Notes · ChemistryUK · A-Levels

Energetics

Every chemical reaction is accompanied by an energy change, released to or taken from the surroundings as bonds break and form. This chapter defines enthalpy change and reaction profiles, measures enthalpy changes by calorimetry, and calculates them by two routes that never require the reaction to be performed: Hess's law with enthalpy cycles, and mean bond enthalpies.

4 sections~13 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 1

T·0444 / 18

Exam profile

AO1 · Define enthalpy change, the standard enthalpy changes, Hess's law and mean bond enthalpyAO2 · Calculate enthalpy changes from calorimetry data, Hess cycles and bond enthalpiesAO3 · Evaluate experimental enthalpy determinations and account for heat loss and non-standard conditions

Operators:statedefinecalculatedetermineexplainevaluate

basic level

AS-Level requires enthalpy changes, reaction profiles, q = mc(delta T) calorimetry, Hess's law and mean bond enthalpies.

higher level

The full A-Level builds on this foundation with Born-Haber cycles, entropy and free energy in the thermodynamics topic, so a secure understanding of enthalpy cycles here is essential.

Depth

Text

Contents · 4 sections

Energetics

Enthalpy change and reaction profiles#

FoundationAQA 7405 3.1.4.1DfE GCE Chemistry - energetics

Exothermic reaction profile

Fig. 1Exothermic: the products lie below the reactants, so delta H is negative; the barrier is the activation energy.

The enthalpy change

\Delta H

of a reaction is the heat energy transferred at constant pressure, and it is defined as the enthalpy of the products minus the enthalpy of the reactants:

\Delta H = H_{\text{products}} - H_{\text{reactants}}

. It is measured in

\text{kJ mol}^{-1}

. In an exothermic reaction the system releases energy to the surroundings, the products have lower enthalpy than the reactants, and

\Delta H

is negative (the surroundings warm up). In an endothermic reaction the system absorbs energy, the products have higher enthalpy, and

\Delta H

is positive (the surroundings cool down).

Energy is required to break bonds (an endothermic process) and released when bonds form (an exothermic process). Whether a reaction is overall exothermic or endothermic depends on the balance: if more energy is released making new bonds than is taken in breaking old ones, the reaction is exothermic. This single idea - break bonds in, make bonds out - underlies both the reaction profile and the bond-enthalpy calculation later in the chapter.

A reaction profile plots enthalpy against the reaction pathway (the reaction coordinate). Reactants and products appear as horizontal levels; between them is a hump whose peak is the transition state. The height of the hump above the reactants is the activation energy

E_a

, the minimum energy colliding particles must have to react. The difference in level between reactants and products is

\Delta H

: a drop for exothermic, a rise for endothermic. The activation energy exists even for exothermic reactions, which is why a mixture of petrol and air is stable until a spark provides

E_a

Reading and drawing these profiles correctly is a common exam skill. For an exothermic reaction the products line sits below the reactants line, and

\Delta H

is the (negative) drop; for an endothermic reaction the products line sits above. The activation energy is always measured from the reactants up to the peak, never from the products. Labelling the axes 'enthalpy' and 'reaction pathway', and marking both

E_a

and

\Delta H

, is expected in a full answer.

\Delta H = H_{\text{products}} - H_{\text{reactants}}

Enthalpy change

Negative for exothermic (products lower), positive for endothermic (products higher).

Endothermic reaction profile

Fig. 2Endothermic: the products lie above the reactants, so delta H is positive.

Worked example

Interpreting a profile

A reaction has an activation energy of 75 kJ mol^-1 and the products lie 120 kJ mol^-1 below the reactants. State delta H and describe the reaction.

Sign of delta H
The products are lower than the reactants, so energy is released and delta H is negative.
Value of delta H
The drop is 120 kJ mol^-1, so delta H = -120 kJ mol^-1.
Classify
A negative delta H means the reaction is exothermic; the 75 kJ mol^-1 barrier must still be supplied to start it.

Result: delta H = -120 kJ mol^-1; the reaction is exothermic with an activation energy of 75 kJ mol^-1.

Typical mistakes

Measuring the activation energy from the products instead of from the reactants.
Giving delta H the wrong sign - exothermic is negative, endothermic is positive.

Active revision

Sketch a fully labelled reaction profile for an exothermic reaction with a high activation energy, marking Ea, delta H, reactants, products and the transition state.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Chemistry 7405 specification (AQA)

Standard enthalpy changes and calorimetry#

StandardAQA 7405 3.1.4.1DfE GCE Chemistry - calorimetry

Calorimetry of a burning fuel

Fig. 3Energy from the burning fuel heats a known mass of water; q = mc(delta T) gives the heat transferred.

To make enthalpy changes comparable they are quoted under standard conditions - a pressure of

100\ \text{kPa}

, a stated temperature (usually

298\ \text{K}

) and substances in their standard states - and denoted

\Delta H^{\ominus}

. The key defined changes are: the standard enthalpy of reaction; the standard enthalpy of formation

\Delta H_f^{\ominus}

(one mole of a compound formed from its elements in their standard states, so

\Delta H_f^{\ominus}

of an element is zero); the standard enthalpy of combustion

\Delta H_c^{\ominus}

(one mole of a substance burned completely in oxygen); and the standard enthalpy of neutralisation (one mole of water formed from an acid and an alkali).

