A Level Entropy

© Philip Allan Updates THERMODYNAMICS & KINETICS 48

Entropyand Gibbs

free energy

Gibbsfree energy

What is entropy?

Some applicationsof Gibbs free

energy

Using entropyto predict the

feasibility of a reaction

Teaching AimsTeaching AimsTeaching AimsE N T R O P Y A N DGIBBS FREE ENERGY


Teaching PlanTeaching Plan Teaching PlanE N T R O P Y A N DGIBBS FREE ENERGY

This has traditionally been a difficult topic to get across and requiresclose attention to detail. The teaching plan is therefore quite prescriptive.However, once the basics have been mastered, the more able pupils willusually appreciate the elegance of these concepts.

1 Introduction to entropy (A2)● Entropy may be defined as: ‘The trend in a chemical reaction is from order to disorder’, where

‘order’ means few arrangements of energy and ‘disorder’ means many arrangements of energy.● Therefore for the states of matter:

solids (least entropy) < liquids < gases (highest entropy)

Gases have the highest entropy because they are the most disordered particles (they have the mostways of rearranging the energy within a given number of particles), e.g. ice, liquid water andsteam – look up values of entropy in a data book to show the increase on moving from left to right.

● Thus entropy is a measure of disorder. Symbol = S, units = J K–1mol–1 . Ask students to predictrelative entropy values for a range of compounds and check them in a data book. It will becomeevident that simple molecules have lower entropies than complicated molecules, as there are fewerways of rearranging energy within them.

● Entropy increases with temperature, as the extra heat provides more energy and therefore moreways of rearranging the energy. Whenever there is a change of state there is a large increase inentropy.

Timing: 1 lesson

2 Entropy used to predict the feasibility of reactions (A2)● The Second Law of Thermodynamics states that: ‘In spontaneous change the total entropy must

increase’. Total entropy means that of the system and its surroundings. Entropy can explain whysome reactions occur, even though their enthalpy change is not favourable (i.e. they areendothermic). It is suggested that Practical 1 is performed as this provides a good introductioninto entropy in practice.

● As with enthalpy, it is necessary to take time to ensure that the terms ‘system’ and ‘surroundings’are fully understood, i.e. the ‘system’ comprises the reacting particles and the ‘surroundings’represent anything outside the system (i.e. the surrounding aqueous medium, the atmosphere,etc.).

● The entropy change of the system (∆Ssys) can be calculated from:

∆Ssys = ΣS(products) – ΣS(reactants)

It is worthwhile practising use of this equation with entropy figures from the data book.● The entropy change in the surroundings (∆Ssurr) is given by:

∆Ssurr = –∆H/T (i)

∆H is usually measured in kJ mol–1 which must be multiplied by 1000 to match the units for theentropy (J K–1 mol–1). It can be seen that for exothermic reactions, ∆Ssurr becomes more positive(entropy in the surroundings increases) as the temperature decreases and vice versa.

● The total entropy (∆Stot) is the sum of these two:

∆Stot = ∆Ssys + ∆Ssurr (ii)

And ∆Stot MUST be positive for a reaction to proceed.

● Therefore the total entropy can be found as long as the individual entropies (data book) and theenthalpy change of the reaction are known. Combining (i) and (ii) gives:

∆Stot = ∆Ssys – ∆H/T

Worksheet 1 gives useful practice in the use of these equations together with a revision ofenthalpy. After this worksheet has been corrected, hand out Worksheet 2 which gives moreapplications of the use of entropy (guidance will almost certainly be necessary for Worksheet 2,Questions 2 and 3).

Timing: 4 lessons

3 Gibbs free energy change (A2)● From above: ∆Stot = – ∆H/T + ∆Ssys

Multiply through by T: T∆Stot = −∆H + T∆Ssys

Multiply through by –1: –T∆Stot = ∆H – T∆Ssys (iii)

The term –T∆Stot is called the Gibbs free energy change (∆G). It represents the energy available todo work. A helpful analogy here is the game of squash, i.e. the energy lost through perspiration isnot useful energy, rather it is the energy left to play the shot which is ‘free’ to do useful work.

