Enthalpy change, ΔH

1. Overview

Almost every chemical reaction transfers heat energy to or from its surroundings. The heat transferred at constant pressure is the enthalpy change, $\Delta H$, and its sign tells you which way the energy flows. This section covers exothermic and endothermic reactions, reaction pathway diagrams, the standard enthalpy changes you must define, and two ways to find $\Delta H$: from bond energies and from temperature changes measured in experiments. These ideas are the base for Hess's law and every later energetics calculation.

Key Definitions

• Enthalpy change (ΔH): the heat energy transferred between a reacting system and its surroundings at constant pressure, usually given in kJ mol⁻¹.
• Exothermic reaction: a reaction that transfers heat energy from the system to the surroundings, so the surroundings warm up and ΔH is negative.
• Endothermic reaction: a reaction that takes in heat energy from the surroundings, so the surroundings cool down and ΔH is positive.
• Standard conditions: a temperature of 298 K and a pressure of 101 kPa; an enthalpy change measured under these conditions, with every substance in its standard state, is a standard enthalpy change.
• Standard enthalpy change of formation (ΔHf): the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.
• Standard enthalpy change of combustion (ΔHc): the enthalpy change when one mole of a substance is burned completely in oxygen under standard conditions, with all reactants and products in their standard states.
• Standard enthalpy change of neutralisation (ΔHneut): the enthalpy change when an acid and an alkali react in aqueous solution to form one mole of water under standard conditions.
• Bond energy: the energy required to break one mole of a particular covalent bond in the gaseous state; it is always positive.
• Standard enthalpy change of reaction (ΔHr): the enthalpy change when the amounts of reactants shown in a balanced equation react completely under standard conditions, with all substances in their standard states.
• Activation energy (EA): the minimum energy required for a collision between particles to be effective, that is, to lead to a reaction.

Content

Exothermic and endothermic reactions

The system is the reacting chemicals. The surroundings are everything else: the solvent, the container and the air. Every substance has an enthalpy, $H$. You cannot measure $H$ itself, only the change:

$$\Delta H = H_{\text{products}} - H_{\text{reactants}}$$

	Exothermic	Endothermic
Enthalpy of products	lower than reactants	higher than reactants
Heat flow	system to surroundings	surroundings to system
Temperature of surroundings	rises	falls
Sign of $\Delta H$	negative (−)	positive (+)
Examples	burning fuels, neutralisation	thermal decomposition of calcium carbonate

Always write the sign, including + for an endothermic value. When comparing, say more exothermic or less exothermic, not "bigger" or "smaller": −300 kJ mol⁻¹ is a smaller number than −200 kJ mol⁻¹, but it is the more exothermic change.

Reaction pathway diagrams

A reaction pathway diagram shows how enthalpy changes as reactants become products. Colliding particles need at least the activation energy, $E_{\text{A}}$, to react, so the line climbs to a peak before it reaches the products. To construct one:

1. Label the vertical axis enthalpy and the horizontal axis progress of reaction.
2. Draw a labelled line for the reactants and one for the products: lower for exothermic, higher for endothermic.
3. Join them with a smooth curve with a single hump.
4. Draw a vertical arrow from the reactants line up to the peak and label it $E_{\text{A}}$.
5. Draw a vertical arrow from the reactants line to the products line and label it $\Delta H$. It points down for exothermic and up for endothermic.

$E_{\text{A}}$ is always positive. For an endothermic reaction the peak must be above the products, so $E_{\text{A}}$ is larger than $\Delta H$.

Interpreting a diagram. The height of the peak above the products is the activation energy of the reverse reaction. If $E_{\text{A}} = +68$ kJ mol⁻¹ and $\Delta H = -29$ kJ mol⁻¹, the reverse reaction needs 68 + 29 = 97 kJ mol⁻¹. A catalyst lowers the peak but leaves the reactant and product levels, and so $\Delta H$, unchanged.

Standard conditions and standard states

$\Delta H$ depends on temperature, pressure and physical states, so values are compared as standard enthalpy changes, marked $\Delta H^\ominus$.

• Standard conditions are 298 K (25 °C) and 101 kPa.
• The standard state of a substance is its physical state under those conditions, in its most stable form, for example H₂O(l), O₂(g), Br₂(l), I₂(s) and carbon as graphite, C(s).

Standard conditions describe temperature and pressure; standard states describe the form of each substance. A full definition often needs both. States matter because forming liquid water releases more energy than forming steam, so every equation for an enthalpy change needs state symbols.