Enthalpy changes are measured by calorimetry, using the relationship

q = mc\,\Delta T

, where

q

is the heat energy transferred,

m

the mass of the substance being heated (usually water or the solution),

c

its specific heat capacity (

4.18\ \text{J g}^{-1}\text{K}^{-1}

for water) and

\Delta T

the temperature change. To find a molar enthalpy change, divide

q

by the number of moles reacting and attach the correct sign: a temperature rise means an exothermic reaction, so

\Delta H

is negative.

For a combustion experiment a measured mass of fuel is burned to heat a known mass of water; for a reaction in solution the reactants are mixed in an insulated cup and the temperature change of the solution is recorded. In the latter case it is usual to assume the solution has the density and specific heat capacity of water. Careful practice - a lid, a stirrer, an accurate thermometer, and extrapolating a temperature-time graph back to the moment of mixing - improves accuracy.

Experimental values are almost always less exothermic than data-book values, and evaluating why is a standard exam demand. The main reasons are heat loss to the surroundings (especially in an open combustion experiment), incomplete combustion producing carbon monoxide and soot, evaporation of the fuel, and the reaction not occurring under standard conditions. Naming these specific sources, and suggesting improvements such as insulation, a draught shield or a bomb calorimeter, gains the marks.

q = mc\,\Delta T

Heat transferred

m = mass of water/solution (g), c = 4.18 J g^-1 K^-1 for water, delta T = temperature change (K or C).

\Delta H = -\dfrac{q}{n}

Molar enthalpy change

Divide the heat by the moles reacting; the minus sign makes an exothermic reaction (temperature rise) come out negative.

Worked example

Enthalpy of combustion from calorimetry

Burning 0.46 g of ethanol (Mr = 46.0) raises the temperature of 100 g of water by 30.0 K. Calculate the enthalpy of combustion. (c = 4.18 J g^-1 K^-1.)

Heat gained by the water
$q = mc\,\Delta T = 100 \times 4.18 \times 30.0 = 12540\ \text{J} = 12.54\ \text{kJ}$ .
$q = 100 \times 4.18 \times 30.0 = 12540\ \text{J}$
Moles of ethanol
$n = m/M = 0.46/46.0 = 0.0100\ \text{mol}$ .
Enthalpy per mole
$\Delta H_c = -q/n = -12.54/0.0100 = -1254\ \text{kJ mol}^{-1}$ .

Result: The experimental enthalpy of combustion is -1254 kJ mol^-1; it is less exothermic than the data-book -1367 kJ mol^-1 mainly because of heat loss to the surroundings and incomplete combustion.

Typical mistakes

Using the mass of the fuel instead of the mass of the water in q = mc(delta T).
Forgetting the negative sign for an exothermic change, or leaving q in J when delta H is wanted in kJ.

Active revision

Burning $0.32\ \text{g}$ of methanol ( $M_r = 32.0$ ) raises the temperature of $150\ \text{g}$ of water by $18.0\ \text{K}$ . Calculate the experimental enthalpy of combustion of methanol.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Hess's law and enthalpy cycles#

StandardAQA 7405 3.1.4.2DfE GCE Chemistry - Hess's law

Hess cycle using enthalpies of formation

Fig. 4The direct route (delta Hr) equals the indirect route via the elements; travelling up the formation arrows changes their sign.

Hess's law states that the total enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same. It is a direct consequence of the conservation of energy: because enthalpy is a state function, the change depends only on the start and end points, not on the path. This lets us calculate enthalpy changes that are impossible or dangerous to measure directly, by constructing an alternative route through substances whose enthalpy changes are known.

An enthalpy cycle is drawn as a triangle. The direct reaction (reactants to products) is one side, with the unknown

\Delta H_r

; the other two sides go via a common set of substances (the elements, or the combustion products). Hess's law then says the direct route equals the sum of the changes along the indirect route. In practice you follow the arrows: if you must travel against an arrow, you reverse the sign of that enthalpy change.

Two standard cases give shortcut formulae. When formation data are given, the enthalpy of reaction is the sum of the standard enthalpies of formation of the products minus that of the reactants:

\Delta H_r = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})

. When combustion data are given, the roles reverse:

\Delta H_r = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})

. Recognising which data you have been given tells you at once which formula to use.

Setting the cycle out carefully avoids the commonest errors. Remember to multiply each enthalpy of formation or combustion by the number of moles in the balanced equation, to include every substance, and to take the enthalpy of formation of an element as zero. A quick check is that a sensible answer for a combustion should be strongly negative; a positive value for a combustion is a warning that a sign or a direction has gone wrong.

\Delta H_r = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})

Using formation data

Multiply each by its number of moles; the enthalpy of formation of an element is zero.

\Delta H_r = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})

Using combustion data

Use this form when the data given are enthalpies of combustion.