● As ∆G = –T∆Stot and ∆Stot must be positive, it can be seen that ∆G must be negative for aspontaneous change.

● From (iii) above: ∆G = ∆H – T∆Ssys

This is a very important equation as it gives the full thermodynamic evidence as to whether a reactionwill occur – it takes into account the enthalpy and the entropy. Rearranging this equation gives:

∆H = ∆G + T∆Ssys

This is a good way to help explain free energy, i.e. it is the ‘useful’ part of the enthalpy (T∆Ssys

being the part of enthalpy that is not available for useful work, i.e. the perspiration in the squashgame analogy).

● ∆G follows the same rules as enthalpy in calculations.i.e. ∆GR = Σ∆Gf(products) – Σ∆Gf(reactants)

Resource Sheet 1 summarises the important equations and ideas for this topic.Worksheet 3 gives practice at the use of ∆G in predicting reactions.

Timing: 2 lessons

4 Applications of Gibbs free energy (A2)● The amount of detail here will depend entirely on syllabus requirements. The information is

covered on the following Resource Sheets:2: Free energy and equilibria3: Free energy and electrode potentials4: Ellingham diagrams (also covered under extraction of metals)Worksheet 4 gives practice for these applications.

Timing: Very variable, circa 1–4 lessons

Revision sheet

Test: 40 minutes

Total time for Topic 2: 9–12 lessons

THERMODYNAMICS & KINETICS 50

Teaching PlanTeaching Plan Teaching PlanE N T R O P Y A N DGIBBS FREE ENERGY

© Philip Allan Updates


111E N T R O P Y A N DGIBBS FREE ENERGY

Essential equations for entropy and free energy

● To calculate the entropy of the system in a reaction:

∆Ssys = ΣSproducts – ΣSreactants

● To calculate the entropy of the surroundings in a reaction:

∆Ssurr = –∆H/T

The units of entropy are J K–1 mol–1, therefore ∆H must be in Joules (not kJ).

● The total entropy change is given by:

∆Stotal = ∆Ssys + ∆Ssurr

● ∆G for a reaction can be calculated in the same way as ∆H:

i.e. ∆GR = Σ∆Gf (products) – Σ∆Gf (reactants)

● ∆G, ∆H and ∆S are related by the following equation:

∆G = ∆H – T∆Ssys

● A reaction is only spontaneous at a given temperature when:

∆G ≤ 0

● Also for a reaction to proceed, ∆Stotal must be positive.

● ∆H is often negative, but there is no law regarding the sign of ∆H and spontaneous change.

● However, ∆G, ∆H and ∆S give no indication as to the kinetic feasibility of a reaction, i.e. the rateat which it occurs.


© Philip Allan Updates THERMODYNAMICS & KINETICS

111E N T R O P Y A N DGIBBS FREE ENERGY

Entropy

Consider the following reactions:(i) N2(g) + 3H2(g) → 2NH3(g)(ii) SO2(g) + 2H2S(g) → 3S(s) + 2H2O(l)(iii) MgO(s) + 2HCl(g) → MgCl2(s) + H2O(g)(iv) CO2(g) + Mg(s) → 2MgO(s) + C(s) (v) C2H6(g) + 3 O2(g) → 2CO2(g) + 3H2O(l)

1 Look up enthalpy of formation values in your data book and use the expression:

Enthalpy change (∆HR) = Σ∆Hf(products) – Σ∆Hf(reactants)

to calculate ∆HR for each reaction, giving your answers in kJ.

2 By inspection of the equations only, state whether you would expect the entropy change in the system(∆Ssys) to be positive (indicating that the entropy of the system increases as the reaction moves from leftto right) or negative (indicating that the entropy of the system decreases as the reaction proceeds).Explain each choice by referring to the equation in question.

3 In your data book, look up the entropy values of each participant and calculate the value of ∆Ssys for eachreaction, using the expression:

∆Ssys = ΣSproducts – ΣSreactants

Give your answers in J K–1 mol–1.

Do your calculations confirm your predictions in Question 2?

4 Use your values of ∆HR from Question 1 to calculate the enthalpy of the surroundings for each reactionat 298 K, using

∆Ssurr = – ∆H/T

Give your answers in J K–1 mol–1 (remember to convert your values of ∆HR to J mol–1).

