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1.4 Energetics

1.4.1 Enthalpy Change 1.4.2 Calorimetry 1.4.3 Hess's Law 1.4.4 Bond Enthalpies

Hess’s Law

Specification Reference Physical Chemistry, Energetics 3.1.4.3

Quick Notes

  • Hess’s Law states that the total enthalpy change of a reaction is independent of the route taken.
    • If a reaction can take multiple pathways, the enthalpy change for each route will be the same, as long as the initial and final conditions are the same.
  • This allows the calculation of unknown enthalpy changes using enthalpy cycles.
  • Key equations:
    ΔH = ΣΔH (route 1) = ΣΔH (route 2)
    Enthalpy of formation cycle: ΔHr = ΣΔHf(products) − ΣΔHf(reactants)

Full Notes

Hess's Law has been outlined with more background theory here.

This page is just what you need to know for AQA A-level Chemistry :)

Hess’s Law is based on the Law of Conservation of Energy, which states that energy cannot be created or destroyed, only transferred.

Definition of Hess’s Law:
“The total enthalpy change for a reaction is the same, regardless of the route taken, provided the initial and final conditions are the same.”

Hess’s Law and Enthalpy Cycles

Hess’s Law is useful when directly measuring an enthalpy change is difficult. Instead, we use enthalpy cycles to calculate it indirectly.

AQA A-Level Chemistry Hess’s Law enthalpy cycle diagram with multiple reaction pathways.

Hess cycles are commonly constructed for:

Hess’s Law for Enthalpy of Formation

The standard enthalpy of formation (ΔHf°) is the energy change when 1 mole of a compound forms from its constituent elements in their standard states at 298 K (25°C) and 1 atm.

For elements in their standard states, ΔHf° = 0.

Examples:

We can calculate the enthalpy change (ΔH) of a reaction by using the standard enthalpies of formation of the reactants and products.

Use the formula:

AP Chemistry diagram showing ΔH°reaction equals the sum of standard enthalpies of formation of products minus the sum for reactants.

Steps

Worked Example

Calculate ΔH° for this reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

Standard enthalpies of formation (in kJ/mol):
CH4(g) = −74.8; O2(g) = 0; CO2(g) = −393.5; H2O(l) = −285.8

  1. Products
    1 mol CO2: 1 × (−393.5) = −393.5
    2 mol H2O: 2 × (−285.8) = −571.6
    Total: −965.1 kJ
  2. Reactants
    1 mol CH4: 1 × (−74.8) = −74.8
    2 mol O2: 2 × 0 = 0
    Total: −74.8 kJ
  3. ΔH°reaction
    ΔH° = (−965.1) − (−74.8) = −890.3 kJ

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Matt’s exam tip

Always base your calculations on a balanced chemical equation and account for the mole ratios of all substances involved. For example, if the balanced equation shows 2 moles of H2O are formed, you must multiply the enthalpy of formation of H2O by 2 when calculating the total enthalpy change.

Hess’s Law for Enthalpy of Combustion

The enthalpy of combustion (ΔHc) is the enthalpy change when one mole of a substance is completely burned in oxygen.

Example: Using enthalpies of combustion to find enthalpy of formation of propane ( C3H8 ):

AQA A-Level Chemistry Hess cycle using enthalpies of combustion for carbon, hydrogen, and propane.

Given data:
ΔHc(C(s)) = −394 kJ mol−1
ΔHc(H2(g)) = −286 kJ mol−1
ΔHc(C3H8(g)) = −2220 kJ mol−1

AQA A-Level Chemistry Hess’s Law cycle showing two routes to form CO2 and H2O from C3H8 or from elements.

Using Hess’s Law:
ΔH (elements → CO2 + H2O) = −2326 kJ mol−1
ΔH (C3H8 → CO2 + H2O) = −2220 kJ mol−1
So: ΔHf(C3H8) = −2326 − (−2220) = −106 kJ mol−1

Common Exam Mistakes and How to Avoid Them

Mistake Why it Happens How to Avoid
Forgetting ΔHf of elements is 0 Students include values for O2(g), H2(g), C(s) Always remember: ΔHf = 0 for elements in standard states
Wrong direction in cycle Arrows misread in enthalpy diagram Follow the arrows carefully; keep consistent direction
Mixing up combustion vs formation Using ΔHc when ΔHf required (or vice versa) Underline the type of enthalpy in the question
Incorrect sign Not applying negative for exothermic reactions Check whether bonds broken (positive) or bonds formed (negative)

Summary