Aim for FiveChemistry

Unit 6 · Topic 6.9

6.9 Hess's Law

Hess's law says that if you can write a reaction as a series of steps, its ΔH is the sum of the steps' ΔH values. Reversing a step flips the sign of its ΔH and multiplying a step multiplies its ΔH, so you can combine known reactions to find the ΔH of one that is hard to measure.

Key terms

  • Hess's law
  • reversing a reaction
  • multiplying a reaction
  • adding reactions
  • first law of thermodynamics

The idea

Many changes can be broken into a series of steps, and each step has its own energy change. Because energy is conserved (the first law of thermodynamics), the total heat transferred by the whole sequence must equal the sum of the heat transferred in each step. At constant pressure, that means the ΔH of the overall process is the sum of the ΔH values of the steps.

Think of hiking from a valley to a ridge. Whether you take a gentle trail or a steep one, your net change in height is the same. Hess's law lets you pick whichever path has known values.

You won't be tested on the term 'state function', but you will be tested on using the rules below.

The three rules

  • Reverse a reaction: keep the size of ΔH and flip its sign. If A → B has ΔH = −50 kJ, then B → A has ΔH = +50 kJ.
  • Multiply a reaction by a factor: multiply ΔH by the same factor. Doubling every coefficient doubles ΔH; halving them halves ΔH.
  • Add reactions: add their ΔH values. Species that appear on both sides of the combined equation cancel.

A reliable strategy

  • Write the target equation at the top of your work.
  • For each substance in the target, find a given equation that contains it, ideally one where it appears only once.
  • Reverse that equation if the substance is on the wrong side, and multiply it so the coefficient matches the target.
  • Apply the same changes to each equation's ΔH as you go.
  • Add the equations, cancel anything that appears on both sides, and confirm the result is exactly the target. Then add the ΔH values.

ΔH rules versus K rules

In Unit 7 you'll combine equilibrium constants in a similar way, but the math is different. Students mix these up, so learn them side by side.

Change to the equationWhat happens to ΔHWhat happens to K
Reverse itFlip the sign: −ΔHTake the reciprocal: 1/K
Multiply coefficients by nMultiply: n × ΔHRaise to a power: Kⁿ
Add two equationsAdd: ΔH₁ + ΔH₂Multiply: K₁ × K₂

Picturing the steps

You can draw a Hess's law problem as an energy diagram with more than two levels. For carbon monoxide, start at C(s) + O₂(g). Burning carbon all the way to CO₂ drops the energy by 393.5 kJ. From CO₂, going 'back up' to CO + ½O₂ rises by 283.0 kJ. The net drop from the start to CO + ½O₂ is 110.5 kJ, which is the ΔH you calculated. Any route between the same starting and ending levels gives the same overall ΔH.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1

    Forming carbon monoxide

    Find ΔH for C(s) + ½O₂(g) → CO(g), given (1) C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ and (2) CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ.

    Show the solution
    1. Step 1: C(s) is a reactant in the target and in equation 1, so keep equation 1 as written: ΔH = −393.5 kJ.
    2. Step 2: CO(g) is a product in the target but a reactant in equation 2, so reverse equation 2: CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0 kJ.
    3. Step 3: Add: C(s) + O₂(g) + CO₂(g) → CO₂(g) + CO(g) + ½O₂(g). Cancel CO₂ from both sides and ½O₂ from both sides, leaving C(s) + ½O₂(g) → CO(g). That's the target.
    4. Step 4: ΔH = −393.5 + 283.0 = −110.5 kJ.

    Answer: ΔH = −110.5 kJ

  2. Example 2Calculator allowed

    Three equations, with reversing and multiplying

    Find ΔH for C(s) + 2H₂(g) → CH₄(g), given (1) C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ; (2) H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8 kJ; (3) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890.3 kJ.

    Show the solution
    1. Step 1: C(s): keep equation 1. ΔH = −393.5 kJ.
    2. Step 2: 2H₂(g): multiply equation 2 by 2, giving 2H₂(g) + O₂(g) → 2H₂O(l). ΔH = 2(−285.8) = −571.6 kJ.
    3. Step 3: CH₄(g) must be a product, so reverse equation 3: CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g). ΔH = +890.3 kJ.
    4. Step 4: Add the three. Reactants: C + O₂ + 2H₂ + O₂ + CO₂ + 2H₂O. Products: CO₂ + 2H₂O + CH₄ + 2O₂. CO₂, 2H₂O and 2O₂ cancel, leaving C(s) + 2H₂(g) → CH₄(g).
    5. Step 5: ΔH = −393.5 + (−571.6) + 890.3 = −74.8 kJ. This is the enthalpy of formation of methane.