Worked example

Enthalpy of reaction from formation data

Calculate the enthalpy of combustion of methane, CH4 + 2O2 -> CO2 + 2H2O, given delta Hf (kJ mol^-1): CH4 -74.8, CO2 -393.5, H2O(l) -285.8, O2 0.

Sum for products
$\sum \Delta H_f(\text{products}) = (-393.5) + 2(-285.8) = -965.1\ \text{kJ mol}^{-1}$ .
Sum for reactants
$\sum \Delta H_f(\text{reactants}) = (-74.8) + 2(0) = -74.8\ \text{kJ mol}^{-1}$ .
Apply Hess's law
$\Delta H_r = -965.1 - (-74.8) = -890.3\ \text{kJ mol}^{-1}$ .
$\Delta H_r = -965.1 - (-74.8) = -890.3\ \text{kJ mol}^{-1}$

Result: The enthalpy of combustion of methane is -890.3 kJ mol^-1, strongly exothermic as expected.

Typical mistakes

Mixing up the formation and combustion formulae (products - reactants for formation, reactants - products for combustion).
Forgetting to multiply an enthalpy of formation or combustion by the number of moles, or omitting a substance.

Active revision

Using enthalpies of formation (kJ mol^-1): C2H5OH(l) -278, CO2(g) -394, H2O(l) -286, calculate the enthalpy of combustion of ethanol, C2H5OH + 3O2 -> 2CO2 + 3H2O.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Mean bond enthalpies and their limitations#

AdvancedAQA 7405 3.1.4.3DfE GCE Chemistry - bond enthalpies

Bond enthalpy balance for a reaction

Fig. 5For H2 + Cl2 -> 2HCl, more energy is released making bonds than is taken in breaking them, so the reaction is exothermic.

A bond enthalpy is the energy needed to break one mole of a particular covalent bond in the gaseous state. Because the same type of bond has slightly different strengths in different molecules (a

\text{C-H}

bond in methane is not identical to one in ethanol), the value tabulated is a mean bond enthalpy, averaged over many compounds. Breaking bonds is always endothermic (positive) and forming bonds is exothermic (negative), which is the basis of the calculation.

To estimate an enthalpy of reaction from bond enthalpies, add up the energy to break all the bonds in the reactants and subtract the energy released forming all the bonds in the products:

\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds made})

. Everything is treated in the gas phase. Because breaking bonds is the positive term, a reaction that makes stronger bonds than it breaks comes out negative (exothermic). Care with the number of each bond in the balanced equation is essential.

Values calculated from mean bond enthalpies are only estimates, and comparing them with the more accurate values from Hess cycles is a favourite discussion point. The discrepancy arises because mean bond enthalpies are averages taken across many different molecules, whereas the actual bonds in the specific reactants and products differ slightly from those averages. Bond-enthalpy calculations also assume all species are gaseous, so they ignore the enthalpy changes involved in melting, vaporising or dissolving.

In practice, therefore, use a Hess cycle with formation or combustion data whenever precise data are available, and reserve mean bond enthalpies for quick estimates or for reactions where formation data are missing. Being able to state clearly why the two methods differ - averaging over many compounds, and the gas-phase assumption - and which is more reliable, is exactly the kind of AO3 evaluation the examiners reward.

\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds made})

Enthalpy from bond enthalpies

Breaking bonds is endothermic (positive), forming bonds exothermic; use the numbers of each bond from the balanced equation.

Worked example

Enthalpy of formation of HCl from bond enthalpies

Estimate delta H for H2(g) + Cl2(g) -> 2HCl(g) using bond enthalpies (kJ mol^-1): H-H 436, Cl-Cl 243, H-Cl 432.

Bonds broken (reactants)
One H-H and one Cl-Cl: $436 + 243 = 679\ \text{kJ mol}^{-1}$ (endothermic, positive).
Bonds made (products)
Two H-Cl bonds: $2 \times 432 = 864\ \text{kJ mol}^{-1}$ (exothermic, released).
Combine
$\Delta H = 679 - 864 = -185\ \text{kJ mol}^{-1}$ .
$\Delta H = 679 - 864 = -185\ \text{kJ mol}^{-1}$

Result: delta H = -185 kJ mol^-1; the reaction is exothermic because the two H-Cl bonds released more energy than was needed to break H-H and Cl-Cl.

Typical mistakes

Reversing the formula to bonds made minus bonds broken and getting the wrong sign.
Miscounting the number of a given bond, for example forgetting there are four C-H bonds in methane.

Active revision

Using mean bond enthalpies (kJ mol^-1) C-H 412, O=O 498, C=O 805, O-H 463, estimate the enthalpy of combustion of methane, CH4 + 2O2 -> CO2 + 2H2O.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Chemistry 7405 specification

Energetics

4 sections~13 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 1

Energetics

Key points

Interpreting a profile

Key points

Enthalpy of combustion from calorimetry

Key points

Enthalpy of reaction from formation data

Key points

Enthalpy of formation of HCl from bond enthalpies

Sources

Energetics

Key points

Interpreting a profile

Key points

Enthalpy of combustion from calorimetry

Key points

Enthalpy of reaction from formation data

Key points

Enthalpy of formation of HCl from bond enthalpies

Sources