Question Bank (Pg 1-58)

1© Oxford Fajar Sdn. Bhd. (008974-T) 2008

Objective Questions

1. Alpha particles, 4

2He2+, are emitted when a radioactive nucleus disintegrates. Which of the

following species has the same number of electrons as one alpha particle?A H+ C Li+

B H– D H2

2. In a mass spectrum, the peak at m/e 22 is caused byA CO

2+ C CO+

B CO22+ D CO2+

3. A sample of oxygen gas which contains atoms of two isotopes 16O and 18O is analysed in a mass spectrometer. Which of the ions below will be deflected most in the magnetic field?A 16O

2+ C 16O – 18O+

B 18O2+ D 16O

22+

4. The mass spectrum of a metal Q is shown below.

1 Matter

What is the relative atomic mass of Q? (85 � 23) + (87 � 7.5) (85 � 23) + (87 � 7.5)A ——————————————— C ——————————————— 100 23 + 7.5 (85 � 21.5) + (87 � 8.5) (85 � 21.5) + (87 � 8.5)B ———————————————— D ———————————————— 100 21.5 + 8.5

5. When potassium chlorate(V) is heated strongly at its melting point, a mixture of potassium chlorate(VII), KClO

4 and potassium chloride is produced. The equation for the reaction is

4KClO3(s) → 3KClO

4(s) + KCl(s)

What amount of potassium chlorate(V) must be used to produce 0.075 mol of potassium chlorate(VII)?A 0.05 molB 0.10 molC 0.15 molD 0.20 mol

Ace Ahead Chemistry (Question Bank) 5th (19/2/08).indd 1Ace Ahead Chemistry (Question Bank) 5th (19/2/08).indd 1 2/20/08 11:30:12 AM2/20/08 11:30:12 AM

2 © Oxford Fajar Sdn. Bhd. (008974-T) 2008

6. The reaction between sulphur dichloride dioxide, SCl2O

2 and water is represented by the

chemical equation:SCl

2O

2 + 2H

2O → H

2SO

4 + 2HCl

What is the volume (in dm3) of 2.0 mol dm–3 NaOH required to neutralise completely the reaction products when 2.0 mol of sulphur dichloride dioxide reacts with excess water?A 4 C 8B 6 D 12

A B C D

1 only is correct 1 and 2 onlyare correct

2 and 3 onlyare correct

1, 2 and 3are correct

7. Which of the formulae given below can be used to calculate the value of the Avogadro constant, L? Mass of one mole of hydrogen gas1 L = ——————————————————————————— Mass of two atoms of hydrogen

Volume of one mole of hydrogen gas (cm3)2 L = —————————————————————————————————— Volume of one molecule of hydrogen (cm3)

Mass of one mole of carbon–123 L = ————————————————————————— Mass of one atom of carbon–12

8. Which of the following compounds contain(s) the same number of atoms as 1.0 mol of oxygen gas?1 1.0 mol of nitrogen gas2 0.5 mol of ammonia gas3 0.5 mol of carbon monoxide

Structured Questions

1. The mass spectrum for an element is shown below.

(a) Calculate the relative atomic mass of this element.(b) With the help of the Periodic Table, identify the element.

2. A sample of hydrogen chloride molecules consists of the isotopes 1H, 35Cl and 37Cl. The relative abundance of 35Cl and 37Cl isotopes are 75% and 25% respectively.(a) Write down the formulae of all the different ions produced in the ionisation chamber of the

mass spectrum.

50

40

30

20

10

90 91 92 93 94 m/e

rela

tive

abun

danc

e %



(b) Sketch the mass spectrum of this sample of hydrogen chloride molecules, ignoring the peak at m/e 1.

(c) Comment on your answer in (b).(d) Calculate the relative molecular mass of the hydrogen chloride in this sample.

Essay Question

1. The mass spectrum of rhombic sulphur is shown below.

(a) (i) Explain why the mass spectrum of rhombic sulphur consists of a number of peaks. (ii) Identify the species which is most abundant in the sample of sulphur vapour. (iii) From the mass spectrum, deduce the formula of rhombic sulphur.(b) Cisplatin, Pt(NH

3)

2Cl

2, is used to treat certain types of cancer. It is prepared by the following

reactionK

2PtCl

4(aq) + 2NH

3(aq) → 2KCl(aq) + Pt(NH

3)

2Cl

2(aq)

The percentage yield for the reaction is 93.2% where

actual yield from experimentpercentage yield = ———————————————————————————— � 100

theoretical yield

If a student wants to prepare 5.72 g of Pt(NH3)

2Cl

2, what is the mass of K

2PtCl

4 he should

use?



Objective Questions

1. A Alpha-particles are helium nuclei, 42He2+

2. (a) 1H+, 35Cl+, 37Cl+

1H – 35Cl+, 1H – 37Cl+

(b)

ANSWER

CHAPTER 1 Matter

2. B

Ion

m/e

CO2+

44/1 = 44

CO2

2+

44/2= 22

CO+

28/1= 28

CO2+

28/2= 14

3. D 16O22+ has the smallest mass and highest charge.

4. D 5. B From the equation,

3 mol of KClO4 is obtained from 4 mol of KClO

3

0.075 mol KClO4 is obtained from

40.075 � — = 0.10 mol KClO

3 3

6. A SCl2O

2 + 2H

2O → H

2SO

4 + 2HCl

2 mol of SCl2O

2 produces 2 mol of H

2SO

4 and

4 mol of HCl, that is, 8 mol of H+ ions.H+(aq) + OH–(aq) → H

2O(l)

Number of moles of NaOH required = 8.08.0 = MV = 2.0 � VV = 4 dm3

7. D 1 mol of H2 gas = L × 2 hydrogen atoms

(1 correct)1 mol of H

2 gas = L molecules of H

2

Volume of 1 mol of H2 = L � volume of

1 molecule of H2 (2 correct)

1 mol of carbon = L atoms of carbon (3 correct) 8. B 1.0 mol of O

2 contains 2 � (6.02 � 1023)

oxygen atoms.1.0 mol of N

2 contains 2 � (6.02 � 1023)

nitrogen atoms. (1 correct)0.5 mol of NH

3 contains 0.5 x (6.02 � 1023)

nitrogen atoms and (3 � 0.5) � (6.02 � 1023) hydrogen atoms. (2 correct)0.5 mol of CO contains 0.5 � (6.02 � 1023) carbon atoms and 0.5 � (6.02 � 1023) oxygen atoms. (3 incorrect)


1. (a) Relative (90 � 52) + (91 � 9) + (92 � 12)atomic + (93 � 14) + (94 � 13)mass

= ——————————————————————————— 52 + 9 + 12 + 14 + 13

= 91.3

(b) Zirconium (Zr)

SpeciesNumber of

protonsNumber of electrons

Helium nuclei

H+

H–

Li+

H2

2

1

1

3

2

0

0

2

2

2

(c) m/e 35 : 35Cl+; m/e 37 : 37Cl+

Relative heights of these two lines: 3 : 1m/e 36 : 1H – 35Cl+; m/e 38 : 1H – 37Cl+.Relative heights of these two lines: 3 : 1

(36 × 3) + (38 × 1)(d) M

r of HCl = ——————————————— = 36.5

3 + 1

Essay Question

1. (a) (i) The mass spectrum of sulphur molecules is caused by the formation of molecular ions and fragmented molecular ions. (Note: The mass spectrum of sulphur as shown in the question is not caused by the isotopes of sulphur).

(ii) The most abundant species is the fragmented ion, 32S

2+ (or 32S – 32S+) which

produces the peak at m/e 64. (iii) From the mass spectrum, M

r of sulphur

= 256The relative atomic mass of sulphur = 32Number of atoms in 1 molecule of rhombic sulphur

256= ——— = 8

32

Thus, the structure of rhombic sulphur is made up of simple molecules. Each molecule contains 8 atoms, that is, the molecular formula of sulphur is S

8.

5.72(b) 92.3 = ———————————— � 100

theoretical yield

Theoretical yield = 6.20 gM

r of Pt(NH

3)

2Cl

2 = 195 + 2 � (17) + 2 �

(35.5) = 300M

r of K

2PtCl

4 = 2 � (39) + 195 + 4 � (35.5)

= 415 6.20

Number of moles of Pt(NH3)

2Cl

2 = ———

300= 0.0207

1 mol of K2PtCl

4 → 1 mol of Pt(NH

3)

2Cl

2

Number of moles of K2PtCl

4 needed = 0.0207

Mass of K2PtCl

4 needed = 0.0207 � 415

= 8.59 g

100

60

20

30 35 40 m/e

rela

tive

abun

danc

e 80

40



Objective Questions

1. A gaseous hydride of nitrogen, X, contains 87.5 % by mass of nitrogen. The relative molecular mass of X is 32. When X is heated, it gives off a mixture of nitrogen gas and hydrogen gas. What is the total volume of the mixture of gases produced when 100 cm3 of X is heated?A 200 cm3 C 400 cm3 B 300 cm3 D 600 cm3

2. A closed container at s.t.p. contains 11 g of P (carbon dioxide) and 8 g of Q (oxygen). Which of the following statements is true?A The average kinetic energy of P is higher than the average kinetic energy of Q.B The partial pressure of P is higher than the partial pressure of Q.C The partial pressure for each gas becomes 0.5 atm when 1 dm3 of P is mixed with 1 dm3 of

Q.D P and Q do not contain the same number of molecules.

ρ 3. The graph of — against p for a gas, X, at 273 K is shown below. p

2 The Gaseous State

(ρ = density of gas in g dm–3; p = gas pressure in atm; R = gas constant)What is the relative molecular mass of X at a pressure of p = 0 and a temperature T = 273 K?

273YA 273 RY C ————

R 273Y RY

B ———— D ——— R 273

4. A gaseous mixture contains 64 g of methane, 64 g of oxygen and 64 g of sulphur dioxide. The gaseous mixture exerts a pressure of 210 kPa. What is the partial pressure of methane?[Relative atomic mass: C, 12, O, 16, S, 32]A 30 kPaB 60 kPaC 70 kPaD 120 kPa

5. A 1.0 dm3 flask contains oxygen at 114 kPa and a 2.0 dm3 flask contains nitrogen at 67.5 kPa. The two flasks are connected and the two gases allowed to mix. What is the total pressure of the gaseous mixture?[The temperature is assumed to remain constant.]A 7 kPaB 38 kPaC 45 kPaD 83 kPa

Y



6. Which of the following formulae represent(s) the properties of an ideal gas?(ρ = density; m = mass of gas; M = relative molecular mass)

ρ1 p = —— RT 3 pV = M

rRT

Mr

RT2 pV = ——

ρ

7. A 10.0 dm3 glass bulb contains 2.0 g of hydrogen, 9.98 g of argon and 3.42 g of oxygen.We can deduce that1 the amount of moles of oxygen in the mixture is 0.214 mol.2 the total amount of gases in the mixture is 1.357 mol.3 the total pressure of the gaseous mixture is 3.34 atm.(Relative atomic mass: H, 1.0; Ar, 39.9; O, 16; temperature = 27 °C)

8. A gaseous mixture contains 0.8 mol of Cl2, 0.8 mol of PCl

3 and 0.2 mol of PCl

5. At 423 K, the

pressure of the mixture is 96 000 Pa. Which of the following statements is/are true?1 The partial pressure of chlorine is 10 667 kPa.2 The partial pressure of PCl

3 is 42 667 kPa.

3 The partial pressure of PCl5 is 10 667 kPa.

9. Which of the following graphs represent(s) the behaviour of an ideal gas?1 2 3

A B C D




p

V 0

p

01v

p

t (°C)0


1. (a) Explain why a decrease in the volume of a gas causes an increase in its pressure.(b) State the law that shows the relationship between the pressure and volume of a gas.(c) A sample of chlorine gas occupies a volume of 45.2 dm3 at a pressure of 340 kPa and a

temperature of 57 °C. Calculate the mass of chlorine gas in the sample.

