# Enthalpy changes

> CIE A-Level Chemistry · 9701
> Source: https://www.owlsprep.com/study/cie-9701-u5-enthalpy-changes/

This module covers core definitions of enthalpy and standard conditions, calculation of enthalpy changes from calorimetry data, application of Hess's Law, and enthalpy calculations from average bond enthalpies for CIE A-Level Chemistry.

**Prerequisites:** [Basic exothermic/endothermic reaction classification](https://www.owlsprep.com/study/cie-9701-u2-energy-changes/)

## Learning objectives

- Define enthalpy change and related standard state terms
- Calculate enthalpy changes from experimental calorimetry data
- Apply Hess's Law to construct enthalpy cycles and calculate unknown enthalpy changes
- Calculate reaction enthalpy changes from average bond enthalpy data

## Key Definitions and Standard Conditions

Enthalpy is a measure of the total heat content of a system. We almost always measure changes in enthalpy ($\Delta H$) rather than absolute enthalpy values. Standard conditions are defined as 1 atm pressure (100 kPa), a stated temperature (usually 298 K for CIE), and all substances in their most stable (standard) state.

**Standard Enthalpy Change** — Enthalpy change measured when reaction occurs under standard conditions, with all reactants and products in their standard states.

*Notation:* $\Delta H^\ominus$

**Worked example:** State whether each of the following is in its standard state at 298 K and 1 atm: (a) $Cl_2(g)$, (b) $C(\text{diamond})$, (c) $H_2O(l)$, (d) $Br_2(g)$

1. Recall: standard state is the most stable form of an element/compound at 1 atm and 298 K.
2. (a) Chlorine is a gas at 298 K, so $Cl_2(g)$ is the standard state:
3. $$\text{Answer: Yes}$$
4. (b) The most stable form of carbon at 298 K is graphite, not diamond:
5. $$\text{Answer: No}$$
6. (c) Water is liquid at 298 K and 1 atm, so this is the standard state:
7. $$\text{Answer: Yes}$$
8. (d) Bromine is liquid at 298 K, so gaseous bromine is not the standard state:
9. $$\text{Answer: No}$$

*Calculator:* allowed

## Enthalpy Calculations from Calorimetry

For reactions that can be carried out in an insulated container (calorimeter), we measure the temperature change of the surroundings (usually the reaction solution) to calculate the heat released or absorbed by the reaction. The formula for heat change of the surroundings is:

$$q = mc\Delta T$$

Where $q$ = heat gained by the surroundings (J), $m$ = mass of solution (g), $c$ = specific heat capacity (usually $4.18\ J\ g^{-1}\ ^\circ C^{-1}$ for aqueous solutions), $\Delta T$ = change in temperature ($^\circ C$ or K). To get the molar enthalpy change:

$$\Delta H = -\frac{q}{n}$$

Where $n$ = moles of the limiting reactant, and the negative sign accounts for the direction of heat flow: exothermic reactions have negative $\Delta H$, endothermic have positive $\Delta H$.

**Worked example:** 50 cm³ of 1.0 mol dm⁻³ HCl is mixed with 50 cm³ of 1.0 mol dm⁻³ NaOH in a calorimeter. The temperature increases from 22°C to 28.5°C. Calculate the enthalpy change of neutralisation. Assume density of solution = 1 g cm⁻³, $c = 4.18\ J\ g^{-1}\ ^\circ C^{-1}$.

1. Calculate total mass of solution and temperature change:
2. $$m = 50 + 50 = 100\ g, \quad \Delta T = 28.5 - 22 = 6.5^\circ C$$
3. Calculate heat gained by the surroundings:
4. $$q = mc\Delta T = 100 \times 4.18 \times 6.5 = 2717\ J = 2.717\ kJ$$
5. Calculate moles of limiting reactant (both are 0.05 mol here):
6. $$n = \frac{1.0 \times 50}{1000} = 0.05\ mol$$
7. Calculate molar enthalpy change (temperature increases, so exothermic, negative):
8. $$\Delta H = -\frac{2.717}{0.05} = -54.3\ kJ\ mol^{-1}$$

*Calculator:* allowed

## Hess's Law and Enthalpy Cycles

Many reactions cannot be carried out directly in a calorimeter, so we use Hess's Law to calculate the unknown enthalpy change from known values. For enthalpy of formation values, the general formula is:

**Hess's Law** — The total enthalpy change for a reaction is independent of the route taken from reactants to products.

$$\Delta H^\ominus_r = \sum \Delta H^\ominus_f(\text{products}) - \sum \Delta H^\ominus_f(\text{reactants})$$

**Worked example:** Given $\Delta H^\ominus_f(C_2H_4(g)) = +52\ kJ\ mol^{-1}$, $\Delta H^\ominus_f(H_2O(l)) = -286\ kJ\ mol^{-1}$, $\Delta H^\ominus_f(C_2H_5OH(l)) = -278\ kJ\ mol^{-1}$. Calculate $\Delta H^\ominus_r$ for the reaction: $C_2H_4(g) + H_2O(l) \rightarrow C_2H_5OH(l)$

