Enthalpy changes

In a nutshell

Almost every reaction either releases heat to its surroundings or takes in heat from them. The enthalpy change, ΔH, measures that heat change, in kJ mol^-1.

This subtopic covers the sign of ΔH, the four named enthalpy changes, measuring ΔH in the lab with q = mcΔT, calculating it from average bond enthalpies, and using Hess' law cycles to find enthalpy changes you cannot measure directly.

Assumed knowledge: Amount of substance: moles from mass and from concentration and volume; Bonding and structure: average bond enthalpy as a measure of bond strength.

Core content

Enthalpy changes: exothermic and endothermic reactions

Enthalpy, H, is the heat energy stored in a chemical system. You cannot measure it directly, but you can measure the enthalpy change, ΔH, when a reaction happens at constant pressure.

$\Delta H = H(\text{products}) - H(\text{reactants})$

Chemists split the world in two:

• the system: the chemicals that are reacting
• the surroundings: everything else (the water, the beaker, the air, the thermometer)

Still don't get it? · why a temperature rise means a negative ΔH

Imagine you hand a friend £10. Your friend's wallet goes up by £10, but yours goes down by £10. If you only ever look at your friend's wallet, you have to remember that the money came out of yours.

The thermometer is your friend's wallet. It sits in the water, which is part of the surroundings, not part of the reacting chemicals.

1. The reaction happens and the water gets hotter.
2. That heat had to come from somewhere: it came out of the chemicals.
3. So the chemicals (the system) now have less enthalpy than before.
4. Less enthalpy at the end than at the start means ΔH = H(products) - H(reactants) is negative.

Exam version: a temperature rise means heat is released from the system to the surroundings, so the reaction is exothermic and ΔH is negative. A temperature fall means the reaction is endothermic and ΔH is positive.

Enthalpy changes: enthalpy profile diagrams and activation energy

An enthalpy profile diagram plots enthalpy against the progress of the reaction. It shows at a glance whether the products sit above or below the reactants.

When you draw one, the marks are in the labels:

• label the reactants and products levels (the formulae if you know them)
• draw ΔH as a vertical arrow from the reactants level to the products level (down for exothermic, up for endothermic)
• draw the activation energy, E_a, as a vertical arrow from the reactants level up to the top of the hump

Activation energy is the minimum energy required for a reaction to take place. Even an exothermic reaction needs this energy put in to break bonds and get started, which is why a fuel does not catch fire until it is lit.

Enthalpy changes: standard conditions and standard states

ΔH depends on temperature, pressure and the physical states of the chemicals. To compare values fairly, they are quoted under standard conditions:

• pressure of 100 kPa
• a stated temperature, usually 298 K (25 °C)
• (for solutions, a concentration of 1 mol dm^-3)

The standard state of a substance is its physical state under standard conditions. For example, water is H₂O(l), oxygen is O₂(g) and magnesium is Mg(s).

An enthalpy change measured under standard conditions, with everything in its standard state, gets the symbol ^⦵: for example Δ_rH^⦵.

Enthalpy changes: the four named enthalpy changes

Enthalpy change of reaction, Δ_rH, is the enthalpy change for a stated equation, in the molar amounts shown in that equation. Double the equation and you double Δ_rH. (You need to use this term, not give a formal definition.)

The other three are each fixed to 1 mole of something. These three definitions are required, word for word:

Standard enthalpy change of formation, Δ_fH^⦵: the enthalpy change that takes place when one mole of a compound is formed from its elements, under standard conditions, with all reactants and products in their standard states.

• The elements must be in their standard states, so you get fractions: ½Cl₂(g), not Cl(g).
• Δ_fH^⦵ of an element in its standard state is zero (nothing is formed).

Standard enthalpy change of combustion, Δ_cH^⦵: the enthalpy change that takes place when one mole of a substance reacts completely with oxygen, under standard conditions, with all reactants and products in their standard states.

• "Completely" matters: carbon ends up as CO₂, not CO, and hydrogen as H₂O(l).
• Combustion is always exothermic, so Δ_cH^⦵ is always negative.