2. Nitric oxide reacts rapidly with oxygen as represented by the following equation

2NO(g) + O2(g) → 2NO

2(g)

The apparatus shown below can be used to carry out this reaction.

NO O2

4 dm3

0.5 atm

2 dm3

1.0 atm

When the tap T is opened, a fast reaction occurs. Calculate (a) the amount (in moles) of NO and O

2 present before the reaction.

(b) the amount (in moles) of NO2 produced in the reaction.

(c) the number of moles of gases remaining after the experiment.(d) the partial pressure for every gas in the resultant mixture.



Essay Question

1. (a) Describe qualitatively in terms of kinetic theory the essential difference between the gaseous, liquid and solid states.

(b) Explain why the behaviour of a real gas differs from that predicted by the ideal gas behaviour.

(c) When a gas is collected over water, it is saturated with water vapour. A sample of hydrogen (45.5 cm3) was collected over water at 25 °C and 0.992 atm. The vapour pressure of water at 25 °C is 0.0313 atm.What is the amount (in mol) of hydrogen collected in this experiment?



Objective Questions

1. B

2.0 7. C n

hydrogen =

— —— — = 1.0

2.0

9.98n

argon =

— —— — — = 0.25

39.9

3.42n

oxygen =

— —— — — = 0.107 (1 incorrect)

32

Total number of moles = 1.0 + 0.25 + 0.107= 1.357 (2 correct)

nRT 1.357 � 0.0821 � 300P

T =

— —— — — =

— ———— — ——— — ———— — — ——— — — — —

V 10.0= 3.34 atm (3 correct)

8. C Total number of moles = 0.8 + 0.8 + 0.2 = 1.8

0.8Partial pressure of Cl

2 = ——— � 96 000

1.8= 42 667 kPa (1 incorrect)

0.8Partial pressure of PCl

3 = ——— � 96 000

1.8= 42 667 kPa

(2 correct)

0.2Partial pressure of PCl

5 = ——— � 96 000

1.8= 10 667 kPa

(3 correct) 9. D Graphs 1 and 2 are different representations of

Boyle’s law. Graph 3 represents the effect of temperature on pressure. As the temperature increases, the kinetic energy of the gaseous particles increases and the frequency of collisions on the walls of the container increases. Thus, the pressure increases. (1, 2, 3 correct)


1. (a) According to the kinetic theory of gases, the pressure of a gas is caused by the collisions of gaseous particles on the walls of the container. When the volume of the gas decreases, the number of molecules per unit volume increases. This increase causes the molecules to collide more frequently with the walls and therefore increases the pressure.

(b) Boyle’s law: The volume of a given mass of gas is inversely proportional to its pressure if the temperature of the gas is kept constant.

(c) pV = nRT

pV (340 � 103) � (45.2 � 10–3)n = ——— = ———————————————————————

RT 8.31 � (273 + 57)= 5.60 mol

Mass of chlorine = 5.60 � 71= 397.6 g

2. (a) Before reactionNumber of moles of nitrogen oxide (NO)

pV 0.5 � 4 2= — — = —————— = ——

RT RT RT

ANSWER

CHAPTER 2 The Gaseous State

Empirical formula of X is NH2.

nNH2 = 32

n(14 + 2) = 32n = 2Molecular formula of X is N

2H

4.

N2H

4 → N

2 + 2H

2

1 mol → 1 mol + 2 mol = 3 mol 1 vol → 3 vol100 cm3 → 300 cm3

2. C 11

11 g of P (CO2) = —— = 0.25 mol

44 8

8 g of Q (O2) = —— = 0.25 mol

32

mRT ρRT 3. A M

r = ———— = ————

pV p ρ

where p = 0; T = 273, — = Y p

Mr = YRT = 273RY

4. D Relative molecular mass:CH

4 = 16, O

2 = 32, SO

2 = 64

64Number of moles of CH

4 = —— = 4

16

64Number of moles of O

2 = —— = 2

32

64Number of moles of SO

2 = —— = 1

64

7 mol of gases exert a pressure of 210 kPa. 4

4 mol of CH4 exert a pressure of — � 210

7= 120 kPa.

5. D p1V

1 = p

2V

2

114 � 1.0Partial pressure of O

2 = ———— —— —— — — = 38 kPa

3

67.5 � 2.0Partial pressure of N

2 = ———— —— —— — — = 45 kPa

3

Total pressure = 38 + 45 = 83 kPa 6. A mRT ρRT

Mr = — —— — — = — —— — —

pV p ρ

p = (— —— )RT (1 correct) M

r

Element

Mass

No. of moles

Mole ratio

87.5

87.5——— = 6.25

14

6.25——— = 1

6.25

100 – 87.5 = 12.5

12.5——— = 12.5

1

12.5——— = 2

6.25

N H



causes an increase in the average kinetic energy of the particles.

Gaseous stateMolecules are in constant motion (vibrational, rotational and translational motion).Molecules are more widely separated.Molecules have very high energy content.

Liquid state Molecules have vibrational, rotational and translational motion. Molecules in the liquid state are much closer together than in gases.Greater intermolecular forces restrict molecular movement.

Solid stateMolecules have vibrational and rotational motion but no translational motion. The particles are close together and are arranged in a defi nite pattern, held together by strong intermolecular forces. Molecules have lowest energy content.

(b) The behaviour of a real gas differs from the behaviour of an ideal gas because of• the existence of attractive forces between

the gaseous molecules and• the volume occupied by the gas molecules.

At high pressures, the volume is decreased.The volume occupied by the molecules and the attractive forces cannot be ignored. Hence real gases do not obey the ideal gas equation,pV = nRT.

(c) Total pressure = 0.992 atm0.992 = P

hydrogen + P

water vapour

Phydrogen

= 0.992 – 0.0313= 0.961

Volume = 45.5 cm3

= 0.0455 dm3

pV = nRT 0.961 � 0.0455

n = ————————————— 0.0821 � 298

= 1.79 � 10–3 mol

Number of moles of oxygen (O2)

1.0 � 2 2= ——————— = ———

RT RT(b) Amount of NO

2 produced

2NO(g) + O2(g) → 2NO

2(g)

2 ——— mol of nitrogen oxide (NO) reacts with

RT 1

——— mol of oxygen (O2) to produce

RT 2

——— mol of nitrogen dioxide (NO2).

RT (c) After the experiment

Number of moles of O2 remained

2 1 1

= ——— – ——— = ——— … (1) RT RT RT

Number of moles of NO2 produced

2= ——— … (2)

RT(d) pV = nRT

For oxygenVolume of oxygen after mixing = 2 + 4 = 6 dm3

1Partial pressure of oxygen � 6 = —— � RT

RTFrom equation (1)

1 Partial pressure of oxygen = — = 0.167 atm

6

For nitrogen dioxideVolume of NO

2 after mixing = 6 dm3

Partial pressure of nitrogen dioxide � 6From equation (2)

2= —— � RT

RT 2

Partial pressure of nitrogen dioxide = — 6

= 0.33 atm

Essay Question

1. (a) Kinetic theory: all matter is composed of tiny particles in constant motion.The essential distinction between the gaseous, liquid and solid states is the energy content of the molecules. An increase in temperature



Objective Questions

1. The phase diagram for iodine is shown below.

3 The Liquid and Solid States

Which of the following statements is true?A The line OC represents the effect of pressure on the melting point of iodine.B Point C is the critical temperature.C At point B, solid iodine and liquid iodine exist in equilibrium. D Iodine sublimes when heated from room temperature to 120 °C at a constant pressure of 14 kPa.

2. A “permanent” gas is a gas that cannot be liquefied at 25 °C. A “non-permanent gas” is a gas that can be liquefied at 25 °C. Which of the following gas is a permanent gas?

A

B

C

D

3. Which of the following does not exhibit allotropy?A Chlorine C PhosphorusB Ozone D Carbon

4. Diamond and graphiteA have the same bond length.B have delocalised electrons in their lattice structure.C produce carbon dioxide and water on combustion.D contain covalent bonds between carbon atoms.

A B C D




5. When a few crystals of ice are added to supercooled water at –5 °C,1 more ice is precipitated. 3 the temperature of water increases.2 the vapour pressure increases.

6. Dry ice is used as refrigerant in the food industry because1 its melting point is well below 0 °C.2 it is non-toxic and non-carcinogenic.3 it does not melt to form a liquid at room conditions.

Substance

Methane

Hydrogen chloride

Sulphur dioxide

Water

Critical temperature (K)

191

324

431

647

Critical pressure (atm)

46

82

78

218

100 110 120 130 140temperature (°C)

pres

sure

(kP

a)16

15

14

13

12

11

A

B

C

O



7. Which of the following statements is/are true?1 The combustion of fullerene in excess oxygen produces carbon dioxide only. 2 Fullerene and graphite have similar bonding and structure.3 Fullerene is a macromolecule with thousands of carbon atoms.

8. The crystal structure of an iodine crystal is shown below.

(a) Name the points O and A.(b) Name the phases in equilibrium

(i) at the point O, (ii) along OA, (iii) within AOC.(c) What is the signifi cance of

(i) the point A, (ii) the point D.(d) Explain why the line OC slopes to the left.(e) Explain what happens if ice at point X is gently heated, keeping the pressure constant.

Essay Question

1. (a) (i) Defi ne freezing point. (ii) Explain freezing using simple kinetic molecular theory.

(b) Describe the structure of fullerene.

What conclusion(s) can be made about the iodine crystal?1 The iodine crystal has a face-centred cubic lattice.

12 Each unit cell has — I

2 molecule at each corner.

8 1

3 Each unit cell has — I2 molecule at each face.

2

Structured Question

1. The phase diagram for water is shown below.

C

X

D

B

A

O612

0.0076

P (

N m

–2)



Objective Questions

1. A When solid iodine at room temperature is heated to 120 °C at a constant pressure of 14 kPa, it changes from solid to liquid.

2. A A gas can be liquefi ed only at temperatures below its critical temperature. For methane, its critical temperature is 191 K (–82 °C).

3. A 4. D Diamond and graphite contain carbon atoms

only. Therefore, the combustion of diamond and graphite gives carbon dioxide only.

5. D A supercooled liquid is metastable. The presence of ice-crystals or even dust particles will cause ice-crystals to be precipitated. This process is called seeding. (1 correct)The vapour pressure and temperature increase as more ice-crystals are formed. (2 and 3 correct).

6. D (1, 2, 3 correct) 7. A Fullerene is an allotrope of carbon. (1 correct) 8. D Each of the six faces has I

2 molecules at the

centre. (1correct) Each I

2 molecule at the corner of the unit cell

is shared between eight unit cells. (2 correct)Each I

2 molecule on the face of the unit cell is

shared between two unit cells. (3 correct)

Structured Question

1. (a) O : triple pointA : critical point

(b) (i) Solid, liquid and vapour (ii) Liquid and vapour (iii) Liquid (c) (i) Point A corresponds to the critical

temperature and pressure of the substance.At higher temperatures and pressures beyond this point, the liquid phase and vapour phase are indistinguishable.

(ii) The H2O(s) H

2O(l) system at point

D is metastable.(d) OC shows the effect of increasing pressure on the

melting point of ice. The line OC slopes towards the left because there is a decrease in volume when ice (solid) changes into water (liquid).

Hence, increase in pressure favours the formation of water.

(e)

ANSWER

CHAPTER 3 The Liquid and Solid States

At condition X – water exists in the solid state (in the form of ice).At condition (I) – ice melts. Equilibrium exists between ice and water.Between (I) and (II) – only water is present.At (II) – water evaporates. Equilibrium exists between water and water vapour.At temperatures above (II), only water vapour is present.