5 Hence calculate the value of ∆Stotal for each reaction by using:

∆Stotal = ∆Ssys + ∆Ssurr

According to your results, which reactions should be spontaneous at 298 K? Comment on any unexpectedanswers.

6 Now summarise your results in the following table, taking great care with signs.

Reaction

(i)

(ii)

(iii)

(iv)

(v)

∆HR(kJ)

∆Ssys(J K–1 mol–1) (J K–1 mol–1) (J K–1 mol–1)

∆Ssurr ∆Stotal Spontaneous?

12

56


E N T R O P Y A N DGIBBS FREE ENERGY

Worksheet 1The method for this worksheet is explained within the questions. The answers given will depend onthe data book used. The table in Question 6 summarises the results; the only qualitative answersrequired are for Questions 2 and 3.

2 For all the reactions, there is more disorder on the left-hand side of the equation; therefore ∆Ssys

would be expected to be negative in all cases.3 This prediction is found to be correct by the figures.6

There are no surprising results, but the more violent reactions (iv and v) are shown to proceed witha large increase in total entropy.

Worksheet 21 (a) When NaCl dissolves there is a large increase in disorder, therefore entropy.

(b) ∆Ssys = (320.9 + 56.5) – 72.4 = +305 J K–1mol–1

∆Ssurr = −3900/298 = –13.1 J K–1mol–1

Therefore, ∆Stot = +292 J K–1mol–1

Despite the unfavourable enthalpy change, the reaction is spontaneous because of the favourablechange in entropy.

2 (a) (i) The following values for ∆Ssys are produced:C2H6 = 87.3; CCl4 = 87.1; CS2 = 85.0. All in J K–1mol–1.

(ii) ∆Ssys is always positive as there is an increase in entropy when a solid is changed into a gas.(b) (i) Water: ∆Ssys = 110; ethanoic acid ∆Ssys = 60.5. All in J K–1mol–1.

(ii) Water is higher than is usual as it hydrogen bonds in the liquid state, giving it more orderthan the other liquids quoted, and the increase in entropy on changing to vapour is greater.Ethanoic acid is less than expected as it forms a dimer in the gas phase: the gas is moreordered than is usual and the increase in entropy is less.

3 (a) C(graphite) + 2H2(g) + O2(g) → CH3OH(l)(b)

(c) Values are in the Hess’s triangle.∆Hat (CH3OH(l)) – 239 = 715 + 872 + 248∆Hat (CH3OH(l)) = +2074 kJ mol–1

(d) ∆Ssurr = –∆H/T = –2074000/298 = –6960 J K–1 mol–1

(e) ∆Ssys will be positive at 298 K (increase in disorder in the reaction). ∆Hat (CH3OH(l)) will occur ata temperature that is high enough to reduce ∆Ssurr to a value that is less than ∆Ssys, thereforemaking ∆Stotal a positive value, which must be the case for a spontaneous change.

CH3OH(l)∆Hat CH3OH(l)

C(g) + 4H(g) + O(g)

C(graph) (g) (g) + 2H2 + 12 O2

+715 ∆Hat (elements)4 × +218 +248

∆Hf= –239

12

Reaction

(i)

(ii)

(iii)

(iv)

(v)

∆HR(kJ)

–92

–235

–97.4

–810

–1560

∆Ssys(J k–1 mol–1)

–199

–424

–123

–187

–310

∆Ssurr(J k–1 mol–1)

310

787

327

2720

5239

∆Stotal(J k–1 mol–1)

111

363

204

2533

4929

Spontaneous?

Yes

Yes

Yes

Yes

Yes

ANSWERS


Teacher/Technician Notes Teacher/Technician Notes Teacher/Technician Notes E N T R O P Y A N DGIBBS FREE ENERGY

Practical 1 Investigating entropy changesEach pupil or pair will need:● Test tubes● Magnesium ribbon (HIGHLY FLAMMABLE)● Tongs● Sodium hydrogencarbonate (solid)● Limewater● Glacial ethanoic acid● Ammonium carbonate (solid)● Zinc granules● Copper sulphate solution (0.5 M)● 2 mol dm–3 HCl(aq)● Splint and access to a Bunsen burner● Eye protection

Results

Equations1 NaCl(s) + aq → NaCl(aq)2 2Mg(s) + O2(g) → 2MgO(s)3 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)4 2CH3COOH(l) + (NH4)2CO3(s) → 2CH3COONH4(aq) + CO2(g) + H2O(l)5 Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

Cu2+(aq) is a catalyst.

