Hess's law — energy cycles — Chemistry

Before this

This leads to

Bond enthalpies
Calorimetry
The Born–Haber cycle · coming soon

A heat that nobody can measure directly

When carbon burns in plenty of oxygen, it forms carbon dioxide. With less oxygen, some carbon monoxide forms as well. Now suppose you want the enthalpy change for making carbon monoxide alone:

C(s, graphite) + ½O₂(g) \to CO(g) ΔH = ?

You cannot simply burn carbon in a measured amount of oxygen and read the thermometer. Some of the carbon always goes on to carbon dioxide, and some stays unburned. The heat you measure belongs to a mixture of reactions. There is no clean experiment for this one.

Yet every data book gives a value: −110.5 kJ/mol. Nobody measured it directly. It was calculated, from two reactions that can be measured cleanly. The last lecture promised that a table of "sea-level heights" would give ΔH for reactions nobody has ever carried out. This lecture keeps that promise. The tool is Hess's law.

Two routes, one answer

Start with a prediction. Carbon can reach carbon dioxide in two ways.

Route 1: burn the carbon completely, in one step, straight to CO₂.
Route 2: first make CO, then burn the CO to CO₂.

Predict, then reveal — which route gives out more heat?

1 mol of graphite ends up as 1 mol of CO₂ by route 1 (one step) or by route 2 (through CO). Which route releases more heat in total?

Both routes release exactly the same total heat, 393.5 kJ per mole of carbon, because they start in the same state and end in the same state. Route 2 releases 110.5 kJ in its first step and 283.0 kJ in its second, which adds up to 393.5 kJ.

The two routes must give the same total. Here is why.

Suppose route 2 released more heat than route 1, say 10 kJ more. Then you could run route 2 forwards and route 1 backwards, turning the CO₂ back into carbon and oxygen. Going backwards takes in exactly what going forwards gave out. So each time round the loop you would end with the same chemicals you started with, plus 10 kJ of extra heat, made from nothing. You could repeat it for ever. That would break the conservation of energy, which no experiment has ever found broken. So the two totals must be equal.

Think of enthalpy as height above sea level. Two climbers start at the same village and finish at the same hut. One takes the steep path, the other goes the long way round, over a ridge and down again. They walked different distances, but they gained exactly the same height. Height depends only on where you are, not on how you got there. Enthalpy is the same.

That is it. That is the whole idea.

Hess's law

The total enthalpy change for a reaction is the same, whatever route is taken, provided the starting and finishing states are the same.

So if a reaction can be written as a sum of steps, its ΔH is the sum of the ΔH values of the steps.

Germain Hess, a chemist born in Geneva in 1802 who worked in St Petersburg, published this result in 1840, from careful measurements of heats of reaction. At that time the law of conservation of energy had not yet been stated in general form. Hess's law turned out to be one special case of it.

Building a cycle

A Hess cycle is a diagram with two routes between the same start and finish. One of the routes contains the reaction you want. Every other arrow is a reaction you know.

A Hess cycle for making carbon monoxide. Burning graphite straight to CO₂ (left arrow) and going through CO (top arrow, then right arrow) start and finish in the same place, so the two routes have the same total ΔH. The ½O₂ is carried along the top and used up on the right.

Now walk around the cycle. To get from the top left to the top right without using the unknown arrow, go down the left arrow, then up the right arrow.

Down the left arrow, in its own direction: −393.5 kJ/mol.
Up the right arrow, against its direction: the reverse of burning CO, so the sign changes. +283.0 kJ/mol.

ΔH = -393.5 + 283.0 = -110.5 kJ/mol

This uses the two rules from the first lecture on enthalpy. Reversing a reaction changes the sign of ΔH, because energy given out going forwards is taken in going backwards. And doubling a reaction doubles ΔH, because twice the chemicals react. In a cycle, the first rule becomes a simple habit: with an arrow, keep its sign; against an arrow, change its sign.