    Answer: ΔH = −74.8 kJ

  3. Example 3

    Trap: reverse and scale together

    Given CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ, find ΔH for 2CO₂(g) → 2CO(g) + O₂(g).

    Show the solution
    1. Step 1: The target is the given equation reversed and then doubled.
    2. Step 2: Reversing flips the sign: +283.0 kJ.
    3. Step 3: Doubling multiplies by 2: 2 × (+283.0) = +566.0 kJ.
    4. Step 4: Students often do only one of the two changes and answer −566.0 kJ or +283.0 kJ. Apply every change you make to the equation to its ΔH too.

    Answer: ΔH = +566.0 kJ

Common mistakes

  • Reversing or multiplying an equation but forgetting to change its ΔH to match.
  • Leaving species uncanceled so the sum isn't actually the target equation. Always check the final equation before adding ΔH values.
  • Using K rules for ΔH: ΔH values are added and multiplied by factors, not multiplied together or raised to powers.
  • Getting the sign of a reversed step wrong when it was already positive. Reversing +50 kJ gives −50 kJ.

On the exam

  • Hess's law problems give two to four equations with ΔH values and a target. Show each manipulated equation and its new ΔH so a grader can follow your reasoning.
  • You may be asked to justify the method: total energy is conserved, so the enthalpy change for the overall process equals the sum of the enthalpy changes for the steps.

Connected topics

Videos

  • Hess's Law and ΔH - Let's Practice! AP Chemistry Unit 6, Topic 9

    Jeremy Krug (krugslist)Watch on YouTube (opens in a new tab)

  • Unit 6.9 - Hess's Law

    Abigail GiordanoWatch on YouTube (opens in a new tab)

  • Hess's law | Thermodynamics | AP Chemistry | Khan Academy

    Khan AcademyWatch on YouTube (opens in a new tab)

  • Hess's Law Problems & Enthalpy Change - Chemistry

    The Organic Chemistry TutorWatch on YouTube (opens in a new tab)

  • Hess's Law and Heats of Formation

    Professor Dave ExplainsWatch on YouTube (opens in a new tab)

  • Hess's Law Common Test Question

    Melissa MaribelWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 6.9 Hess's Law. Pick an answer to see if you got it, and why.

Reaction 1: C(s) + O₂(g) → CO₂(g) ΔH° = −393.5 kJ/mol

Reaction 2: CO(g) + ½O₂(g) → CO₂(g) ΔH° = −283.0 kJ/mol

Thermochemical data at 25 °C

Question 1 of 4Calculator allowed

Use the reactions above to find ΔH° for C(s) + ½O₂(g) → CO(g).

Question 2 of 4Calculator allowed

For the reaction 2H₂(g) + O₂(g) → 2H₂O(l), ΔH° = −571.6 kJ/mol. What is ΔH° for H₂O(l) → H₂(g) + ½O₂(g)?

Reaction 1: N₂(g) + O₂(g) → 2NO(g) ΔH° = +180.6 kJ/mol

Reaction 2: 2NO(g) + O₂(g) → 2NO₂(g) ΔH° = −114.1 kJ/mol

Thermochemical data at 25 °C

Question 3 of 4Calculator allowed

What is ΔH° for 2NO₂(g) → N₂(g) + 2O₂(g)?

Reaction 1: C(s) + O₂(g) → CO₂(g) ΔH° = −393.5 kJ/mol

Reaction 2: H₂(g) + ½O₂(g) → H₂O(l) ΔH° = −285.8 kJ/mol

Reaction 3: C₂H₆(g) + 7/2 O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH° = −1,560.7 kJ/mol

Thermochemical data at 25 °C

Question 4 of 4Calculator allowed

Use the reactions above to find ΔH° for 2C(s) + 3H₂(g) → C₂H₆(g).

0 of 4 answered