D. Energy Changes and Rates of Reaction · Energy Changes and Rates of Reaction

Hess's Law: Adding Thermochemical Equations

D3.4 D2.5 A2.2About 70 minutes

Learning goals

I can state Hess's law and explain it using the idea of a state function.
I can reverse and multiply thermochemical equations and add them to match a target equation.
I can calculate the enthalpy change of a reaction from two or three given equations.
I can explain why Hess's law lets chemists find ΔH for reactions that are hard to measure.

Words to know

Hess's law
The enthalpy change of a reaction is the same whether it occurs in one step or several; the ΔH values of steps that add to the overall reaction add to its ΔH.

state function
A property, like enthalpy or elevation, whose change depends only on the initial and final states, not the path.

target equation
The equation whose ΔH you want to find in a Hess's law problem.

additivity of reaction enthalpies
Another name for Hess's law: enthalpy changes can be added like the equations they belong to.

Measuring the unmeasurable

Carbon burning in a limited supply of oxygen makes carbon monoxide: C(s) + ½O₂(g) → CO(g). Chemists would love to measure ΔH for this reaction directly, but they cannot: whenever carbon burns, some CO₂ always forms too. So how do tables list ΔH = −110.5 kJ for it? The answer is a law discovered in 1840 by Germain Henri Hess, a Swiss-born chemist working in St. Petersburg, Russia. It lets us calculate the enthalpy change of reactions we cannot measure, from reactions we can.

Hess's law

Hess's law (the law of additivity of reaction enthalpies): the enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps. If two or more thermochemical equations add up to give a target equation, their ΔH values add up to give the target ΔH.

Why it works: enthalpy is a state function. Its change depends only on the starting state (the reactants) and the final state (the products), not on the path between them. Think of elevation on a hike: whether you take the steep route over the ridge or the gentle trail, your change in elevation from the parking lot to the lookout is the same.

Two routes from the same start to the same end. Route A goes up and down, route B is gentle (drawn to scale), but the overall change in elevation is +350 m either way. Enthalpy behaves the same way: ΔH depends only on the start and end states.

Carbon can burn to CO₂ in one step (−393.5 kJ) or in two steps through CO (−110.5 kJ, then −283.0 kJ). Both paths give the same total, so ΔH for the hard-to-measure first step can be calculated. (Drawn to scale.)

How to solve a Hess's law problem

Write the target equation.
Look at each given equation and find a substance in it that appears only once among the given equations. Arrange that equation so the substance is on the same side as in the target, with the same coefficient.

If it is on the wrong side, reverse the equation and change the sign of ΔH.
If its coefficient is wrong, multiply the whole equation (and ΔH) by the needed factor.

Add the arranged equations. Cancel any substance that appears on both sides (in the same state). You should be left with exactly the target equation.
Add the arranged ΔH values.

Check your work: every substance that does not appear in the target must cancel completely.

Worked example

Example 1: Carbon monoxide

Find ΔH for C(s) + ½O₂(g) → CO(g), given:

(1) C(s) + O₂(g) → CO₂(g) ΔH₁ = −393.5 kJ

(2) CO(g) + ½O₂(g) → CO₂(g) ΔH₂ = −283.0 kJ

Target has C(s) as a reactant (1 mol). Equation (1) has C(s) as a reactant: keep it. ΔH = −393.5 kJ.
Target has CO(g) as a product. In equation (2) CO is a reactant: reverse it: CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0 kJ.
Add: C(s) + O₂(g) + CO₂(g) → CO₂(g) + CO(g) + ½O₂(g). Cancel CO₂(g) from both sides and ½O₂ from the right (leaving ½O₂ on the left): C(s) + ½O₂(g) → CO(g). ✓
Add the ΔH values: −393.5 + 283.0 = −110.5 kJ.

