Energetics

In a nutshell

Energetics is about the heat energy exchanged in a chemical reaction, called the enthalpy change (ΔH).

You need to measure it by experiment (calorimetry), and calculate it indirectly two ways: with Hess's law and with mean bond enthalpies.

Reactions that release heat are exothermic (ΔH negative); reactions that absorb heat are endothermic (ΔH positive).

Assumed knowledge: Amount of substance, Bonding.

Core content

Enthalpy change, exothermic and endothermic

Enthalpy change (ΔH) is the heat energy change measured at constant pressure.

Energy is transferred between the system (the chemicals reacting) and the surroundings (everything else).

• Exothermic: energy is transferred from the system to the surroundings. The products have less energy than the reactants, so ΔH is negative. Examples: combustion, neutralisation.
• Endothermic: energy is transferred from the surroundings to the system. The products have more energy than the reactants, so ΔH is positive. Example: thermal decomposition of calcium carbonate.

A reaction (energy) profile shows this. The hump is the activation energy (E_a); the gap between the reactant and product levels is ΔH.

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

Think of the reacting chemicals (the system) as a bank account, and ΔH as the change in its balance.

In an exothermic reaction the chemicals give heat away to the surroundings. The account has paid money out, so its balance goes down: the change is a negative number. That is why ΔH carries a minus sign even though the surroundings get hotter.

In an endothermic reaction the chemicals take heat in from the surroundings. Money has come in, the balance goes up, so ΔH is positive, even though the surroundings get colder.

In the exam: if the temperature of the mixture rises, heat has left the chemicals, the reaction is exothermic, and you must write ΔH with a minus sign. Leaving that minus sign off is the single most common lost mark in this topic.

Standard enthalpy changes and standard conditions

Enthalpy changes depend on conditions, so they are quoted under standard conditions, shown by the symbol ^⦵ (for example ΔH^⦵):

• a pressure of 100 kPa
• a stated temperature, usually 298 K
• every substance in its standard state (its physical state at 100 kPa and that temperature)
• any solutions at a concentration of 1 mol dm^-3

Two standard enthalpy changes are defined in this subtopic.

Standard enthalpy of combustion (Δ_cH^⦵) is the enthalpy change when 1 mole of a substance is burned completely in oxygen under standard conditions, with all reactants and products in their standard states. It is always negative (combustion is exothermic).

$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$

Standard enthalpy of formation (Δ_fH^⦵) is the enthalpy change when 1 mole of a compound is formed from its elements, with all substances in their standard states under standard conditions.

$2\text{Fe}(s) + \tfrac{3}{2}\text{O}_2(g) \rightarrow \text{Fe}_2\text{O}_3(s)$

The equation is balanced to give exactly 1 mole of the product, which is why fractions such as 3/2 are allowed. The standard enthalpy of formation of any element in its standard state is zero.

Calorimetry: q = mcΔT

The heat change in a reaction is found from the temperature change it causes:

$q = mc\Delta T$

where q is the heat change (J), m is the mass of the substance being heated (g), c is its specific heat capacity (J g^-1 K^-1) and ΔT is the temperature change (K or °C, which are the same size).

Two rules decide what m is, and they are where marks are won and lost:

• For a reaction in solution (neutralisation, dissolving, displacement), m is the total mass of the solution, not the mass of any solid added. Assume the solution has the density of water (1 g cm^-3) and the specific heat capacity of water (4.18 J g^-1 K^-1).
• For burning a fuel to heat water, m is the mass of the water in the calorimeter, not the mass of fuel burned.

q is only the heat for the quantity you used. To get the molar enthalpy change you divide q by the moles of the reactant that is not in excess, then add a sign and convert J to kJ.

Required practical 2: measuring an enthalpy change

A reaction in solution is carried out in an insulated polystyrene cup and the temperature change measured.

Method:

1. Use a measuring cylinder or pipette to put a measured volume of one solution into the cup.
2. Record its temperature every minute for about 3 minutes to get a steady starting value.
3. At the next minute, add the second measured reagent (or a solid weighed on a balance), stir, and keep recording the temperature every minute for several more minutes.
4. If a solid is used, find its mass by weighing by difference (weigh the container before and after adding it).

Because the mixture starts cooling as soon as it has warmed, the true maximum temperature is higher than any value you actually read. You find it by plotting temperature against time and extrapolating the cooling line back to the time of mixing.

Here the baseline is 20.0 °C and the cooling line extrapolated back to the mixing time reads 34.0 °C, so the true temperature change is ΔT = 14.0 °C, larger than any single reading.

The method carries built-in errors, almost all causing heat loss so the measured enthalpy change is less exothermic than the true value:

• heat is lost to (or gained from) the surroundings
• the specific heat capacity and density of the solution are assumed equal to water
• the heat absorbed by the cup and thermometer is ignored
• reaction may be slow or incomplete

Improvements reduce heat loss: add a lid, insulate the cup, and use the extrapolation above to correct for cooling.

Hess's law

Some enthalpy changes cannot be measured directly. Hess's law lets you calculate them from ones that can.