Essay Question

1. (a) (i) The freezing point is the temperature at which a solid and a liquid exist in equilibrium.

(ii) When the temperature of a liquid is lowered, • the kinetic energy of the particles

decreases,• the motion of particles in the liquid is

slowed down. At the freezing point, the attractive forces can hold the particles together in a fi xed position and the liquid freezes (change from liquid state to solid state).

(b) Fullerenes: carbon allotropes, cage-like hollow carbon molecules whose atoms are bonded into structures having hexagonal and pentagonal faces.Carbon atom uses sp2 hybridised orbitals to form C – C bonds. Hence, fullerenes contain π electrons.

pressure

temperature

X(I) (II)



Objective Questions

1. The figure below shows the lines in the Lyman series of the hydrogen spectrum.

4 The Electronic Structure of Atoms

The line M is caused by the transition of electronsA from n = 3 to n = 1B from n = 6 to n = 1C from n = 4 to n = 1D from n = 3 to n = 2

2. The electron configuration of a species which is isoelectronic with H2O is

A Li+ C N3–

B S2– D Cl–

3. Which of the following elements have all the electrons in the p valence shell orbitals unpaired?A NitrogenB OxygenC SulphurD Fluorine

4. An element forms a stable ion of charge +2. Which of the following electronic configuration represents this element?A 1s22s22p4

B 1s22s22p63s23p63d14s2

C 1s22s22p63s23p2

D 1s22s22p63s23p63d104s2

5. In the emission spectrum of hydrogen, how many lines are formed due to electron transitions involving energy levels n = 1 and n = 6?A 5 C 10B 6 D 15

Structured Question

1. (a) Draw the non-directional orbital which are found in an atom with the proton number 6. (b) Write the electronic configuration for the following:

(i) Fe in FeSO4

(ii) Mn in MnO2

(c) Write the valence electronic configuration of S in SF6 by using the box diagram and arrows

to repesent the electrons.

Essay Question

1. (a) The frequencies of the first six lines in the Lyman series (1014 Hz) are:24.7, 29.3, 30.9, 31.6, 32.0 and 32.2. Sketch a labelled energy level diagram to show the electron transitions responsible for forming the lines. By using the above frequencies, draw a suitable line graph that will enable you to calculate the ionisation energy of hydrogen.

(b) By using an energy level diagram of orbitals, show the electronic configuration of Fe in Fe

2O

3. Explain why it is difficult to find Fe(IV) compounds in nature.

M



1. C The Lyman line series are formed when electrons return to the lowest energy leveln = 1. M is the third line. Therefore it involves electron transition from n = 4 to n = 1.

2. C The total number of electrons in H2O

= 2(1) + 8 = 10Nitrogen has 7 electrons and it receives 3 more electrons to form N3–. Therefore there are 10 electrons in N3–.

3. A The valence shell of nitrogen has the electronic confi guration 2s22p3. Therefore each of the three 2p orbitals have a single unpaired electron.

Essay Question

1. (a)

ANSWER

CHAPTER 4 The Electronic Structure of Atoms

2s 2p

4. D The element must be from Period 4, Group 12 as shown by the electronic confi guration. It is zinc. Zinc only has a +2 charge.

5. D Total number of lines = 5 + 4 + 3 + 2 + 1 = 15

Structured Question

1. (a) Carbon has proton number 6. Electronic configuration of carbon is 1s22s22p2. The non-directional orbitals are 1s and 2s.

(b) (i) Fe in FeSO4 has +2 charge. The

electronic configuration of Fe2+ is 1s22s22p63s23p63d 6.

(ii) Mn in MnO2

has +4 charge. The electronic configuration of Mn4+ is 1s22s22p63s23p63d 3.

(c) S has electronic confi guration 1s22s22p63s23p 4. In SF

6, the 3s and 3p electrons are excited

to the empty 3d orbitals. Hence there are six unpaired electrons which will share electrons with an electron of each of the six fl uorine atoms. The valence electronic confi guration of S in SF

6 is:

3d

3s 3p

f (x1014 Hz)

∆f

24.7

4.6

29.3

1.6

30.9

0.7

31.6

0.4

32.0

0.2

32.2

–

To form iron(IV), the electron has to be removed from the Fe3+ ion.Fe3+ → Fe4+ + e–

Fe3+ has the electronic configuration 1s22s22p6

3s23p63d 5. It is stable because it has half-filled 3d orbitals. So it is difficult to remove an electron from the stable Fe3+ ion.

From the graph above, the frequency of the convergence limit is 33.1 � 1014 Hz.∆E = hfN

A

= 6.63 � 10–34 � 33.1 � 1014 � 6.02 � 1023

= 1321.1 � 103 J/mol–1

= 1321.1 � 103 /1000 kJ mol–1

= 1.32 � 103 kJ mol–1

(b) Fe in Fe2O

3 has a charge of +3. The electronic

configuration of Fe in Fe2O

3 is

f (x1014 Hz)

33.1

1

2

3

4

1s 2s

z

y

z

y

x x

n = 6n = 7

n = 5

n = 4

n = 3

n = 2

n = 1

24.7 29.3 30.9 31.6 32 32.2

frequency (X 1014 Hz)

ener

gy (

kJ m

ol–1

)

2p6

3p6

3s2

2s2

1s2

3d5

ener

gy



Objective Questions

1. A metal X has only one oxidation state in all its compounds. Atom X is most likely to have the electronic configuration ofA 1s2 2s2 2p3 C 1s2 2s2 2p6 3s2

B 1s2 2s2 2p5 D 1s2 2s2 2p6 3s2 3p6 3d5 4s1

2. Beryllium (Be), carbon (C), fluorine (F) and lithium (Li) are Period 2 elements. Which element has the lowest melting point and which element has the highest melting point?

A

B

C

D 3. In which of the following pairs is the atomic size of the first atom larger than the second atom?

A Na, Mg C Li, NaB Al, Mg D F, Cl

4. Which of the following pairs of elements will form a compound with the highest ionic character?A Cs and F C Ba and FB Li and I D Ba and I

5. The successive energies (in kJ mol–1) of an element, X, are shown below.

1020 (1st), 2150, 3400, 4650, 7080, 27 300, 32 100

This element is expected to form an ion with the formulaA X 2+ C X 2–

B X 3+ D X 3–

6. Which of the following pairs of elements has giant structures?A Carbon and phosphorusB Boron and siliconC Silicon and sulphurD Boron and phosphorus

5 The Periodic Table

Lowest melting point

Be

Be

F

F

Highest melting point

C

Li

C

Li

A B C D




7. Which of the following properties increases on going across Period 3 (sodium to chlorine)?1 Atomic size 3 First ionisation energy2 Electronegativity

8. On descending from beryllium to barium in Group 2 of the Periodic Table,

1 the first ionisation energy decreases.2 the metallic character increases.3 the metallic radius increases.



9. Which of the following graphs correctly represent(s) the variation of electronegativity against proton number?1 2 3


1. The physical properties of four elements, P, Q, X and Y are shown below.

Element

Melting point (°C)

First ionisation energy (kJ mol–1)

Electrical conductivity (25 °C)

P

937

762

Poor

Q

650

736

Good

X

1535

762

Good

Y

–7.2

1140

Non-conductor

(a) Which element is a (i) Group 2 element? (ii) Group 15 element? (iii) Group 17 element? (iv) d-block element?Give a reason for your choice.

(b) Identify the s and p elements from the list.

2. (a) The electronic configuration of an ion, X2+, is shown below.

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6

(i) What is the proton number of element X? (ii) Write the electronic configuration of element X. (iii) Classify element X as s-, p-, d- or f-block element.

(b) (i) Using aluminium as an example, explain what is meant by first ionisation energy. (ii) State the factors that influence ionisation energy.

(c) Explain the trend in the first ionisation energy on going down Group 1 elements.

Essay Question

1. (a) Explain the term first ionisation energy using chlorine as an example. (b) Explain the factors that influence the ionisation energy of an atom. (c) Explain the variation of first ionisation energy for the first row d-block elements.

0proton number

elec

tron

egat

ivity

3 4 5 6

Li

Be

B

C

0proton number

elec

tron

egat

ivity

5 10 15 20

LiNa

K

0proton number

elec

tron

egat

ivity

4 5 6 7

C

N

O

F



Objective Questions

1. C Group 1 and Group 2 metals have only one oxidation state.

2. C Fluorine: Simple covalent molecule(lowest melting point)

Carbon: Giant covalent molecule(highest melting point)

3. A Atomic size decreases across Periods 2 and 3. Atomic size increases down a group.

4. A Cs is the most electropositive, compared to Li and Ba. F is the most electronegative, compared to I. Hence, the ionic compound, CsF has the highest ionic character.

5. D 1st 2nd 3rd 4th 5th 6th 7th 1020 2150 3400 4650 7080 27 300 32 100 1130 1250 1250 2430 20 220 4800 ↑ It is Group 15 element.

Since the 1st I.E. is very high, it is expected to be a non-metal. A Group 15 non-metal will tend to form the X3– ion.

6. B The elements, phosphorus and sulphur have simple covalent structures.

7. C 8. D 9. B Electronegativity increases across periods but

decreases down a group.


1. (a) (i) QReason: Q is a Group 2 element because Group 2 elements are metals, with low fi rst ionisation energy, good conductor in the solid state and relatively low melting point.

(ii) P Reason: P is a metalloid (poor conductivity and fairly high melting point).P is Group 15 element. On going down Group 15, the metallic character changes from non-metals, to metalloid to metals.

(iii) YReason: Y is a non-metal (high fi rst ionisation energy and non-conductor).Y is a Group 17 element such as chlorine, which has a very low melting point.

(iv) XX is a metal (low fi rst ionisation energy, good conductor in the solid state).

X is a d-block element because it has a very high melting point.

(b) s-block element: Element Q p-block elements: Elements P and Y

2. (a) (i) 38There are 36 electrons in X2+ ion. Thus there are 38 electrons in element X.

(ii) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s2

(iii) X is a s-block element.(b) (i) The fi rst ionisation energy is the

energy required to remove one mole of electrons from one mole of atoms in the gaseous state.Al(g) → Al+(g) + e–;∆H = First ionisation energy of

aluminium (ii) The nuclear charge, that is, the number

of protons in the nucleus.The screening effect, that is, the effect of shielding of the outermost electrons from the attraction of the nucleus by the repelling effect of the electrons in the inner shells.

(c) On going down Group 1 elements, the fi rst ionisation energy decreases. This is because of• the increase in atomic radius. The further

the electron is from the nucleus, the weaker is the force of attraction of the nucleus on the electron and the more easily the electron can be removed.

• the increase in the shielding effect. This causes the electron to be held less strongly by the nucleus.

Essay Question

1. (a) The first ionisation energy is the minimum energy required to remove one mole of electrons from one mole of atoms in the gaseous state.

Cl(g) → Cl+(g) + e–

(b) Ionisation energy is influenced by the atomic size, the nuclear charge and the screening effect.The larger the atomic size, the lower the ionisation energy.The higher the nuclear charge, the higher the ionisation energy.The higher the screening effect, the lower the ionisation energy.

(c) The first ionisation energy remains fairly constant.The slight increase in ionisation energy is due to the fact that the effect of increased nuclear charge is roughly cancelled by the increasing shielding effect of the inner 3d electrons.

ANSWER

CHAPTER 5 The Periodic Table



Objective Questions

1. The diagram shows the effect of a charged rod on a liquid flowing out a burette.

6 Chemical Bonding

polar liquid

charged rod (positive or negative)

jet of deflected liquid

Which of the following liquids will be deflected by the charged rod?A Br

2 C SiCl

4

B BCl3

D PCl3

2. The formula of sodium peroxide is Na2O

2. Which of the ‘dot-and-cross’ diagrams represents the

structure of the peroxide ion?

