DP Chemistry HL R1.4 Entropy and Spontaneity

DP Chemistry HL R1.4 Entropy and Spontaneity - Full notes

RevisionPrep

DP Chemistry HL | R1.4 Entropy and Spontaneity

DP CHEMISTRY · REACTIVITY

R1.4

Entropy and Spontaneity

Full notes for HL · IB-style

Contents

1	Spontaneous change
2	Entropy as a measure of disorder
3	Predicting the sign of ΔS
4	Calculating standard entropy changes
5	Gibbs energy and spontaneity
6	The effect of temperature on spontaneity
7	Gibbs energy and equilibrium
8	Entropy as dispersal of energy and matter
9	Predicting the sign of ΔS
10	Calculating ΔS from standard entropies
11	Gibbs energy: the criterion for spontaneity
12	The temperature at which a reaction becomes feasible
13	The four combinations of signs
14	Gibbs energy of formation
15	Gibbs energy and equilibrium
16	Kinetic versus thermodynamic stability
17	Entropy and the second law
18	Mixed review: worked problems
19	Additional exam-style practice
20	Entropy of dissolving
21	Entropy changes for phase changes
22	Thermodynamics in biology and industry
23	Thermodynamics checklist
24	Common misconceptions and links
25	Exam-style question with markscheme
26	Check your understanding
27	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
explain what is meant by a spontaneous reaction
define entropy and relate it to the number of ways energy and particles can be arranged
predict the sign of ΔS from changes in state and numbers of gas particles
calculate ΔS⊖ from standard entropy values
calculate ΔG using ΔG = ΔH − TΔS and use it to predict spontaneity
determine the temperature at which a reaction becomes spontaneous
relate ΔG to the position of equilibrium

R1.4 is Additional Higher Level (AHL) content. It is studied by HL students only.

Guiding questions

What determines whether a reaction happens on its own? How do enthalpy and entropy together decide the direction of change?

How this topic is examined

R1.4 is an HL-only subtopic, examined through calculations of ΔS and ΔG, temperature at which ΔG = 0, and qualitative predictions of entropy changes.

Convert ΔS from J K−1 mol−1 to kJ K−1 mol−1 before combining with ΔH in kJ mol−1, and use temperature in kelvin.

Command term	What the examiner wants
State / Identify	a brief answer, no explanation
Outline / Describe	a brief or detailed account, without reasons
Explain	a detailed account with causes, reasons or mechanisms
Calculate / Determine	a numerical answer with working, correct significant figures and units
Deduce / Predict	reach a conclusion from given information
Draw / Sketch	a clear labelled diagram, structure or graph
Evaluate / Discuss	weigh strengths and limitations; reach a judgement

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer comes in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the chemistry shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Spontaneous change

Some changes happen on their own; their reverse does not.

Think about it

Ice melts at room temperature, even though melting is endothermic. So why does it happen? Why doesn't water freeze spontaneously at 25 °C?

A spontaneous process is one that, once started, occurs without continuous outside intervention. Examples: a gas spreading through a room, ice melting at 25 °C, iron rusting, combustion of fuels once ignited. 'Spontaneous' says nothing about speed: diamond turning into graphite is spontaneous but extremely slow (kinetics, R2.2).

Many spontaneous reactions are exothermic, but not all: dissolving ammonium nitrate and melting ice are endothermic yet spontaneous. Enthalpy alone cannot predict spontaneity. A second factor is needed: entropy.

2 Entropy as a measure of disorder

Energy and matter tend to spread out.

Entropy (S) is a measure of the dispersal or distribution of energy and matter in a system: the number of possible ways (microstates) in which the particles and their energy can be arranged. The greater the number of possible arrangements, the higher the entropy. Informally, entropy is often described as disorder.

The second law of thermodynamics states that the total entropy of the universe increases in a spontaneous process. A reaction can decrease the entropy of the system, provided it increases the entropy of the surroundings by more (for example, by releasing heat).

