conventions and calculation methods you learned here are required for Hess's law, which allows you to calculate ΔH for reactions that cannot be measured directly experimentally. You will later apply these concepts to calculate enthalpy changes from experimental calorimetry data, and eventually explore entropy and the spontaneity of chemical reactions. Mastery of this sub-topic is critical for all higher level energetics questions in your final exam.

- [Hess's Law](https://www.owlsprep.com/study/ib-chemistry-sl-u5-hess-s-law/)
- [Collision theory and reaction rates](https://www.owlsprep.com/study/ib-chemistry-sl-u5-collision-theory-and-reaction-rates/)

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