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

Enthalpy and Thermochemical Equations

D2.2 D2.1 D3.1 A1.12About 70 minutes
Learning goals
  • I can define enthalpy change and relate its sign to exothermic and endothermic changes.
  • I can write thermochemical equations with a ΔH value or with a heat term.
  • I can scale and reverse thermochemical equations correctly.
  • I can calculate the heat released or absorbed for a given amount of a substance, and work backward to a molar enthalpy.

Words to know

enthalpy (H)
The total energy stored in a system at constant pressure. Only its change can be measured.
enthalpy change (ΔH)
H(products) − H(reactants); equal to the heat transferred at constant pressure. Negative for exothermic, positive for endothermic.
thermochemical equation
A balanced chemical equation that includes the enthalpy change, either as ΔH or as a heat term.
molar enthalpy
The enthalpy change per mole of a named substance, in kJ/mol, for example ΔH_comb.
standard enthalpy change (ΔH°)
An enthalpy change measured at standard conditions: 25 °C and 100 kPa.
enthalpy diagram
A diagram with an enthalpy axis showing reactants and products as levels, with an arrow for ΔH.

Reading the label on a propane tank

A barbecue propane cylinder holds about 9 kg of propane. How much energy is that? To answer, you need two things: the amount of propane (in moles) and how much energy each mole releases when it burns. Chemists record the second number in a thermochemical equation. By the end of this lesson you will be able to write one and use it to calculate the energy from any amount of fuel.

Enthalpy and enthalpy change

Every substance stores energy in the motion of its particles, in the attractions between them and in its chemical bonds. Chemists use a quantity called enthalpy, symbol H, for the total stored energy of a system at constant pressure. We cannot measure an absolute enthalpy, but we can measure how much it changes:

ΔH = H(products) − H(reactants)

For a reaction in an open container (constant pressure), the enthalpy change ΔH equals the heat absorbed or released by the system.

  • Exothermic: the products have less enthalpy than the reactants, so ΔH is negative. The difference is released to the surroundings.
  • Endothermic: the products have more enthalpy, so ΔH is positive. The difference is absorbed from the surroundings.

The symbol ΔH° ("delta H naught") means the value was measured at standard conditions, which in this course means SATP: 25 °C and 100 kPa, with solutions at 1 mol/L.

HCH₄(g) + 2O₂(g)CO₂(g) + 2H₂O(l)ΔH = −890.3 kJenergy releasedExothermic: products lowerHCaCO₃(s)CaO(s) + CO₂(g)ΔH = +178.3 kJenergy absorbedEndothermic: products higher
Enthalpy diagrams. Burning methane: the products are lower in enthalpy, the arrow points down and ΔH is negative. Decomposing limestone: the products are higher, the arrow points up and ΔH is positive.

Four ways to show an energy change

1. ΔH notation (the usual way): write ΔH after the balanced equation.

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = −890.3 kJ

2. Heat term in the equation: put the energy in as a product (exothermic) or as a reactant (endothermic). Note that the number is written without a sign.

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 890.3 kJ

CaCO₃(s) + 178.3 kJ → CaO(s) + CO₂(g)

3. Molar enthalpy: the enthalpy change per mole of one named substance, in kJ/mol. For example, the molar enthalpy of combustion of methane is ΔH_comb = −890.3 kJ/mol. Common types: combustion (ΔH_comb), solution (ΔH_soln), neutralization (ΔH_neut), vaporization (ΔH_vap), fusion or melting (ΔH_fus) and formation (ΔH_f).

4. Enthalpy (potential energy) diagram: a vertical enthalpy axis with reactants and products as horizontal lines and an arrow labelled with ΔH, as in the diagram above.

Three rules for thermochemical equations

  1. ΔH is proportional to the amounts. Double every coefficient and you double ΔH. Fractional coefficients such as ½O₂ are allowed, because coefficients now mean moles, not molecules.
  2. Reverse the equation, reverse the sign. If A → B has ΔH = −50 kJ, then B → A has ΔH = +50 kJ.
  3. States matter. Always include (s), (l), (g) or (aq). Forming liquid water releases more energy than forming water vapour.
HH₂(g) + ½O₂(g)H₂O(g)H₂O(l)−241.8 kJ−285.8 kJ−44.0 kJSame reactants, different product state, different ΔH
Burning hydrogen to make water vapour releases 241.8 kJ per mole; making liquid water releases 285.8 kJ, because condensing the vapour releases another 44.0 kJ. That is why state symbols are required. (Arrows not to scale.)

