Energetics

Every chemical reaction takes in or gives out energy, because bonds have to be broken and new ones made. This topic shows you how to define, measure and calculate these enthalpy changes: with a thermometer and a polystyrene cup, with Hess cycles that find values nobody can measure directly, and with bond enthalpies.

Enthalpy changes

In energetics we split the world into two parts. The system is the chemicals that are reacting. The surroundings are everything else: the water they are dissolved in, the beaker, the thermometer, the air. When a reaction happens, energy moves between the two, and the total amount of energy stays the same.

The heat content of a system is called its enthalpy, \(H\). Nobody can measure the enthalpy of a substance on its own, but we can measure how much it changes. The enthalpy change, ΔH, is the heat energy transferred between the system and the surroundings when a reaction takes place at constant pressure. Most reactions in a lab happen in open containers, at the pressure of the atmosphere, so this is the quantity we actually measure.

• Exothermic: the system gives heat energy out to the surroundings, so the surroundings warm up. The products have less enthalpy than the reactants, so ΔH is negative. Combustion, neutralisation and respiration are all exothermic.
• Endothermic: the system takes heat energy in from the surroundings, so the surroundings cool down (or the reaction needs continuous heating). The products have more enthalpy than the reactants, so ΔH is positive. Thermal decomposition, such as CaCO₃(s) → CaO(s) + CO₂(g), and photosynthesis are endothermic.

Why do reactions transfer energy at all? Breaking a bond always takes in energy, because you have to pull the atoms apart against the attraction between the bonding electrons and the two nuclei. Making a bond always gives out energy. If the bonds made in the products are stronger overall than the bonds broken in the reactants, more energy comes out than goes in and the reaction is exothermic. If the bonds broken are stronger overall, the reaction is endothermic.

Enthalpy level diagrams and reaction profiles

An enthalpy level diagram puts the reactants and products at heights that show their enthalpy. The ΔH arrow always starts at the reactants and points to the products, so it points down for an exothermic reaction and up for an endothermic one. A reaction profile adds the path between them. Even an exothermic reaction has to climb a hump first: some bonds must start to break before new ones can form. The height of that hump above the reactants is the activation energy, E_a, the minimum energy that colliding particles need for a reaction to happen.

Standard conditions

The size of an enthalpy change depends on the temperature, the pressure, the concentrations and the physical states of the substances. Water vapour condensing to liquid water gives out extra energy, for example, so burning methane to make H₂O(l) gives a bigger ΔH than burning it to make H₂O(g). To compare values fairly, chemists quote them under standard conditions and mark them with the symbol ⦵, as in ΔH^⦵.

• Pressure: 100 kPa.
• Temperature: a stated temperature, normally 298 K (25 °C).
• Solutions: concentration 1.00 mol dm⁻³.
• Standard states: every substance is in the physical state it normally has at 100 kPa and 298 K. Water is H₂O(l), oxygen is O₂(g), bromine is Br₂(l), iodine is I₂(s), carbon is C(s) in the form of graphite.

ΔH values are quoted in kJ mol⁻¹. The "per mole" refers to the equation as written: if you double every number in the equation, you double ΔH.

Standard enthalpy change definitions

You need to know five definitions word for word in meaning. Each one fixes three things: what reacts, how much (usually exactly one mole of something), and the conditions and states.

The equation must make exactly one mole of the compound, which often means fractions in front of the elements:

2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)    Δ_fH^⦵ = −277 kJ mol⁻¹
Na(s) + ½Cl₂(g) → NaCl(s)    Δ_fH^⦵ = −411 kJ mol⁻¹

The standard enthalpy of formation of an element in its standard state is zero. Forming O₂(g) from O₂(g) is no change at all. This is not true of other forms of the element: O₃(g) or Br₂(g) have non-zero values.

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)    Δ_cH^⦵ = −1367 kJ mol⁻¹

Combustion is always exothermic. Complete combustion of a compound of carbon and hydrogen gives only CO₂ and H₂O. With too little oxygen you get incomplete combustion, making carbon monoxide and soot (carbon). Less energy is released, because the carbon has not been fully oxidised. Notice also that the combustion of hydrogen, H₂(g) + ½O₂(g) → H₂O(l), is the same reaction as the formation of water, so Δ_cH^⦵(H₂) = Δ_fH^⦵(H₂O(l)) = −286 kJ mol⁻¹. In the same way, Δ_cH^⦵(C, graphite) = Δ_fH^⦵(CO₂) = −394 kJ mol⁻¹.

