What is enthalpy — heat at constant pressure — Chemistry

Before this

Lattice structures
Combustion · coming soon
Kinetic energy

This leads to

A hot pack, a cold pack, and a sign that looks backwards

Some first-aid kits contain two kinds of pack. Squeeze one, and it gets warm. Squeeze the other, and it gets cold, cold enough to calm a twisted ankle. Inside each pack, a chemical change happens. One change gives energy to your hand. The other takes energy from your hand.

Chemists describe both with one quantity, the enthalpy change, $Δ H$ . And here is the part that confuses almost everyone at first. The warm pack, which gives energy out, has a negative $Δ H$ . The cold pack, which takes energy in, has a positive $Δ H$ .

That looks backwards. Surely giving something out should be the positive one? This lecture shows why the sign is not backwards at all. Once you see whose point of view the sign takes, every sign in this unit makes sense.

First decide: what is the system?

Every energy question in chemistry starts by dividing the world into two parts.

The system is the chemicals that are reacting: the particles whose bonds break and form. The surroundings are everything else: the water the chemicals are dissolved in, the container, the air, your hand.

This split is a choice, and it matters. In a beaker of acid and alkali, the reacting ions are the system. The water around them is part of the surroundings. So when the thermometer in the beaker rises, it is measuring the surroundings, not the system. The thermometer tells you what happened to the surroundings, and you work out what happened to the system from that.

Energy moves between the system and the surroundings as heat. It can go in either direction.

Exothermic and endothermic

A change that transfers heat from the system to the surroundings is exothermic. The surroundings get warmer. Burning fuel, neutralising an acid with an alkali, and respiration in your cells are all exothermic. So is the warm pack.

A change that transfers heat from the surroundings into the system is endothermic. The surroundings get cooler, or the change needs continuous heating to keep going. Dissolving ammonium nitrate in water (the cold pack), photosynthesis, and breaking down limestone (calcium carbonate) by strong heating are all endothermic.

The words carry their meaning: exo means "out" and endo means "in", and thermic means "heat". Physical changes count too. Ice melting is endothermic, because the ice takes heat from its surroundings. Steam condensing is exothermic.

Stop and think

Kofi holds a cold pack. His hand gets cold. He says: "The pack is giving out cold, so the reaction is exothermic." What is wrong with his reasoning?

Show the answer

There is no such thing as "giving out cold". Cold is only less energy. Heat flows from Kofi's warmer hand into the pack, where the dissolving chemicals take it in. Heat flows into the system, so the change is endothermic. His hand cools because it is the surroundings, and it has lost energy. Always ask which way the heat flows between the system and the surroundings. Never ask whether something "feels hot".

Enthalpy, and why the sign takes the system's side

Every substance stores energy in its particles: in the motion of the particles, and in the bonds and forces between them. Chemists use a quantity called enthalpy, $H$ , for this stored energy when the pressure stays constant. (Older books call it "heat content".) Its full definition includes a small extra term for the work done pushing back the atmosphere. At this level, you can think of enthalpy as the energy stored in a substance under ordinary lab conditions.

Nobody can measure the total enthalpy of a substance. There is no zero to measure it from. But we can measure a change in enthalpy, and that is all chemistry needs:

Δ H = H_{products} - H_{reactants}

Now the sign is easy. In an exothermic change, the system gives heat to the surroundings, so the system ends with less enthalpy than it started with. The products are lower than the reactants, and $H_{products} - H_{reactants}$ is negative. In an endothermic change, the system takes in heat, so the products end higher, and $Δ H$ is positive.

That is it. That is the whole idea. The sign of $Δ H$ always describes the system. A negative sign means "the system lost enthalpy", which means the surroundings gained heat. Think of a bank account. When you pay a shop, your account shows a negative number, even though the shop received money. The account belongs to you, the system.

Enthalpy change

The enthalpy change $Δ H$ is the heat transferred between a system and its surroundings at constant pressure.

Exothermic: heat transferred to the surroundings, $Δ H$ negative. Endothermic: heat taken from the surroundings, $Δ H$ positive.

Why "at constant pressure"? Most reactions in a lab happen in open containers, under the pressure of the atmosphere. Under those conditions, the heat transferred is exactly the enthalpy change. This is why chemists use enthalpy rather than some other measure of energy. The idea was first described by Josiah Willard Gibbs in 1875, as "a heat function for constant pressure". The word "enthalpy", from the Greek for "to put heat in", first appeared in print in 1909, and is credited to the Dutch physicist Heike Kamerlingh Onnes.