Standard enthalpy change of neutralisation, Δ_neutH^⦵: the enthalpy change that takes place when an acid and a base react to form one mole of water, under standard conditions, with all reactants and products in their standard states.

• It is per mole of water, not per mole of acid. One mole of H₂SO₄ fully neutralised makes two moles of water.

Enthalpy changes: measuring ΔH with q = mcΔT

In a calorimetry experiment the reaction heats (or cools) a known mass of water or solution. The heat energy transferred to or from that water is:

$q = m\,c\,\Delta T$

Turning q into ΔH always takes the same four moves:

1. Calculate q = mcΔT in J, using the mass of the solution (assume 1.00 cm³ has a mass of 1.00 g).
2. Calculate the moles of the substance the ΔH is per mole of (the reactant not in excess, the fuel, or the water formed).
3. Divide: q ÷ n, then ÷ 1000 to get kJ mol^-1.
4. Add the sign: temperature rose, so ΔH is negative; temperature fell, so ΔH is positive.

$\Delta H = \frac{-q}{n} \quad \text{(for a temperature rise)}$

Bond enthalpies: average bond enthalpy

Average bond enthalpy is the energy needed to break one mole of a given type of bond in gaseous molecules, averaged over many different compounds. (You need to explain and use the term; a formal definition is not required.)

• Breaking a bond always takes in energy, so bond enthalpies are always positive.
• The larger the value, the stronger the bond.
• It is an average: a C-H bond in methane is not exactly the same strength as a C-H bond in ethanol. The actual bond enthalpy in a particular molecule may differ from the average value.

Bond enthalpies: breaking and making bonds

Every reaction breaks bonds in the reactants and makes new bonds in the products.

• Bond breaking is endothermic: energy is taken in.
• Bond making is exothermic: energy is released.

So the sign of ΔH comes down to which is bigger:

• Exothermic (ΔH negative): more energy is released making bonds in the products than is taken in breaking bonds in the reactants.
• Endothermic (ΔH positive): more energy is taken in breaking bonds than is released making bonds.

Bond enthalpies: calculating ΔH from average bond enthalpies

$\Delta_r H = \sum(\text{bond enthalpies of bonds broken}) - \sum(\text{bond enthalpies of bonds made})$

Method:

1. Write the balanced equation and draw out every bond (displayed formulae) so you can count them.
2. Add up the bond enthalpies of all bonds broken in the reactants, multiplying by the number of each bond and by the balancing numbers.
3. Add up the bond enthalpies of all bonds made in the products in the same way.
4. Broken minus made.

A ΔH from average bond enthalpies will not exactly match an experimental or data-book value, because:

• the bond enthalpies used are averages, not the actual values in these molecules
• bond enthalpies apply to gases, so the method ignores the energy for any state change (for example, H₂O(g) condensing to H₂O(l) in a standard combustion)

The same equation can be rearranged to find an unknown bond enthalpy when ΔH and the other bond enthalpies are given (see Worked example 5).

Hess' law and enthalpy cycles

Hess' law: the enthalpy change for a reaction is the same whatever route is taken from the reactants to the products, as long as the start and end are the same. It follows from conservation of energy. (You need to apply it, not define it.)

This lets you find a ΔH you cannot measure by going round an enthalpy cycle: the direct route and the indirect route must give the same total.

Rules for going round any cycle:

• Follow the arrows from start to finish by the indirect route.
• An arrow you travel with its direction: add its ΔH.
• An arrow you travel against its direction: subtract its ΔH (reverse its sign).
• Multiply each ΔH by the number of moles in the equation.

Still don't get it? · going round a Hess cycle without sign errors

Think of driving from home to school. The direct road is closed, so you go via the shop. The height you climb from home to school is the same either way: it only depends on where home and school are, not on the road.

Now treat each arrow in a cycle as a road with a height change written on it.

1. Put your finger on the start of the arrow you want (the reactants).
2. Trace the other way round to the end of that arrow (the products).
3. Each time you move along an arrow in the direction it points, write down its value as it is.
4. Each time you move along an arrow backwards, write down its value with the sign flipped.
5. Add up what you wrote. That total is the ΔH you wanted.