[Key: o = electron from first oxygen atom, x = electron from second oxygen atom, • = electron from sodium atom]

A C

B D

3. Which of the following pairs contains both molecules that have linear shape?A CS

2 and XeF

2

B SCl2 and SO

2

C CS2 and SCl

2

D XeF2 and SO

2

4. Which of the following molecules has six bonding electrons?A NF

3 C C

2H

4

B CO2

D H2O

5. Chloromethane forms a carbonium ion R+ when it reacts with aluminium chloride as shown in the chemical equation below.

CH3Cl + AlCl

3 → CH

3+ + AlCl

4–

This reaction occurs becauseA AlCl

3 is a simple covalent molecule.

B the aluminium atom in AlCl3 does not have eight electrons in the valence shell.

C the chlorine atom in AlCl3 does not have eight electrons in the valence shell.

D the chlorine atom in CH3Cl has an empty p orbital.



6. Which of the following compounds contain(s) ionic bonds?1 Sodium chloride, NaCl2 Aluminium chloride, AlCl

3

3 Silicon(IV) chloride, SiCl4

7. Iron forms a complex with the formula, [Fe(CN)6]Cl

3. This complex contains

1 electrovalent bond2 covalent bond3 coordinate bond

8. Which of the following molecules contain(s) both sigma (σ) and pi (π) bonds?

1 NH3

2 CO2

3 C2H

4

9. Tetraethyllead has the formula (C2H

5)

4Pb. Which of the following statements is/are correct?

1 It is soluble in water.2 It is a covalent compound.3 It has a tetrahedral shape.


1. (a) Show the ‘dot-and-cross’ structures for the following species: (i) CO (ii) IF

3

(iii) H3O+

(b) Describe and explain the molecular shape of (i) IF3 and (ii) H

3O+?

(c) The carbonyl group, C = O, contains a σ-bond and a π-bond. With the help of a diagram,

explain the terms (i) σ-bond and (ii) π-bond.

2. Consider the elements and compounds given below:

potassium, silicon, helium, trichloromethane, sodium bromide

Select from this list, an element or a compound that shows the following properties:(a) A monatomic gas (b) A giant molecule that consists of atoms covalently bonded(c) A polyatomic molecule with low boiling point(d) A solid that conducts electricity in molten state but not in solid state(e) A substance that contains delocalised electrons

Essay Question

1. (a) The structure of propanoic acid is shown below.

OH ⏐

aCH3 – bCH

2 – cC = O

Describe the hybridised orbitals used by the carbon atoms marked a, b and c to form covalent bonds.

(b) Explain the effect of temperature on the electrical conductivity of (i) iron, (ii) silicon.

A B C D






xxx xC O

Objective Questions

1. D Br2

(linear), BCl3

(trigonal planar) and SiCl4

(tetrahedral) have symmetrical shapes. PCl3

(pyramidal shape) has resultant dipole. 2. D In the peroxide ion, each oxygen atom carries

a negative charge, –O – O–.Hence each oxygen atom receives one electron from each sodium atom.

3. A

(ii)

ANSWER

CHAPTER 6 Chemical Bonding

xxx x

x

x xxxx

x

x

S SC

Linear shape

The XeF2 molecule

Xe : 8e– (Group 18)2F : 2e–

Total : 10e–

The geometry of the electron-pairs and the shape of the molecule is shown below:

4. A xxx x

x

xx

xxxx

x

x

x xF

F

FN

5. B The central Al atom in AlCl3 is surrounded

by six electrons. So, it can accept 2 electrons from Cl– ion to complete the octet.

+

++

++

+ +

+ +++ +

+

+++

+ +++

+ +or

••

••

••

6. A AlCl3 and SiCl

4 are covalent compounds.

7. D Electrovalent bond between [Fe(CN)6l3+ and

Cl– ions.Covalent bond between C and N in CN– ion.Coordinate bond between Fe3+ and CN– ion.

8. C σ- and π-bonds are found in a double bond.O = C = O and CH

2 = CH

2 contain double bonds.

9. C (C2H

5)

4Pb is a covalent compound. The C – H

and C – Pb bonds are covalent bonds.Covalent compounds are insoluble in water.


1. (a) (i)

(iii)x

xH

H

H

+

O

T – shaped

Explanation: IF3 is T – shape because

the iodine atom is surrounded by three bond pairs of electrons and two lone pairs of electrons.

(ii)

H H H

O

+

pyramidal structure

Explanation: H3O+ has a pyramidal

structure because the oxygen atom is surrounded by three bond pairs of electrons and one lone pair of electrons.

(c) (i) A σ-bond is a covalent bond formed due to the head-on overlapping of two atomic orbitals.

(ii) A π-bond is a covalent bond formed by the sideways overlapping of atomic orbitals.

2. (a) Helium(b) Silicon(c) Trichloromethane

(d) Sodium bromide (e) Potassium

(b) (i)

F

F

F

I

xxx x

x

xx

xxxx

x

x

x xF F

xx

xx

x xF

I



Essay Question

1. (a) aCH3 – sp3 hybrid orbitals

bCH2 – sp3 hybrid orbitals.

Carbon atom in the excited state:

2s2 2s1 2p1x 2p1

y 2p1

z

The sp3 hybrid orbitals are formed when the one 2s orbital combines with three 2p orbitals.

H ⏐

cC = O – sp2 hybrid orbitals

Carbon atom in the excited state:

2s2 2s1 2p1x 2p1

y 2p1

z.

The sp2 hybrid orbitals are formed when the one 2s orbital combines with two 2p orbitals.

(b) (i) Iron is a metal. Electrical conductivity decreases with increase in temperature.Reason: Vibration of ions in the metal lattice impede the free movement of electrons in the conduction band.

(ii) Silicon is a metalloid. Electrical conductivity increases with increase in temperature.Reason: More electrons in the valence band gain thermal energy and are able to move into the conduction band.



Objective Questions

1. The compound that has the highest boiling point isA CH

3CH

2CH

2OH C CH

3 – O – CH

3

B (CH3)

2CHOH D CH

3CH

2CH

2Cl

2. Which of the following organic liquids dissolves most readily in water?A CCl

4 C CH

3(CH

2)

4COOH

B CH3CH

2CH

2OH D C

6H

5OH

3. What is the intermolecular bonding that is present in dry ice (solid carbon dioxide)?A Ionic bondB Covalent bondC Van der Waals forcesD Hydrogen bond

4. The structure of an amino acid, H2NCH

2COOH is shown below. Which of the hydrogen atoms,

labelled as I, II or III can form intermolecular hydrogen bonds.

H H O ⏐ ⏐ �

H – N – C – C – O – H (I) ⏐ (III)

H(II)

A I only C I and II onlyB II only D I and III only

7 Intermolecular Forces of Attraction

A B C D




5. Which of the following pairs of liquids are miscible?1 Water with methanol2 Benzene (C

6H

6) with methylbenzene (C

6H

5CH

3)

3 Benzene with water

6 Hydrogen bonding exists in1 liquid ammonia 3 liquid methane2 liquid hydrogen chloride

7. Which of the following can be explained in terms of hydrogen bonding?1 The density of ice is less than water.2 NH

3 has a higher boiling point than PH

3.

3 HI has a higher boiling point than HCl.

8. Which of the following organic compounds are polar molecules?1 Chloromethane, CH

3Cl

2 Trichloromethane, CHCl3

3 Tetrachloromethane, CCl4




1. The boiling points of some organic compounds are shown below.

Compound

Structural formula

Boiling point (oC)

I

CH3(CH

2)

3CH

3

36.3

II

CH3(CH

2)

4CH

3

68.7

III

CH3

⏐CH

3 – C – CH

3

⏐ CH

3

9.5

IV

CH3

⏐CH

3 – C – CH

3

⏐ OH

82.5

Explain why the boiling point of(a) compound II is higher than compound I.(b) compound I is higher than compound III.(c) compound IV is higher than compound II.

2. Water, ethanol, CH3CH

2OH and 1-propanol, CH

3CH

2CH

2OH are compounds containing the

–OH group and can form intermolecular hydrogen bonding. (a) Which of these compounds forms the strongest hydrogen bonds. (b) Explain your choice in (a).

Essay Question

1. (a) Which of the molecules, Br2 or ICl, is expected to have a higher boiling point? Explain your

answer. (b) The standard enthalpy of vaporisation, ∆H

vap for four compounds are shown below.

Compound

∆Hvap

(kJ mol–1)

H2Se

19.9

H2S

18.7

H2O

41.1

RbCl

154

Explain the trend of the standard enthalpy of vaporisation of these four compounds in terms of chemical bonding and intermolecular forces.



Objective Questions

1. A CH3CH

2CH

2OH is a straight chain alcohol

with hydrogen bonding. 2. B The –OH group is hydrophilic and causes organic

compounds to be soluble in water through hydrogen bonding. The hydrocarbon chain is hydrophobic which decreases the solubility of organic compounds in water. The organic compound becomes less soluble as the length of the carbon chain increases.

3. C The bond within CO2 is covalent bond. The bond

between CO2 molecules is van der Waals forces.

4. D A hydrogen atom attached to a carbon atom cannot form hydrogen bond.

5. B Water and methanol molecules have the same type of intermolecular forces, hydrogen bonding. Hence they are miscible.Benzene and methylbenzene molecules have the same type of intermolecular forces, van der Waals forces. Hence they are miscible.Benzene and water molecules have different types of intermolecular forces. Hence they are immiscible.

6. A 7. B Hydrogen bonds do not exist in HI and HCl. 8. B CCl

4 has symmetrical shape. It has polar bonds

but the molecule is non-polar.


1. (a) Compounds I and II are simple covalent molecules. The van der Waals forces in compound II is stronger because it has a larger molecular size.

(b) Compounds I and III are isomers. Compound III is a branched isomer. Hence it has a

smaller molecular size and therefore weaker van der Waals forces.

(c) Compound IV contains the –OH group. Hence, intermolecular hydrogen bonding is present in compound IV.

2. (a) Water molecules(b) Water has the highest boiling point compared

to ethanol and 1-propanol. Hence it has the strongest hydrogen bond.Water molecules do not contain the alkyl groups, whereas ethanol and 1-propanol contains the alkyl groups. The van der Waals forces of attraction between ethanol molecules or between 1-propanol molecules weakens the hydrogen bonds in these molecules.

Essay Question

1. (a) Mr of Br

2 = 160

Mr of ICl = 162.5

They have similar relative molecular mass. But Br

2 is non-polar and ICl is polar and has

a dipole moment. Hence, ICl should have a higher boiling point.

(b) RbCl has the highest ∆Hvap

because it is an ionic compound. Ionic bond is a very strong bond.H

2Se, H

2S and H

2O are simple covalent

molecules. The intermolecular forces are much weaker than ionic bond. Hence, the values of ∆H

vap are much smaller.

The molecular size of H2Se > H

2S. Hence the

van der Waals forces in H2Se are stronger.

Hence, H

2Se has a higher value of ∆H

vap.

Hydrogen bonds exist between water molecules. Hydrogen bonds are stronger than van der Waals forces. Hence, H

2O has a much

higher value of ∆Hvap

than H2Se or H

2S.