Entropy has units of J K−1 mol−1. Unlike enthalpy, absolute entropies can be measured; a perfect crystal at 0 K has zero entropy. Standard entropy values (S⊖) are always positive.

Substance	State	S⊖ / J K−1 mol−1
C (diamond)	solid	2.4
H2O	liquid	70
H2O	gas	189
CO2	gas	214
O2	gas	205

3 Predicting the sign of ΔS

Gases dominate entropy.

Entropy increases (ΔS positive) when: a substance changes from solid to liquid to gas; the number of moles of gas increases; a solid dissolves in a liquid; temperature increases; particles become more spread out.

Entropy decreases (ΔS negative) when the number of moles of gas decreases, when a gas dissolves or condenses, or when a precipitate forms from solution. Changes in the number of moles of gas usually dominate.

WORKED EXAMPLE 1 Predicting ΔS

Predict the sign of ΔS for (a) CaCO3(s) → CaO(s) + CO2(g); (b) N2(g) + 3H2(g) → 2NH3(g); (c) H2O(l) → H2O(s).

Solution

(a) gas produced from solid: positiveA1

(b) 4 mol gas → 2 mol gas: negativeA1

4 Calculating standard entropy changes

Products minus reactants.

ΔS⊖ = ΣS⊖(products) − ΣS⊖(reactants)

Remember to multiply each S⊖ by its coefficient in the balanced equation. Unlike ΔHf, elements have non-zero standard entropies.

WORKED EXAMPLE 2 Entropy change for the Haber process

Calculate ΔS⊖ for N2(g) + 3H2(g) → 2NH3(g). S⊖: N2 192, H2 131, NH3 193 J K−1 mol−1.

Solution

Σ products = 2(193) = 386M1

Σ reactants = 192 + 3(131) = 585M1

ΔS⊖ = 386 − 585 = −199 J K−1 mol−1: negative, as predictedA1

Common mistake

Assuming elements have zero entropy, as with enthalpy of formation. All substances have positive absolute entropies.

5 Gibbs energy and spontaneity

Combining enthalpy and entropy.

ΔG⊖ = ΔH⊖ − TΔS⊖

The Gibbs energy change (ΔG) combines the enthalpy change of the system with its entropy change at temperature T (in kelvin). It reflects the total entropy change of the universe. A reaction is spontaneous (thermodynamically feasible) when ΔG is negative. When ΔG is positive, the reaction is not spontaneous (the reverse reaction is). When ΔG = 0, the system is at equilibrium.

Units: ΔH is usually in kJ mol−1 and ΔS in J K−1 mol−1, so divide ΔS by 1000 before substituting.

WORKED EXAMPLE 3 Calculating ΔG

For the Haber process at 298 K: ΔH⊖ = −92 kJ mol−1, ΔS⊖ = −199 J K−1 mol−1. Calculate ΔG⊖ and comment.

Solution

ΔS = −0.199 kJ K−1 mol−1M1

ΔG = −92 − 298 × (−0.199) = −92 + 59.3 = −32.7 kJ mol−1A1

Negative: spontaneous at 298 K (though slow without a catalyst)R1

Quick check

1. What sign of ΔG indicates a spontaneous reaction?

2. Why must ΔS be divided by 1000?

Answers: 1. negative 2. to convert J to kJ to match ΔH

6 The effect of temperature on spontaneity

The TΔS term grows with temperature.

Because ΔG = ΔH − TΔS, temperature can change the sign of ΔG. There are four cases:

ΔH	ΔS	ΔG	Spontaneous?
negative	positive	always negative	at all temperatures
positive	negative	always positive	never
negative	negative	negative at low T	only at low temperatures
positive	positive	negative at high T	only at high temperatures

The temperature at which a reaction becomes (or stops being) spontaneous is where ΔG = 0: T = ΔH ÷ ΔS.