Energy level diagrams

A picture makes the sign visible. An energy level diagram draws enthalpy up the side. The reactants are a line on the left, and the products are a line on the right, at their own heights.

Top: an exothermic reaction. The products are lower than the reactants, so

Δ H

is negative and the arrow points down. Bottom: an endothermic reaction. The products are higher, so

Δ H

is positive and the arrow points up. The dashed curve is the path the reaction takes. In both cases it first rises by the activation energy,

E_{a}

The dashed hump shows that reactants must first gain some energy before they can react. This is the activation energy, $E_{a}$ . It is why methane does not burn until a spark lights it, even though burning releases a lot of energy. The activation energy is always needed, whether the overall change is exothermic or endothermic. You will meet it again in the lectures on rates of reaction.

Numbers: kJ/mol, as written

An enthalpy change is written beside a balanced equation. For example, for burning methane:

CH₄(g) + 2O₂(g) \to CO₂(g) + 2H₂O(l) ΔH = -890 kJ/mol

The unit kJ/mol means kilojoules for each mole of the reaction exactly as written. Here that means: when 1 mol of methane reacts with 2 mol of oxygen to make 1 mol of carbon dioxide and 2 mol of water, 890 kJ of heat is released. The equation and the value belong together. Change the equation, and the value changes with it.

Two rules follow from this, and both come from the conservation of energy.

Scaling. Twice as much reaction transfers twice as much heat. If the equation is doubled, $Δ H$ doubles. If you burn 3 mol of methane, $3 \times 890 = 2670$ kJ is released.

Reversing. If a reaction releases 890 kJ going forwards, then going backwards it must take in exactly 890 kJ. Otherwise you could run it forwards and backwards and create energy from nothing. So reversing an equation reverses the sign of $Δ H$ .

Explore both rules below.

Explore — the sign, the size and the direction of ΔH

Choose a reaction. Change the amount of reaction, and press "Reverse the equation". Watch the arrow and the sign of ΔH, and read what happens to the surroundings.

Reaction:

Amount of reaction: 1 mol

Four reactions, each as written, with their enthalpy changes. Reversing an equation reverses the sign. Doubling the amount doubles the heat.

Equation	ΔH in kJ/mol	Type
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)	−890	exothermic
H₂(g) + ½O₂(g) → H₂O(l)	−286	exothermic
N₂(g) + O₂(g) → 2NO(g)	+181	endothermic
CaCO₃(s) → CaO(s) + CO₂(g)	+178	endothermic
H₂O(l) → H₂(g) + ½O₂(g)	+286	endothermic (the second row reversed)

For example, 2 mol of the methane reaction releases 2 × 890 = 1780 kJ to the surroundings.

What should you have seen? The size of the heat transfer grows with the amount of reaction, but the sign never changes when you only change the amount. Only reversing the equation changes the sign, and it never changes the size. An exothermic reaction run backwards is always endothermic, by exactly the same amount.

The promised sign of lattice energy

The lecture on lattice structures gave lattice energy as a size only: 786 kJ/mol for sodium chloride. It said that signs would come later, in the lectures on enthalpy. Here they are, and the reversing rule explains everything.

When gaseous sodium ions and chloride ions come together to form solid sodium chloride, energy is released:

Na⁺(g) + Cl⁻(g) \to NaCl(s) ΔH = -786 kJ/mol

Reverse the equation, and separating the solid into gaseous ions takes the same energy in:

NaCl(s) \to Na⁺(g) + Cl⁻(g) ΔH = +786 kJ/mol

Some textbooks define lattice energy using the first equation, so they give it as negative. Others use the second, so they give it as positive. Neither book is wrong. They describe the same energy in opposite directions. When you meet a lattice energy, look for the equation, or the definition, that it belongs to.

Worked examples

Worked Example 1 — From a mass to a heat

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$ , $Δ H = - 286$ kJ/mol. How much heat is released when 10 g of hydrogen burns? ( $M$ of $H_{2}$ = 2 g/mol.)

Moles of hydrogen: $n = \frac{10}{2} = 5.0$ mol.

The equation uses 1 mol of $H_{2}$ , so 5.0 mol of $H_{2}$ is 5.0 mol of the reaction as written.

Heat released: $5.0 \times 286 = 1430$ kJ.

