DP Chemistry HL R2.3 How Far? The Extent of Chemical Change

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DP Chemistry HL | R2.3 How Far? The Extent of Chemical Change

DP CHEMISTRY · REACTIVITY

R2.3

How Far? The Extent of Chemical Change

Full notes for HL · IB-style

Contents

1	Reversible reactions and dynamic equilibrium
2	The equilibrium constant, K
3	The magnitude of K
4	The reaction quotient, Q
5	Le Chatelier's principle
6	Equilibrium in industry: the Haber and Contact processes
7	Calculating K and equilibrium concentrations (HL only)
8	Gibbs energy and K (HL only)
9	Reversible reactions and dynamic equilibrium
10	Concentration–time graphs at equilibrium
11	The equilibrium constant expression
12	Magnitude of K
13	Manipulating K
14	The reaction quotient Q
15	Le Châtelier's principle: concentration
16	Le Châtelier's principle: pressure
17	Le Châtelier's principle: temperature
18	Catalysts and equilibrium
19	Industrial compromise: the Haber process
20	The Contact process
21	Equilibrium calculations: ICE tables (HL)
22	Mixed review: worked problems
23	Additional exam-style practice
24	Equilibria in nature
25	Common misconceptions and links
26	Exam-style question with markscheme
27	Check your understanding
28	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
describe the characteristics of dynamic equilibrium
write equilibrium constant expressions for homogeneous reactions
interpret the magnitude of K
use the reaction quotient Q to predict the direction of change
apply Le Chatelier's principle to changes in concentration, pressure and temperature
explain the effect of a catalyst and evaluate conditions used in industry
calculate K and equilibrium concentrations using ICE tables
relate ΔG⊖ to K

Sections marked HL only (Sections 7 and 8) are Additional Higher Level content. Every other section is common to SL and HL.

Guiding questions

How can the extent of a reversible reaction be predicted and controlled? Why are industrial conditions often a compromise?

How this topic is examined

R2.3 is examined through K expressions, Le Chatelier predictions with explanations, and evaluation of industrial conditions. HL questions include ICE table calculations and ΔG = −RT ln K.

When applying Le Chatelier's principle, state the change, the direction the equilibrium shifts, and why.

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 Reversible reactions and dynamic equilibrium

Forward and backward reactions at the same rate.

Think about it

In a sealed bottle of fizzy water, CO2 is constantly dissolving and escaping, yet the pressure stays the same. How can both happen at once?

A reversible reaction can go in both directions: reactants form products, and products re-form reactants. It is shown with the symbol ⇌: N2(g) + 3H2(g) ⇌ 2NH3(g).

In a closed system (nothing enters or leaves), a reversible reaction reaches dynamic equilibrium. At first the forward rate is high and the reverse rate zero; as products build up, the reverse rate increases and the forward rate decreases, until the rates of the forward and reverse reactions are equal. Characteristics of dynamic equilibrium: the forward and reverse reactions continue (it is dynamic); the rates are equal; the concentrations of reactants and products remain constant (but are not usually equal); macroscopic properties (colour, pressure) are constant; it can be reached from either direction.

Figure 1. Concentrations become constant (not equal) at equilibrium.

Common mistake

Saying concentrations of reactants and products are equal at equilibrium. They are constant, but usually different.

Quick check

1. What is meant by dynamic equilibrium?

2. Why must the system be closed?

Answers: 1. forward and reverse rates equal; concentrations constant; reactions continue 2. so no substances can escape or enter

2 The equilibrium constant, K

A ratio of concentrations at equilibrium.

For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant expression is:

K = [C]c[D]d ÷ [A]a[B]b

Square brackets mean equilibrium concentration in mol dm−3. Products are on top, reactants below, each raised to the power of its coefficient. K is constant at a given temperature; it changes only with temperature.

For homogeneous equilibria (all species in the same phase), all species are included. For N2(g) + 3H2(g) ⇌ 2NH3(g): K = [NH3]2 ÷ [N2][H2]3.