Check with route 2 from the prediction. Making CO gives out 110.5 kJ. Burning it gives out 283.0 kJ. In total: 110.5 + 283.0 = 393.5 kJ, exactly as route 1. ✓

Stop and think

If ΔH depends only on the start and the finish, does the route make any difference at all?

Show the answer

Yes, a great deal, but not to ΔH. The route decides how fast the reaction goes, whether it happens at all at room temperature, and how hot things get at each stage. Graphite does not turn into diamond on a shelf, even though ΔH for the change is tiny. Hess's law says only one thing: the total enthalpy change is fixed by where you start and where you finish.

Formation cycles: back to sea level

The last lecture gave every compound a "height": its standard enthalpy of formation, ΔH_f^⦵, measured from the elements in their standard states at zero. That table makes a ready-made cycle for any reaction.

Put the elements at the bottom. Both the reactants and the products can be made from those same elements. So there are two routes from the elements to the products: straight there, or through the reactants.

Elements → products: ΣΔH_f^⦵(products).
Elements → reactants → products: ΣΔH_f^⦵(reactants) + ΔH(reaction).

By Hess's law the two are equal, and rearranging gives:

ΔH from formation enthalpies

ΔH(reaction) = ΣΔH_f^⦵(products) − ΣΔH_f^⦵(reactants)

Each ΔH_f^⦵ is multiplied by its coefficient in the balanced equation. Elements in their standard states count as zero.

Do not learn this as "products minus reactants". Learn the picture. It is the height of the finish minus the height of the start, both measured from sea level.

The burning of methane on an enthalpy "map". The elements are at zero. Methane and oxygen sit 74.8 kJ/mol below them, and carbon dioxide and water sit 965.1 kJ/mol below them. The reaction drops from one level to the other: −965.1 − (−74.8) = −890.3 kJ/mol.

Worked Example 1 — The combustion of methane from formation enthalpies

Find ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), using ΔH_f^⦵ values (kJ/mol): CH₄(g) −74.8, CO₂(g) −393.5, H₂O(l) −285.8.

Products: −393.5 + 2 × (−285.8) = −393.5 − 571.6 = −965.1 kJ/mol.

Reactants: −74.8 + 2 × 0 = −74.8 kJ/mol. Oxygen is an element in its standard state, so it counts as zero.

ΔH = -965.1 - (-74.8) = -890.3 kJ/mol

Check: the standard enthalpy of combustion of methane, from the first lecture on enthalpy, is −890 kJ/mol. ✓

The same method explains the puzzle from the start of the last lecture. Burning hydrogen gives −285.8 kJ/mol if the water is liquid, and −241.8 kJ/mol if it is steam. The difference is the reaction H₂O(l) → H₂O(g). Its ΔH is the height of steam minus the height of liquid water: −241.8 − (−285.8) = +44.0 kJ/mol. That is the energy needed to turn 1 mol of liquid water into steam at 298 K. Two burning experiments tell you something about boiling.

Combustion cycles: everything burns to the same place

For most organic compounds, the formation reaction cannot be carried out, but the compound can be burned cleanly and the heat measured. So the other common cycle puts the combustion products at the bottom. Reactants and products both burn to the same CO₂ and H₂O.

Now the arrows point down, from each substance to its combustion products. To go from reactants to products the long way, go down from the reactants (with the arrows), then up to the products (against the arrows, so change the sign):

ΔH from combustion enthalpies

ΔH(reaction) = ΣΔH_c^⦵(reactants) − ΣΔH_c^⦵(products)

The order is the opposite of the formation formula. This is not a new rule to memorise. It comes from the direction of the arrows. Formation arrows point up from the elements to the substances, so the products' arrow is followed and the reactants' arrow is reversed. Combustion arrows point down from the substances, so the reactants' arrow is followed and the products' arrow is reversed. If you ever forget which way round, draw the arrows.

A combustion cycle for adding hydrogen to ethene. Both sides burn in oxygen (not shown) to the same products. Going down the left (−1696.8 kJ/mol) and up the right (+1561 kJ/mol) gives the unknown top arrow.