ANSWER

CHAPTER 7 Intermolecular Forces of Attraction



Objective Questions

1. The rate equation for a reaction is given as follows

Rate = k [A]2 [B]

Which of the following is the unit of the rate constant, k?A s–1 C mol–1 dm3 s–1

B mol dm3 s–1 D mol–2 dm6 s–1

2. The table below shows the result of an experiment for the reaction at 25 °C.

X(aq) + Y(aq) → Z(aq)

8 Kinetics

No. of experiment

1

2

3

[X]

x

2x

2x

[Y]

y

y

3y

Initial rate of reaction

r

8r

24r

What is the rate equation for the reaction?A Rate = k [X] [Y] C Rate = k [X]3 [Y]B Rate = k [X]2 [Y] D Rate = k [X]3 [Y]2

3. The mechanism of the reaction between diazonium ions with iodide ions is given below.

slowC

6H

5N

2+ ⎯⎯→ C

6H

5+ + N

2

fastC

6H

5+ + I– ⎯⎯→ C

6H

5 – I

Which of the following is true of the reaction between diazonium ions with iodide ions? A The unit of the rate constant is time–1.B The reaction is first order with respect to diazonium ions and iodide ions.C The rate is doubled when the concentration of iodide ions is doubled.D The half-life of the reaction is inversely proportional to the initial concentration of

diazonium ions.

4. Iodine reacts with propanone in the presence of dilute acids as represented by the following equation: I

2 + CH

3COCH

3 → HI + ICH

2COCH

3

The graph of the concentration of iodine in the reaction mixture measured at regular time intervals is given as follows

Based on the graph, which of the following statements is true of the reaction above?A The half-life of the reaction is a constant.B The rate of reaction is inversely proportional to [I

2].

C The rate-determining step in the reaction mechanism is independent of I2.

D The reaction is first order with respect to I2.

time

[I2]



A B C D




5. Which of the following statements is true about the rate constant, k in the equation of a reaction?1 k will increase when the temperature is increased.2 k will increase when the activation energy is increased.3 k will increase when the concentration of the reactants is increased.

6. Which of the following statement is true about a catalysed reaction?1 A catalyst provides a different mechanism for the reaction.2 A catalyst changes the order of the reaction in the rate equation.3 A catalyst reaction speeds up the forward reaction more than the backward reaction.


1. The table below shows the results of an investigation of the rate of reaction between hydroxide ion, OH– and phosphinate ion, PH

2O

2– at 80 °C. The overall reaction is

PH2O

2–(aq) + OH–(aq) → PHO

32–(aq) + H

2(g)

No ofexperiment

1

2

3

4

Initial concentration

0.1

0.3

0.6

0.6

Initial rate for the production of hydrogen gas (dm3 min–1)

14.4

43.2

2.4

9.6

[PH2O

2–] (mol dm–3)

6.0

6.0

1.0

2.0

[OH–] (mol dm–3)

(a) What is the order of reaction with respect to phosphinate ion?(b) What is the order of reaction with respect to hydroxide ion?(c) What is the overall order of reaction?(d) Write down the rate equation for the reaction above.(e) Calculate the value of the rate constant at 80 °C.(f) Predict the rate of production of hydrogen gas when the initial rate of phosphinate ions and

hydroxide ions are 0.6 mol dm–3 and 3.0 mol dm–3 respectively.

2. Trichloromethane, CHCl3 reacts with sodium hydroxide solution as represented by the following

equation: 2CHCl3 + 7OH– → CO + HCOO– + 6Cl– + 4H

2O

The reaction is first order with respect to each reactant.(a) Write the rate equation for the reaction between trichloromethane and sodium hydroxide.(b) Determine the rate of production of chloride ions at 28 °C when the rate of loss of

trichloromethane is 2.0 � 10–5 mol s–1.(c) If the rate of reaction is r when the concentrations of both trichloromethane and sodium

hydroxide are both 2.0 mol dm–3, what is the rate of reaction in terms of r when half of the hydroxide ion is reacted?

(d) Sketch a graph of the rate of reaction against the concentration of trichloromethane if sodium hydroxide is in excess such that the hydroxide ion concentration remains practically constant in the reaction mixture.



Essay Question

1. (a) Explain what is meant by the rate-determining step of a reaction mechanism. (b) Nitrogen monoxide reacts with fluorine to produce the compound nitrosyl fluoride, NOF as

represented the equation

2NO(g) + F2(g) → 2NOF(g)

It is found the reaction is first order with respect to both nitrogen monoxide and fluorine. Suggest a two-step mechanism for the reaction if the free radical fluorine, F is identified as an intermediate for the reaction.Justify your answer in terms of reaction kinetics.

(c) Describe and explain how the initial rate of reaction in (b) is affected by (i) an increase in temperature, (ii) an increase of 100 % of partial pressure of each reactant.



Objective Questions

unit of rate 1. D Unit of k = —————————————— [unit of reactant]3

mol dm–3 s–1

= —————————— (mol dm–3)3

= mol–2 dm6 s–1

Experiment 2 8r k (2x)n(y)m

2. C ——————————— = —— = ———————— Experiment 1 r k (x)n (y)m

8 = 2n

n = 3

Experiment 3 24r k (2x)n(3y)m

——————————— = ——— = ————————— Experiment 2 8r k (2x)n(y)m

3 = 3m

m = 1Rate = k [X]3 [Y]1

3. A Since the fi rst step is the slowest step, it is the rate-determining step. Hence the rate is fi rst order with respect to diazonium ions only (B and C incorrect). The unit of the rate constant of a fi rst order reaction is time–1 (A correct). The half-life of a fi rst order reaction is a constant (D incorrect).

4. C The graph of [I2] against time is a straight

line indicating that the gradient (which is the rate of reaction) is constant. Hence the rate is independent of [I

2]. The rate is zero order with

respect to I2. (C correct; A, B and D incorrect).

5. A k will increase when the temperature is increased (1 correct) and when the activation energy is decreased (2 incorrect) according to the Arrhenius equation:

k = A e–Ea/ RT

k is the proportional constant in the rate equation and is independent of the concentration of the reactants (3 incorrect).

6. B A catalyst will provide an alternate mechanism which involves a lower activation energy (1 correct). The rate-determining step in the reaction mechanism is different, hence the order of reaction will be different (2 correct). A catalyst speeds up the forward reaction and the backward reaction by the same amount (3 incorrect).


1. (a) Rate = k[PH2O

2–]n[OH–]m

By substituting the values of rate, [PH2O

2–]

and [OH–] from the results of experiments 1 and 2:

43.2 k[0.3]n[6.0]m

———— = —————————— 14.4 k[0.1]n[6.0]m

3 = 3n

n = 1∴ The reaction is fi rst order with respect to

phosphinate ion.(b) Rate = k[PH

2O

2–]n[OH–]m

By substituting the values of rate, [PH2O

2–]

and [OH–] from the results of experiments 3 and 4:

9.6 k[0.6]n[2.0]m

—— = —————————— 2.4 k[0.6]n[1.0]m 4 = 2m

m = 2∴ The reaction is second order with respect

to hydroxide ion.(c) The overall order of reaction is third order.(d) The rate equation is rate = k[PH

2O

2–] [OH–]2

(e) By substituting the values of rate, [PH2O

2–]

and [OH–] from the results of experiment 1 in the rate equation

14.4 = k[0.1]1 [6.0]2

14.4 dm3 min–1

k = ————————————————— (0.1 � 36) mol3 dm–9

= 4 mol–3 dm12 min–1

(f) By substituting the values of k, [PH2O

2–] and

[OH–] into the rate equation:Rate = 4 � 0.6 � (3.0)2

= 21.6 dm3 min–1

2. (a) The rate equation is: Rate = k[CHCl3] [OH–]

1 d[CHCl3] 1 d[Cl–]

(b) Rate = — ——————— = — ————— 2 dt 6 dt

d[Cl–]————— = 3 � 2.0 � 10–5 mol s–1

dt= 6.0 � 10–5 mol s–1

(c) When half of the hydroxide ion is reacted, [OH–]

remaining = 2.0 � ½ = 1.0 mol dm–3

According to the equation, 7 mol of OH– ions react with 2 mol of CHCl

3.

When 1.0 mol of OH– ions has reacted, 2

number of moles of CHCl3 reacted = — � 1.0

7

2 12[CHCl

3]

remaining = 2.0 – �—� 1.0� = ——

7 7

By substituting the values of r, [CHCl3] and

[OH–] into the rate equation:Rate = k [CHCl

3] [OH–]

and assuming the rate of reaction when half of the hydroxide ion is reacted is r’

r’ k (12/7) (1.0)— = ———————————

r k (2.0) (2.0)

3rr’ = ——

7(d) Graph of the rate of reaction against the

concentration of trichloromethane.

ANSWER

CHAPTER 8 Kinetics



Essay Question

1. (a) The rate-determining step is the slowest step of a reaction mechanism. The order of reaction with respect to the reactant in the rate equation is determined by the number of moles of reactants in the slowest step.

(b) The two-step mechanism: slow

NO + F2 ⎯→ NOF + F

fastF + NO ⎯→ NOFThe slowest step of the mechanism involves one mole of NO and one mole of F

2. This is

consistent with the rate equation in which the order of reaction is fi rst order with respect to NO and also fi rst order with respect to F

2:

Rate = k [NO] [F2].

(c) (i) When temperature increases, the rate constant, k increases as represented by Arrhenius equation:

k = A e–Ea/RT

rate of reaction(mol dm-3 time-1)

[CHCI ] (mol dm )3-3

The rate of reaction increases when the rate constant, k is increased.According to the Boltzman distribution curve, the fraction of molecules having a kinetic energy equal or greater than the activation energy, E

a is very much

larger at the higher temperature.

Shaded area = fraction of molecules with E > Ea

At higher temperature, T2, fraction of molecules with E > Ea increases signifi cantly

(ii) Partial pressure is directly proportional to the number of moles of the reactant.An increase in 100% of partial pressure of each reactant will result in a doubling of the concentration of each reactant.According to the rate equation: Rate = k [NO] [F

2]; when [NO] increases by 2

times and [F2] also increases 2 times,

the rate will increase by 4 times.

Graph of fraction of molecules with a particular energy attwo different temperatures



Objective Questions

1. The graph below shows the yield of gas X at different temperatures and pressures.

9 Chemical Equilibrium

Based on the graph only, what conclusion can be drawn?A There is a decrease in volume when the forward reaction proceeds.B The forward reaction is endothermic.C The reaction rate increases in the presence of a catalyst.D Pressure has no effect on the position of equilibrium.

2. A mixture of iron and steam is allowed to reach equilibrium at 800 K:

3Fe(s) + 4H2O(g) Fe

3O

4(s) + 4H

2(g)

The equilibrium pressures for hydrogen and steam are 3.5 kPa and 2.4 kPa respectively. The equilibrium constant, K

p, for the reaction is

A 1.46B 4.5C 6.0D 9.0

3. Hydrogen and carbon dioxide react as represented by the equation:

H2(g) + CO

2(g) CO(g) + H

2O(g)

When 4 mol of hydrogen and 0.9 mol of carbon dioxide were used for this reaction, 0.1 mol of CO

2 was found in the equilibrium mixture. Calculate K

c for the experiment.

A 0.4B 0.8C 2.0D 2.4

4. Consider the following equilibrium:

2NO2(g) N

2O

4(g)

At 350 K and 7000 kPa pressure, an equilibrium mixture contains 48 % by volume of N2O

4.

What is the value of Kp for the equilibrium?

A 2.54 � 10–3 (kPa)–1

B 2.54 � 10–4 (kPa)–1

C 3.92 � 102 kPaD 3.92 � 103 kPa



5. 2.0 mol of sulphur dioxide was allowed to react with 1.0 mol of oxygen at 10 atm pressure and at a fixed temperature. When equilibrium was achieved, 1.0 mol of sulphur dioxide was converted to sulphur trioxide.

2SO2(g) + O

2(g) 2SO

3(g)

Calculate Kp for the reaction.

A 0.5 atm–1

B 1.0 atm–1

C 3.0 atm–1

D 4.0 atm–1

A B C D




6. For the reaction:N

2(g) + O

2(g) 2NO(g)

the equilibrium constant, Kp, is 1.3 � 10–15 at 298 K and 5.5 � 10–2 at 2000 K. These data show that

1 the forward reaction is endothermic.2 the value of K

p depends on the amount (in mol) of nitrogen used.