If the equation is reversed, the new K is 1/K; if the coefficients are doubled, the new K is K2.

WORKED EXAMPLE 1 Writing and using K

For H2(g) + I2(g) ⇌ 2HI(g), equilibrium concentrations are [H2] = 0.020, [I2] = 0.020, [HI] = 0.14 mol dm−3. Calculate K.

Solution

K = [HI]2 ÷ [H2][I2]M1

= 0.142 ÷ (0.020 × 0.020) = 0.0196 ÷ 0.00040 = 49A1

Quick check

1. Write K for 2SO2 + O2 ⇌ 2SO3.

2. What is the only factor that changes K?

Answers: 1. K = [SO3]2 ÷ [SO2]2[O2] 2. temperature

3 The magnitude of K

Large K, products; small K, reactants.

The size of K shows the position of equilibrium. If K ≫ 1, the equilibrium lies to the right: mainly products; the reaction goes almost to completion. If K ≪ 1, the equilibrium lies to the left: mainly reactants; little reaction occurs. If K is close to 1, significant amounts of both are present.

K tells us nothing about the rate at which equilibrium is reached.

4 The reaction quotient, Q

Is the system at equilibrium yet?

The reaction quotient (Q) has the same expression as K but uses the concentrations at any moment, not necessarily at equilibrium. Comparing Q with K predicts the direction of change: if Q < K, there are too few products, so the reaction proceeds forward (to the right); if Q > K, too many products, so it proceeds in reverse (to the left); if Q = K, the system is at equilibrium.

WORKED EXAMPLE 2 Using Q

For the HI equilibrium above (K = 49), a mixture contains [H2] = 0.10, [I2] = 0.10, [HI] = 0.50 mol dm−3. Predict the direction of change.

Solution

Q = 0.502 ÷ (0.10 × 0.10) = 25M1

Q < K, so the reaction proceeds forward, forming more HIA1

5 Le Chatelier's principle

Equilibrium shifts to oppose a change.

Le Chatelier's principle: if a system at equilibrium is disturbed by a change in conditions, the position of equilibrium shifts to oppose the change and restore equilibrium.

Concentration: adding a reactant (or removing a product) shifts equilibrium to the right, using up the added substance; K is unchanged. Pressure (gases): increasing pressure shifts equilibrium to the side with fewer moles of gas, reducing the pressure; if both sides have equal moles of gas, there is no effect. K is unchanged. Temperature: increasing temperature shifts equilibrium in the endothermic direction, absorbing heat. K changes: for an exothermic forward reaction, K decreases as temperature rises. Catalyst: no effect on the position of equilibrium or K; equilibrium is reached faster because forward and reverse rates increase equally.

Change	Shift for N2 + 3H2 ⇌ 2NH3, ΔH = −92 kJ mol−1	Effect on K
add N2	right	none
remove NH3	right	none
increase pressure	right (4 mol → 2 mol gas)	none
increase temperature	left (endothermic direction)	decreases
add catalyst	none; faster to reach equilibrium	none

Common mistake

Saying a catalyst increases the yield at equilibrium. It only increases the rate at which equilibrium is reached.

Quick check

1. State Le Chatelier's principle.

2. Predict the effect of increasing pressure on H2 + I2 ⇌ 2HI.

Answers: 1. equilibrium shifts to oppose a change 2. no shift: equal moles of gas

6 Equilibrium in industry: the Haber and Contact processes

Compromise between yield, rate and cost.

The Haber process makes ammonia: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol−1. High pressure favours ammonia (fewer moles of gas) and increases rate, but is expensive and dangerous: about 200 atm is used. Low temperature favours ammonia (exothermic) but makes the reaction too slow: a compromise temperature of about 450 °C is used. An iron catalyst increases the rate. Ammonia is removed by cooling and liquefying it, and unreacted gases are recycled.

The Contact process makes sulfuric acid: 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = −196 kJ mol−1. Conditions: about 450 °C, 1 to 2 atm (the yield is already high, so high pressure is not worth the cost), vanadium(V) oxide catalyst.