The commonest mistake is to use "products minus reactants" with combustion data. Find it below.

Spot the error — adding hydrogen to ethene

A student finds ΔH for C₂H₄(g) + H₂(g) → C₂H₆(g) from combustion data. Tap the line that contains the mistake.

The working below uses the formation formula with combustion data, so every sign comes out reversed. The mistake is in line 3.

Data, ΔH_c^⦵ in kJ/mol: C₂H₄(g) −1411, H₂(g) −285.8, C₂H₆(g) −1561.
Both sides burn to the same products: 2CO₂(g) + 3H₂O(l).
ΔH = ΣΔH_c^⦵(products) − ΣΔH_c^⦵(reactants). (Wrong: with combustion data it is reactants minus products.)
ΔH = −1561 − (−1411 − 285.8)
ΔH = +135.8 kJ/mol, so adding hydrogen to ethene is endothermic.

The correct answer is ΔH = (−1411 − 285.8) − (−1561) = −135.8 kJ/mol. The reaction is exothermic.

More worked examples

Worked Example 2 — Adding hydrogen to ethene

Find ΔH for C₂H₄(g) + H₂(g) → C₂H₆(g), using ΔH_c^⦵ values (kJ/mol): C₂H₄(g) −1411, H₂(g) −285.8, C₂H₆(g) −1561.

Reactants burn: −1411 + (−285.8) = −1696.8 kJ/mol. Products burn: −1561 kJ/mol.

ΔH = -1696.8 - (-1561) = -135.8 kJ/mol

Check the balance: on the left, 2 C and 6 H burn to 2CO₂ and 3H₂O. On the right, ethane also has 2 C and 6 H, and burns to the same 2CO₂ + 3H₂O. The bottom of the cycle really is the same place. ✓

Worked Example 3 — Turning graphite into diamond

Graphite does not turn into diamond at room temperature, so ΔH for C(s, graphite) → C(s, diamond) cannot be measured directly. Both forms of carbon burn to CO₂: ΔH_c^⦵ is −393.5 kJ/mol for graphite and −395.4 kJ/mol for diamond. Find ΔH.

Use the combustion cycle. Down from graphite: −393.5. Up from CO₂ to diamond, against the arrow: +395.4.

ΔH = -393.5 + 395.4 = +1.9 kJ/mol

Check: this is the formation value of diamond from the last lecture, +1.9 kJ/mol. That makes sense: forming diamond from graphite is forming diamond from the element carbon in its standard state. ✓ Diamond sits slightly higher than graphite. It still does not change into graphite on a shelf, because the route is extremely slow. That is the Stop-and-think in action.

Three classic mistakes

Trap 1 — Forgetting the coefficients

Every ΔH_f^⦵ or ΔH_c^⦵ must be multiplied by its coefficient in the balanced equation. In Worked Example 1, water appears as 2H₂O(l), so its value is counted twice: 2 × (−285.8). Leaving out the 2 gives −604.5 instead of −890.3.

Trap 2 — The formation formula with combustion data

"Products minus reactants" is only for formation enthalpies. With combustion enthalpies it is reactants minus products. Using the wrong one reverses the sign of the answer. If you are unsure, draw the cycle and follow the arrows.

Trap 3 — Mixing up the state of water

H₂O(l) and H₂O(g) have different formation values: −285.8 and −241.8 kJ/mol. A cycle closes only if the bottom is exactly the same on both routes. Read the state symbol in the equation, and use the matching value.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Find ΔH for C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), using ΔH_f^⦵ (kJ/mol): C₂H₅OH(l) −277.7, CO₂(g) −393.5, H₂O(l) −285.8.

Hint 1
Products minus reactants, each times its coefficient. O₂ counts as zero.

Hint 2
Products: 2 × (−393.5) + 3 × (−285.8).