Using a thermochemical equation in calculations

Because ΔH is proportional to the amount that reacts, you can use the same mole methods you learned in SCH3U (Quantities in Chemical Reactions):

ΔH = n × ΔH_molar

where n is the amount (mol) of the substance the molar enthalpy refers to. For a mass, first find n = m ÷ M.

Worked example

Example 1: Writing a thermochemical equation

The molar enthalpy of combustion of ethanol, C₂H₅OH(l), is −1367 kJ/mol (water produced as a liquid). Write the thermochemical equation in two ways.

  1. Balance the combustion: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). (Check: C 2 = 2; H 6 = 6; O 1 + 6 = 7 = 4 + 3.)
  2. The equation has 1 mol of ethanol, so ΔH = 1 mol × (−1367 kJ/mol) = −1367 kJ.
  3. Exothermic, so as a heat term the energy goes on the product side, without a sign.

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH = −1367 kJ, or C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) + 1367 kJ

Worked example

Example 2: Scaling and reversing

Given 2H₂(g) + O₂(g) → 2H₂O(l) ΔH = −571.6 kJ, find ΔH for H₂O(l) → H₂(g) + ½O₂(g).

  1. The target equation is the given one reversed (water is now a reactant): the sign changes to +571.6 kJ.
  2. It is also halved (1 mol of water instead of 2): +571.6 kJ ÷ 2 = +285.8 kJ.

ΔH = +285.8 kJ. Splitting water is endothermic; that is why electrolysis needs a steady supply of electrical energy.

Worked example

Example 3: Energy from a mass of fuel

How much heat is released when 25.0 g of propane, C₃H₈(g), burns completely? ΔH_comb = −2220 kJ/mol.

  1. M(C₃H₈) = 3(12.01) + 8(1.01) = 44.11 g/mol.
  2. n = 25.0 g ÷ 44.11 g/mol = 0.5668 mol.
  3. ΔH = n × ΔH_comb = 0.5668 mol × (−2220 kJ/mol) = −1258 kJ.

ΔH = −1.26 × 10³ kJ: about 1260 kJ of heat is released.

Worked example

Example 4: Working backward to a molar enthalpy

Burning 3.00 g of magnesium ribbon releases 74.2 kJ of heat. Find the molar enthalpy of combustion of magnesium and write the thermochemical equation for 2Mg(s) + O₂(g) → 2MgO(s).

  1. n(Mg) = 3.00 g ÷ 24.31 g/mol = 0.1234 mol.
  2. Heat was released, so ΔH is negative: ΔH_comb = ΔH ÷ n = −74.2 kJ ÷ 0.1234 mol = −601 kJ/mol.
  3. The equation has 2 mol of Mg: ΔH = 2 mol × (−601 kJ/mol) = −1202 kJ.

ΔH_comb(Mg) = −601 kJ/mol; 2Mg(s) + O₂(g) → 2MgO(s) ΔH = −1202 kJ

Hands-on activity

Fuel labels at home

You need: a calculator and one or more fuel containers you can read without opening or moving them: a barbecue propane cylinder (about 9 kg of propane when full), a camping-stove butane canister, or a butane lighter. A grown-up should be with you near any fuel; never open a valve or squeeze a lighter.

  1. Record the mass of fuel printed on the label (in grams or kilograms).
  2. Use these molar enthalpies of combustion: propane C₃H₈, −2220 kJ/mol; butane C₄H₁₀, −2877 kJ/mol. Calculate the molar mass of the fuel and the amount in moles.
  3. Calculate the total energy the full container would release, in kilojoules and in megajoules (1 MJ = 1000 kJ).
  4. Write the thermochemical equation for the fuel in both ΔH notation and heat-term form.

What to look for (sample answer for 9.0 kg of propane): n = 9000 g ÷ 44.11 g/mol = 204 mol; energy = 204 mol × 2220 kJ/mol ≈ 4.5 × 10⁵ kJ = 450 MJ. Equation: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l), ΔH = −2220 kJ. A real barbecue delivers less than this to the food, because much of the heat escapes.

Just for fun

The word enthalpy comes from Greek words meaning "to heat within". So ΔH is literally "the change in the heat within". Chemists are very good at naming things after what they mean, and not so good at making the names easy to spell.

Your turn

1

Write the thermochemical equation, in ΔH notation, for the combustion of methane, CH₄(g), given ΔH_comb = −890.3 kJ/mol and liquid water as a product.