For any strong acid with any strong alkali the value is about −57 kJ mol⁻¹. The reason is that strong acids and alkalis are fully ionised, so the only reaction that actually happens is the same every time:

H⁺(aq) + OH⁻(aq) → H₂O(l)

The other ions (Na⁺, Cl⁻, K⁺, NO₃⁻ and so on) are spectators. A weak acid such as ethanoic acid gives a slightly less exothermic value, because some energy is used up in ionising the acid as it reacts. The definition is per mole of water: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O makes two moles of water, so the enthalpy change for that equation is about twice Δ_neutH^⦵.

Na(s) → Na(g)    Δ_atH^⦵ = +107 kJ mol⁻¹
½Cl₂(g) → Cl(g)    Δ_atH^⦵ = +122 kJ mol⁻¹

Atomisation is always endothermic, because bonds (metallic bonds in sodium, the covalent Cl–Cl bond in chlorine) have to be broken. For chlorine, the definition is per mole of Cl atoms, so it is half the energy needed to break one mole of Cl–Cl bonds.

Measuring enthalpy changes: calorimetry

You can't measure the heat a reaction gives out directly. What you can measure is its effect: the temperature change of a known mass of water or solution that absorbs the heat. The energy needed to warm a substance depends on its mass, on how far its temperature changes, and on how much energy that substance needs per gram per degree, its specific heat capacity, \(c\).

\[ Q = mc\,\Delta T \]

• Q: heat energy transferred, in J.
• m: the mass of the water or solution whose temperature changes, in g. Not the mass of the reactants, and not the mass of any solid added.
• c: specific heat capacity. For water and dilute solutions, use 4.18 J g⁻¹ K⁻¹.
• ΔT: temperature change. A change of 1 K is the same size as a change of 1 °C, so either scale works for ΔT.

\(Q\) is the energy for the amounts you actually used. To turn it into ΔH you divide by the number of moles of the substance that sets the amount of reaction, then convert J into kJ and add the sign: a temperature rise means the reaction gave out heat, so ΔH is negative.

\[ \Delta H = \frac{-Q}{n} \]

Reactions in solution

For neutralisation, displacement, dissolving and similar reactions, the solution itself absorbs the heat. The reaction is done in an expanded polystyrene cup, which is a good insulator and has a very small heat capacity, so almost all the energy stays in the solution.

• Stand the polystyrene cup in a beaker for support and extra insulation, and use a lid with a hole for the thermometer.
• Measure the solution into the cup with a pipette or burette. Weigh any solid reactant accurately, preferably by weighing the container before and after adding the solid.
• Record the temperature every minute for 3 minutes to get a steady starting value. If two solutions are mixed, measure both and use the average.
• At the fourth minute add the second reactant, stir, and keep recording every half-minute or minute for several more minutes. Stirring makes sure all of the solution is at the same temperature.
• Plot temperature against time and extrapolate the cooling line back to the moment of mixing to find ΔT.

Why extrapolate? While the reaction is still giving out heat, the warm solution is already losing some to the room. The highest temperature you read is therefore lower than it would have been if all the heat had been released instantly. The steady cooling afterwards shows how fast heat is being lost; following that line back to the time of mixing estimates the temperature the solution would have reached with no losses. The slower the reaction, the more this matters.

Sources of error and assumptions

• Heat loss (or gain) to the surroundings, the biggest error. It makes the measured ΔH less exothermic (or less endothermic) than the true value.
• Heat capacity of the apparatus is ignored: the cup and thermometer absorb a little energy too.
• The solution is treated as water: density 1.00 g cm⁻³ and c = 4.18 J g⁻¹ K⁻¹.
• The reaction may be incomplete or slow, or a solid may not fully dissolve.
• Uncertainty in ΔT: a thermometer reading has an uncertainty of half a division, and ΔT uses two readings. A larger temperature change (more concentrated solutions, or more solid) makes the percentage uncertainty in ΔT smaller.

Enthalpy change of combustion

To measure Δ_cH of a liquid fuel such as an alcohol, burn it in a spirit burner under a copper can containing a known mass of water. Weigh the burner (with its cap on, to stop the fuel evaporating) before and after, measure the water's temperature rise, and use Q = mcΔT with the mass of the water.

Combustion calorimetry is much less accurate than the polystyrene-cup method. Most of the error comes from:

• Heat loss: hot gases from the flame escape around the can and warm the air instead of the water. A draught shield, a lid on the can and a short gap between flame and can all help.
• Incomplete combustion: soot on the bottom of the can shows that some carbon was not burned to CO₂, so less energy was released.
• Evaporation of the fuel from the wick between weighings, making the mass "burned" seem too large.
• The heat capacity of the can is ignored.
• Non-standard conditions: the water formed escapes as steam, not as liquid, so the energy it would release on condensing is lost.