ΔH = −110.5 kJ

Worked example

Example 2: Making acetylene (three equations)

Find ΔH for the formation of acetylene (ethyne), 2C(s) + H₂(g) → C₂H₂(g), given:

(1) C₂H₂(g) + 5/2O₂(g) → 2CO₂(g) + H₂O(l) ΔH₁ = −1299.5 kJ

(2) C(s) + O₂(g) → CO₂(g) ΔH₂ = −393.5 kJ

(3) H₂(g) + ½O₂(g) → H₂O(l) ΔH₃ = −285.8 kJ

C₂H₂ is a product in the target but a reactant in (1): reverse (1): 2CO₂(g) + H₂O(l) → C₂H₂(g) + 5/2O₂(g), ΔH = +1299.5 kJ.
The target needs 2C(s): multiply (2) by 2: 2C(s) + 2O₂(g) → 2CO₂(g), ΔH = −787.0 kJ.
The target needs 1 H₂(g) as a reactant: keep (3): ΔH = −285.8 kJ.
Add and cancel: 2CO₂ cancels; H₂O(l) cancels; O₂: 2 + ½ = 5/2 on the left cancels 5/2 on the right. Left: 2C(s) + H₂(g) → C₂H₂(g). ✓
ΔH = +1299.5 + (−787.0) + (−285.8) = +226.7 kJ.

ΔH = +226.7 kJ. Acetylene is formed endothermically, which is one reason it releases so much energy (and such a hot flame) when it burns in a welding torch.

Worked example

Example 3: Hydrogenating ethene

Use combustion data to find ΔH for C₂H₄(g) + H₂(g) → C₂H₆(g):

(1) C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l) ΔH₁ = −1411.0 kJ

(2) H₂(g) + ½O₂(g) → H₂O(l) ΔH₂ = −285.8 kJ

(3) C₂H₆(g) + 7/2O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH₃ = −1559.7 kJ

Keep (1) (C₂H₄ is a reactant): −1411.0 kJ.
Keep (2) (H₂ is a reactant): −285.8 kJ.
Reverse (3) (C₂H₆ must be a product): 2CO₂(g) + 3H₂O(l) → C₂H₆(g) + 7/2O₂(g), +1559.7 kJ.
Cancel: 2CO₂; 3H₂O (2 + 1 on the left, 3 on the right); O₂ (3 + ½ = 7/2 on each side). ✓
ΔH = −1411.0 − 285.8 + 1559.7 = −137.1 kJ.

ΔH = −137.1 kJ. This trick, using heats of combustion that are easy to measure, is how many reaction enthalpies were first found.

Hess's law and the mole

Coefficients in thermochemical equations mean moles, so fractions like 5/2O₂ are perfectly fine. When you multiply an equation by 2 to clear fractions, remember to multiply its ΔH by 2 as well.

Hands-on activity

Equation-card puzzle

You need: six index cards (or squares of paper), a pen, and a calculator.

Write each of these on its own card, with ΔH on the back:

Card A: S(s) + O₂(g) → SO₂(g), ΔH = −296.8 kJ
Card B: 2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = −197.8 kJ
Card C: C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ
Card D: CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ
Card E: H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8 kJ
Card F: H₂O(l) → H₂O(g), ΔH = +44.0 kJ

Rules: you may turn a card upside down to reverse it (write the new sign on a sticky note), or put "× 2" or "× ½" beside it.
Build each target from the cards: (a) S(s) + 3/2O₂(g) → SO₃(g); (b) C(s) + ½O₂(g) → CO(g); (c) H₂(g) + ½O₂(g) → H₂O(g).
Ask someone at home to give you a target from the cards and race the clock.

Answers: (a) A + ½B: −296.8 + (−98.9) = −395.7 kJ. (b) C + reversed D: −393.5 + 283.0 = −110.5 kJ. (c) E + F: −285.8 + 44.0 = −241.8 kJ.

Just for fun

Germain Hess published his law in 1840. That was a few years before scientists such as Mayer, Joule and Helmholtz clearly stated the law of conservation of energy, which is the deeper reason his law works. Hess's law is, in a sense, conservation of energy wearing a chemistry lab coat.

Your turn

State Hess's law in your own words.

Find ΔH for S(s) + 3/2O₂(g) → SO₃(g) from: S(s) + O₂(g) → SO₂(g), ΔH = −296.8 kJ; 2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = −197.8 kJ.

