Unit 1 · Spec 3.1.4

Energetics, Calorimetry & Hess Law Cycles

Standard enthalpy definitions, solution calorimetry, cooling curve extrapolation, Hess Law cycles, and mean bond enthalpies.

1. Standard Enthalpy Definitions

Enthalpy Change (ΔH)

The heat energy transferred in a chemical reaction at constant pressure. Standard conditions (°): 100 kPa (1 bar) pressure and a stated temperature (usually 298 K / 25°C), with substances in their standard physical states.

Exothermic Reaction

Releases heat energy to the surroundings. ΔH is negative. The surroundings get warmer.

Examples: combustion of fuels, neutralisation.

Endothermic Reaction

Takes in heat energy from the surroundings. ΔH is positive. The surroundings get cooler.

Example: thermal decomposition of calcium carbonate.

Standard Enthalpy of Formation (\(\Delta_f H^\circ\))

The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions.

Rule: The standard enthalpy of formation of any element in its standard physical state is zero by definition (e.g. \(\Delta_f H^\circ[\text{O}_2(g)] = 0\)).

Standard Enthalpy of Combustion (\(\Delta_c H^\circ\))

The enthalpy change when one mole of a substance burns completely in excess oxygen under standard conditions, with all reactants and products in their standard states.

Rule: Enthalpies of combustion are always negative (exothermic).

2. Calorimetry Principles

Calorimetry measures thermal energy changes by tracking temperature changes in an aqueous solution or calorimeter:

Heat Energy Transferred Formula

\[ q = mc\Delta T \]

  • \(q\) = heat energy absorbed or released in joules (J).
  • \(m\) = mass of solution being heated or cooled in grams (g). We assume aqueous solutions have a density of 1.00 g cm⁻³ (so 50 cm³ = 50 g).
  • \(c\) = specific heat capacity of water (4.18 J g⁻¹K⁻¹).
  • ΔT = temperature change in K or °C (\(T_{\text{final}} - T_{\text{initial}}\)).

Molar Enthalpy Change Formula

\[ \Delta H = -\frac{q}{n \times 1000} \]

Where \(n\) is the amount of limiting reactant in moles (mol), dividing by 1000 converts joules to kilojoules (kJ mol⁻¹). The negative sign ensures that an exothermic temperature rise produces a negative ΔH.

3. Cooling Curve Temperature Extrapolation

In simple polystyrene cup calorimetry, heat is inevitably lost to the laboratory surroundings during the reaction. The recorded peak temperature is therefore lower than the true theoretical maximum.

Calorimetry cooling curve extrapolation 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Time / minutes Temperature / °C steady start: T₀ reactants mixed at 4 min 5 min: still rising, not used for the line line of best fit through cooling points corrected ΔT Tmax (extrapolated)

Experimental Protocol (RP2 Alignment)

  1. Record the temperature of the initial solution every minute for 3 minutes to establish a stable baseline.
  2. At minute 4, add the second reactant and stir. Do not record a temperature at minute 4.
  3. Resume recording temperature at minute 5, taking readings every minute until minute 15.
  4. Plot temperature against time on graph paper.
  5. Draw a straight line of best fit through the cooling points, from the highest temperature onwards, and extrapolate this line back to minute 4 (the exact moment of mixing). The minute 5 reading is often still rising, so do not include it if it is below the highest temperature.
  6. Determine ΔT as the vertical difference between the extrapolated temperature at minute 4 and the initial baseline temperature.
Heat loss and RP2
  • Heat is lost to the surroundings, so the measured temperature change is too small. The calculated ΔH is then smaller in size than the true value.
  • Reduce heat loss with a polystyrene cup, a lid and insulation around the cup.
  • Extrapolating the cooling line back to the time of mixing corrects for the heat lost while the reaction takes place.

The full method, apparatus and errors are on RP2: Measurement of an Enthalpy Change.

4. Hess's Law Cycles

Hess's Law

The total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final states are identical.

Enthalpies of Formation Route

\[ \Delta H_r^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants}) \]

Constituent elements in standard states form the bottom baseline. Arrows point upwards from elements to products and reactants.

Enthalpies of Combustion Route

\[ \Delta H_r^\circ = \sum \Delta_c H^\circ(\text{reactants}) - \sum \Delta_c H^\circ(\text{products}) \]

Combustion products (CO₂,H₂O) form the bottom baseline. Arrows point downwards from reactants and products to combustion products.

Worked Example: Hess Cycle Calculation
Calculate the standard enthalpy of formation of ethanol, C₂H₅OH(l), given: ΔcH(C) = -393.5 kJ mol⁻¹; ΔcH(H₂) = -285.8 kJ mol⁻¹; ΔcH(C₂H₅OH) = -1367.3 kJ mol⁻¹.

