Enthalpy changes

CIE A-Level Chemistry· Unit 5: Chemical energetics, Sub-topic 1: Enthalpy changes· 25 min read

1. Key Definitions and Standard Conditions★★☆☆☆⏱ 5 min

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Enthalpy is a measure of the total heat content of a system. We almost always measure changes in enthalpy ( $Δ H$ ) 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.

📘 Definition

Standard Enthalpy Change

$Δ H^{⊖}$

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

📐 Worked Example

State whether each of the following is in its standard state at 298 K and 1 atm: (a) $C l_{2} (g)$ , (b) $C (diamond)$ , (c) $H_{2} O (l)$ , (d) $B r_{2} (g)$

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 $C l_{2} (g)$ 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
$\text{Answer: No}$

2. Enthalpy Calculations from Calorimetry★★★☆☆⏱ 7 min

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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), $Δ T$ = 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 $Δ H$ , endothermic have positive $Δ H$ .

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

3. Hess's Law and Enthalpy Cycles★★★☆☆⏱ 7 min

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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:

📘 Definition

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 $Δ H_{f}^{⊖} (C_{2} H_{4} (g)) = + 52 k J m o l^{- 1}$ , $Δ H_{f}^{⊖} (H_{2} O (l)) = - 286 k J m o l^{- 1}$ , $Δ H_{f}^{⊖} (C_{2} H_{5} O H (l)) = - 278 k J m o l^{- 1}$ . Calculate $Δ H_{r}^{⊖}$ for the reaction: $C_{2} H_{4} (g) + H_{2} O (l) \to C_{2} H_{5} O H (l)$

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

4. Enthalpy Changes from Bond Enthalpies★★★☆☆⏱ 6 min

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Bond enthalpy is a measure of the strength of a covalent bond. Breaking bonds requires energy (endothermic, positive $Δ H$ ) and forming bonds releases energy (exothermic, negative $Δ H$ ). 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 $Δ H$ for the reaction: $H_{2} (g) + C l_{2} (g) \to 2 H C l (g)$ . Bond enthalpies (kJ mol⁻¹): $H - H = + 436$ , $C l - C l = + 242$ , $H - C l = + 431$ .

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:
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$\sum \text{bonds formed} = 2 \times 431 = 862\ kJ\ mol^{-1}$
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Substitute into the formula to get $Δ H$ :
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$\Delta H = 678 - 862 = -184\ kJ\ mol^{-1}$

5. Common Pitfalls

Wrong move:

Forgetting the negative sign in calorimetry $Δ H$ calculations

Why:

$q$ measures heat gained by the surroundings, so exothermic reactions (temperature increase) have negative $Δ H$

Correct move:

Always use $Δ H = - q / n$ and confirm the sign matches if the reaction is exothermic/endothermic

Wrong move:

Using moles of excess reactant instead of limiting reactant

Why:

Enthalpy change is quoted per mole of reaction, which is limited by the reactant that is fully consumed

Correct move:

Always identify the limiting reactant and use its number of moles in the $Δ H$ calculation

Wrong move:

Reversing the order in bond enthalpy calculations

Why:

Common sign error from mixing up bond breaking vs bond making contributions

Correct move:

Always use: $Δ H = \sum (bonds broken) - \sum (bonds formed)$

Wrong move:

Forgetting to flip the sign of $Δ H$ when reversing a reaction in Hess's Law

Why:

Reversing a reaction reverses the direction of heat flow, so the enthalpy change sign must change

Correct move:

Always change the sign of $Δ H$ whenever you reverse a chemical equation in an enthalpy cycle

Wrong move:

Using non-gaseous species in bond enthalpy calculations

Why:

Bond enthalpies are only defined for gaseous species, extra enthalpy changes for state changes are ignored

Correct move:

Always confirm all reactants and products are gaseous when using average bond enthalpies

6. Quick Reference Cheatsheet

Concept	Formula/Rule	Key Exam Notes
Calorimetry	$q = m c Δ T$ , $Δ H = - \frac{q}{n}$	Use moles of limiting reactant
ΔH from formation	$Δ H_{r}^{⊖} = \sum Δ H_{f}^{⊖} (p r o d u c t s) - \sum Δ H_{f}^{⊖} (r e a c t an t s)$	ΔH⊖f of elements = 0
ΔH from combustion	$Δ H_{r}^{⊖} = \sum Δ H_{c}^{⊖} (r e a c t an t s) - \sum Δ H_{c}^{⊖} (p r o d u c t s)$	Reverse order vs formation
ΔH from bond enthalpies	$Δ H = \sum (b o n d s b r o k e n) - \sum (b o n d s f or m e d)$	Average values = approximate result
Hess's Law	Total ΔH is independent of route	Flip sign of ΔH when reversing reactions

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 · 2
Hess cycle enthalpy calculation
2023 · 1
Calorimetry enthalpy MCQ
2021 · 4
Bond enthalpy calculation

Going deeper

revision noteCIE 9701 Energetics Syllabus OverviewOfficial syllabus document

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.