Check: the answer is in kJ, not kJ/mol, because it is for a stated amount. The sign is negative, so the heat is released, and the surroundings warm up. ✓

Worked Example 2 — Reading the thermometer

Amira mixes sodium hydroxide solution with hydrochloric acid in a polystyrene cup. The temperature rises by 6.2 °C. Is the reaction exothermic or endothermic, and what is the sign of $Δ H$ ?

The system is the reacting ions. The water around them is the surroundings, and so is the thermometer. The surroundings got warmer, so they gained heat. That heat came from the system.

So the reaction is exothermic, and $Δ H$ is negative.

Check: neutralisation of an acid by an alkali is a standard exothermic reaction. ✓ The lecture on calorimetry shows how to turn the 6.2 °C into a value of $Δ H$ .

Worked Example 3 — The same reaction, written two ways

$N_{2} (g) + O_{2} (g) \to 2NO (g)$ , $Δ H = + 181$ kJ/mol. Find $Δ H$ for $\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) \to NO (g)$ , and for $2NO (g) \to N_{2} (g) + O_{2} (g)$ .

The second equation is the first one halved, so $Δ H$ halves: $\frac{181}{2} \approx + 90.5$ kJ/mol.

The third equation is the first one reversed, so the sign reverses: $Δ H = - 181$ kJ/mol.

Check: "kJ/mol" always means per mole of the reaction as written. That is why one reaction can be described by three different numbers. ✓ This reaction happens in the very hot gases inside car engines, which is one source of nitrogen oxide pollution.

Worked Example 4 — Per mole or per gram?

Burning methane releases 890 kJ per mole of methane, and burning hydrogen releases 286 kJ per mole of hydrogen. Which fuel releases more heat per gram? ( $M$ : $C H_{4}$ = 16 g/mol, $H_{2}$ = 2 g/mol.)

Methane: $\frac{890}{16} \approx 55.6$ kJ/g. Hydrogen: $\frac{286}{2} = 143$ kJ/g.

Per mole, methane releases more. Per gram, hydrogen releases more than twice as much.

Check: a hydrogen molecule is very light, so a gram of hydrogen contains many more molecules than a gram of methane. ✓ Fuels are compared per gram or per kilogram, because fuel is bought and carried by mass. This ratio is the fuel's energy per unit mass. The size of $Δ H$ per mole alone cannot tell you which fuel is better.

Now sort some changes by the direction of their heat flow.

Sort — exothermic or endothermic?

Put each change in its box. For each one, decide what the system is, and ask: does heat flow out of it, or into it?

Eight changes, sorted by the direction of heat flow.

Change	Type	Reason
Methane burning on a gas hob	Exothermic	The kitchen warms up.
Ammonium nitrate dissolving in a cold pack	Endothermic	It takes heat from your hand.
Ice melting in a drink	Endothermic	The ice takes heat from the drink.
Steam condensing on a cold window	Exothermic	The steam gives heat to the glass.
Heating limestone to make lime	Endothermic	It stops when the heating stops.
Hydrochloric acid neutralising sodium hydroxide	Exothermic	The solution warms up.
Photosynthesis in a leaf	Endothermic	It takes in energy from light.
Respiration in your cells	Exothermic	It keeps your body warm.

Look at the ice and the steam. Melting and condensing are the same change in opposite directions. So one is endothermic and the other exothermic, by the same amount, exactly as the reversing rule predicts.

Three classic mistakes

Trap 1 — "Energy is given out, so ΔH is positive"

The sign of $Δ H$ describes the system, not the surroundings. When heat is given out, the system loses enthalpy, so $Δ H$ is negative. Write the sign every time, including the + for an endothermic change: $Δ H = + 178$ kJ/mol.

Trap 2 — Treating the thermometer as the system

The thermometer and the water measure the surroundings. If the temperature falls, the surroundings lost heat to the system, so the change is endothermic. Decide what the system is before you decide the sign.

Trap 3 — Using ΔH without its equation

"kJ/mol" means per mole of the reaction as written. If the equation is doubled, halved or reversed, $Δ H$ changes too. And the heat for a stated amount of substance is in kJ, not kJ/mol.

Try it yourself

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

$C (s) + O_{2} (g) \to C O_{2} (g)$ , $Δ H = - 394$ kJ/mol. Is the reaction exothermic or endothermic? How much heat is released when 60 g of carbon burns completely? ( $A_{r}$ of C = 12.)