Figure 2. Yield of ammonia increases with pressure and decreases with temperature.

Quick check

1. Why is 450 °C used in the Haber process?

2. Why is low pressure used in the Contact process?

Answers: 1. compromise: acceptable rate and yield 2. yield already high; high pressure costly

7 Calculating K and equilibrium concentrations (HL only)

ICE tables.

HL only

This section is Additional Higher Level content. SL students do not need it.

An ICE table (Initial, Change, Equilibrium) organizes equilibrium calculations. Write initial amounts or concentrations; use the stoichiometry to write the changes; add to find equilibrium values; substitute into the K expression. If amounts are in moles, convert to concentrations by dividing by the volume (unless the volumes cancel).

WORKED EXAMPLE 3 ICE table

0.200 mol of H2 and 0.200 mol of I2 are placed in a 1.00 dm3 flask. At equilibrium, 0.040 mol of I2 remains. Calculate K for H2 + I2 ⇌ 2HI.

Solution

Initial: H2 0.200, I2 0.200, HI 0M1

Change: −0.160, −0.160, +0.320M1

Equilibrium: 0.040, 0.040, 0.320M1

K = 0.3202 ÷ (0.040 × 0.040) = 64A1

WORKED EXAMPLE 4 Finding an equilibrium concentration

For the same reaction, K = 64. If 0.100 mol each of H2 and I2 start in 1.00 dm3, calculate [HI] at equilibrium.

Solution

Let x react: [HI] = 2x; [H2] = [I2] = 0.100 − xM1

√64 = 8 = 2x ÷ (0.100 − x), so 0.800 − 8x = 2xM1

x = 0.0800; [HI] = 0.160 mol dm−3A1

8 Gibbs energy and K (HL only)

Thermodynamics predicts the position of equilibrium.

HL only

This section is Additional Higher Level content. SL students do not need it.

The standard Gibbs energy change is related to K: ΔG⊖ = −RT ln K (R1.4). A negative ΔG⊖ gives K > 1 (products favoured); positive ΔG⊖ gives K < 1; ΔG⊖ = 0 gives K = 1.

WORKED EXAMPLE 5 K from ΔG

Calculate K at 298 K when ΔG⊖ = +10.0 kJ mol−1.

Solution

ln K = −ΔG ÷ RT = −10 000 ÷ (8.31 × 298) = −4.04M1

K = e−4.04 = 0.018: reactants favouredA1

9 Reversible reactions and dynamic equilibrium

Forward and backward at equal rates.

In a closed system, a reversible reaction reaches dynamic equilibrium when the forward and reverse rates are equal. Concentrations then stay constant, but reactions continue in both directions. Macroscopic properties (colour, pressure) no longer change.

WORKED EXAMPLE 6 Characteristics (standard)

State three characteristics of a system at dynamic equilibrium.

Solution

Forward and reverse rates are equalR1

Concentrations constant; closed systemR1

Can be approached from either directionR1

Common mistake

Saying that at equilibrium 'the concentrations of reactants and products are equal'. They are constant, not necessarily equal.

Quick check

1. Why must the system be closed?

2. Sketch how forward and reverse rates change as equilibrium is reached.

Answers: 1. so no substance can escape 2. forward rate falls, reverse rate rises, until they meet

10 Concentration–time graphs at equilibrium

Levelling off.

When reactants are mixed, their concentrations fall and product concentrations rise until both level off at equilibrium. The changes are in the ratio of the coefficients.

Figure 3. N2O4 ⇌ 2NO2: NO2 rises twice as fast as N2O4 falls.

WORKED EXAMPLE 7 Reading the graph (standard)

Using the graph, explain why the change in [NO2] is twice the change in [N2O4].

Solution

1 mol N2O4 gives 2 mol NO2R1

11 The equilibrium constant expression

Products over reactants.

For aA + bB ⇌ cC + dD, Kc = [C]c[D]d/([A]a[B]b), using equilibrium concentrations. Pure solids and liquids are omitted for heterogeneous equilibria.