For a combustion cycle, you go down from the reactants (with the arrow) and then up to the products (against the arrow), which is why the formula is Σ Δ_cH(reactants) - Σ Δ_cH(products). For a formation cycle you go down to the elements (against the arrow) then up to the products (with the arrow), giving Σ Δ_fH(products) - Σ Δ_fH(reactants).

Exam version: show the cycle, label every arrow with its ΔH multiplied by the moles, and write the line "Δ_rH = ..." with each term and its sign before you calculate.

Hess' law: ΔrH from enthalpy changes of combustion

Both the reactants and the products can be burned to the same combustion products (CO₂ and H₂O). The combustion arrows point down from both.

$\Delta_r H = \sum \Delta_c H(\text{reactants}) - \sum \Delta_c H(\text{products})$

• Use this when you are given Δ_cH data, for example to find the Δ_fH of an organic compound, which cannot be made directly from its elements.
• O₂ has no enthalpy change of combustion (it does not burn in oxygen), so it never appears in the sum.

Hess' law: ΔrH from enthalpy changes of formation

Both the reactants and the products can be formed from the same elements. The formation arrows point up from the elements to both.

$\Delta_r H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})$

• Any element in its standard state (O₂(g), Fe(s), H₂(g)) has Δ_fH^⦵ = 0, but still write it in so you do not lose track.

Notice the two formulas are opposite ways round. Do not memorise them blindly: if you can draw the cycle, you can always work out which way round to subtract.

Hess' law: unfamiliar enthalpy cycles

An exam can give you a cycle you have never seen (it will be provided). The rules are the same: find the direct arrow you want, trace the indirect route, add with the arrows and subtract against them.

Typical reasons a ΔH must be found indirectly:

• the reaction needs strong heating, so the temperature change cannot be measured (for example, CaCO₃(s) → CaO(s) + CO₂(g))
• the reaction is too slow, or other reactions happen at the same time
• the compound cannot be made directly from its elements (for example, most organic compounds)

PAG3: enthalpy determination

Direct method 1: reactions in solution (neutralisation, displacement, dissolving a solid in acid).

1. Measure a known volume of solution into a polystyrene cup standing in a beaker, with a lid.
2. Record the temperature every minute for a few minutes to get a steady starting value.
3. Add the second reactant at a recorded time (for a solid, weigh it before and weigh the empty container after). Stir.
4. Keep recording the temperature every minute for several minutes after mixing.
5. Plot temperature against time and extrapolate the cooling line back to the time of mixing to find ΔT.

Why extrapolate? While the mixture is still heating up, it is already losing heat to the surroundings, so the highest reading is lower than the true maximum. Extending the cooling line back to the moment of mixing estimates the temperature the mixture would have reached with no heat loss.

Here ΔT = 31.5 - 21.0 = 10.5 K, even though the highest reading was only 31.2 °C. This graph is used in Worked example 2.

Direct method 2: enthalpy change of combustion of a liquid fuel.

1. Weigh the spirit burner with its cap on (cap stops the fuel evaporating).
2. Measure a known volume of water into the copper can and record its starting temperature.
3. Light the burner, stir the water, and let the flame heat it by about 15 to 20 K.
4. Put out the flame, record the highest temperature, then reweigh the burner to find the mass of fuel burned.

Why experimental values are less exothermic than data-book values

• heat lost to the surroundings (the biggest error in combustion experiments), instead of all going into the water
• incomplete combustion of the fuel (soot on the can is the clue)
• evaporation of the fuel from the wick, so the mass loss is not all burned fuel
• heat absorbed by the calorimeter itself is ignored
• the solution is assumed to have the density and specific heat capacity of water
• the experiment is not under standard conditions (for example, the water formed in combustion leaves as a gas)

Improvements: a lid and insulation, a draught shield, a shorter gap between flame and can, stirring, and the extrapolation method for slow reactions in solution.

Indirect method: using Hess' law. When ΔH cannot be measured directly, measure two reactions that can be measured and link them in a cycle.