2

Rewrite N₂(g) + O₂(g) → 2NO(g) ΔH = +180.6 kJ with a heat term.

3

Given N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = −92.2 kJ, find ΔH for 2NH₃(g) → N₂(g) + 3H₂(g).

4

Using the equation in the previous question, find ΔH for ½N₂(g) + 3/2H₂(g) → NH₃(g).

5

How much heat is released when 10.0 g of methane burns? (ΔH_comb = −890.3 kJ/mol, M = 16.05 g/mol)

6

Is the reaction CaO(s) + H₂O(l) → Ca(OH)₂(s) + 65.2 kJ exothermic or endothermic? Write its ΔH.

7

Why does H₂(g) + ½O₂(g) → H₂O(l) release more energy than H₂(g) + ½O₂(g) → H₂O(g)?

More practice: warm-up

1

What is the sign of ΔH for an exothermic reaction?

2

What does the symbol ΔH° tell you?

3

If ΔH for A → B is −120 kJ, what is ΔH for B → A?

4

If ΔH for A → B is −120 kJ, what is ΔH for 3A → 3B?

5

In a heat-term equation for an endothermic reaction, on which side is the energy written?

More practice: on your own

1

The molar enthalpy of combustion of octane, C₈H₁₈(l), is −5470 kJ/mol. Write the balanced thermochemical equation using whole-number coefficients.

2

What mass of methane must burn to release 5.00 × 10⁴ kJ? (ΔH_comb = −890.3 kJ/mol)

3

Dissolving 1.00 mol of NaOH(s) in water releases 44.5 kJ. How much heat is released when 5.00 g of NaOH dissolves? (M = 40.00 g/mol)

4

Photosynthesis: 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g) ΔH = +2803 kJ. How much energy must a plant absorb to make 1.00 g of glucose? (M = 180.18 g/mol)

5

Draw (describe) an enthalpy diagram for N₂(g) + O₂(g) → 2NO(g), ΔH = +180.6 kJ.

Challenge

1

A barbecue burns propane (ΔH_comb = −2220 kJ/mol). How many grams of propane are needed to heat a 2.0 L pot of water from 10.0 °C to 100.0 °C if only 30.0% of the heat reaches the water? (Water needs 4.18 kJ per kilogram per °C, so the pot needs 2.0 × 4.18 × 90.0 = 752 kJ.)

2

Given H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8 kJ and H₂(g) + ½O₂(g) → H₂O(g), ΔH = −241.8 kJ, find ΔH_vap of water at 25 °C, and explain why it differs from 40.7 kJ/mol at 100 °C.

3

A 9.0 kg propane cylinder and a 9.0 kg block of paraffin wax (taken as C₂₅H₅₂, ΔH_comb ≈ −16 500 kJ/mol, M = 352.8 g/mol) are compared. Which stores more energy per kilogram? Show your work.

Check your understanding

Quiz

1.

Which thermochemical equation shows an endothermic change?

2.

If 2SO₂(g) + O₂(g) → 2SO₃(g) has ΔH = −197.8 kJ, what is ΔH for SO₃(g) → SO₂(g) + ½O₂(g)?

3.

ΔH for a reaction is negative. This means

4.

How much heat is released when 0.250 mol of ethanol burns (ΔH_comb = −1367 kJ/mol)?

5.

Why must state symbols be included in thermochemical equations?

6.

What does "standard conditions" (the ° in ΔH°) mean in this course?

For parents and teachers

Big ideas (D2.2, D2.1): enthalpy change, ΔH, is the heat absorbed or released at constant pressure. Negative ΔH means exothermic and positive means endothermic. A thermochemical equation can show the energy as a ΔH value or as a heat term, and ΔH scales with the coefficients and changes sign when the equation is reversed.

How to help: Ask your teen to explain the two forms of the same equation to you, and to say which side the heat term goes on for an exothermic reaction (products) and why there is no minus sign in that form. Give a quick "double it, reverse it" drill: "If A → B is −50 kJ, what is 2B → 2A?" (+100 kJ).

Common difficulties:

  • Writing a negative number in a heat-term equation (the side tells you the sign, so the number has none).
  • Forgetting to scale ΔH when coefficients change, or forgetting to change the sign when reversing.
  • Leaving out state symbols, especially (l) versus (g) for water.
  • Using grams instead of moles in ΔH = nΔH_molar.

Activity safety: read fuel labels only; a grown-up stays nearby, and no valves are opened or lighters used.