WORKED EXAMPLE 4 Temperature for spontaneity

For CaCO3(s) → CaO(s) + CO2(g): ΔH = +178 kJ mol−1, ΔS = +161 J K−1 mol−1. Calculate the minimum temperature for spontaneous decomposition.

Solution

T = ΔH ÷ ΔS = 178 ÷ 0.161M1

= 1106 K (about 833 °C)A1

Above this temperature TΔS exceeds ΔH, so ΔG is negativeR1

Figure 1. ΔG for the decomposition of CaCO3 falls with temperature and becomes negative above about 1106 K.

7 Gibbs energy and equilibrium

ΔG tells us the position of equilibrium.

The value of ΔG⊖ is related to the equilibrium constant K (R2.3) by ΔG⊖ = −RT ln K. If ΔG⊖ is large and negative, K is much greater than 1 and the equilibrium lies far to the right (mainly products). If ΔG⊖ is positive, K is less than 1 (mainly reactants). If ΔG⊖ = 0, K = 1.

As a reaction proceeds, the Gibbs energy of the reaction mixture falls until it reaches a minimum at equilibrium, where ΔG = 0 for the forward and reverse reactions.

WORKED EXAMPLE 5 ΔG and K

Calculate K at 298 K for a reaction with ΔG⊖ = −32.7 kJ mol−1.

Solution

ln K = −ΔG ÷ RT = 32 700 ÷ (8.31 × 298) = 13.2M1

K = e13.2 = 5.4 × 105: equilibrium lies far to the rightA1

Quick check

1. What does a negative ΔG⊖ imply about K?

2. State the relationship between ΔG⊖ and K.

Answers: 1. K > 1: products favoured 2. ΔG⊖ = −RT ln K

8 Entropy as dispersal of energy and matter

Counting arrangements.

Entropy S measures the dispersal (spreading out) of energy and particles: the number of ways the particles and their energy can be arranged. Systems tend to move towards greater dispersal. Entropy has units J K−1 mol−1.

WORKED EXAMPLE 6 Predicting (basic)

Which has higher entropy: 1 mol of H2O(l) or 1 mol of H2O(g) at 100 °C?

Solution

H2O(g)A1

Gas particles are more dispersed and have more possible arrangementsR1

Quick check

1. Order by entropy: NaCl(s), NaCl(aq), Cl2(g).

2. Why does a solid have low entropy?

Answers: 1. NaCl(s) < NaCl(aq) < Cl2(g) 2. particles fixed in an ordered lattice

9 Predicting the sign of ΔS

Count moles of gas.

ΔS is positive when: a solid or liquid becomes a gas; a solid dissolves; the number of moles of gas increases. ΔS is negative when the number of moles of gas decreases or a gas condenses or solidifies.

WORKED EXAMPLE 7 Signs (standard)

Predict the sign of ΔS for (a) CaCO3(s) → CaO(s) + CO2(g); (b) N2(g) + 3H2(g) → 2NH3(g).

Solution

(a) positive: gas formedA1

(b) negative: 4 mol gas → 2 mol gasA1

Quick check

1. Predict the sign of ΔS for 2H2(g) + O2(g) → 2H2O(l).

2. Predict the sign for dissolving KNO3.

Answers: 1. negative 2. positive

10 Calculating ΔS from standard entropies

Products minus reactants.

Unlike ΔHf, standard entropies of elements are not zero (only a perfect crystal at 0 K has S = 0).

WORKED EXAMPLE 8 Ammonia (exam level)

Calculate ΔS for N2 + 3H2 → 2NH3. S⦵: N2 192, H2 131, NH3 193 J K−1 mol−1.

Solution

2(193) − [192 + 3(131)]M1

= −199 J K−1 mol−1A1

Quick check

1. Calculate ΔS for CaCO3 → CaO + CO2 (S: CaCO3 93, CaO 40, CO2 214).

Answers: 1. +161 J K−1 mol−1

11 Gibbs energy: the criterion for spontaneity

ΔG = ΔH − TΔS.