1. React a weighed sample of CaCO₃(s) with excess dilute HCl(aq) in a polystyrene cup; use q = mcΔT to find ΔH₁ per mole of CaCO₃.
2. Repeat with a weighed sample of CaO(s) to find ΔH₂ per mole of CaO.
3. Go round the cycle: Δ_rH = ΔH₁ - ΔH₂ (see Worked example 8).

Worked examples

Worked example 1: enthalpy change of neutralisation from two solutions

30.0 cm³ of 2.00 mol dm^-3 HCl(aq) at 19.8 °C is mixed with 30.0 cm³ of 2.00 mol dm^-3 NaOH(aq) at 20.2 °C in a polystyrene cup. The highest temperature reached is 33.6 °C. Calculate Δ_neutH.

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Step 1: ΔT. Two solutions start at different temperatures, so use their mean starting temperature.

$T_{\text{start}} = \frac{19.8 + 20.2}{2} = 20.0\ ^\circ\text{C}$ $\Delta T = 33.6 - 20.0 = 13.6\ \text{K}$

Step 2: q. The mass is the total volume of both solutions: 30.0 + 30.0 = 60.0 cm³, so 60.0 g.

$q = m\,c\,\Delta T = 60.0\ \text{g} \times 4.18\ \text{J g}^{-1}\,\text{K}^{-1} \times 13.6\ \text{K}$ $q = 3410.88\ \text{J} = 3.41088\ \text{kJ}$

Step 3: moles of water formed.

$n(\text{HCl}) = n(\text{NaOH}) = 2.00\ \text{mol dm}^{-3} \times \frac{30.0}{1000}\ \text{dm}^3 = 0.0600\ \text{mol}$ $n(\text{H}_2\text{O}) = 0.0600\ \text{mol}$

Step 4: ΔH, with sign. The temperature rose, so the reaction is exothermic.

$\Delta_{neut} H = \frac{-3.41088\ \text{kJ}}{0.0600\ \text{mol}} = -56.848\ \text{kJ mol}^{-1}$ $\Delta_{neut} H = -56.8\ \text{kJ mol}^{-1}\ \text{(3 s.f.)}$

Where marks go: using 30.0 g instead of 60.0 g halves the answer; leaving off the minus sign loses the final mark.

Worked example 2: a solid reacting with a solution (excess reagent)

1.00 g of zinc powder is added to 50.0 cm³ of 0.200 mol dm^-3 CuSO₄(aq). The temperature readings are plotted in the PAG3 graph above. Calculate Δ_rH for:

Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

Step 1: ΔT from the graph (extrapolated, not the highest reading).

$\Delta T = 31.5 - 21.0 = 10.5\ \text{K}$

Step 2: q. The mass is the solution only (50.0 g). Do not add the 1.00 g of zinc.

$q = 50.0\ \text{g} \times 4.18\ \text{J g}^{-1}\,\text{K}^{-1} \times 10.5\ \text{K} = 2194.5\ \text{J} = 2.1945\ \text{kJ}$

Step 3: find the reactant not in excess (the reaction is 1 : 1).

$n(\text{Zn}) = \frac{1.00\ \text{g}}{65.4\ \text{g mol}^{-1}} = 0.0153\ \text{mol}$ $n(\text{CuSO}_4) = 0.200\ \text{mol dm}^{-3} \times \frac{50.0}{1000}\ \text{dm}^3 = 0.0100\ \text{mol}$

Zinc is in excess, so only 0.0100 mol reacts. Divide by 0.0100 mol.

Step 4: ΔH, with sign.

$\Delta_r H = \frac{-2.1945\ \text{kJ}}{0.0100\ \text{mol}} = -219.45\ \text{kJ mol}^{-1}$ $\Delta_r H = -219\ \text{kJ mol}^{-1}\ \text{(3 s.f.)}$

Where marks go: dividing by the moles of zinc (the reagent in excess) gives -144 kJ mol^-1, which is wrong.