3 the value of Kp increases as the pressure exerted on the system increases.

7. For the reversible reaction,

2SO2(g) + O

2(g) 2SO

3(g) ; ∆H = –198 kJ mol–1

the proportion of sulphur trioxide in the equilibrium mixture can be increased by1 increasing the total pressure on the system.2 increasing the temperature of the system.3 using a catalyst.


1. In the Contact process for the production of sulphuric acid(VI), sulphur dioxide and oxygen are converted into sulphur trioxide.

2SO2(g) + O

2(g) 2SO

3(g) ; ∆H = –198 kJ

(a) How would you expect the equilibrium position to vary with (i) increasing temperature? (ii) increasing pressure? (iii) the presence of a catalyst?

(b) (i) What are the industrial conditions for this process? (ii) Explain why these conditions are used.

2. The reaction as shown by the equation below represents the first step in the manufacture of nitric(V) acid from ammonia.

4NH3(g) + 5O

2(g) 4NO(g) + 6H

2O(g) ; ∆H = –909 kJ mol–1

(a) How would the yield of NO(g) be affected if (i) the pressure is increased at constant temperature? (ii) the temperature is increased at constant pressure?

(b) In the industrial process, a temperature of 900 °C is used. Based on your answer in (a)(ii), suggest a reason for using this temperature.

(c) In the industrial process, the mixture of NH3(g) and O

2(g) is allowed to pass through

platinum gauze. (i) Suggest a reason why platinum is used in this reaction. (ii) Give one advantage of using platinum in the form of platinum gauze.



Essay Question

1. (a) What is the effect of (i) temperature, (ii) pressure and (iii) catalyst on the equilibrium constant for the reaction:

N2O

4(g) 2NO

2(g) ; ∆H° = +58 kJ mol–1

(b) At 25 °C, the equilibrium constant for the reaction,

CH3COOH(aq) + C

2H

5OH(aq) CH

3COOC

2H

5(aq) + H

2O(l)

is 4.0. Calculate the molar composition of the mixture at equilibrium at 25 °C starting from (i) 1 mol of CH

3COOH and 2 mol of C

2H

5OH,

(ii) 1 mol of CH3COOH, 1 mol of H

2O and 2 mol of C

2H

5OH.

Comment on the results in terms of position of equilibrium.



Objective Questions

1. A The graph shows that the yield is increased by low temperatures and high pressures.Hence, the forward reaction is exothermic and accompanied by a decrease in the number of molecules (that is, a decrease in volume).

pH2

4 3.5 2. B Kp = �———— � = (——— )4 = 4.5

pH2O

2.4

3. C Amount of CO2 reacted = 0.9 – 0.1 = 0.8 mol

Amount of CO/H2O formed = 0.8 mol

Amount of H2 at equilibrium = 4 – 0.8

= 3.2 mol

H2(g) + CO

2(g) CO(g) + H

2O(g)

Initial 4 0.9 0 0(mol)Eqm 3.2 0.1 0.8 0.8(mol)

0.8 � 0.8K

c = ——————— = 2.0

3.2 � 0.1 4. B 2NO

2(g) N

2O

4(g)

48p

N2O4 = ——— � 7000 = 3360 kPa

100

100 – 48p

NO2 = ——————— � 7000 = 3640 kPa

100 p

N2O4 3360Kp = ————— = ————

(pNO2

)2 36402

= 2.54 � 10–4 (kPa)–1

5. A 2SO2(g) + O

2(g) 2SO

3(g)

Initally(mol) 2.0 1.0 0Equilibrium 2.0 – 1.0 1.0 – 0.5 1.0(mol)Total number of moles at equilibrium= (2.0 – 1.0) + (1.0 – 0.5) + 1.0 = 2.5

1.0p

SO3 = ——— � 10 = 4.0 atm

2.5

1.0p

SO2 = ——— � 10 = 4.0 atm

2.5

0.5p

O2 = ——— � 10 = 2.0 atm

2.5 (p

SO3)2 4.02

Kp = —————————— = ———————— = 0.5 atm–1

(pSO2

)2 � pO2

4.02 � 2.0

6. A Kp increases as the temperature increases.

So, the forward reaction is endothermic. 7. A


1. (a) (i) Equilibrium position shifts to the left (ii) Equilibrium position shifts to the right (iii) No effect

(b) (i) Temperature: 450 °CPressure: 1 or 2 atmCatalyst: V

2O

5

(ii) Increase in pressure will increase the yield of SO

3 but the effect is small for

this reaction. Hence, a pressure of 1 or 2 atm is used to cut down the cost of maintaining high pressures.

2. (a) (i) The number of moles of products is more than the number of moles of the reactants.Thus an increase in pressure will cause the equilibrium position to shift to the left. That is, the yield of NO(g) will decrease.

(ii) The reaction is an exothermic reaction. Thus, an increase in temperature will cause the equilibrium position to shift to the left, that is, the yield of NO(g) will decrease.

(b) If the temperature is lowered, the yield of NO(g) will increase. However, at low temperatures, the rate of reaction is very low. A temperature of 900 °C is used so that a reasonable yield of NO(g) is obtained at a relatively high rate of reaction.

(c) (i) Platinum is used as a catalyst in this reaction to speed up the reaction.

(ii) Platinum gauze is used because it has a larger surface area and this increases the rate of catalytic reaction.

Essay Question

1. (a) (i) In a reversible reaction, a change of temperature will affect the rates of both forward and reverse reactions, but not necessarily to the same extent. Thus, equilibrium constant varies with temperature.The given reaction is an endothermic reaction. For endothermic reaction, the equilibrium constant increases as temperature increases.

(ii) A change in the total pressure on the system does not change the equilibrium constant, even though the composition of the equilibrium mixture is changed.

(iii) A catalyst increases the rates of both forward and reverse reactions to the same extent. Hence, the catalyst has no effect on the equilibrium constant.

(b) (i) CH3COOH(aq) + C

2H

5OH(aq)

Initial 1 2(mol)Eqm(mol) 1 – x 2 – x

CH3COOC

2H

5(aq) + H

2O(l)

Initial(mol) 0 0Eqm(mol) x x

ANSWER

CHAPTER 9 Chemical Equilibrium



x2

4 = —————————— (1 – x)(2 – x)

x = 0.845Amount of CH

3COOC

2H

5 = 0.845 mol

Amount of water = 0.845 molAmount of CH

3COOH = 1 – 0.845

= 0.155 molAmount of alcohol = 2 – 0.845

= 1.16 mol

(ii) CH3COOH(aq) + C

2H

5OH(aq)

Initial 1 2(mol)Eqm(mol) 1 – x 2 – x

CH3COOC

2H

5(aq) + H

2O(l)

Initial(mol) 0 1Eqm(mol) x 1 + x

[CH3COOC

2H

5] [H

2O]

Kc = ———————————————————

[CH3COOH] [C

2H

5OH]

x(1 + x)4 = ——————————

(1 – x)(2 – x)x = 0.7427Amount of CH

3COOC

2H

5 = 0.743 mol

Amount of water = 1 + 0.743= 1.743 mol

Amount of CH3COOH = 1 – 0.743

= 0.257 molAmount of C

2H

5OH = 2 – 0.743

= 1.26 molComment: In the presence of H

2O,

equilibrium shifts to the left. Hence amount of CH

3COOC

2H

5 decreases

while the amounts of CH3COOH and

C2H

5OH increase.



Objective Questions

1. In the following reaction,

H2O(l) + NH

3(g) NH

4+(aq) + OH–(aq)

water acts as aA Bronsted-Lowry acidB Bronsted-Lowry baseC conjugate acidD conjugate base

2. The acid dissociation constant of propanoic acid, CH3CH

2COOH is 1.3 � 10–5 mol dm–3. What

is the concentration of H3O+ ions in 0.010 mol dm–3 propanoic acid?

1.3 � 10–5 2

A (———————— ) mol dm–3

102

1.3 � 10–5 ½

B (———————— ) mol dm–3

102

1.3 � 10–5

C (———————— ) mol dm–3

102

3 � 10–5 –2

D (———————— ) mol dm–3

102

3. When a monoprotic acid (in the conical flask) is titrated with a base, the pH changes abruptly from 3 to 7 at the equivalence point. This shows that it is a A strong acid-strong base titrationB strong acid-weak base titrationC weak acid-strong base titrationD weak acid-weak base titration

4. 25.0 cm3 of 1.0 mol dm–3 of ammonia solution is added to 25.0 cm3 of 1.0 mol dm–3 of ammonium chloride solution. What is the pH of the mixture?(K

b of ammonia is 1.75 � 10–5 mol dm–3)

A 3.5B 6.4C 9.2D 12.0

10 Acid-base Equilibria and Buffer Solutions

A B C D




5. Name of indicator

Congo red

Methyl yellow

Diphenol purple

pH range of indicator

3.0 – 5.0

2.9 – 4.0

7.0 – 9.0

Based on the information given above, we can infer that1 Congo red is suitable for the titration of aqueous ammonia with ethanoic acid.2 methyl yellow is suitable for the titration of aqueous ammonia with hydrochloric acid.3 diphenol purple is suitable for the titration of aqueous sodium hydroxide with ethanoic acid.



6. In the following reaction,

8NH3 + 3Cl

2 → N

2 + 6NH

4Cl

ammonia acts as a1 Lewis acid2 Lewis base3 reducing agent

7. The graph below shows the changes in pH for the titration between a monoprotic acid and aqueous sodium hydroxide.

Which of the following statements is/are correct?1 Addition of 10.0 cm3 of sodium hydroxide produces an acidic buffer solution in the conical

flask 2 Addition of 20.0 cm3 of sodium hydroxide produces a basic buffer solution in the conical

flask3 Addition of 40.0 cm3 of sodium hydroxide produces an acidic buffer solution in the conical

flask

8. Which of the following pairs of substance can act as a buffer solution?1 CH

3COOH and CH

3COO–

2 NH4+ and NH

3

3 HCO3– and CO

32–

9. When a little acid is added separately to the following mixtures, which mixture will show a marked change in pH?1 A mixture of NaHSO

4 and Na

2SO

4

2 A mixture of NaH2PO

4 and Na

2HPO

4

3 A mixture of CH3CH

2COOH and CH

3CH

2COONa


1. (a) The values of Kw for water at 10 °C, 30 °C and 40 °C are shown below.

Temperature (°C)

Kw � 1014 (mol2 dm–6)

10

0.29

30

1.5

40

2.9

(i) From the data given, what inference can you make regarding the dissociation of water? (ii) Calculate the pH of water at 40 °C. (iii) Calculate K

c at 30 °C for the following equilibrium.

H2O H+ + OH–

(Assume the density of water at 30 °C = 1 g cm–3)



(b) The formula of an acid-base indicator can be represented by HIn.

HIn(aq) H+(aq) + In–(aq)

[In–] (i) Show that pH = pK

In + lg ————

[HIn]

(ii) An acid-base indicator has pKIn = 4.0. The colour of HIn(aq) is yellow and the colour

of In–(aq) is blue. Calculate the pH of a solution if the colour of the indicator in the solution is green.

2. The graph below shows the changes in pH when 20.0 cm3 of saturated sulphur dioxide solution are added to 1.0 mol dm–3 sodium hydroxide solution in a titration.

(a) Write the chemical equation for the reaction that occurs at (i) the first equivalence point and (ii) the second equivalence point of the titration.

(b) Suggest a suitable indicator that can be used to show the first equivalence point.(c) Write the chemical equation for the complete reaction between sulphur dioxide and sodium

hydroxide.(d) Calculate the concentration in mol dm–3 for sulphur dioxide in the original solution.

