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Hess’ Law and Cycles

Learn and apply Hess’ Law and Cycles in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Hess’ Law and Cycles: Orientation

Hess questions are bookkeeping questions: the chemistry is in choosing the right route (formation vs combustion), and the marks are in clean sign/scale handling so your equations add to the target.

Keep Enthalpy Changes and Energy Profiles and the Energetics and Thermodynamics hub in view, because most questions mix definitions with cycle reasoning.

Definitions (Must Know)

A. Hess’ Law

The enthalpy change of a reaction is independent of route, because enthalpy is a state function.

B. Standard enthalpy change, Δ H⦵

A standard enthalpy change, Δ H⦵, is measured under standard conditions (typically 298 K, 100 kPa; solutions at 1.00 mol dm⁻³ when relevant).

C. Standard state

A standard state is the most stable physical form of an element under standard conditions (e.g. O₂(g), H₂(g), Cl₂(g), C(graphite)).

D. Standard enthalpy of formation, Δ H_f⦵

Δ H_f⦵ is the enthalpy change when 1 mol of a compound is formed from its elements in their standard states.

For elements in their standard states, Δ H_f⦵ = 0.

E. Standard enthalpy of combustion, Δ H_c⦵

Δ H_c⦵ is the enthalpy change when 1 mol of a substance is completely burned in oxygen under standard conditions, with products in their standard states.

Detailed Explanations

A. Hess cycle workflow (always works)

  1. Write the target equation clearly.
  2. Choose the route based on the data provided:
    • formation data → formation cycle (elements as the common baseline)
    • combustion data → combustion cycle (common products are CO₂ and H₂O)
  3. Write the equations with correct coefficients.
  4. Apply the sign + scaling rules so that the equations sum to the target.
  5. Write the final algebra line before substituting numbers.

Mini example (formation method idea):

  • If Δ H_f⦵ is given for every species, then “products − reactants” with coefficients gives Δ H_reaction⦵ directly.

B. Why “products − reactants” works for formation data

Because forming the reactants from elements and forming the products from elements are two alternative routes between the same initial and final states, Hess’ Law says the difference between those routes equals the reaction enthalpy.

Therefore: Δ H_reaction⦵ = ∑ ν Δ H_f⦵(products) - ∑ ν Δ H_f⦵(reactants)

C. Combustion cycles (the common trap)

In combustion cycles, everything is taken to the same “common products” (usually CO₂ and H₂O).

Because the cycle goes reactants → products either directly or via combustion, the reaction enthalpy becomes: Δ H_reaction⦵ = ∑ ν Δ H_c⦵(reactants) - ∑ ν Δ H_c⦵(products)

Worked Examples

Modelled example 1

Write a formation equation

Core

Problem

Write the standard formation equation for ethanol, C₂H₅OH(l).
Study the worked solution

  1. Choose the starting substances

    Method

    Use each element in its standard state.

    Reason

    A standard enthalpy of formation starts from the constituent elements under standard conditions.

    Working

    C(graphite), H₂(g) and O₂(g).

  2. Form exactly one mole

    Method

    Balance atoms while keeping one mole of ethanol as the sole product.

    Reason

    The definition is tied to formation of one mole, so fractional coefficients are allowed.

    Working

    2C(graphite) + 3H₂(g) + 1/2 O₂(g) → C₂H₅OH(l)

Guided practice 2

Use formation enthalpies

About 7 min

Problem

Calculate Δ H⦵ for CO(g) + 1/2O₂(g) → CO₂(g) using Δ H_f⦵(CO₂) = -394, Δ H_f⦵(CO) = -111, and Δ H_f⦵(O₂) = 0 kJ mol⁻¹.

Try this before viewing the solution

Hints

Hint 1: route relationship
For formation data, use products minus reactants.
Hint 2: substitute coefficients
Write (-394)-[(-111) + (1/2)(0)] before calculating.
View solution step by step

  1. Write the route relationship

    Method

    Sum coefficient-weighted formation values for products and subtract the reactant sum.

    Reason

    Both formation routes share the elements in their standard states as a common baseline.

    Working

    Δ H_reaction⦵ = ∑νΔ H_f⦵(products)-∑νΔ H_f⦵(reactants).

  2. Substitute and calculate

    Method

    Retain signs and the one-half coefficient.

    Reason

    The standard formation enthalpy of elemental oxygen is zero, but its term still shows the balanced route.

    Working

    Δ H⦵ = (-394)-[(-111) + (1/2)(0)] = -283 kJ mol⁻¹.

Common misconception 3

Transform an equation and its enthalpy together

Find and correct the mistake

Learner step

For H₂ + 1/2O₂ → H₂O(l), Δ H = -286 kJ mol⁻¹. A learner writes 2H₂O(l) → 2H₂ + O₂ but keeps Δ H = -286 kJ mol⁻¹. Diagnose the transformation.