Hint 1
The sign tells you the type. For the heat, first find the moles of carbon.

Hint 2
$n = 60 \div 12 = 5$ mol, and the equation uses 1 mol of C.

Full answer
$Δ H$ is negative, so the reaction is exothermic. Moles of carbon: $n = \frac{60}{12} = 5.0$ mol, which is 5.0 mol of the reaction as written. Heat released: $5.0 \times 394 = 1970$ kJ. Check: the answer is in kJ because it is for a stated amount, and 5 times the value for 1 mol is sensible. ✓
Using $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$ , $Δ H = - 286$ kJ/mol, write $Δ H$ for: (a) $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$ ; (b) $2 H_{2} O (l) \to 2 H_{2} (g) + O_{2} (g)$ .

Hint 1
Compare each equation with the given one: is it multiplied, reversed, or both?

Hint 2
(a) is the given equation doubled. (b) is (a) reversed.

Full answer
(a) Doubled: $Δ H = 2 \times (- 286) = - 572$ kJ/mol. (b) The reverse of (a): $Δ H = + 572$ kJ/mol. This is the energy needed to split water into hydrogen and oxygen, for example by electrolysis. Check: (a) and (b) have the same size and opposite signs, as reversing requires. ✓
Book A says: "The lattice energy of sodium chloride is −786 kJ/mol." Book B says: "The lattice energy of sodium chloride is +786 kJ/mol." Explain how both books can be right. Then find the energy needed to separate 0.50 mol of solid sodium chloride into gaseous ions.

Hint 1
Write the equation that each book must be using.

Hint 2
Separating the ions is the direction that takes energy in.

Full answer
Book A uses forming the solid: $N a^{+} (g) + C l^{-} (g) \to NaCl (s)$ , which releases energy, so $Δ H = - 786$ kJ/mol. Book B uses separating the solid: $NaCl (s) \to N a^{+} (g) + C l^{-} (g)$ , the reverse, so $Δ H = + 786$ kJ/mol. Both describe the same energy in opposite directions. To separate 0.50 mol: $0.50 \times 786 = 393$ kJ must be supplied. Check: separating ions that attract each other must take energy in, so the positive direction is the separating one. ✓
Dev's camping stove burns methane: $Δ H = - 890$ kJ/mol. To boil a pan of water, the water must gain 2000 kJ. Only half of the heat released by the flame reaches the water. What mass of methane must he burn? ( $M$ of $C H_{4}$ = 16 g/mol.) Where does the rest of the heat go?

Hint 1
If only half reaches the water, the flame must release twice 2000 kJ.

Hint 2
Moles of methane = 4000 ÷ 890. Then mass = moles × 16.

Full answer
The flame must release $2 \times 2000 = 4000$ kJ. Moles of methane: $n = \frac{4000}{890} \approx 4.49$ mol. Mass: $4.49 \times 16 \approx 71.9$ g, so about 72 g. The rest of the heat goes into the surroundings that are not the water: the air around the flame, the pan and the stove. Check: 4.49 mol × 890 kJ/mol ≈ 4000 kJ, and half of that is 2000 kJ. ✓ The same idea, that some heat always escapes, is the main source of error in the lecture on calorimetry.

The bigger idea

Enthalpy changes are energy accounts. Like any account, they balance. If a reaction releases 890 kJ going forwards, it needs 890 kJ going backwards, and the total never changes. This simple rule, the conservation of energy, is what the lecture on Hess's law turns into a powerful tool. It lets chemists find enthalpy changes that nobody can measure directly, from the difference between ones that they can.

The same accounts explain why life needs food, and why the world burns fuels. Photosynthesis stores the energy of sunlight in glucose, an endothermic change. Respiration and burning release it again, an exothermic one. Much of human energy use moves energy between these two directions.

Key concepts in this lesson

The system is the reacting chemicals and the surroundings are everything else, including the water and the thermometer.
An exothermic change transfers heat to the surroundings and has a negative ΔH, and an endothermic change takes heat in and has a positive ΔH.
The sign of ΔH describes the system, because ΔH = H(products) − H(reactants).
On an energy level diagram, the products are below the reactants for an exothermic change and above them for an endothermic one, with the activation energy as a hump in both.
ΔH in kJ/mol is per mole of the reaction as written, so it scales with the equation, and reversing the equation reverses its sign.
Lattice energy is negative for forming a solid from gaseous ions and positive for separating it, because the two are the same change in opposite directions.