Worked example 3: enthalpy change of combustion from a spirit burner

A spirit burner of ethanol, C₂H₅OH, has a mass of 62.814 g before and 62.112 g after heating 150.0 cm³ of water from 18.5 °C to 34.0 °C. Calculate Δ_cH of ethanol.

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

Step 1: q. The mass is the water being heated (150.0 g), not the ethanol.

$\Delta T = 34.0 - 18.5 = 15.5\ \text{K}$ $q = 150.0\ \text{g} \times 4.18\ \text{J g}^{-1}\,\text{K}^{-1} \times 15.5\ \text{K} = 9718.5\ \text{J} = 9.7185\ \text{kJ}$

Step 2: moles of ethanol burned.

$m = 62.814 - 62.112 = 0.702\ \text{g}$ $M(\text{C}_2\text{H}_5\text{OH}) = (2 \times 12.0) + (6 \times 1.0) + 16.0 = 46.0\ \text{g mol}^{-1}$ $n = \frac{0.702\ \text{g}}{46.0\ \text{g mol}^{-1}} = 0.015261\ \text{mol}$

Step 3: ΔH, with sign.

$\Delta_c H = \frac{-9.7185\ \text{kJ}}{0.015261\ \text{mol}} = -636.8\ \text{kJ mol}^{-1}$ $\Delta_c H = -637\ \text{kJ mol}^{-1}\ \text{(3 s.f.)}$

The data-book value is about -1367 kJ mol^-1. The experimental value is far less exothermic, mainly because of heat lost to the surroundings and incomplete combustion.

Worked example 4: ΔrH from average bond enthalpies

Use the average bond enthalpies (kJ mol^-1) to calculate Δ_rH for the combustion of methane, with all species as gases:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Step 1: count the bonds. CH₄ has 4 C-H; 2O₂ has 2 O=O; CO₂ has 2 C=O; 2H₂O has 2 × 2 = 4 O-H.

Step 2: bonds broken.

$\sum(\text{broken}) = (4 \times 413) + (2 \times 498) = 1652 + 996 = +2648\ \text{kJ mol}^{-1}$

Step 3: bonds made.

$\sum(\text{made}) = (2 \times 805) + (4 \times 464) = 1610 + 1856 = 3466\ \text{kJ mol}^{-1}$

Step 4: broken minus made.

$\Delta_r H = 2648 - 3466 = -818\ \text{kJ mol}^{-1}$

More energy is released making bonds (3466) than is taken in breaking them (2648), so the reaction is exothermic. The data-book Δ_cH^⦵ of methane is about -890 kJ mol^-1: the difference comes from using average bond enthalpies and from water being a liquid, not a gas, under standard conditions.

Worked example 5: finding an unknown bond enthalpy

H₂(g) + Cl₂(g) → 2HCl(g)    Δ_rH = -184 kJ mol^-1

Bond enthalpies: H-H = 436 kJ mol^-1, Cl-Cl = 243 kJ mol^-1. Calculate the H-Cl bond enthalpy.

Step 1: write the broken minus made equation, with x for H-Cl. Two H-Cl bonds are made.

$-184 = (436 + 243) - 2x$

Step 2: rearrange.

$2x = 436 + 243 + 184 = 863$ $x = \frac{863}{2} = 431.5\ \text{kJ mol}^{-1}$ $\text{H-Cl bond enthalpy} = +432\ \text{kJ mol}^{-1}\ \text{(3 s.f.)}$

Where marks go: forgetting the "2" in front of HCl gives 863 kJ mol^-1. A bond enthalpy is always positive; a negative answer means a sign has slipped.

Worked example 6: ΔrH from enthalpy changes of formation

Calculate Δ_rH for the reduction of iron(III) oxide in a blast furnace:

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

Step 1: formation cycle formula (both arrows point up from the elements).

$\Delta_r H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})$

Step 2: products, multiplied by moles.

$\sum \Delta_f H(\text{products}) = (2 \times 0) + (3 \times -393.5) = -1180.5\ \text{kJ mol}^{-1}$

Step 3: reactants, multiplied by moles.