Full answer
Products: −787.0 − 857.4 = −1644.4 kJ/mol. Reactants: −277.7 kJ/mol. ΔH = −1644.4 − (−277.7) = −1366.7 kJ/mol. Check: the data-book combustion value for ethanol from the last lecture is −1367 kJ/mol. ✓
Ethene cannot be made cleanly from carbon and hydrogen. Find its ΔH_f^⦵ from combustion data (kJ/mol): C(s, graphite) −393.5, H₂(g) −285.8, C₂H₄(g) −1411.

Hint 1
The target is 2C(s, graphite) + 2H₂(g) → C₂H₄(g). Everything burns to 2CO₂ + 2H₂O.

Hint 2
Reactants minus products, with combustion data.

Full answer
Reactants burn: 2 × (−393.5) + 2 × (−285.8) = −787.0 − 571.6 = −1358.6 kJ/mol. Product burns: −1411 kJ/mol. ΔH = −1358.6 − (−1411) = +52.4 kJ/mol. Ethene sits above "sea level": its formation is endothermic. Check: going down from the elements (−1358.6), then up from the products to ethene (+1411), gives +52.4. ✓
Propane burns with ΔH_c^⦵ = −2220 kJ/mol. Using the values in question 2, find ΔH_f^⦵ of propane, C₃H₈(g).

Hint 1
Write the formation equation: 3C(s, graphite) + 4H₂(g) → C₃H₈(g).

Hint 2
Elements burn: 3 × (−393.5) + 4 × (−285.8).

Full answer
Elements burn: −1180.5 − 1143.2 = −2323.7 kJ/mol. Propane burns: −2220 kJ/mol. ΔH = −2323.7 − (−2220) = −103.7 kJ/mol. Check the balance at the bottom of the cycle: 3C → 3CO₂, 8H → 4H₂O on both sides. ✓
Find ΔH_f^⦵ of ethane from its combustion value (−1561 kJ/mol) and the values in question 2. Then use your formation values for ethene (question 2) and ethane to find ΔH for C₂H₄(g) + H₂(g) → C₂H₆(g). Compare with Worked Example 2, and explain why the two answers must agree.

Hint 1
Ethane formation: 2C(s, graphite) + 3H₂(g) → C₂H₆(g).

Hint 2
Then use products minus reactants, with H₂ as zero.

Full answer
Elements burn: 2 × (−393.5) + 3 × (−285.8) = −787.0 − 857.4 = −1644.4 kJ/mol. So ΔH_f^⦵(ethane) = −1644.4 − (−1561) = −83.4 kJ/mol. Then ΔH = −83.4 − (52.4 + 0) = −135.8 kJ/mol, the same as Worked Example 2. They must agree, because both are routes between the same start (ethene and hydrogen) and the same finish (ethane). The formation route goes via the elements, and the combustion route goes via CO₂ and H₂O, but Hess's law says the route cannot change the total. ✓

The bigger idea

Hess's law works because enthalpy depends only on the state a system is in, not on its history. Quantities like this are called state functions, and they are what make chemistry's tables possible. A few hundred measured values can give ΔH for millions of reactions, including dangerous ones, slow ones, and ones that have never been carried out.

The same idea of building a cycle returns again and again. The next lecture estimates ΔH by breaking and making bonds, which is another route between the same start and finish. At Pre-University, the Born–Haber cycle uses Hess's law to find lattice energies that no experiment can measure directly.

Key concepts in this lesson

Hess's law: the total enthalpy change is the same whatever route is taken, provided the starting and finishing states are the same.
The law follows from conservation of energy: if two routes differed, a loop could make energy from nothing.
In a Hess cycle, keep the sign of ΔH when you follow an arrow, and change it when you go against an arrow.
With formation enthalpies, ΔH = ΣΔH_f^⦵(products) − ΣΔH_f^⦵(reactants), because formation arrows point up from the elements.
With combustion enthalpies, ΔH = ΣΔH_c^⦵(reactants) − ΣΔH_c^⦵(products), because combustion arrows point down to the same products.