Select the corrected value

Correct transformed ΔH

View solution step by step

  1. Reverse the equation

    Method

    Change the sign of the enthalpy.

    Reason

    The energy change for the reverse route is the negative of the forward route.

    Working

    H₂O(l) → H₂ + 1/2O₂ has Δ H = +286 kJ mol⁻¹.

  2. Scale the equation

    Method

    Multiply the enthalpy by the same factor as every coefficient.

    Reason

    Two reaction amounts require twice the energy change.

    Working

    2H₂O(l) → 2H₂ + O₂ has Δ H = +572 kJ mol⁻¹.

Examiner practice 4

Find an unknown formation enthalpy

4 marks

Problem

For CaCO₃(s) → CaO(s) + CO₂(g), Δ H⦵_reaction = +178 kJ mol⁻¹. Given Δ H_f⦵(CaO) = -635 and Δ H_f⦵(CO₂) = -394 kJ mol⁻¹, calculate Δ H_f⦵(CaCO₃). [4 marks]

Try this before viewing the solution

View solution step by step

  1. Write the relationship

    1 mark

    Method

    Place the unknown inside the reactant sum.

    Reason

    The target reaction has calcium carbonate as its only reactant.

    Working

    Δ H⦵_reaction = [Δ H_f⦵(CaO) + Δ H_f⦵(CO₂)]-Δ H_f⦵(CaCO₃).

  2. Substitute data

    1 mark

    Method

    Keep brackets around the product sum.

    Reason

    This preserves the products-minus-reactants structure.

    Working

    178 = [(-635) + (-394)]-Δ H_f⦵(CaCO₃).

  3. Collect known values

    1 mark

    Method

    Add the product values before rearranging.

    Reason

    Separating arithmetic from algebra reduces sign errors.

    Working

    178 = -1029-Δ H_f⦵(CaCO₃).

  4. Rearrange

    1 mark

    Method

    Isolate the unknown formation enthalpy.

    Reason

    The negative before the unknown must be retained through rearrangement.

    Working

    Δ H_f⦵(CaCO₃) = -1207 kJ mol⁻¹.

Challenge 5

Transfer to combustion data

Minimal support

Problem

Given Δ H_c⦵(C₂H₄) = -1411, Δ H_c⦵(H₂) = -286, and Δ H_c⦵(C₂H₆) = -1560 kJ mol⁻¹, calculate Δ H⦵ for C₂H₄ + H₂ → C₂H₆.

Try this before viewing the solution

Hints

Hint 1: common destination
Both sides combust completely to the same totals of carbon dioxide and liquid water.
Hint 2: combustion relationship
Use the reactant combustion total minus the product combustion total.
View solution step by step

  1. Form the reactant route total

    Method

    Add the combustion enthalpies of ethene and hydrogen.

    Reason

    Both reactants must reach the shared combustion products.

    Working

    -1411 + (-286) = -1697 kJ mol⁻¹.

  2. Subtract the product route

    Method

    Subtract the combustion enthalpy of ethane.

    Reason

    Reactants-to-common-products equals target reaction plus product-to-common-products.

    Working

    Δ H⦵ = -1697-(-1560) = -137 kJ mol⁻¹.

Mind Stretchers

Mind stretcher 1Extension

Use the data below to find Δ H for: C(s) + 2H₂(g) → CH₄(g)

Data:

  • C(s) + O₂(g) → CO₂(g) Δ H = -394 kJ mol⁻¹
  • 2H₂(g) + O₂(g) → 2H₂O(l) Δ H = -572 kJ mol⁻¹
  • CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) Δ H = -890 kJ mol⁻¹
Show Hint

Choose a common intermediate or use the products-minus-reactants formation relationship; preserve coefficients and signs.

Show Answer

Mark scheme (one valid route):

  • Add (1) and (2): C + 2H₂ + 2O₂ → CO₂ + 2H₂O Δ H = -394 + (-572) = -966
  • Reverse (3): CO₂ + 2H₂O → CH₄ + 2O₂ Δ H = +890
  • Add them and cancel CO₂ and 2H₂O to get C + 2H₂ → CH₄
  • Δ H = -966 + 890 = -76 kJ mol⁻¹

Mind stretcher 2: Combining formation and combustion evidenceExtension

Question. Given Δ H_f°(CO₂) = -394, Δ H_f°(H₂O(l)) = -286 and Δ H_f°(CH₃OH(l)) = -239 kJ mol⁻¹, calculate Δ H_c° for methanol.

Show Hint

Write the balanced combustion equation before using products minus reactants.

Show Answer

For CH₃OH(l) + 3/2O₂(g) → CO₂(g) + 2H₂O(l), Δ H_c° = [-394 + 2(-286)]-[-239] = -727 kJ mol⁻¹.