Hess's law states that the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.

This works because energy is conserved: going from the same reactants to the same products must involve the same total enthalpy change whichever path you follow.

You will use Hess's law with two kinds of data, and the direction of the indirect route is different for each.

From enthalpies of formation (the indirect route goes up from the elements to both reactants and products):

$\Delta H_r = \sum \Delta_f H^{\ominus}(\text{products}) - \sum \Delta_f H^{\ominus}(\text{reactants})$

From enthalpies of combustion (the indirect route goes down from both reactants and products to the combustion products):

$\Delta H_r = \sum \Delta_c H^{\ominus}(\text{reactants}) - \sum \Delta_c H^{\ominus}(\text{products})$

Still don't get it? · which way round to subtract in a Hess cycle

Imagine two routes up a hill from the same car park to the same café: a short direct path, and a long way round through a lower valley. The change in height from car park to café is fixed, so (direct climb) must equal (down into the valley, then up the other side). That fixed height change is ΔH; the valley is the shared set of elements or combustion products.

Build the cycle as a triangle. Put your reaction along the top (reactants → products, that is ΔH_r), and join both corners to the shared box at the bottom with the data you are given.

• With formation data, the arrows point up from the elements into both top corners. Following the cycle, ΔH_r = (go up to products) − (go up to reactants) = ΣΔ_fH(products) − ΣΔ_fH(reactants).
• With combustion data, the arrows point down from both top corners to the combustion products, so it flips: ΔH_r = ΣΔ_cH(reactants) − ΣΔ_cH(products).

Drawing the arrows and reading round the cycle is safer than memorising the formulae, because the arrow directions tell you which term is subtracted. The "products minus reactants" formula only applies to formation data; using it with combustion data gives the wrong sign.

Mean bond enthalpies

Breaking a bond always needs energy (endothermic, positive); making a bond always releases energy (exothermic, negative).

Mean bond enthalpy is the enthalpy change needed to break one mole of a given covalent bond into gaseous atoms, averaged over a range of different compounds.

Every individual bond of one type has a slightly different strength depending on the molecule it is in (the four C–H bonds in methane are not all identical), so a single mean value is used across many compounds.

For a reaction in which all species are gaseous:

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

Why bond enthalpy and Hess's law values differ

The spec asks you to explain this, so give the reasons, not "experimental error":

• Mean bond enthalpies are averaged over many different compounds, so they are not the exact bond enthalpies for the particular molecules in this reaction.
• Mean bond enthalpy calculations assume all species are gaseous, whereas Δ_cH^⦵ and Δ_fH^⦵ apply to substances in their standard states (for example water is a liquid, not a gas).

A Hess's law value from formation or combustion data is therefore more accurate than one from mean bond enthalpies.

Worked examples

Example 1 (calorimetry, a reaction in solution). 50.0 cm³ of 1.00 mol dm^-3 hydrochloric acid is neutralised by 50.0 cm³ of 1.00 mol dm^-3 sodium hydroxide. The temperature rises from 19.50 °C to 26.32 °C. Calculate the enthalpy change per mole of HCl.

Use the total mass of the solution (50.0 + 50.0 = 100 g), not the acid alone:

$q = mc\Delta T = 100 \times 4.18 \times 6.82 = 2850.76 \text{ J}$

Moles of HCl (neither reactant is in excess, 1:1):

$n = c \times V = 1.00 \times \frac{50.0}{1000} = 0.0500 \text{ mol}$

$\Delta H = -\frac{q}{n} = -\frac{2850.76}{0.0500} = -57015 \text{ J mol}^{-1}$

The temperature rose, so the reaction is exothermic and ΔH is negative:

$\Delta H = -57.0 \text{ kJ mol}^{-1} \text{ (3 sf)}$

Example 2 (calorimetry, burning a fuel). Burning 0.320 g of methanol, CH₃OH, raises the temperature of 150 g of water by 11.4 °C. Calculate the enthalpy of combustion of methanol.

Use the mass of the water being heated, not the mass of fuel:

$q = mc\Delta T = 150 \times 4.18 \times 11.4 = 7147.8 \text{ J}$

Moles of methanol (M = 32.0 g mol^-1):

$n = \frac{m}{M} = \frac{0.320}{32.0} = 0.0100 \text{ mol}$

$\Delta_c H = -\frac{q}{n} = -\frac{7147.8}{0.0100} = -714780 \text{ J mol}^{-1} = -715 \text{ kJ mol}^{-1} \text{ (3 sf)}$

This is less exothermic than the data-book value (about −726 kJ mol^-1), mainly because of heat loss to the surroundings.

Example 3 (Hess's law from enthalpies of formation). Calculate the enthalpy of combustion of methane.

$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$

Data (Δ_fH^⦵ / kJ mol^-1): CH₄(g) = −74.8, CO₂(g) = −393.5, H₂O(l) = −285.8. For O₂, an element, Δ_fH^⦵ = 0.