1. Substitute into the standard enthalpy of reaction formula:
2. $$\Delta H^\ominus_r = \Delta H^\ominus_f(\text{product}) - \left[\Delta H^\ominus_f(C_2H_4) + \Delta H^\ominus_f(H_2O)\right]$$
3. Plug in the values:
4. $$\Delta H^\ominus_r = (-278) - \left[(+52) + (-286)\right]$$
5. Simplify to get the final answer:
6. $$\Delta H^\ominus_r = -278 - (-234) = -44\ kJ\ mol^{-1}$$

*Calculator:* allowed

## Enthalpy Changes from Bond Enthalpies

Bond enthalpy is a measure of the strength of a covalent bond. Breaking bonds requires energy (endothermic, positive $\Delta H$) and forming bonds releases energy (exothermic, negative $\Delta H$). Average bond enthalpies are mean values taken from many different compounds, so calculations using them are approximate. The formula for reaction enthalpy is:

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

**Worked example:** Calculate $\Delta H$ 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. Calculate the total enthalpy required to break all bonds in reactants:
2. $$\sum \text{bonds broken} = 436 + 242 = 678\ kJ\ mol^{-1}$$
3. Calculate the total enthalpy released when all bonds form in products:
4. $$\sum \text{bonds formed} = 2 \times 431 = 862\ kJ\ mol^{-1}$$
5. Substitute into the formula to get $\Delta H$:
6. $$\Delta H = 678 - 862 = -184\ kJ\ mol^{-1}$$

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting the negative sign in calorimetry $\Delta H$ calculations
  - Why it fails: $q$ measures heat gained by the surroundings, so exothermic reactions (temperature increase) have negative $\Delta H$
  - Correct: Always use $\Delta H = -q/n$ and confirm the sign matches if the reaction is exothermic/endothermic
- **Wrong:** Using moles of excess reactant instead of limiting reactant
  - Why it fails: Enthalpy change is quoted per mole of reaction, which is limited by the reactant that is fully consumed
  - Correct: Always identify the limiting reactant and use its number of moles in the $\Delta H$ calculation
- **Wrong:** Reversing the order in bond enthalpy calculations
  - Why it fails: Common sign error from mixing up bond breaking vs bond making contributions
  - Correct: Always use: $\Delta H = \sum (\text{bonds broken}) - \sum (\text{bonds formed})$
- **Wrong:** Forgetting to flip the sign of $\Delta H$ when reversing a reaction in Hess's Law
  - Why it fails: Reversing a reaction reverses the direction of heat flow, so the enthalpy change sign must change
  - Correct: Always change the sign of $\Delta H$ whenever you reverse a chemical equation in an enthalpy cycle
- **Wrong:** Using non-gaseous species in bond enthalpy calculations
  - Why it fails: Bond enthalpies are only defined for gaseous species, extra enthalpy changes for state changes are ignored
  - Correct: Always confirm all reactants and products are gaseous when using average bond enthalpies

## Cheatsheet

| Concept | Formula/Rule | Key Exam Notes |
| --- | --- | --- |
| Calorimetry | $q = mc\Delta T$, $\Delta H = -\frac{q}{n}$ | Use moles of limiting reactant |
| ΔH from formation | $\Delta H^\ominus_r = \sum \Delta H^\ominus_f(products) - \sum \Delta H^\ominus_f(reactants)$ | ΔH⊖f of elements = 0 |
| ΔH from combustion | $\Delta H^\ominus_r = \sum \Delta H^\ominus_c(reactants) - \sum \Delta H^\ominus_c(products)$ | Reverse order vs formation |
| ΔH from bond enthalpies | $\Delta H = \sum (bonds\ broken) - \sum (bonds\ formed)$ | Average values = approximate result |
| Hess's Law | Total ΔH is independent of route | Flip sign of ΔH when reversing reactions |

## What's next

Enthalpy changes are the foundational concept for all further energetics topics in CIE A-Level Chemistry. The calculation skills you have practiced here (Hess cycles, enthalpy arithmetic) are repeated in more advanced topics like lattice enthalpy and Born-Haber cycles, and are required for understanding entropy and Gibbs free energy, which are core to predicting reaction spontaneity. These skills are also frequently tested in combination with organic chemistry topics, where enthalpy changes of reaction are commonly asked.

- [Hess' Law](https://www.owlsprep.com/study/cie-9701-u5-hess-law/)
- [Bond Enthalpies](https://www.owlsprep.com/study/cie-9701-u5-bond-enthalpies/)
- [Redox reactions and electrolysis](https://www.owlsprep.com/study/cie-9701-u6-overview/)

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