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Hess Law From Given Cycles

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

  • GCE A-Level H1 Chemistry 8873-2027
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H1 Hess’ Law from Given Simple Cycles: Orientation

Hess’ Law questions are route-accounting problems. H1 supplies the simple cycle: your task is to trace equivalent routes, preserve direction and coefficients, and isolate the unknown enthalpy change.

H1 8873 scope
  • 8873 requires calculations involving given simple cycles using formation, combustion and neutralisation terms.
  • Constructing an energy cycle, including a Born–Haber cycle, is not required.

Definitions (Must Know)

  • Hess’ Law states that the enthalpy change of a reaction depends only on the initial and final states, not on the route taken.
  • A standard enthalpy change of formation, Δ H_f°, forms one mole of a compound from its elements in their standard states.
  • A standard enthalpy change of combustion, Δ H_c°, is the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions.
  • A standard enthalpy change of neutralisation is the enthalpy change when an acid and base form one mole of water under standard conditions.

Detailed Explanations

A. Route-equation method

Write an equation for each complete route between the same starting and finishing states. Hess’ Law makes the route totals equal, so solve the resulting algebraic equation.

B. Formation-data method

Formation values share the elements in their standard states as a common origin:

Δ H°_reaction = ∑ νΔ H_f°(products)-∑ νΔ H_f°(reactants)

C. Combustion-data method

When reactants and products combust to the same final substances, the combustion route from reactants equals the direct reaction plus the combustion route from products. This gives reactant combustion totals minus product combustion totals.

D. Why H1 does not require cycle construction

The supplied cycle is evidence, not a drawing prompt. Annotate and use it accurately; do not add atomisation, ionisation or electron-affinity steps that the question does not supply.

Worked Examples

Modelled example 1

Formation values

Core

Problem

Calculate Δ H° for CH₄ + 2O₂ → CO₂ + 2H₂O(l) using Δ H_f°(CH₄) = -75, Δ H_f°(CO₂) = -394, Δ H_f°(H₂O(l)) = -286, and Δ H_f°(O₂) = 0 kJ mol⁻¹.
Study the worked solution

  1. Build the product total

    Method

    Multiply each supplied formation value by its equation coefficient.

    Reason

    Two moles of liquid water are formed for each reaction amount.

    Working

    ∑Δ H_f°(products) = -394 + 2(-286) = -966 kJ mol⁻¹.

  2. Subtract the reactant total

    Method

    Subtract the methane and oxygen formation terms.

    Reason

    Formation values share the standard elements as a common origin.

    Working

    Δ H° = -966-[-75 + 2(0)] = -891 kJ mol⁻¹.

Guided practice 2

A supplied alternative route

About 6 min

Problem

A direct route has enthalpy change x. The supplied alternative route between the same starting and finishing states has consecutive steps of -394 and + 283 kJ mol⁻¹. Find x.

Try this before viewing the solution

Hints

Hint 1: same endpoints
Hess’ Law makes the total enthalpy change independent of route.
Hint 2: signed sum
Add the two supplied alternative-route values with their signs.
View solution step by step

  1. Equate the routes

    Method

    Set the direct-route value equal to the complete alternative-route total.

    Reason

    Both routes connect identical initial and final states.

    Working

    x = (-394) + (+283).

  2. Add signed values

    Method

    Preserve each supplied direction and sign.

    Reason

    No sign is changed unless the corresponding chemical step is reversed.

    Working

    x = -111 kJ mol⁻¹.

Common misconception 3

Separate arrow geometry from reaction direction

Find and correct the mistake

Learner claim

A supplied diagram includes the combustion of hydrogen, tabulated as H₂ + 1/2O₂ → H₂O(l), Δ H_c° = -286 kJ mol⁻¹. A learner uses + 286 solely because the route arrow is drawn upward. Identify and correct the error.

Choose the sign rule

The sign changes when

View solution step by step

  1. Anchor the tabulated direction

    Method

    Match the value to the combustion equation as written.

    Reason

    Combustion of hydrogen to liquid water is the defined forward chemical change.