Going up a homologous series, each extra CH₂ group means breaking one more C–C bond, two more C–H bonds and 1½ more O=O bonds, and making two more C=O bonds and two more O–H bonds. The extra amount of energy is the same each time, so Δ_cH becomes more negative by a roughly constant amount, about 650 kJ mol⁻¹ per CH₂: methanol −726, ethanol −1367, propan-1-ol −2021, butan-1-ol −2676 kJ mol⁻¹.

Hess's law

Hess's law follows from the conservation of energy. Imagine two routes from A to B with different enthalpy changes. You could go from A to B by the more exothermic route and come back by the reverse of the other, ending where you started with energy left over. That would create energy from nothing, which never happens.

The law is useful because many enthalpy changes can't be measured directly. You can't make carbon and hydrogen react to give methane in a calorimeter, and you can't heat calcium carbonate while measuring a temperature change. Instead, you build an enthalpy cycle: the direct route from reactants to products, plus an indirect route through some other substances whose enthalpy changes are known. The rule for going round a cycle is simple: follow an arrow and you use its ΔH; go against an arrow and you change its sign.

Using enthalpies of formation

Every compound can be made from its elements, so the elements make a convenient bottom corner for the cycle. Going from reactants to products the indirect way means first going backwards down the left arrow (breaking the reactants back into elements, so the sign changes) and then up the right arrow (forming the products). That gives:

\[ \Delta_\mathrm{r}H = \sum \Delta_\mathrm{f}H(\text{products}) - \sum \Delta_\mathrm{f}H(\text{reactants}) \]

Using enthalpies of combustion

Enthalpies of combustion are easy to measure for anything that burns, and every organic compound burns to the same products, CO₂ and H₂O. So the combustion products make the bottom corner. This time the indirect route goes down the left arrow (burning the reactants) and then backwards up the right arrow (un-burning the products):

\[ \Delta_\mathrm{r}H = \sum \Delta_\mathrm{c}H(\text{reactants}) - \sum \Delta_\mathrm{c}H(\text{products}) \]

Finding enthalpy changes that can't be measured directly

Some reactions are impossible to carry out cleanly in a calorimeter, but you can reach the same products by two reactions that can be measured.

Hydrating copper(II) sulfate. White anhydrous copper(II) sulfate turns into blue hydrated crystals when it takes up water: CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s). You can't add exactly five moles of water and measure the temperature of a solid. Instead, dissolve each solid separately in plenty of water. Both give the same solution, CuSO₄(aq), so they share a corner of the cycle.

Decomposing calcium carbonate. CaCO₃(s) → CaO(s) + CO₂(g) needs strong heating, so you can't measure its heat change with a thermometer. Both calcium carbonate and calcium oxide react with hydrochloric acid to give the same products, and those two reactions can be done in a polystyrene cup. ΔH₁ and ΔH₂ are measured, and ΔH = ΔH₁ − ΔH₂. (The CO₂ from the decomposition simply ends up in the bottom box alongside the CO₂ from the acid reaction.)

Bond enthalpies

Bond enthalpies are always positive, because breaking a bond always needs energy. The strength of a bond depends a little on its surroundings. Breaking the four C–H bonds in methane one after another needs four different amounts of energy, because each time the remaining fragment is different. The C–H bond in ethanol is slightly different again. Data books therefore list a mean value for each type of bond, such as 413 kJ mol⁻¹ for C–H, which is close to, but not exactly, the value in any one molecule.

To estimate ΔH for a reaction between gases, imagine it happening in two stages: break every bond in the reactants to make separate gaseous atoms (energy in), then join the atoms up into the products (energy out). By Hess's law that is the same as the real reaction:

\[ \Delta H = \sum(\text{bond enthalpies of bonds broken}) - \sum(\text{bond enthalpies of bonds made}) \]

Values from mean bond enthalpies are less accurate than values from enthalpies of formation or combustion, for two reasons. The mean values are averages, not the true values for the bonds in these particular molecules. And the method only works for gases: if a substance is a liquid or solid in its standard state, extra energy is involved in separating its molecules, which bond enthalpies leave out. That is why the methane example uses H₂O(g); the standard Δ_cH, which makes H₂O(l), is −890 kJ mol⁻¹.

Working out a bond enthalpy from other data

You can turn the method round and use a known ΔH to find a bond enthalpy. If you need a substance as gaseous atoms and it is not a simple diatomic gas, use its enthalpy of atomisation instead of a bond enthalpy: carbon, C(s) → C(g), needs +717 kJ mol⁻¹.

Bond enthalpies also give a feel for how reactive a substance will be. A reaction usually has to start by breaking a bond, so weak bonds (such as the I–I bond, 151 kJ mol⁻¹) break more easily and lead to faster reactions at room temperature than very strong bonds like N≡N (945 kJ mol⁻¹). That is one reason nitrogen is so unreactive.

Quick checks

Interactive diagrams