A reaction is thermodynamically spontaneous (feasible) when ΔG < 0. Remember to convert ΔS from J to kJ before combining with ΔH in kJ.

WORKED EXAMPLE 9 Feasibility (exam level)

For CaCO3 → CaO + CO2, ΔH = +178 kJ mol−1, ΔS = +161 J K−1 mol−1. Find ΔG at 298 K.

Solution

ΔG = 178 − 298(0.161)M1

= +130 kJ mol−1: not spontaneousA1 R1

Common mistake

Forgetting to convert ΔS into kJ K−1 mol−1. Dividing by 1000 is essential.

Quick check

1. Find ΔG at 298 K for ΔH = −92 kJ, ΔS = −199 J K−1.

Answers: 1. −92 + 59.3 = −32.7 kJ mol−1: spontaneous

12 The temperature at which a reaction becomes feasible

Setting ΔG = 0.

When ΔH and ΔS have the same sign, feasibility depends on temperature. The changeover temperature is T = ΔH/ΔS (with consistent units).

WORKED EXAMPLE 10 Decomposition temperature (exam level)

Find the minimum temperature for CaCO3 decomposition.

Solution

T = 178 000/161M1

= 1106 K (833 °C)A1

WORKED EXAMPLE 11 Upper temperature (standard)

Above what temperature is the Haber process no longer feasible?

Solution

T = 92 000/199M1

= 462 KA1

Quick check

1. Explain why the Haber process runs at about 700 K despite the result above.

Answers: 1. rate is too slow at lower temperatures; equilibrium and kinetics are compromised

13 The four combinations of signs

A summary table.

ΔH	ΔS	ΔG	Spontaneous?
−	+	always −	at all temperatures
+	−	always +	never
−	−	− at low T	at low temperatures
+	+	− at high T	at high temperatures

WORKED EXAMPLE 12 Applying (standard)

Ice melting has ΔH > 0 and ΔS > 0. Explain why it is spontaneous above 0 °C.

Solution

At high T, TΔS exceeds ΔHR1

ΔG < 0 above 273 KR1

14 Gibbs energy of formation

Products minus reactants.

ΔG can also be calculated from standard Gibbs energies of formation: ΔG = ΣΔGf(products) − ΣΔGf(reactants). ΔGf of elements in their standard states is zero.

WORKED EXAMPLE 13 From Gibbs energies of formation (standard)

Find ΔG for CH4 + 2O2 → CO2 + 2H2O(l). ΔGf: CH4 −51, CO2 −394, H2O(l) −237 kJ mol−1.

Solution

−394 + 2(−237) + 51M1

= −817 kJ mol−1A1

15 Gibbs energy and equilibrium

ΔG = −RT ln K.

ΔG⦵ is related to the equilibrium constant: ΔG⦵ = −RT ln K. A large negative ΔG gives K ≫ 1 (products favoured); a positive ΔG gives K ≪ 1.

WORKED EXAMPLE 14 Finding K (exam level)

ΔG⦵ = −32.7 kJ mol−1 at 298 K. Find K.

Solution

ln K = 32 700/(8.31 × 298) = 13.2M1

K = 5.4 × 105A1

Quick check

1. Find ΔG⦵ at 298 K for a reaction with K = 1.0 × 10−5.

Answers: 1. +28.5 kJ mol−1

16 Kinetic versus thermodynamic stability

Spontaneous does not mean fast.

A negative ΔG means a reaction is feasible, but it says nothing about rate. Diamond converting to graphite has ΔG < 0 but is extremely slow because the activation energy is very high: diamond is kinetically stable.

WORKED EXAMPLE 15 Explaining (standard)

Explain why petrol does not ignite spontaneously in air although combustion has ΔG < 0.

Solution

High activation energyR1

Kinetically stable at room temperature until a spark is suppliedR1

17 Entropy and the second law

The universe's entropy increases.

