Lesson 6 – Hess’ law and enthalpy cycles using

enthalpy of combustionFriday, April 21, 2023

Objectives

•Recall Hess’ Law •Interpret an enthalpy cycle using Hess’ Law•Construct an enthalpy cycle and use it to calculate an enthalpy change of reaction from combustion data

Specification link2.3.1 Enthalpy Changes, Candidates should be able to:• (a) explain that some chemical reactions are accompanied by enthalpy changes,

exothermic (ΔH, negative) or endothermic ΔH, positive);• (b) describe the importance of oxidation as an exothermic process in the combustion

of fuels and the oxidation of carbohydrates such as glucose in respiration;• (c) describe that endothermic processes require an input of heat energy, i.e., the

thermal decomposition of calcium carbonate;• (d) construct a simple enthalpy profile diagram for a reaction to show the difference

in the enthalpy of the reactants compared with that of the products;• (e) explain qualitatively, using enthalpy profile diagrams, the term activation energy;• (f) define and use the terms:

– (i) standard conditions,– (ii) enthalpy change of reaction,– (iii) enthalpy change of formation,– (iv) enthalpy change of combustion;

• (g) calculate enthalpy changes from appropriate experimental results directly, including use of the relationship: energy change = mcΔT;

• Bond Enthalpies• (h) explain exothermic and endothermic reactions in terms of enthalpy changes

associated with the breaking and making of chemical bonds;• (i) define and use the term average bond enthalpy (ΔH positive; bond breaking of one

mole of bonds);• (j) calculate an enthalpy change of reaction from average bond enthalpies;• Hess’ Law and Enthalpy Cycles• (k) use Hess’ law to construct enthalpy cycles and carry out calculations to

determine:– (i) an enthalpy change of reaction from enthalpy changes of combustion,– (ii) an enthalpy change of reaction from enthalpy changes of formation,– (iii) an enthalpy change of reaction from an unfamiliar enthalpy cycle;

Starter - Test1. What are standard conditions?2. Define ΔHc

3. Define ΔHf

4. Define ΔHr

Describe the following equations as ΔHc, ΔHr or ΔHf

5. 3C(s) + 4H2(g) C3H8(g)6. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)7. C2H4(g) + H2(g) C2H6(g)8. 2C2H6(l) + 7O2(g) 4CO2(g) + 6H2O(l)

Write the equation for ΔHc for9. H2

10. CH3OH11. C

Starter - Answers

Starter - Answers

Measuring enthalpy changes

Problems

Solution

Measuring enthalpy of reaction indirectly• We can use Hess’ law to measure the

energy change of reactions indirectly.• We use enthalpy changes that we can

measure to calculate ones that we can’t.

• This means we need to work out a “cycle” of reactions.

Using ΔHC Ө

Many things burn and chemists love burning things!So we have huge tables of ΔHC

Ө including very many of the most common chemicals.In a reaction which we cannot measure directly, ΔHC

Ө gives us a link between reactants and products.

Hess’ Law

Definition (on sheet)Hess’ law states that, if a reaction can take place by more than one route and the initial and final conditions are the same, the total enthalpy change is the same.

Reactants Products

Intermediate

ΔH(Route A) = ΔH(Route B) – ΔH(Route C)

Using combustion data

The combustion of reactants and combustion of products is a useful way to complete the cycle in a way that can be experimentally determined.

ΔHr = Σ ΔHc(reactants) – Σ ΔHc(products)

Reactants Products

Combustion products

• Example on worksheet

Practice question 1aYou are provided with the following enthalpy changes of combustion

Determine the enthalpy change for the following reaction:

4C(s) + 5H2(g) C4H10(g)

Substance

C(s) H2(g) C4H10(g) C2H5OH(l)

ΔHc / kJmol-1

-394 -286 -2877 -1367

Substance C(s) H2(g) C4H10(g) C2H5OH(l)

ΔHc / kJmol-1 -394 -286 -2877 -1367

4C(s) + 5H2 (g) C4H10(g)

Combustion products

ΔHf = (4 x -394) + (5 x -286) – (-2877) = -129 kJmol-1

Practice question 1bYou are provided with the following enthalpy changes of combustion

Determine the enthalpy change for the following reaction:

2C(s) + 3H2(g) + ½O2(g) C2H5OH(l)

Substance

C(s) H2(g) C4H10(g) C2H5OH(l)

ΔHc / kJmol-1

-394 -286 -2877 -1367

Substance

C(s) H2(g) C4H10(g) C2H5OH(l)

ΔHc / kJmol-1

-394 -286 -2877 -1367

2C(s) + 3H2 (g) + ½ O2(g) C2H5OH(l)

Combustion products

ΔHr = (2 x -394) + (3 x -286) – (-1367) = -279 kJmol-1

Examination question

Mark scheme

Mark scheme