Find ΔH for N₂(g) + 2O₂(g) → 2NO₂(g) from: N₂(g) + O₂(g) → 2NO(g), ΔH = +180.6 kJ; 2NO(g) + O₂(g) → 2NO₂(g), ΔH = −114.2 kJ.

Find ΔH for C(graphite) → C(diamond) from: C(graphite) + O₂(g) → CO₂(g), ΔH = −393.5 kJ; C(diamond) + O₂(g) → CO₂(g), ΔH = −395.4 kJ.

Find ΔH for H₂O(l) → H₂O(g) from: H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8 kJ; H₂(g) + ½O₂(g) → H₂O(g), ΔH = −241.8 kJ.

Why must every substance not in the target equation cancel completely?

More practice: warm-up

If you reverse an equation, what happens to ΔH?

If you multiply an equation by 3, what happens to ΔH?

What is a state function?

Who stated the law of additivity of reaction heats, and in what year?

Two steps have ΔH = −50 kJ and +20 kJ. What is ΔH for the overall reaction?

More practice: on your own

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

Find ΔH for C₂H₄(g) + H₂O(l) → C₂H₅OH(l) from: C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l), −1411.0 kJ; C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), −1366.7 kJ.

Find ΔH for 2C(s) + 2H₂(g) → C₂H₄(g) using C(s) + O₂(g) → CO₂(g), −393.5 kJ; H₂(g) + ½O₂(g) → H₂O(l), −285.8 kJ; C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l), −1411.0 kJ.

Given A → 2B, ΔH = +40 kJ and B → C, ΔH = −30 kJ, find ΔH for A → 2C.

Find ΔH for 2NO₂(g) → N₂O₄(g) from: N₂(g) + 2O₂(g) → 2NO₂(g), +66.4 kJ; N₂(g) + 2O₂(g) → N₂O₄(g), +11.1 kJ.

Challenge

Find ΔH for NO(g) + O(g) → NO₂(g) from: (1) NO(g) + O₃(g) → NO₂(g) + O₂(g), −198.9 kJ; (2) O₃(g) → 3/2O₂(g), −142.3 kJ; (3) O₂(g) → 2O(g), +495.0 kJ.

Diamond → graphite has ΔH = −1.9 kJ, yet diamonds do not turn into graphite in a jewellery box. What does this tell you about what Hess's law can and cannot predict?

Find ΔH for 2Fe(s) + 3/2O₂(g) → Fe₂O₃(s) from: 2Al(s) + 3/2O₂(g) → Al₂O₃(s), −1675.7 kJ; 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s), −851.5 kJ.

Check your understanding

Quiz

Hess's law works because enthalpy is

Given A → B, ΔH = −100 kJ and C → B, ΔH = −40 kJ, what is ΔH for A → C?

To use the equation 2H₂O(l) → 2H₂(g) + O₂(g), ΔH = +571.6 kJ in a target that needs H₂(g) + ½O₂(g) → H₂O(l), you must

Given C(s) + O₂(g) → CO₂(g), −393.5 kJ and CO(g) + ½O₂(g) → CO₂(g), −283.0 kJ, ΔH for C(s) + ½O₂(g) → CO(g) is

When added equations are correct for a Hess's law problem, which statement is true?

Hess's law is most useful for finding ΔH of a reaction that

For parents and teachers

Big ideas (D3.4, D2.5): Hess's law says the enthalpy change of a reaction is the same whether it happens in one step or many, because enthalpy is a state function. By reversing (change the sign) and multiplying (multiply ΔH) known equations so that they add up to a target, we can calculate ΔH for reactions we cannot measure.

How to help: Use the equation-card puzzle as a game. Ask your teen to explain each move aloud: "I reversed this because CO has to be a product, so the sign changes." Check the final sum by listing what cancels.

Common difficulties:

Reversing an equation but forgetting to change the sign of ΔH, or multiplying the equation but not ΔH.
Starting with a substance that appears in several given equations (start with one that appears only once).
Not checking that every unwanted substance cancels, including O₂ with fractional coefficients.
Thinking Hess's law says anything about the speed of a reaction (it does not).

Activity: no safety concerns.