The four standard enthalpy changes

Enthalpy change	What the value is per	Example equation	Sign
Reaction, $\Delta H_{\text{r}}^\ominus$	the amounts in the equation	2H₂(g) + O₂(g) → 2H₂O(l)	either
Formation, $\Delta H_{\text{f}}^\ominus$	one mole of compound, from its elements	H₂(g) + ½O₂(g) → H₂O(l)	usually negative
Combustion, $\Delta H_{\text{c}}^\ominus$	one mole of substance, burned completely	CH₃CH(OH)CH₃(l) + 4½O₂(g) → 3CO₂(g) + 4H₂O(l)	always negative
Neutralisation, $\Delta H_{\text{neut}}^\ominus$	one mole of water, from acid and alkali	H⁺(aq) + OH⁻(aq) → H₂O(l)	always negative

• Reaction. The value belongs to the equation as written. The table's equation has $\Delta H_{\text{r}}^\ominus = -572$ kJ mol⁻¹; halve the equation and the value halves to −286 kJ mol⁻¹.
• Formation. Fractions such as ½O₂ are fine because the product must be exactly one mole. For an element in its standard state, such as O₂(g), $\Delta H_{\text{f}}^\ominus$ is zero.
• Combustion. "Completely" means the carbon ends as CO₂(g) and the hydrogen as H₂O(l). Burning hydrogen is the same equation as forming water, so $\Delta H_{\text{c}}^\ominus$(H₂) $= \Delta H_{\text{f}}^\ominus$(H₂O) $= -286$ kJ mol⁻¹.
• Neutralisation. The value is per mole of water, not of acid. Every strong acid with a strong alkali is really H⁺(aq) + OH⁻(aq) → H₂O(l), so the value is close to −57 kJ mol⁻¹. A weak acid, such as ethanoic acid, gives a slightly less exothermic value, because some energy is used to ionise the acid fully as it reacts. For sulfuric acid the equation is ½H₂SO₄(aq) + NaOH(aq) → ½Na₂SO₄(aq) + H₂O(l).

Why energy is transferred: breaking and making bonds

• Breaking a bond needs energy: bond breaking is endothermic.
• Making a bond releases energy: bond making is exothermic.

Every reaction breaks bonds in the reactants and makes bonds in the products. If making releases more energy than breaking takes in, the reaction is exothermic; otherwise it is endothermic. Energy is never "released when a bond breaks".

Calculating ΔH from bond energies

Bond energies, $E$, are listed in the Data Booklet. They refer to gases and are always positive, because they describe breaking.

$$\Delta H_{\text{r}} = \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})$$

1. Write the balanced equation. Bond energies are for gases, so your answer is $\Delta H$ for the reaction with every substance as a gas.
2. Count every bond in each molecule, using displayed formulae.
3. Multiply by the number in front of each molecule.
4. Total the bonds broken (reactants) and the bonds formed (products), then subtract.

Choose values carefully. Oxygen contains O=O (496), not O-O (150). Carbon dioxide has its own value, "C=O in CO₂" (805), not the general C=O value (740).

Worked example 1: combustion of propan-2-ol from bond energies

Question: Estimate the enthalpy change for the combustion of propan-2-ol vapour.

CH₃CH(OH)CH₃(g) + 4½O₂(g) → 3CO₂(g) + 4H₂O(g)

Bond energies / kJ mol⁻¹: C-C 350, C-H 410, C-O 360, O-H 460, O=O 496, C=O in CO₂ 805.

Step 1: count the bonds. Propan-2-ol has 2 C-C, 7 C-H, 1 C-O and 1 O-H. Each CO₂ has two C=O and each H₂O has two O-H.

Bonds broken	Energy / kJ mol⁻¹	Bonds formed	Energy / kJ mol⁻¹
2 C-C	2 × 350 = 700	6 C=O (in CO₂)	6 × 805 = 4830
7 C-H	7 × 410 = 2870	8 O-H	8 × 460 = 3680
1 C-O	360
1 O-H	460
4½ O=O	4.5 × 496 = 2232
Total	6622	Total	8510

Step 2: subtract. $$\Delta H = 6622 - 8510 = -1888\ \text{kJ mol}^{-1}$$

Answer: $\Delta H = -1888$ kJ mol⁻¹. It is negative because forming the bonds in CO₂ and H₂O releases more energy than breaking the bonds in the reactants takes in.

Exact and average bond energies

• A bond in a diatomic molecule, such as H-H or H-Cl, exists in only that molecule. It can be measured directly, so its value is exact.
• A bond such as C-H or O-H occurs in many different molecules, and its strength varies slightly with the neighbouring atoms. Even in methane, removing the first hydrogen atom takes a different energy from removing the last. The Data Booklet gives an average.

With exact values and only gases, the answer is reliable. For H₂(g) + Cl₂(g) → 2HCl(g), $\Delta H = (436 + 242) - (2 \times 431) = -184$ kJ mol⁻¹, which agrees closely with measured data.