WORKED EXAMPLE 8 Writing equilibrium expressions (basic)

Write Kc for N2(g) + 3H2(g) ⇌ 2NH3(g) and for CaCO3(s) ⇌ CaO(s) + CO2(g).

Solution

Kc = [NH3]2/([N2][H2]3)A1

Kc = [CO2]A1

Quick check

1. Write Kc for 2SO2 + O2 ⇌ 2SO3.

2. Write Kc for H2 + I2 ⇌ 2HI.

Answers: 1. [SO3]2/([SO2]2[O2]) 2. [HI]2/([H2][I2])

12 Magnitude of K

How far does it go?

K ≫ 1: the equilibrium lies to the right (mainly products). K ≪ 1: lies to the left (mainly reactants). K ≈ 1: significant amounts of both. K depends only on temperature.

WORKED EXAMPLE 9 Interpreting (standard)

For H2 + I2 ⇌ 2HI, Kc = 54 at 700 K. Describe the position of equilibrium.

Solution

K > 1: products favouredR1

Significant amounts of reactants remain (not ≫ 1)R1

Quick check

1. Kc = 1 × 10−30 for N2 + O2 ⇌ 2NO at 298 K. Interpret.

Answers: 1. almost no NO forms at room temperature

13 Manipulating K

Reversing and multiplying equations.

Reversing an equation gives K' = 1/K. Multiplying the coefficients by n gives K' = Kn. Adding equations multiplies their K values.

WORKED EXAMPLE 10 Manipulation (exam level)

For 2SO2 + O2 ⇌ 2SO3, K = 280. Find K for SO3 ⇌ SO2 + ½O2.

Solution

Reverse: 1/280; halve: square rootM1

K' = (1/280)1/2 = 0.0598A1

Quick check

1. K for A ⇌ B is 4.0. Find K for 2B ⇌ 2A.

Answers: 1. 1/16 = 0.0625

14 The reaction quotient Q

Predicting direction.

Q has the same form as K but uses current (non-equilibrium) concentrations. If Q < K, the reaction proceeds forwards; if Q > K, backwards; if Q = K, the system is at equilibrium.

WORKED EXAMPLE 11 Direction (exam level)

For H2 + I2 ⇌ 2HI, K = 54. A mixture has [H2] = [I2] = 0.10, [HI] = 0.50 mol dm−3. Predict the direction.

Solution

Q = 0.25/0.010 = 25M1

Q < K: proceeds to the rightA1

Quick check

1. If [HI] were 1.00, which way would it shift?

Answers: 1. Q = 100 > 54: to the left

15 Le Châtelier's principle: concentration

Opposing the change.

If the concentration of a substance is increased, the equilibrium shifts to reduce it (away from that substance). Removing a product shifts the equilibrium to the right. K is unchanged.

WORKED EXAMPLE 12 Colour change (standard)

Fe3+ + SCN− ⇌ FeSCN2+ (blood red). Predict the effect of adding more SCN−.

Solution

Equilibrium shifts rightR1

Deeper red colourA1

Quick check

1. Predict the effect of removing NH3 in the Haber process.

Answers: 1. shifts right, more NH3 formed

16 Le Châtelier's principle: pressure

Count moles of gas.

Increasing pressure shifts the equilibrium towards the side with fewer moles of gas. If both sides have the same number of gas moles, pressure has no effect on position. K is unchanged.

WORKED EXAMPLE 13 Pressure (standard)

Predict the effect of increasing pressure on 2NO2(g) ⇌ N2O4(g).

Solution

Left side 2 mol gas, right side 1 molM1

Shifts right (fewer gas moles)A1

Quick check

1. Predict the effect of pressure on H2 + I2 ⇌ 2HI.

Answers: 1. no effect: 2 mol gas each side

17 Le Châtelier's principle: temperature

The only factor that changes K.

Increasing temperature shifts the equilibrium in the endothermic direction, decreasing it shifts in the exothermic direction. This changes the value of K: for an exothermic forward reaction, K decreases as temperature increases.