The second law of thermodynamics: in a spontaneous process, the total entropy of the universe increases. ΔStotal = ΔSsystem + ΔSsurroundings, where ΔSsurroundings = −ΔH/T. This leads to ΔG = −TΔStotal.

WORKED EXAMPLE 16 Total entropy (exam level)

For a reaction with ΔH = −50 kJ mol−1 and ΔS = −50 J K−1 mol−1 at 298 K, find ΔStotal.

Solution

ΔSsurr = 50 000/298 = +168 J K−1 mol−1M1 A1

ΔStotal = +118 J K−1 mol−1: spontaneousA1

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 17 Mixed problem (exam level)

For 2SO2(g) + O2(g) → 2SO3(g), ΔH = −198 kJ mol−1, ΔS = −188 J K−1 mol−1. (a) Find ΔG at 700 K. (b) Find the temperature above which it is not feasible.

Solution

(a) −198 + 700(0.188) = −66.4 kJ mol−1M1 A1

(b) 198 000/188 = 1053 KA1

Quick check

1. Explain why ΔS for the reaction is negative.

Answers: 1. 3 mol gas → 2 mol gas

19 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 18 Paper 1 question (standard)

Which change has the largest positive ΔS? A. H2O(l) → H2O(s) B. NH4Cl(s) → NH3(g) + HCl(g) C. 2NO2(g) → N2O4(g) D. CO2(g) → CO2(s)

Solution

B: solid to two moles of gasA1

WORKED EXAMPLE 19 Paper 2 question (exam level)

The reaction C(s) + H2O(g) → CO(g) + H2(g) has ΔH = +131 kJ mol−1 and ΔS = +134 J K−1 mol−1. (a) Explain the sign of ΔS. (b) Find the minimum temperature for feasibility.

Solution

(a) 1 mol gas → 2 mol gasR1

(b) 131 000/134 = 978 KM1 A1

20 Entropy of dissolving

Why some salts dissolve endothermically.

Many salts (NH4NO3, KCl) dissolve endothermically because the large increase in entropy (ions spreading through the solution) makes ΔG negative even though ΔH is positive.

WORKED EXAMPLE 20 Dissolving (standard)

Explain why ammonium nitrate dissolves spontaneously although ΔHsol = +26 kJ mol−1.

Solution

ΔS is large and positive (ordered lattice → dispersed ions)R1

TΔS > ΔH, so ΔG < 0R1

21 Entropy changes for phase changes

ΔS = ΔH/T at the transition.

At a melting or boiling point, the two phases are in equilibrium (ΔG = 0), so ΔS = ΔH/T. For many liquids, ΔSvap ≈ 85 J K−1 mol−1 (Trouton's rule); water is higher because hydrogen bonding orders the liquid.

WORKED EXAMPLE 21 Boiling water (standard)

ΔHvap(water) = 40.7 kJ mol−1 at 373 K. Find ΔSvap.

Solution

40 700/373M1

= 109 J K−1 mol−1A1

22 Thermodynamics in biology and industry

Coupling reactions.

Living cells drive non-spontaneous reactions (ΔG > 0) by coupling them to spontaneous ones such as ATP hydrolysis (ΔG ≈ −30 kJ mol−1). Industry chooses temperatures and pressures that make reactions both feasible and fast enough.

In the real world

Metal extraction from oxides is analysed with Ellingham diagrams, plots of ΔG against temperature that show which reducing agents work at which temperatures.

Quick check

1. How can a reaction with ΔG = +20 kJ mol−1 occur in a cell?

Answers: 1. coupled to ATP hydrolysis so that the combined ΔG is negative

23 Thermodynamics checklist

Avoiding common errors.

Convert °C to K. Convert ΔS to kJ. Remember elements have non-zero S but zero ΔHf and ΔGf. State whether the reaction is spontaneous from the sign of ΔG. Distinguish feasibility (thermodynamics) from rate (kinetics).