Average values give only estimates, and states matter too. The accepted $\Delta H_{\text{c}}^\ominus$ of propan-2-ol is about −2006 kJ mol⁻¹, against −1888 kJ mol⁻¹ from Worked example 1.

• States (the main cause here). Bond energies apply to gases, so Worked example 1 forms steam. In the standard reaction the 4 mol of water are liquid, and turning steam into liquid water releases extra energy. Vaporising the liquid propan-2-ol takes in some energy, but much less than the water gives out.
• Averages (a small effect here). An average can make an estimate too exothermic or not exothermic enough. For propan-2-ol the estimate happens to be close to the true value for the gases.

Measuring enthalpy changes by experiment

The reaction heats (or cools) a known mass of water or solution, and you measure its temperature change.

$$q = mc\Delta T$$

• $q$: heat energy transferred, in J.
• $m$: mass of the water or solution that changes temperature, in g (not the solid or the fuel).
• $c$: specific heat capacity, 4.18 J g⁻¹ K⁻¹ for water and dilute solutions.
• $\Delta T$: temperature change, in K. A change of 1 K equals a change of 1 °C, so never add 273.

Then divide by $n$, the amount in moles that $\Delta H$ refers to: the limiting reactant, the water formed for neutralisation, or the fuel burned for combustion.

$$\Delta H = -\frac{mc\Delta T}{n}$$

A temperature rise means the system lost heat, hence the minus sign. With $\Delta T$ taken as final minus initial, a rise gives a negative $\Delta H$ and a fall a positive one. Divide by 1000 for kJ mol⁻¹.

	Reactions in solution	Burning a liquid fuel
Apparatus	polystyrene cup, lid, thermometer, stirring	spirit burner under a copper can of water
$m$	volume of solution, taking 1.00 g cm⁻³	mass of water in the can
$n$	moles of limiting reactant	mass lost by the burner ÷ $M$

Why experimental values are less exothermic than expected. Heat escapes to the air and apparatus, so $\Delta T$ is too small, $q$ is too small and $\Delta H$ is less exothermic than the true value. Fuels lose far more heat than cups. Incomplete combustion (soot) and fuel evaporating from the wick make the value less exothermic still. Lids, insulation and a draught shield reduce the error. For a reaction in a cup, plotting temperature against time and extending the cooling line back to the moment of mixing gives a better $\Delta T$ than the highest reading.

Worked example 2: enthalpy change of neutralisation

Question: 35.0 cm³ of 1.80 mol dm⁻³ nitric acid at 20.8 °C is mixed in a polystyrene cup with 35.0 cm³ of 2.00 mol dm⁻³ potassium hydroxide, also at 20.8 °C. The highest temperature reached is 32.6 °C. Calculate $\Delta H_{\text{neut}}$. Take the density as 1.00 g cm⁻³ and $c = 4.18$ J g⁻¹ K⁻¹.

HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O(l)

Step 1: heat released. $m = 35.0 + 35.0 = 70.0$ g and $\Delta T = 32.6 - 20.8 = 11.8$ K. $$q = 70.0 \times 4.18 \times 11.8 = 3452.68\ \text{J}$$

Step 2: moles of water. $$n(\text{HNO}_3) = \frac{35.0}{1000} \times 1.80 = 0.0630\ \text{mol} \qquad n(\text{KOH}) = \frac{35.0}{1000} \times 2.00 = 0.0700\ \text{mol}$$ The ratio is 1 : 1, so the acid is limiting and 0.0630 mol of water forms.

Step 3: molar enthalpy change. $$\Delta H_{\text{neut}} = -\frac{3452.68}{0.0630} = -54\,804\ \text{J mol}^{-1} = -54.8\ \text{kJ mol}^{-1}$$

Answer: $\Delta H_{\text{neut}} = -54.8$ kJ mol⁻¹ (3 s.f.). It is less exothermic than about −57 kJ mol⁻¹ because some heat escaped.

Using 70.0 cm³ of each solution would give the same $\Delta T$: twice the heat, but twice the mass to warm.

Worked example 3: enthalpy change of combustion from a spirit burner

Question: A burner of propan-2-ol ($M = 60.0$ g mol⁻¹) heats 250 g of water in a copper can. The burner's mass falls from 74.62 g to 73.54 g and the water warms from 19.2 °C to 41.8 °C. Calculate the enthalpy change of combustion.

Step 1: heat absorbed by the water. $m = 250$ g of water and $\Delta T = 41.8 - 19.2 = 22.6$ K. $$q = 250 \times 4.18 \times 22.6 = 23\,617\ \text{J}$$

Step 2: moles of fuel. Mass burned $= 74.62 - 73.54 = 1.08$ g. $$n = \frac{1.08}{60.0} = 0.0180\ \text{mol}$$

Step 3: molar enthalpy change. $$\Delta H_{\text{c}} = -\frac{23\,617}{0.0180} = -1\,312\,056\ \text{J mol}^{-1} = -1310\ \text{kJ mol}^{-1}$$

Answer: $\Delta H_{\text{c}} = -1310$ kJ mol⁻¹ (3 s.f.).