WORKED EXAMPLE 14 Temperature (exam level)

N2O4 ⇌ 2NO2, ΔH = +57 kJ mol−1. Predict the colour change and the effect on K when heated.

Solution

Shifts right (endothermic direction)R1

Browner colour; K increasesA1 A1

Quick check

1. For an exothermic reaction, what happens to K at higher temperature?

Answers: 1. K decreases

18 Catalysts and equilibrium

Faster, not further.

A catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached faster, but the position of equilibrium and K are unchanged.

WORKED EXAMPLE 15 Catalyst (standard)

Explain the role of iron in the Haber process.

Solution

Lowers Ea for both directionsR1

Equilibrium reached faster; yield unchangedR1

19 Industrial compromise: the Haber process

Yield versus rate versus cost.

N2 + 3H2 ⇌ 2NH3, ΔH = −92 kJ mol−1. Low temperature favours yield but is slow; about 450 °C is a compromise. High pressure favours yield and rate but is expensive; about 200 atm is used. An iron catalyst speeds up the reaction; unreacted gases are recycled.

WORKED EXAMPLE 16 Compromise (exam level)

Explain why the Haber process does not use a very low temperature.

Solution

Low T increases yield (exothermic)R1

but rate would be too slowR1

In the real world

About 1% of the world's energy is used for the Haber process, which provides the ammonia for fertilizers feeding roughly half of the world's population.

Quick check

1. Why is ammonia removed by condensation?

Answers: 1. shifts equilibrium right and gives pure product

20 The Contact process

Making sulfuric acid.

2SO2 + O2 ⇌ 2SO3, ΔH = −198 kJ mol−1, with a V2O5 catalyst at about 450 °C and 1–2 atm. High conversion (over 99%) is achieved at modest pressure, so high pressure is not worth the cost.

WORKED EXAMPLE 17 Pressure choice (standard)

Explain why only low pressure is used in the Contact process.

Solution

Equilibrium already lies far to the rightR1

High pressure costs more than the small gain in yieldR1

21 Equilibrium calculations: ICE tables (HL)

Initial, change, equilibrium.

Set up a table of initial amounts, changes (in the ratio of coefficients) and equilibrium amounts, convert to concentrations and substitute into Kc.

WORKED EXAMPLE 18 Finding K (exam level)

1.00 mol H2 and 1.00 mol I2 in 1.00 dm3 reach equilibrium with 1.56 mol HI. Find Kc.

Solution

H2 and I2: 1.00 − 0.78 = 0.22 molM1 A1

Kc = 1.562/0.222 = 50.3A1

22 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 19 Mixed problem (exam level)

For CO(g) + 2H2(g) ⇌ CH3OH(g), ΔH = −91 kJ mol−1: predict the effect on the yield of methanol of (a) increasing pressure; (b) increasing temperature; (c) adding a catalyst.

Solution

(a) increases: 3 mol → 1 mol gasR1

(b) decreases: exothermicR1

Quick check

1. Write Kc for the methanol equilibrium.

Answers: 1. [CH3OH]/([CO][H2]2)

23 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 20 Paper 1 question (standard)

Which changes the value of Kc? A. pressure B. concentration C. catalyst D. temperature

Solution

DA1

WORKED EXAMPLE 21 Paper 2 question (exam level)

For 2NO(g) + O2(g) ⇌ 2NO2(g), ΔH = −114 kJ mol−1. (a) Write Kc. (b) Predict the effect of increasing temperature on Kc. (c) Predict the effect of decreasing pressure on the amount of NO2.

Solution

(a) [NO2]2/([NO]2[O2])A1

(b) decreasesA1

24 Equilibria in nature

Oceans and blood.

CO2 dissolving in seawater sets up equilibria: CO2(aq) + H2O ⇌ H+ + HCO3−. Increasing atmospheric CO2 shifts these to the right, lowering ocean pH (ocean acidification), which harms shell-forming organisms. Haemoglobin–oxygen binding is also an equilibrium.

In the real world

At high altitude, the lower partial pressure of oxygen shifts the haemoglobin–oxygen equilibrium to the left; the body responds by producing more haemoglobin over a few weeks.