Question 2 Entropy allows endothermic reactions to occur spontaneously.

Expt Equation(s) Is ∆H + or –? Entropy of reactants Surroundingsand products (more or less ordered)

1 + Products higher More

2 See – Reactants higher Less

3 + Products higher More

4 below + Products higher More

5 – Products higher Less


111Investigating entropy changes

Safety Notes

Hydrochloric acid is an irritant. Ethanoic acid is flammable and corrosive. If either acid is spilt onto theskin, wash off with plenty of water. If they get into the eyes, irrigate immediately until medical help canbe obtained. Magnesium is highly flammable. Wear eye protection.

TheoryReactions which are endothermic (∆HR is positive) are unlikely to occur in terms of enthalpy change –they are said to have an unfavourable enthalpy change. The aim of this practical is to show why suchreactions do in fact occur (at 298 K or other temperatures) because of a favourable entropy change (∆S).

Experiments1 Place about 5 cm3 of water in a test tube and add a small spatula measure of sodium chloride.

Carefully stir the mixture with a thermometer and measure the temperature change of the mixingprocess.

2 Take a small length of magnesium ribbon and clean it thoroughly with emery paper. Hold the ribbonwith some tongs and observe if there is any sign of reaction without heat. Now hold the ribbon brieflyin a bunsen flame and observe carefully.

3 Place approximately 2 cm depth of sodium hydrogencarbonate in a hard glass test tube. Warm thetube gently at first, and then more strongly, testing to see at which point carbon dioxide starts to bedriven off by ‘pouring’ the gases evolved onto 2 cm3 of lime water and observing for the formation ofa cloudy precipitate.

4 Measure approximately 4 cm3 of ethanoic acid into a test tube and measure its temperature. Now add2.5 g of solid ammonium carbonate and stir the contents of the tube carefully with a thermometer.Note the final temperature.

5 Place two granules of zinc metal in a test tube and cover them with copper sulphate solution.Carefully add 3 cm depth of 2 mol dm–3 HCl(aq) and test for the evolution of hydrogen gas by seeingif a lighted splint pops in the presence of the gas.

Results and questions1 For each experiment:

(a) Write a balanced equation for any reaction(s) occurring.(b) Were the experiments endo- or exothermic? What does this suggest about the energetic

favourability of the reactions?(c) Consider the reactants and products. Which have the highest degree of disorder (entropy)? Are

the reactions favourable in terms of entropy?(d) Consider the surroundings. Will they become more or less disordered as the reactions proceed?



111Show your results and answers to the questions by drawing out and completing the following table:

2 Summarise your understanding of how entropy influences the spontaneity of these reactions.

3 Further work After completing Topic 2 (Entropy and Gibbs free energy), return to this practical andperform calculations (where data allow) to determine ∆Ssys, ∆Ssurr, ∆Stot and ∆G and see if your figuresback up your earlier qualitative work.



Expt Equation(s) Is ∆H + or –? Entropy of reactants Surroundingsand products (more or less ordered)

1

2

3

4

5


In spontaneous change, the total entropy increases.

It is a gas – more energy.

More ways of rearrangingenergy.

∆Ssys = ΣS(products) – ΣS(reactants)

∆Ssurr =

∆Stot = ∆Ssys + ∆Ssurr

∆G = ∆H – T∆S

∆GR = Σ∆Gf(products) –Σ∆Gf(reactants)

If ∆G ≤ 0, then the reactionis feasible.

J K–1 mol–1

No information about rate.

It is the energy that isavailable, i.e ‘free’, to dowork.