$\sum \Delta_f H(\text{reactants}) = -824.2 + (3 \times -110.5) = -1155.7\ \text{kJ mol}^{-1}$

Step 4: subtract.

$\Delta_r H = -1180.5 - (-1155.7) = -24.8\ \text{kJ mol}^{-1}$

Worked example 7: ΔfH from enthalpy changes of combustion

Calculate Δ_fH^⦵ of ethanol. Ethanol cannot be made directly from carbon, hydrogen and oxygen, so it must be found indirectly.

2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)

Step 1: combustion cycle formula (both arrows point down to CO₂ and H₂O). O₂ is not included: it has no enthalpy change of combustion.

$\Delta_f H = \sum \Delta_c H(\text{reactants}) - \sum \Delta_c H(\text{products})$

Step 2: reactants, multiplied by moles.

$\sum \Delta_c H(\text{reactants}) = (2 \times -394) + (3 \times -286) = -788 + (-858) = -1646\ \text{kJ mol}^{-1}$

Step 3: subtract the product.

$\Delta_f H = -1646 - (-1367) = -279\ \text{kJ mol}^{-1}$

Worked example 8: an unfamiliar cycle (indirect determination)

A student cannot measure Δ_rH for CaCO₃(s) → CaO(s) + CO₂(g) directly. Using Figure 7, she finds:

• ΔH₁ = -16 kJ mol^-1 for CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
• ΔH₂ = -194 kJ mol^-1 for CaO(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l)

Calculate Δ_rH.

Step 1: trace the indirect route. From CaCO₃, go down ΔH₁ (with the arrow), then up ΔH₂ (against the arrow, so subtract it).

$\Delta_r H = \Delta H_1 - \Delta H_2$

Step 2: substitute.

$\Delta_r H = -16 - (-194) = +178\ \text{kJ mol}^{-1}$

The answer is positive, as expected: thermal decomposition is endothermic.

Common exam mistakes

• Using the wrong mass in q = mcΔT. The mass is the water or solution being heated. Students use the mass of the solid added, the mass of fuel burned, or add the solid to the solution.
• Forgetting both volumes when two solutions are mixed: 30.0 cm³ + 30.0 cm³ means m = 60.0 g.
• Dividing by the moles of the reagent in excess. Work out the moles of both reactants first and divide by the one that runs out.
• Leaving off the sign. A temperature rise means ΔH is negative: "-219 kJ mol^-1", not "219 kJ mol^-1". The sign carries its own mark.
• Mixing J and kJ. q comes out in J; ΔH is quoted in kJ mol^-1, so divide by 1000.
• Adding 273 to ΔT. A temperature change of 13.6 °C is a change of 13.6 K. Do not convert it.
• Using the highest reading instead of the extrapolated temperature when a graph is given.
• Definition of Δ_fH^⦵ missing a condition. It needs one mole of compound, from its elements, standard conditions and standard states. The two most often left out are "from its elements" and "all reactants and products in their standard states".
• Definition of Δ_cH^⦵ without "completely" or "in oxygen", or per mole of oxygen instead of per mole of the substance.
• Defining Δ_neutH^⦵ per mole of acid. It is per mole of water formed.
• Writing equations that do not fit the definition: 2Na(s) + Cl₂(g) → 2NaCl(s) for formation (must be 1 mol NaCl, so ½Cl₂), or 2C₂H₆ + 7O₂ for combustion (must be 1 mol C₂H₆, so 3½O₂), or leaving off state symbols.
• Standard conditions written as "room temperature and pressure". Quote the values: 100 kPa and 298 K.
• Activation energy as "the energy needed to start a reaction". It must be the minimum energy.
• Saying bond breaking releases energy. Breaking bonds is endothermic; making bonds is exothermic.
• Bond counting slips: not multiplying by the balancing number (2O₂ is two O=O bonds), missing bonds (CO₂ has two C=O), or calculating "made minus broken".
• Hess' law slips: forgetting to multiply by the moles in the equation, putting a non-zero Δ_fH in for an element, including O₂ in a combustion sum, or using "products minus reactants" with combustion data.
• Vague evaluation. "Human error" or "heat loss" alone is weak; write "heat lost to the surroundings", "incomplete combustion" or "evaporation of the fuel", and link it to the value being less exothermic.
• Wrong sig figs. Give the final answer to the same number of significant figures as the least precise data (usually 3).