$\Delta H = \sum \Delta_f H^{\ominus}(\text{products}) - \sum \Delta_f H^{\ominus}(\text{reactants})$

$\Delta H = [(-393.5) + 2(-285.8)] - [(-74.8) + 0]$

$\Delta H = (-965.1) - (-74.8) = -890.3 \text{ kJ mol}^{-1}$

Example 4 (Hess's law from enthalpies of combustion). Calculate the enthalpy of formation of ethanol.

$2\text{C}(s) + 3\text{H}_2(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l)$

Data (Δ_cH^⦵ / kJ mol^-1): C(s) = −393.5, H₂(g) = −285.8, C₂H₅OH(l) = −1367. O₂ is an element and does not combust, so it has no Δ_cH^⦵.

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

$\Delta_f H = [2(-393.5) + 3(-285.8)] - [(-1367)]$

$\Delta_f H = (-1644.4) - (-1367) = -277.4 \text{ kJ mol}^{-1}$

Example 5 (mean bond enthalpies). Calculate ΔH for the formation of hydrogen chloride, all species gaseous.

$\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g)$

Data (mean bond enthalpy / kJ mol^-1): H–H = 436, Cl–Cl = 243, H–Cl = 432.

Bonds broken: 1 H–H and 1 Cl–Cl. Bonds made: 2 H–Cl.

$\Delta H = \sum(\text{broken}) - \sum(\text{made}) = (436 + 243) - (2 \times 432)$

$\Delta H = 679 - 864 = -185 \text{ kJ mol}^{-1}$

More energy is released making the bonds than is taken in breaking them, so the reaction is exothermic.

Common exam mistakes

• Leaving the minus sign off an exothermic ΔH. If the temperature rises, the reaction is exothermic and ΔH must be negative. This is the most frequently lost mark in the topic.
• In q = mcΔT, using the mass of the solid or the fuel instead of the mass of the solution (or the water heated).
• Stopping at q and quoting it as the enthalpy change. You must divide q by the moles of the reactant not in excess to get the molar ΔH.
• Adding 273 to the temperature rise. A rise of 6.82 °C is a rise of 6.82 K: ΔT is already correct, so do not convert it.
• For neutralisation, thinking the moles reacting are moles of acid plus moles of alkali. Use the moles of one reactant (they react 1:1).
• Using ΔH = products − reactants with combustion data. That form is only for formation data; combustion data is reactants − products. Getting this backwards inverts the sign.
• Defining mean bond enthalpy as breaking one bond rather than one mole of bonds, or referring to bond making instead of bond breaking.
• Forgetting to count bonds correctly, for example not halving an O=O or missing that one mole of a reactant contains several of the same bond.
• Omitting "in their standard states" from the definitions of Δ_fH^⦵ and Δ_cH^⦵.
• Explaining the Hess-vs-bond-enthalpy difference as "heat loss" or "experimental error". The marks are for mean bond enthalpies being averaged over different compounds (not exact) and the calculation assuming gaseous species.
• Defining ΔH as "energy change" alone, dropping "heat" and "at constant pressure".

Key definitions

• Enthalpy change (ΔH) - the heat energy change measured at constant pressure.
• Exothermic reaction - a reaction in which energy is transferred from the system to the surroundings (ΔH is negative).
• Endothermic reaction - a reaction in which energy is transferred from the surroundings to the system (ΔH is positive).
• Standard conditions - a pressure of 100 kPa and a stated temperature (usually 298 K), with all substances in their standard states (and any solution at 1 mol dm^-3).
• Standard enthalpy of combustion (Δ_cH^⦵) - the enthalpy change when 1 mole of a substance is burned completely in oxygen under standard conditions, all reactants and products being in their standard states.
• Standard enthalpy of formation (Δ_fH^⦵) - the enthalpy change when 1 mole of a compound is formed from its elements under standard conditions, all reactants and products being in their standard states.
• Hess's law - the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.
• Mean bond enthalpy - the enthalpy change needed to break one mole of a given covalent bond into gaseous atoms, averaged over a range of different compounds.

Specification

• I can state whether a reaction is exothermic or endothermic and give the sign of ΔH for each.
• I can define enthalpy change and state that it is measured at constant pressure.
• I can state standard conditions (100 kPa and a stated temperature, e.g. 298 K).
• I can define standard enthalpy of combustion (Δ_cH^⦵).
• I can define standard enthalpy of formation (Δ_fH^⦵).
• I can use q = mcΔT to find the heat change and convert it to a molar enthalpy change.
• I can carry out Required practical 2 and process the data, including extrapolating to find the true temperature change.
• I can state Hess's law.
• I can use Hess's law to calculate an enthalpy change from enthalpies of formation.
• I can use Hess's law to calculate an enthalpy change from enthalpies of combustion.
• I can define mean bond enthalpy.
• I can use mean bond enthalpies to calculate an approximate ΔH for a reaction in the gaseous phase.
• I can explain why mean bond enthalpy values differ from values found using Hess's law.

Related notes

• Amount of substance
• Bonding
• Kinetics