– ∆HT



1 What is the Second Law of Thermodynamics?

2 Why does NH3(g) have a higher entropy than NH3(l)?

3 Why does C2H6(g) have a higher entropy than CH4(g)?

4 How can ∆Ssys be calculated?

5 How can ∆Ssurr be calculated?

6 How can ∆Stot be calculated?

7 Write an expression linking ∆S, ∆G and ∆H with temperature (T).

8 How can the free energy change in a reaction be calculated from ∆Gf values?

9 How does ∆G predict if a reaction is feasible?

10 What are the units of entropy?

11 What information do ∆S and ∆G give about the rate of a reaction?

12 What is ‘free’ about free energy?



1 For each of the following reactions, decide whether the entropy of the system increases or decreasesas the reaction proceeds. Write a few words of explanation in each case. (6)(a) 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(l)

(b) CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(aq) + CO2(g)

(c) 3CuO(s) + 2NH3(g) → 3Cu(s) + N2(g) + 3H2O(l)

2 Consider the following reaction at 298 K between nitrogen gas and oxygen gas to form nitrogen monoxide:N2(g) + O2(g) → 2NO(g)

Use the following information to answer the questions which follow:

Entropy (So––)/J K–1 mol–1: N2(g) +191.4; O2(g) +204.9; NO(g) +210.5.Enthalpy of formation (∆Hf)/kJ mol–1: NO(g) +90.4.

(a) Calculate the entropy change of the system (∆Ssys). (2)

(b) Calculate the value of ∆H for the reaction. (2)

(c) Using your answers to parts (a) and (b), calculate the change in free energy of the reaction (∆GR) at298 K. (3)

(d) Indicate, with a reason, whether the reaction is likely to be spontaneous at 298 K. (2)

(e) There are some transition metals that are known to catalyse this reaction. State with a reason whethera catalyst will affect the value of ∆GR. (2)

3 Consider the reaction:

N2O4(g) 2NO2(g)

At what temperature does the reaction become spontaneous?

Relevant information:

So––/J K–1 mol–1 NO2(g) +240; N2O4(g) +304.2∆Hf

o––/kJ mol–1 NO2(g) +33.2; N2O4(g) +9.2 (5)




4 For the reaction: 2SO2(g) + O2(g) 2SO3(g), a graph was plotted of the natural logarithm of theequilibrium constant (ln Kp) against the reciprocal of the absolute temperature (1/T). The graph isshown below:

(a) Using the gradient quoted, calculate a value for the enthalpy change (∆H) stating whether the reactionis exothermic or endothermic.(Gas constant R = 8.314 J K–1 mol–1) (2)

(b) Derive an expression which shows the dependence of ln Kp on temperature T, standard enthalpychange ∆H and the standard entropy change ∆S. (3)

Gradient of line= +2.377 × 104 K

1/T / K–1

In Kp




THERMODYNAMICS & KINETICS

1 (a) Decreases as there are fewer molecules of gas on the right-hand side.(b) Increases as there is a gas on the right-hand side but not the left.(c) Could argue either way as there is a solid and a gas on the left, and a solid, a gas and a liquid on

the right.

2 (a) ∆Ssys = (2 × 210.5) – (191.4 + 204.9) = +24.7 J K–1 mol–1

(b) ∆HR = 2 × 90.4 = +180.8 kJ mol–1

(c) ∆GR = 180.8 – (298 × 0.0247)= 173 kJ mol–1

(d) The positive value of ∆GR suggests that the reaction is not spontaneous at 298 K.(e) No, a catalyst will not affect ∆GR as a catalyst has no effect on the thermodynamics of the reaction; it

acts by lowering the activation energy.

3 ∆Ssys = (2 × 240) – 304.2 = +175.8 J K–1mol–1

∆HR = (2 × 33.2) – 9.2 = +57.2 kJ mol–1

As ∆G = ∆HR – T∆Ssys

When ∆G = 0T = 57.2/0.176 = 325 K

4 (a) As gradient = –∆H/R and the gradient is positive, the reaction must be exothermic (∆H is negative).∆H = –2.377 × 104 × 8.314 = –197624 J = –198 kJ

(b) ∆G = ∆H – T∆S and ∆G = –RT ln Kp

–RT ln Kp = ∆H – T∆S ln Kp = (–∆H/R) × 1/T + ∆S/R

71

ANSWERS



Documents

A Level Entropy