Formation equation: \[ 2\text{C}(s) + 3\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l) \]

Applying Hess combustion formula:

\[ \Delta_f H = \sum \Delta_c H(\text{reactants}) - \sum \Delta_c H(\text{products}) \]

\[ \Delta_f H = [2(-393.5) + 3(-285.8)] - [-1367.3] \]

\[ \Delta_f H = [-787.0 - 857.4] - [-1367.3] = -1644.4 + 1367.3 = \mathbf{-277.1\text{ kJ mol}^{-1}} \]

5. Mean Bond Enthalpies

Mean Bond Enthalpy

The enthalpy change required to break one mole of a specified covalent bond in gaseous molecules, averaged across a wide range of chemical compounds.

\[ \Delta H_r = \sum (\text{Bonds Broken}) - \sum (\text{Bonds Formed}) \]

Why do bond enthalpy calculations differ from experimental Hess cycles?

  • Mean bond enthalpies are averages over a range of compounds. They are not the actual values for the bonds in these molecules. When both values refer to gases, the reason to give is that mean bond enthalpies are averages.
  • Bond enthalpies apply only to the gaseous state. The data-book value of an enthalpy of combustion has water as a liquid, not a gas, so it differs even more.

Breaking bonds is endothermic (energy taken in). Making bonds is exothermic (energy released).

Worked Example: ΔH from Mean Bond Enthalpies
Calculate the enthalpy change for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). Mean bond enthalpies / kJ mol⁻¹: C-H 413, O=O 498, C=O 805, O-H 463.

Step 1: Bonds broken (4 C-H in CH₄, 2 O=O): 4 × 413 + 2 × 498 = 2648 kJ mol⁻¹

Step 2: Bonds made (2 C=O in CO₂, 4 O-H in 2H₂O): 2 × 805 + 4 × 463 = 3462 kJ mol⁻¹

Step 3: ΔH = bonds broken - bonds made = 2648 - 3462 = -814 kJ mol⁻¹

How does this compare with Hess's law? Using enthalpies of formation with water as a gas gives -802 kJ mol⁻¹ ([-393.5 + 2(-241.8)] - [-74.8] = -802.3 kJ mol⁻¹). The 12 kJ difference is because mean bond enthalpies are averages over many compounds. The data-book ΔcH (-890 kJ mol⁻¹) differs more because there the water is a liquid.

Go deeper

These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.

6. Practice Questions

Practice Problem (3 Marks)
In a calorimetry experiment, 50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH. The temperature rises from 19.5 °C to 26.2 °C. Calculate the molar enthalpy of neutralisation. (Assume density = 1.00 g cm⁻³, c = 4.18 J g⁻¹ K⁻¹).
Show Answer & Mark Scheme

Step 1: Total solution mass: \(m = 50.0 + 50.0 = 100.0\text{ g}\).

Step 2: Temperature rise: ΔT = 26.2 − 19.5 = 6.7 K.

Step 3: Heat energy released:

\[ q = mc\Delta T = 100.0 \times 4.18 \times 6.7 = 2800.6\text{ J} = 2.8006\text{ kJ} \]

Step 4: Moles of water formed: \(n = c \times V = 1.00 \times (50.0 / 1000) = 0.0500\text{ mol}\).

Step 5: Molar enthalpy change:

\[ \Delta H = -\frac{2.8006}{0.0500} = \mathbf{-56.0\text{ kJ mol}^{-1}} \]

Practice Problem 2 (3 Marks)
Use the mean bond enthalpies to calculate the enthalpy change for H₂(g) + Cl₂(g) → 2HCl(g). Mean bond enthalpies / kJ mol⁻¹: H-H 436, Cl-Cl 243, H-Cl 432.
Show Answer & Mark Scheme

Mark Scheme:

  • Bonds broken: 436 + 243 = 679 kJ mol⁻¹. (1 mark)
  • Bonds made: 2 × 432 = 864 kJ mol⁻¹. (1 mark)
  • ΔH = 679 - 864 = -185 kJ mol⁻¹. (1 mark)
Practice Problem 3 (4 Marks)
(a) Use the enthalpies of formation to calculate the enthalpy change when methane burns to form gaseous water: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). ΔfH / kJ mol⁻¹: CH₄(g) -74.8; CO₂(g) -393.5; H₂O(g) -241.8. (b) The value from mean bond enthalpies is -814 kJ mol⁻¹. Give the main reason for the difference.
Show Answer & Mark Scheme

Mark Scheme:

  • (a) ΔfH of O₂(g) is zero, because it is an element in its standard state. (1 mark)
  • ΔH = Σ ΔfH(products) - Σ ΔfH(reactants) = [-393.5 + 2(-241.8)] - [-74.8]. (1 mark)
  • ΔH = -877.1 + 74.8 = -802.3 kJ mol⁻¹. (1 mark)
  • (b) Mean bond enthalpies are averages over a range of compounds, not the actual values in these molecules. (1 mark)
Practice Problem 4 (3 Marks)
State what is meant by an exothermic reaction, give the sign of its ΔH, and state the standard conditions of pressure and temperature.
Show Answer & Mark Scheme

Mark Scheme:

  • An exothermic reaction releases heat energy to the surroundings. (1 mark)
  • ΔH is negative. (1 mark)
  • Standard conditions: 100 kPa and a stated temperature, usually 298 K. (1 mark)
Required Practical 2

Measure an enthalpy change

The method, the cooling curve and how to calculate the enthalpy change.

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