    Working

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

  2. Decide whether to reverse

    Method

    Change to + 286 only if the chemical equation is followed from water back to hydrogen and oxygen.

    Reason

    Reversing initial and final chemical states reverses Δ H; page geometry does not.

    Working

    Upward alone is insufficient evidence for a sign change.

Examiner practice 4

Find an unknown from supplied formation data

4 marks

Problem

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

Try this before viewing the solution

View solution step by step

  1. Use the supplied formation relationship

    1 mark

    Method

    Write products minus reactants with the unknown in the reactant term.

    Reason

    The question supplies all the formation values needed, so you can use them directly without constructing another cycle.

    Working

    178 = [Δ H_f°(CaO) + Δ H_f°(CO₂)]-Δ H_f°(CaCO₃).

  2. Substitute

    1 mark

    Method

    Keep the product values in a single bracket.

    Reason

    The bracket makes the route structure and signs visible.

    Working

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

  3. Simplify

    1 mark

    Method

    Add the known product values.

    Reason

    Completing the arithmetic before rearrangement protects the negative unknown term.

    Working

    178 = -1029-Δ H_f°(CaCO₃).

  4. Rearrange

    1 mark

    Method

    Isolate the unknown and include the molar unit.

    Reason

    The enthalpy belongs to formation of one mole of calcium carbonate.

    Working

    Δ H_f°(CaCO₃) = -1207 kJ mol⁻¹.

Challenge 5

Reverse a formation-data result

Minimal support

Problem

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

Try this before viewing the solution

Hints

Hint 1: use the written direction
The products in this question are carbon monoxide and oxygen.
Hint 2: direction check
The forward oxidation CO + 1/2O₂ → CO₂ is -283 kJ mol⁻¹, so reversing it must reverse the sign.
View solution step by step

  1. Use products minus reactants

    Method

    Place the carbon monoxide and oxygen terms in the product sum.

    Reason

    The relationship follows the target equation exactly as written.

    Working

    Δ H° = [(-111) + (1/2)(0)]-[-394].

  2. Calculate and check direction

    Method

    Evaluate the signed difference.

    Reason

    Decomposing carbon dioxide here is the reverse of the exothermic carbon monoxide oxidation.

    Working

    Δ H° = +283 kJ mol⁻¹.

Mind Stretchers

Attempt each unfamiliar application before opening the hint, then compare your reasoning chain with the solution.

Mind stretcher 1: Combustion-cycle deductionExtension

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

Show Hint

Both sides combust to the same carbon dioxide and water products.

Show Answer

Δ H° = [-1411 + (-286)]-[-1560] = -137 kJ mol⁻¹.

Mind stretcher 2: Correcting a sign errorExtension

Question. A student uses + 286 kJ mol⁻¹ for combustion of hydrogen because the cycle arrow is followed upward. Identify and correct the error.

Show Hint

Separate the tabulated reaction direction from the direction followed on the supplied cycle.

Show Answer

The tabulated combustion value is -286 kJ mol⁻¹ for H₂ + 1/2O₂ → H₂O(l). It becomes + 286 only when that chemical equation is reversed. Arrow geometry alone does not decide the sign.

Mind stretcher 3: A formation value that cannot be measured directlyExtension

Question. Carbon and hydrogen do not react cleanly to form only methane, so Δ H_f°(CH₄) cannot be measured directly. Use Δ H_c°(C) = -394, Δ H_c°(H₂) = -286 and Δ H_c°(CH₄) = -890 kJ mol⁻¹ to find it for C(s) + 2H₂(g) → CH₄(g).

Show Hint

Burning the elements and burning methane give the same set of products, CO₂(g) and H₂O(l). Keep the coefficient 2 attached to hydrogen.

Show Answer

Route through the combustion products: Δ H_f°(CH₄) = [Δ H_c°(C) + 2Δ H_c°(H₂)]-Δ H_c°(CH₄) = [-394 + 2(-286)]-(-890) = -76 kJ mol⁻¹.