Method for propan-2-ol	$\Delta H_{\text{c}}$ / kJ mol⁻¹	Why it differs from the accepted value
Spirit burner (Worked example 3)	−1310	heat lost to the air and the can; incomplete combustion; fuel evaporating from the wick
Bond energies (Worked example 1)	−1888	mainly states: water is treated as steam, not liquid; bond energies are also averages
Accepted standard value	about −2006	the reference: propan-2-ol and water are liquids in their standard states

Key Equations

Enthalpy change, in kJ mol⁻¹ (negative = exothermic): $$\Delta H = H_{\text{products}} - H_{\text{reactants}}$$

From bond energies ($E$ = bond energy in kJ mol⁻¹, gases, always positive): $$\Delta H_{\text{r}} = \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})$$

Heat transferred ($q$ in J; $m$ = mass of water or solution in g; $c$ = 4.18 J g⁻¹ K⁻¹; $\Delta T$ in K): $$q = mc\Delta T$$

Molar enthalpy change ($n$ = moles of limiting reactant, of water formed or of fuel burned; divide by 1000 for kJ mol⁻¹): $$\Delta H = -\frac{mc\Delta T}{n}$$

Moles (mass in g, $M$ in g mol⁻¹; concentration in mol dm⁻³, $V$ in cm³): $$n = \frac{\text{mass}}{M} \qquad n = \text{concentration} \times \frac{V}{1000}$$

Reverse activation energy, read from a pathway diagram (kJ mol⁻¹): $$E_{\text{A}}(\text{reverse}) = E_{\text{A}}(\text{forward}) - \Delta H$$

Common Mistakes to Avoid

• Mistake: using the mass of the solid or fuel as $m$. Fix: $m$ is whatever changes temperature. In Worked example 3, $m$ = 250 g of water, not 1.08 g of propan-2-ol.
• Mistake: adding 273 to $\Delta T$. Fix: a rise of 11.8 °C is a rise of 11.8 K. Using 284.8 K would make $q$ about 24 times too big.
• Mistake: dividing by moles of acid with sulfuric acid. Fix: divide by moles of water. 20.0 cm³ of 0.500 mol dm⁻³ H₂SO₄ neutralised by NaOH holds 0.0100 mol of acid but forms 0.0200 mol of water.
• Mistake: dropping the sign or the unit conversion. Fix: a temperature rise means a negative $\Delta H$, and J must become kJ. Write −54.8 kJ mol⁻¹, not 54.8 kJ mol⁻¹ or −54 804 kJ mol⁻¹.
• Mistake: wrong bond energy or a missing multiplier. Fix: in Worked example 1, 4½ O=O bonds need 4.5 × 496 = 2232, not 496 once and not the O-O value of 150.
• Mistake: subtracting the wrong way round. Fix: broken minus formed. In Worked example 1, 8510 − 6622 = +1888 would wrongly make combustion endothermic.
• Mistake: incomplete definitions, or mixing up standard conditions and standard states. Fix: $\Delta H_{\text{f}}^\ominus$ of water is for one mole of H₂O(l) made from H₂(g) and ½O₂(g), elements in their standard states, at 298 K and 101 kPa. The equation 2H₂(g) + O₂(g) → 2H₂O(l) makes two moles, so its value is $\Delta H_{\text{r}}^\ominus$.
• Mistake: pathway arrows at the wrong levels. Fix: start both arrows at the reactants line. An arrow from the products up to the peak is the reverse activation energy (97 kJ mol⁻¹ in the example above), not $E_{\text{A}}$.

Exam Tips

• Define: include one mole of what, standard states, and 298 K and 101 kPa. When asked for an equation, add state symbols, for example 3C(s) + 4H₂(g) + ½O₂(g) → CH₃CH(OH)CH₃(l) for the formation of propan-2-ol.
• Calculate: set out $q$, then $n$, then $\Delta H$ on labelled lines. Round only at the end, and give the sign, kJ mol⁻¹ and as many significant figures as the data (usually 3).
• Explain why an experimental value is less exothermic than expected as a chain: heat lost, so $\Delta T$ smaller, so $q$ smaller, so $\Delta H$ less exothermic. "Human error" explains nothing.
• Sketch a pathway diagram with labelled axes, formulae on both levels, and $E_{\text{A}}$ and $\Delta H$ as vertical arrows with heads.
• Suggest why a bond-energy answer differs from a data value: bond energies are averages from many compounds, and they apply to gases while standard values may involve liquids.
• Compare energy changes as more or less exothermic (or endothermic), not "higher" or "lower".