Essay Question

1. (a) Define an acid based on (i) Arrhenius theory (ii) Bronsted-Lowry theory (iii) Lewis theory. (b) Calculate the pH of a solution formed by mixing 49 cm3 of 0.1 mol dm–3 aqueous sodium

hydroxide with 51 cm3 of 0.1 mol dm–3 hydrochloric acid. (c) Ammonia soluion is 4.2% dissociated at 25 °C in 0.01 mol dm–3 solution. Calculate pK

b for

ammonia at this temperature.



Objective Questions

1. A H2O donates H+ to NH

3(g)

2. B For a weak acid, [H+] = Ka � c

3. B At the equivalence point, an acidic solution is obtained. Hence it is a strong acid-weak base titration.

[salt] 4. C pOH = – lg K

b + lg —————

[base] 1.0 � 25.0

= – lg 1.75 � 10–5 + lg ———————— 1.0 � 25.0

= 4.76pH = 14 – 4.76

= 9.2 5. C The pH range for Congo red and methyl

yellow shows that they are indicators suitable for strong acid-weak base titration. For diphenol purple, the pH range shows that it is suitable for weak acid–strong base titration.

6. C The formation of NH4+ in NH

4Cl shows that

NH3

acts as a Lewis base (an electron-pair donor).NH

3 reduces Cl

2 to Cl–. Thus it is a reducing

agent. 7. A The pH change shows that it is a strong base-

weak acid titration.Before the end point, the solution in the conical fl ask contains excess acid and salt of the weak acid. Thus it is an acidic buffer solution. After the end point, the solution contains excess alkali and salt of the weak acid. Hence, it is not a buffer solution.

8. D 9. A A mixture of NaHSO

4 and Na

2SO

4 is not a

buffer solution.

Structured Question

1. (a) (i) H2O

H+ + OH–

Kw = [H+][OH–]

When the temperature is increased, the value of K

w increases. This shows that

more H+ and OH– ions are produced. That is, equilibrium position shifts to the right. Thus, the dissociation of water is an endothermic process.

(ii) At 40 °C, Kw = 2.9 � 10–14 mol2 dm–6

[H+] = (2.9 � 10–14)

= 1.7 � 10–7 mol dm–3

pH = – lg [H+]= – lg (1.7 � 10–7)= 6.8

[H+][OH–] Kw (iii) K

c = ———————— = ————

[H2O] [H

2O]

1000[H

2O] = ———— = 55.6 mol dm–3

18 1.5 � 10–14

Kc = —————————

55.6= 2.7 � 10–16 mol dm–3

(b) (i) HIn(aq)

H+(aq) + In–(aq)

[H+(aq)][In–(aq)]K

HIn = —————————————

[HIn]By taking negative logarithm (lg) and rearranging, we have

[In–]– lg [H+] – lg ———— = –lg K

HIn [HIn] [In–]

pH = pKHIn

+ lg ———— [HIn]

(ii) A mixture of yellow and blue gives green. Thus, green shows that[HIn] = [In–] and pH = 4.0 + lg 1

= 4.0. 2. (a) (i) SO

2(aq) + NaOH(aq) → NaHSO

3(aq)

(ii) NaHSO3(aq) + NaOH(aq) →

Na2SO

3(aq) + H

2O(l)

(b) Methyl orange(c) SO

2(aq) + 2NaOH(aq) →

Na2SO

3(aq) + H

2O(l)

(M1V

1)

SO2 1(d) ———————— = —

(M2V

2)

NaOH 2

Concentration of aqueous SO2

1.0 � 25.2= ½ � ————————— = 0.63 mol dm–3

20

Essay Question

1. (a) (i) An acid is a substance that dissociates in water to produce H+ ions.

(ii) An acid is a substance that can donate a proton to a base.

(iii) An acid is a substance that accepts a pair of electrons from a base.

(b) 49 cm3 of 0.1 mol dm–3 would neutralise49 cm3 of 0.1 mol dm–3 HCl.2 cm3 of 0.1 mol dm–3 HCl remained in(49 + 51) = 100 cm3 solution.Number of moles of H+ in 100 cm3 solution

0.1 � 2= —————— = 2 � 10–4

1000100 cm3 ≡ 0.1 dm3

(2 � 10–4) [H+] = ———————— = 2 � 10–3 mol dm–3

0.1pH = – lg (2 � 10–3) = 2.7

α2c (c) K

b = ————

1 – α 0.0422 � 0.01

= ——————————— 1 – 0.042

= 1.84 � 10–5 mol dm–3

pKb = – lg (1.84 � 10–5)

= 4.74

ANSWER

CHAPTER 10 Acid-base Equilibria and Buffer Solutions



Objective Questions

1. A metal forms a salt with formula M2X. If the solubility of the salt is s mol dm–3, its solubility

product isA 2s2 mol2 dm–6 C 4s3 mol3 dm–9

B 4s2 mol2 dm–6 D 2s4 mol4 dm–12

2 At 15 °C, the solubility of lead(II) iodide, PbI2 is 0.6 g dm–3. What is the solubility product of

lead(II) iodide at this temperature? (Mr of PbI

2 = 461)

0.6A (——— )3 mol3 dm–9

461

0.6B 2 � (——— )3 mol3 dm–9

461

0.6C 4 � (——— )3 mol3 dm–9

461

461D 4 � (——— )3 mol3 dm–9

0.6

3. A solution containing 7 � 10–7 mol dm–3 Ag+ ions is required for a reaction. Which of the following saturated solution can be used for this purpose?

A

B

C

D

4. 150 cm3 of aqueous silver nitrate is added to 150 cm3 of 0.8 mol dm–3 aqueous sodium chloride. What is the minimum concentration of aqueous silver nitrate needed to cause the precipitation of silver chloride?[K

sp of AgCl = 2.0 � 10–10 mol2 dm–6]

A 1.0 � 10–8 mol dm–3

B 2.0 � 10–8 mol dm–3

C 1.0 � 10–9 mol dm–3

D 2.0 � 10–9 mol dm–3

5. 50 cm3 of a saturated solution of Ca(OH)

2 required 10 cm3 of 0.1 mol dm–3 HCl for complete

neutralisation. The solubility product of Ca(OH)2 is

1 1 1A — � —— � (——— )2 mol3 dm–9

2 50 50

1 1 1B — � —— � (——— )2 mol3 dm–9

2 100 100

1 2C —— � (——— )2 mol3 dm–9

50 50

1 2D —— � (——— )2 mol3 dm–9

50 100

11 Heterogeneous Ionic Equilibrium

Saturated solution of the silver salt

AgCl

AgBr

Ag2SO

4

Ag2CrO

4

Ksp

of the silver salt

2.0 � 10–10 mol2 dm–6

5.0 � 10–13 mol2 dm–6

2.0 � 10–5 mol3 dm–9

1.3 � 10–12 mol3 dm–9



6. Which of the following observations can be explained in terms of common ion effect?1 BaSO

4(s) is less soluble in dilute sulphuric acid than in water.

2 When solid NaCl is added to a saturated solution of silver chloride, AgCl will precipitate out.

3 The concentration of OH– ions is increased when solid NH4Cl is added to ammonia

solution.

7. Silver chloride is sparingly soluble in water. When aqueous ammonia is added to silver chloride in water, it dissolves to form a clear solution because1 soluble silver hydroxide is formed.2 the complex ion, [Ag(NH

3)

2]+ is formed.

3 the ionic product of silver chloride is less than its solubility product.


1. (a) Write an equation to show the equilibrium that exists in a saturated solution of barium fluoride, BaF

2.

(b) Write the solubility product expression for barium fluoride and state its units.(c) A solution is 0.10 mol dm–3 with respect to both Ba2+ and Ca2+ ions. Sodium fluoride is

slowly added to precipitate BaF2.

(i) Calculate the concentration of F– ions at which BaF2 begins to precipitate.

(ii) Hence, calculate the concentration of Ca2+ ion that remains unprecipitated at this point.[K

sp for BaF

2 = 1.7 � 10–6 mol3 dm–9; K

sp for CaF

2 = 1.7 � 10–10 mol3 dm–9]

2. (a) The solubility products at 298 K for AgCl, Fe(OH)2 and Mn(OH)

2 are shown below.

Ksp

for AgCl = 1.0 � 10–10 mol2 dm–6

Ksp

for Fe(OH)2 = 8.0 � 10–16 mol3 dm–9

Ksp

for Mn(OH)2 = 1.9 � 10–13 mol3 dm–9

Which of the following pairs of solution would produce a precipitate when the solutions are mixed?[K

b for ammonia = 1.8 � 10–5 mol dm–3]

(i) 25.0 cm3 of 1.0 � 10–3 mol dm–3 AgNO3(aq) and 50.0 cm3 of 1.0 � 10–5 mol dm–3

KCl(aq) (ii) 25.0 cm3 of 3.0 � 10–3 mol dm–3 FeSO

4(aq) and 75.0 cm3 of 1.0 � 10–4 mol dm–3

NaOH(aq) (iii) 50.0 cm3 of 0.05 mol dm–3 NH

3(aq) and 50.0 cm3 of 0.025 mol dm–3 MnSO

4(aq)

(b) A saturated solution Z contains calcium hydroxide and calcium sulphate. (i) If the pH of solution Z is 12.5, calculate the concentration of Ca2+ in solution Z.

[Ksp

for Ca(OH)2 is 5.5 � 10–6 mol3 dm–9]

(ii) Calculate the concentration of SO42– ions in solution Z.

[Ksp

for CaSO4 is 2.4 � 10–5 mol2 dm-6]

A B C D






Essay Question

1. (a) The values of Ksp

for iron(II) hydroxide and iron(III) hydroxide are shown below.

Compound

Fe(OH)2

Fe(OH)3

Ksp

at 298 K

7.9 � 10–16 mol3 dm–9

2.0 � 10–39 mol4 dm–12

Iron(III) compounds react with potassium iodide as represented by the equation

2Fe3+(aq) + 2I–(aq) 2Fe2+(aq) + I2(s)

When aqueous iodine is added to iron(II) hydroxide solid that contains a few drops of alkali solution, the iodine solution is decolourised and iron(II) hydroxide is converted to iron(III) hydroxide. Explain this observation.

(b) The concentration of Cl–(aq) ion in aqueous solution can be determined by titration with

silver nitrate. The indicator used is potassium chromate(VI). At the end point, a red precipitate of silver chromate(VI), Ag

2CrO

4 is produced.

Ksp

of AgCl = 1.0 � 10–10 mol2 dm–6 ; Ksp

of Ag2CrO

4 = 2.4 � 10–12 mol3 dm–9

(i) Calculate the minimum concentration of Ag+ required to cause precipitation of AgCl. (ii) Calculate the minimum concentration of Ag+ required to cause precipitation of

Ag2CrO

4.

(iii) Based on your answers in (b)(i) and (ii), describe what happens when silver nitrate solution is run into a solution of chloride ions, using potassium chromate(VI) as indicator.



Objective Questions

1. C M2X 2M + X

[M] = 2s, [X] = sK

sp = [M]2[X] = (2s)2(s) mol3 dm–9

2. C Ksp

for PbI2 = [Pb2+][I–]2

0.6 0.6= ——— � (2 � ——— )2 mol3 dm–9

461 461 3. B [Ag+][Br-] = 5 � 10–13 mol2 dm–6

[Ag+] = 5 � 10–13 = 7.1 � 10–7 mol dm–3

Ag2CrO

4 2Ag+ + CrO

42–

Let [Ag+] = 2xK

sp = [Ag+]2[CrO

42–]

1.3 � 10–12 = (2x)2(x)x = 6.9 � 10–5

2x = 1.38 � 10–4

[Ag+] = 1.38 � 10–4 mol dm–3

4. C [Cl–] after mixing = ½ � 0.8 = 0.4 mol dm–3

(as volume is doubled)For precipitation to occur, ionic product of AgCl must exceed the K

sp of AgCl.