Key definitions

• Exothermic: heat energy is transferred from the system to the surroundings; the products have less enthalpy than the reactants; ΔH is negative.
• Endothermic: heat energy is transferred from the surroundings to the system; the products have more enthalpy than the reactants; ΔH is positive.
• Activation energy, E_a: the minimum energy required for a reaction to take place.
• Standard conditions: a pressure of 100 kPa and a stated temperature, usually 298 K.
• Standard state: the physical state of a substance under standard conditions.
• Enthalpy change of reaction, Δ_rH: the enthalpy change for a stated equation, in the molar quantities shown in that equation.
• Standard enthalpy change of formation, Δ_fH^⦵: the enthalpy change that takes place when one mole of a compound is formed from its elements, under standard conditions, with all reactants and products in their standard states.
• Standard enthalpy change of combustion, Δ_cH^⦵: the enthalpy change that takes place when one mole of a substance reacts completely with oxygen, under standard conditions, with all reactants and products in their standard states.
• Standard enthalpy change of neutralisation, Δ_neutH^⦵: the enthalpy change that takes place when an acid and a base react to form one mole of water, under standard conditions, with all reactants and products in their standard states.
• Average bond enthalpy: the energy needed to break one mole of a given type of bond in gaseous molecules, averaged over many compounds (formal definition not required).
• Hess' law (definition not required): the enthalpy change for a reaction is independent of the route taken.

Specification

• I can explain that reactions are exothermic (ΔH negative) or endothermic (ΔH positive), in terms of heat transfer between system and surroundings. (3.2.1 a)
• I can draw and label enthalpy profile diagrams for exothermic and endothermic reactions, showing ΔH. (3.2.1 b)
• I can explain activation energy as the minimum energy required for a reaction to take place, and show it on an enthalpy profile diagram. (3.2.1 c)
• I can state standard conditions (100 kPa and 298 K) and explain what a standard state is. (3.2.1 d i)
• I can use the term enthalpy change of reaction for a stated equation. (3.2.1 d ii)
• I can define the standard enthalpy changes of formation, combustion and neutralisation, and write an equation with state symbols for each. (3.2.1 d iii to v)
• I can calculate an enthalpy change from experimental results using q = mcΔT, with the correct mass, moles, sign, units and significant figures. (3.2.1 e)
• I can explain the term average bond enthalpy, and explain why an actual bond enthalpy may differ from the average value. (3.2.1 f i)
• I can explain exothermic and endothermic reactions in terms of the energy taken in breaking bonds and released making bonds. (3.2.1 f ii)
• I can use average bond enthalpies to calculate an enthalpy change, or an unknown bond enthalpy. (3.2.1 f iii)
• I can construct a Hess' law cycle and calculate an enthalpy change of reaction from enthalpy changes of combustion. (3.2.1 g i)
• I can construct a Hess' law cycle and calculate an enthalpy change of reaction from enthalpy changes of formation. (3.2.1 g ii)
• I can use an unfamiliar enthalpy cycle to calculate an enthalpy change. (3.2.1 g iii)
• I can describe how to measure enthalpy changes directly (solution calorimetry, combustion with a spirit burner, extrapolating a cooling curve) and indirectly (two measured reactions linked by a Hess cycle), and evaluate the sources of error. (3.2.1 e, h; PAG3)

Related notes

• Amount of substance: moles from mass and from concentration and volume
• Acids: neutralisation equations for acids with bases and alkalis
• Bonding and structure: average bond enthalpy as a measure of covalent bond strength
• Reaction rates: activation energy, the Boltzmann distribution and catalysts
• Chemical equilibrium: using the sign of ΔH to predict the effect of temperature on equilibrium position
• Lattice enthalpy (A level): Born-Haber cycles and enthalpy changes of solution and hydration
• Enthalpy and entropy (A level): why some endothermic reactions are still feasible