Quick check

1. Explain how increasing CO2 lowers ocean pH.

Answers: 1. more CO2 dissolves; equilibrium shifts right, producing H+

25 Common misconceptions and links

Students often think…	Actually…
concentrations are equal at equilibrium	they are constant
a catalyst increases yield	it only increases rate
pressure changes K	only temperature changes K
K uses initial concentrations	K uses equilibrium concentrations

How R2.3 connects to the rest of the course

Topic	Connection
R2.2 How fast?	rates and catalysts
R1.4 Entropy (HL)	ΔG and K
R3.1 Proton transfer	acid dissociation equilibria
R2.1 How much?	yield

26 Exam-style question with markscheme

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

Figure 4. Equilibrium yield of ammonia.

(a) Write the equilibrium constant expression for N2O4(g) ⇌ 2NO2(g). [1]

(b) The forward reaction is endothermic. Predict and explain the effect of increasing temperature on the position of equilibrium and K. [3]

(d) Explain why the Haber process uses 450 °C rather than a lower temperature. [2]

(e) 0.500 mol N2O4 in 1.00 dm3 forms 0.200 mol NO2 at equilibrium. Calculate K. [3]

Markscheme

Part	Answer	Marks
(a)	K = [NO2]2 ÷ [N2O4]	A1
(b)	shifts right	A1
	endothermic direction absorbs heat	R1
	K increases	A1
(c)	shifts left	A1
	fewer moles of gas (1 vs 2)	R1
(d)	lower temperature gives higher yield	R1
	but rate too slow; compromise	R1
(e)	N2O4 reacted 0.100; left 0.400	M1
	K = 0.2002 ÷ 0.400	M1
	= 0.100	A1

Examiner's comments. In (b), many candidates forgot that K changes with temperature.

27 Check your understanding

Section A: multiple choice

1. At dynamic equilibrium

A. forward and reverse rates are equal B. reactions stop C. concentrations are equal D. only products remain

2. A large K means

A. the reaction is fast B. mainly reactants C. equilibrium lies to the right D. K is temperature independent

3. If Q < K the reaction

A. proceeds in reverse B. proceeds forward C. is at equilibrium D. stops

4. Adding a catalyst

A. shifts equilibrium right B. increases K C. decreases K D. does not change the position of equilibrium

5. Increasing pressure on 2SO2 + O2 ⇌ 2SO3

A. shifts left B. shifts right C. has no effect D. changes K

6. For an exothermic reaction, increasing temperature

A. decreases K B. increases K C. does not change K D. makes K negative

7. If ΔG⊖ is negative, K is

A. less than 1 B. zero C. negative D. greater than 1

Section B: structured

8. State the characteristics of dynamic equilibrium. [3]

9. Explain the effect of removing ammonia in the Haber process. [2]

10. Evaluate the conditions used in the Contact process. [3]

11. Use an ICE table to find K when 1.0 mol PCl5 in 1.0 dm3 is 40% dissociated: PCl5 ⇌ PCl3 + Cl2. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D

8. Closed system; equal rates; constant concentrations; dynamic.

9. Shifts right to replace NH3; more product.

10. 450 °C compromise; 1–2 atm sufficient as yield high; V2O5 catalyst increases rate.

11. Equilibrium: 0.6, 0.4, 0.4; K = 0.16 ÷ 0.6 = 0.27.

28 Summary and topic map

Idea	Remember
Dynamic equilibrium	equal rates; constant concentrations; closed system
K	products over reactants, powers = coefficients
Magnitude of K	≫1 products; ≪1 reactants
Q	Q < K forward; Q > K reverse
Le Chatelier	concentration, pressure, temperature; catalyst no effect
Industry	Haber 450 °C, 200 atm, Fe; Contact V2O5
HL	ICE tables; ΔG⊖ = −RT ln K

Before the exam, make sure you…

•write K with correct powers
•explain shifts with the opposing change
•state whether K changes
•justify compromise conditions

How Far? 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 5. Topic map for How Far?.

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