A-Level Chemistry Energetics: Enthalpy, Hess's Law and Bond Enthalpy

1. Enthalpy changes: the basics

An enthalpy change (ΔH) is the heat energy transferred between a chemical system and its surroundings at constant pressure. The sign of ΔH tells you which way the energy flows.

• Exothermic: energy is released to the surroundings, ΔH is negative, and the surroundings get warmer (e.g. combustion, neutralisation).
• Endothermic: energy is taken in from the surroundings, ΔH is positive, and the surroundings get colder (e.g. thermal decomposition, dissolving ammonium nitrate).

Enthalpy changes are quoted under standard conditions: a pressure of 100 kPa and a stated temperature (usually 298 K, i.e. 25 C), with solutions at 1 mol/dm³. Each substance is in its standard state, meaning the physical state it exists in under those conditions.

Two definitions you need to know word-for-word:

• Standard enthalpy of formation (ΔfH): the enthalpy change when 1 mole of a compound is formed from its elements in their standard states, under standard conditions. The ΔfH of any element in its standard state is zero.
• Standard enthalpy of combustion (ΔcH): the enthalpy change when 1 mole of a substance is burned completely in oxygen, under standard conditions, with all reactants and products in their standard states.

2. Calorimetry: measuring ΔH with q = mcΔT

In the lab you measure the temperature change of a known mass of solution, then convert that heat into an enthalpy change per mole. Two equations do the work:

• q = mcΔT gives the heat energy transferred to (or from) the solution, in joules (J).
• m = mass of the solution being heated, in grams (assume 1 cm³ of dilute aqueous solution = 1 g).
• c = specific heat capacity = 4.18 J g-1 K-1 for water and dilute aqueous solutions.
• ΔT = temperature change, in K (the value is the same in C).
• ΔH = -q/n, where n = moles of the relevant reactant. The minus sign turns a temperature rise (exothermic) into a negative ΔH.

1. 1Set up the problem: 50 cm³ of 1.0 mol/dm³ HCl is mixed with 50 cm³ of 1.0 mol/dm³ NaOH. The temperature rises from 21.0 C to 27.8 C. Find the enthalpy of neutralisation.
2. 2Find q: Total solution = 100 cm³, so m = 100 g. ΔT = 27.8 - 21.0 = 6.8 K. q = mcΔT = 100 x 4.18 x 6.8 = 2842 J = 2.842 kJ.
3. 3Find moles: Moles of water formed = moles of HCl = 0.050 dm³ x 1.0 mol/dm³ = 0.050 mol (HCl and NaOH react 1:1).
4. 4Find ΔH: ΔH = -q/n = -2.842 / 0.050 = -56.8 kJ/mol. Negative, because the mixture warmed up: neutralisation is exothermic.

3. Hess's law: enthalpy cycles

Hess's law: the total enthalpy change of a reaction is the same whatever route you take, as long as the start and end conditions are identical. This lets you calculate a ΔH you cannot measure directly by going the long way round using data you do have. Which route you use depends on the data you are given.

Route A, using formation data (ΔfH): ΔH = sum of ΔfH(products) - sum of ΔfH(reactants). Picture the elements at the bottom of the cycle, with arrows pointing up to the reactants and to the products.

1. 1Reaction: CaCO3(s) -> CaO(s) + CO2(g), the thermal decomposition of limestone.
2. 2Data (ΔfH, kJ/mol): CaCO3 = -1207, CaO = -635, CO2 = -394.
3. 3Apply: ΔH = [(-635) + (-394)] - [(-1207)] = (-1029) - (-1207) = +178 kJ/mol.
4. 4Check: Positive, so endothermic, which fits: you have to heat limestone strongly to decompose it.

Route B, using combustion data (ΔcH): ΔH = sum of ΔcH(reactants) - sum of ΔcH(products). Here the combustion products (CO2 and H2O) sit at the bottom, with arrows pointing down from the reactants and the products.

1. 1Reaction: C(s) + 2H2(g) -> CH4(g), the formation of methane, which cannot be carried out directly.
2. 2Data (ΔcH, kJ/mol): C = -394, H2 = -286, CH4 = -890.
3. 3Apply: ΔH = [(-394) + 2 x (-286)] - [(-890)] = (-966) - (-890) = -76 kJ/mol.
4. 4Meaning: This is the standard enthalpy of formation of methane, about -76 kJ/mol.

4. Bond enthalpies: bonds broken minus bonds made

A bond enthalpy is the energy needed to break one mole of a particular covalent bond in the gaseous state. Breaking bonds absorbs energy (endothermic, positive); making bonds releases energy (exothermic, negative). The mean bond enthalpies in your data book are averages taken across many different molecules.

The equation to memorise: ΔH = sum of bonds broken (in the reactants) - sum of bonds made (in the products).

1. 1Reaction: H2(g) + Cl2(g) -> 2HCl(g).
2. 2Data (bond enthalpies, kJ/mol): H-H = 436, Cl-Cl = 243, H-Cl = 432.
3. 3Bonds broken: 1 x (H-H) + 1 x (Cl-Cl) = 436 + 243 = 679 kJ/mol.
4. 4Bonds made: 2 x (H-Cl) = 2 x 432 = 864 kJ/mol.
5. 5Apply: ΔH = bonds broken - bonds made = 679 - 864 = -185 kJ/mol. Exothermic, as expected when stable H-Cl bonds form.

Bond-enthalpy answers are only approximate. Mean bond enthalpies are averaged across many molecules, and the method assumes every substance is a gas, so a value from ΔfH or ΔcH data (Hess's law) is more accurate whenever you can use it.

5. Signs, units and the four classic errors

Most lost marks in energetics come from a small number of slips. Get the sign rules straight for each method and the calculations become routine:

• Calorimetry: temperature rises, so exothermic, so ΔH is negative (the minus sign in ΔH = -q/n handles this for you).
• Hess with formation data: ΔH = products - reactants.
• Hess with combustion data: ΔH = reactants - products.
• Bond enthalpy: ΔH = bonds broken - bonds made (broken is positive because breaking bonds absorbs energy).

6. A reliable method for any energetics question

Whatever the numbers, the same five steps get you to the answer. Work through them in order and you will rarely go wrong.

1. 11. Identify the data type: Temperature and mass means calorimetry; ΔfH or ΔcH values means Hess's law; bond enthalpies means the bonds-broken-minus-made method.
2. 22. Write a balanced equation: You need the correct stoichiometry to count moles and to count bonds.
3. 33. Apply the right formula: q = mcΔT then ΔH = -q/n; or the matching Hess expression; or bonds broken - bonds made.
4. 44. Work out ΔH per mole with units: Answers are in kJ/mol, so divide by moles and convert joules to kilojoules.
5. 55. Sanity-check the sign: Does exothermic or endothermic match the physical situation? If it does not, go back and find the sign slip.

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