WORKED EXAMPLE 22 Units (standard)

ΔH = −45 kJ, ΔS = −120 J K−1, T = 25 °C. Find ΔG.

Solution

T = 298 K; ΔS = −0.120 kJ K−1M1

ΔG = −45 + 35.8 = −9.2 kJA1

24 Common misconceptions and links

Students often think…	Actually…
spontaneous means fast	spontaneity is about direction, not rate
all exothermic reactions are spontaneous	ΔS and T matter too
elements have zero entropy	all substances have positive S⊖
ΔS and ΔH can be combined directly	convert J to kJ first

How R1.4 connects to the rest of the course

Topic	Connection
R1.2 Energy cycles	ΔH values
R2.3 How far?	equilibrium constant
R2.2 How fast?	kinetics vs thermodynamics
S1.1 Particulate nature	states of matter

25 Exam-style question with markscheme

Try this under exam conditions (about 15 minutes) before reading the markscheme.

Figure 2. ΔG against temperature for a decomposition.

Text A. For 2NO2(g) → N2O4(g): ΔH⊖ = −57 kJ mol−1; S⊖: NO2 240, N2O4 304 J K−1 mol−1.

(a) Predict the sign of ΔS and explain. [2]

(b) Calculate ΔS⊖. [2]

(d) Calculate the temperature above which the reaction is not spontaneous. [2]

Markscheme

Part	Answer	Marks
(a)	negative	A1
	2 mol gas form 1 mol gas	R1
(b)	304 − 2(240)	M1
	= −176 J K−1 mol−1	A1
(c)	−57 − 298(−0.176)	M1
	= −4.6 kJ mol−1	A1
(d)	T = 57 ÷ 0.176	M1
	= 324 K	A1

Examiner's comments. In (c), forgetting to convert ΔS into kJ gave answers thousands of times too large.

26 Check your understanding

Section A: multiple choice

1. A reaction is spontaneous when

A. ΔG is negative B. ΔH is positive C. ΔS is negative D. ΔG is positive

2. Which has the highest entropy?

A. H2O(l) B. H2O(s) C. H2O(g) D. C (diamond)

3. If ΔH > 0 and ΔS > 0, the reaction is spontaneous

A. at all temperatures B. at high temperatures C. never D. at low temperatures

4. The units of S⊖ are

A. kJ mol−1 B. J mol−1 C. K D. J K−1 mol−1

5. When ΔG⊖ is large and negative, K is

A. less than 1 B. much greater than 1 C. equal to 1 D. zero

6. Forming a precipitate from ions in solution usually

A. decreases entropy B. increases entropy C. has no entropy change D. increases ΔH

Section B: structured

7. Explain why ice melts spontaneously at 25 °C although melting is endothermic. [3]

8. Calculate T at which ΔG = 0 for ΔH = +44 kJ mol−1, ΔS = +119 J K−1 mol−1. [2]

9. Explain the significance of ΔG = 0. [2]

Answers

1. A 2. C 3. B 4. D 5. B 6. A

7. ΔS positive (solid to liquid); at 298 K, TΔS > ΔH; ΔG negative.

8. T = 44 ÷ 0.119 = 370 K.

9. System at equilibrium; neither direction favoured; boundary of spontaneity.

27 Summary and topic map

Idea	Remember
Spontaneous	occurs without outside intervention; not about rate
Entropy	dispersal of energy and matter; J K−1 mol−1
ΔS sign	more gas moles: positive
ΔS⊖	Σ products − Σ reactants
Gibbs energy	ΔG = ΔH − TΔS; negative is spontaneous
Temperature	T = ΔH/ΔS when ΔG = 0
Equilibrium	ΔG⊖ = −RT ln K

Before the exam, make sure you…

•convert ΔS to kJ
•use kelvin
•predict ΔS from moles of gas
•relate ΔG to K

Entropy and Spontaneity on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 3. Topic map for Entropy and Spontaneity.

revisionprep.comIB-style notes · question bank · mock papers