[Ag+] � 0.4 = 2.0 � 10–10

[Ag+] = 5 � 10–10 mol dm–3

M1V

1 before mixing = M

2V

2 after mixing

M1 � 150 = 5 � 10–10 � (150 + 150)

M1 = 5 � 10–10 � 2 = 1.0 � 10–9 mol dm–3

5. A OH– + HCl → H2O + Cl–

M1V

1 = M

2V

2

0.1 � 10Concentration of OH– = ———————

50 1

= —— mol dm–3

50Ca(OH)

2 → Ca2+ + 2OH–

1 1Concentration of Ca2+ = — � —— mol dm–3

2 50K

sp = [Ca2+][OH–]2

1 1 1= — � —— � (—— )2 mol3 dm–9

2 50 50 6. B BaSO

4(s) Ba2+(aq) + SO

42–(aq)

H2SO

4(aq) → 2H+(aq) + SO

42–(aq)

Common ion, SO42– causes BaSO

4 to be less

soluble. (1 correct) AgCl(s) Ag+(aq) + Cl–(aq) NH

4Cl(aq) → NH

4+(aq) + Cl–(aq)

Common ion, Cl– shifts the AgCl equilibrium to the left, hence AgCl precipitates out.(2 correct)

NH3(g) + H

2O(l) NH

4+(aq) + OH–(aq)

NH4Cl(s) + water NH

4+(aq) + Cl–(aq)

The concentration of OH– decreased due to common ion effect. (3 incorrect)

7. C Silver hydroxide is insoluble. (1 incorrect) AgCl(s) Ag+(aq) + Cl–(aq) When aqueous NH

3 is added, it reacts with

silver to form [Ag(NH3)

2]+(aq). (2 correct)

Aqueous NH3 solution added causes [Ag+] and

[Cl–] to decrease hence ionic product [Ag+] [Cl–] < K

sp. (3 correct)


1. (a) BaF2(s) Ba2+(aq) + 2F–(aq)

(b) Ksp

= [Ba2+][F–]2 mol3 dm–9

(c) (i) BaF2 begins to precipitate when ionic

product = Ksp

(saturated solution).[Ba2+][F–]2 = 1.7 × 10–6

0.1 × [F–]2 = 1.7 × 10–6

1.7 × 10–6

∴ [F–] = ———————— 0.1

= 4.1 × 10–3 mol dm–3

(ii) [Ca2+][4.1 × 10–3]2 = 1.7 × 10–10

1.7 × 10–10

[Ca2+] = —————————— (4.1 × 10–3)2

= 1.0 × 10–5 mol dm–3

25 2. (a) (i) [Ag+] = —— × (1 × 10–3) 75

= 3.33 × 10–4 mol dm–3

50[Cl–] = —— × (1 × 10–5)

75= 6.67 × 10–6 mol dm–3

Ionic product = (3.33 × 10–4) ×(6.67 × 10–6)

= 2.22 × 10–9 mol2 dm–6

Ionic product > Ksp

∴ Precipitation of AgCl occurs. 25

(ii) [Fe2+] = ——— × (3.0 × 10–3) 100

= 7.5 × 10–4 mol dm–3

75[OH–] = ——— × (1 × 10–4)

100= 7.5 × 10–5 mol dm–3

Ionic product = (7.5 × 10–4) ×(7.5 × 10–5)2

= 4.2 × 10–12 mol3 dm–9

Ionic product > Ksp

∴Precipitation of Fe(OH)2 occurs.

50 (iii) [Mn2+] = ——— × 0.025

100= 0.0125 mol dm–3

50[NH

3] = ——— × 0.050

100= 0.025 mol dm–3

[OH–] = Kb × c

= (1.8 × 10–5) × 0.025= 6.7 × 10–4 mol dm–3

Ionic product = 0.0125 × (6.7 × 10–4)2

= 5.6 × 10–9 mol3 dm–9

Ionic product > Ksp

∴ Precipitation of Mn(OH)2 occurs.

(b) (i) pH = 14 + lg[OH–]lg[OH–] = 12.5 – 14

= –1.5

ANSWER

CHAPTER 11 Heterogeneous Ionic Equilibrium



[OH–] = 0.032 mol dm-3

Ksp

for Ca(OH)2 = [Ca2+][OH–]2

5.5 � 10–6 = [Ca2+] � (0.032)2

[Ca2+] = 0.0054 mol dm–3

(ii) Ksp

= [Ca2+][SO42–]

2.4 � 10–5

[SO42–] = ——————— = 0.0044 mol dm–3

0.0054

Essay Question

1. (a) In the presence of iodine, equilibrium shifts to the left (le Chatelier’s principle) and Fe3+ ions are produced. Fe3+ ions react with OH– ions from the alkali to form Fe(OH)

3.

The Ksp

for Fe(OH)3 is very much lower than

the Ksp

for Fe(OH)2.

Thus, Fe(OH)3 can be precipitated more

readily than Fe(OH)2.

(b) (i) Ksp

for AgCl = 1.0 � 10–10 mol2 dm–6

[Ag+] = 1.0 � 10–10

= 1.0 � 10–5 mol dm–3

If [Ag+] is higher than 1.0 � 10–5 mol dm–3, precipitation of AgCl will occur.

(ii) Ksp

for Ag2CrO

4 = 2.4 � 10–12 mol3 dm–9

[Ag+]2[CrO42–] = 2.4 � 10–12 mol3 dm–9

Let concentration of Ag+ needed to precipitate Ag

2CrO

4 = x mol dm–3

Ag2CrO

4(aq)

2Ag+(aq) + CrO42–(aq)

[CrO42–] = ½ x mol dm–3

x2 � (½x) = 2.4 � 10–12

x3 = 4.8 � 10–12;x = 1.69 � 10–4 mol dm–3

If [Ag+] is higher than 1.69 � 10–4 mol dm–3, precipitation of Ag

2CrO

4 will

occur. (iii) The answers in (b)(i) and (ii) show that a

higher concentration of Ag+ ions is need to cause the precipitation of Ag

2CrO

4.

When silver nitrate solution is run into a solution of Cl– ions, Ag

2CrO

4 does

not precipitate until all the Cl– ions have precipitated as AgCl. When the precipitation of AgCl has completed, the addition of one drop of silver nitrate will cause the precipitation of Ag

2CrO

4,

which corresponds to the end point of the titration.



Objective Questions

1. Which of the following liquid mixture shows positive deviation from Raoult’s law?A Hexane and heptaneB Ethanol and waterC Hydrogen bromide and waterD Nitric acid and water

2. Two miscible liquids X and Y form an ideal solution, containing 3 mol of X and 1 mol of Y. The

total vapour pressure at 20 °C is 48 kPa. The vapour pressure of pure X is 52 kPa at 20 °C. What is the vapour pressure of pure Y at 20 °C?A 36 kPa C 92 kPaB 48 kPa D 192 kPa

3. The partition coefficient of butanoic acid (CH

3CH

2CH

2COOH) between ether and water is 3.0.

If 1.0 g of butanoic acid is shaken with a mixture of water and ether until equilibrium is established, what is the volume of 1 mol dm–3 sodium hydroxide solution required to neutralise all the acid in the ether layer? [M

r of butanoic acid = 88; the mixture contains equal volumes of water and ether]

A 2.84 cm3

B 8.52 cm3

C 17.1 cm3

D 28.4 cm3

4. When a mixture of compound X and water is steam distilled at a pressure of 97.1 kPa and a

temperature of 98 °C, a distillate containing compound X and water in the ratio of 0.188 : 1 (by mass) is obtained. If the vapour pressure of water at 98 °C is 94.5 kPa, what is the relative molecular mass of X? A 98 C 196B 123 D 246

5. Which of the following solutions has the lowest vapour pressure at 30 °C?

A Distilled waterB 0.1 mol dm–3 MgSO

4 solution

C 0.1 mol dm–3 MgCl2 solution

D A mixture of water and phenylamine

12 Phase Equilibrium

A B C D




6. When propanone (CH3COCH

3) is mixed with chloromethane, heat energy is liberated and the

increase in temperature is ∆T1. If the experiment is repeated using CHCl

3 and CCl

4, the results

as shown in the table below are obtained. In each experiment, the volumes of both the organic compounds are kept constant.

Mixture

CH3COCH

3 + CH

3Cl

CH3COCH

3 + CHCl

3

CH3COCH

3 + CCl

4

Increase in temperature

∆T1

∆T2

∆T3



Which of the following relationships concerning the increase in temperature is/are correct?1 ∆T

1 = ∆T

2 = ∆T

3

2 ∆T1 < ∆T

2

3 ∆T3 < ∆T

2

7. A mixture of two liquids, X and Y shows positive deviation from Raoult’s law. This implies that1 the two liquids are immiscible.2 the vapour pressure-composition curve shows a maximum point.3 the azeotropic mixture has a lower boiling point than pure X or pure Y.

8. The boiling point-composition diagram for a mixture of two miscible liquids, A and B, is shown below.

Based on the diagram above, we can conclude that1 pure A and pure B cannot be obtained from fractional distillation of mixture X. 2 fractional distillation of mixture Y produces pure B.3 the mixture of A and B shows positive deviation from Raoult’s law.


1. (a) A mixture of water and methanol obeys Raoult’s law but a mixture of water and 1-propanol forms an azeotropic mixture. Explain the behaviour of these two mixtures.

(b) Explain why water is immiscible with hexanol, CH3(CH

2)

5OH.

2. Butanedioic acid (HOOCCH2CH

2COOH) is added to a separating funnel which contains equal

volumes of water and ether. When equilibrium is established, the concentrations of the acid in aqueous and ether layers are determined by titration. The results of four experiments are shown below.

Volume of alkali required to neutralise10 cm3 of the ether layer (cm3)

1.4

1.8

2.0

2.5

Volume of alkali required to neutralise10 cm3 of aqueous layer (cm3)

9.0

12.0

13.5

17.5

(a) Name an alkali and a suitable indicator which can be used for the titration above.(b) Is butanedioic acid more soluble in ether or in water? Justify your answer.(c) Plot a graph and determine the partition coefficient of butanedioic acid between ether and

water from the graph.(d) 1.0 g of butanedioic acid is dissolved in 20 cm3 water. The acid in the solution is extracted

by using 40.0 cm3 of ether. Calculate the percentage of butanedioic acid that still remains in the aqueous layer after extraction.

(e) If the acid is extracted twice using two separate portions of 20 cm3 of ether, what is the mass of ethanoic acid still remaining in water after the second extraction?

temperature

liquid

composition0% A

100% B 0% B100% AX Y

vapo

ur



Essay Question

1. (a) The boiling point-composition diagram for hydrogen fluoride and water is shown in the figure below.

120

100

80

60

40

20

0 20

100

40 60 80

80 60 40 20 0 HF

100 H2OX

vapo

ur

vapour

liqui

d

liquid

(i) A solution with composition X is heated until boiling. What is the boiling point of this solution?What is the composition of the vapour at the boiling point?

(ii) Explain why a mixture of hydrogen fl uoride and water shows negative deviation from Raoult’s law.

(b) The liquids P and Q form an ideal mixture. At 30 °C, the vapour pressures of pure P and pure Q are 120 kPa and 40 kPa respectively and the mole fraction of P in the solution is 0.60.

(i) Calculate the total vapour pressure for the liquid mixture at 30 °C. (ii) Calculate the composition of vapour at equilibrium with the liquid mixture at 30 °C.



Objective Questions

1. B Hexane and heptane – ideal solutionHBr/H

2O – negative deviation

HNO3/H

2O – negative deviation

3 2. A Mole fraction of X = — = 0.75 4

Mole fraction of Y = 1 – 0.75 = 0.25Vapour pressure of X = 52 � 0.75 = 39 kPaVapour pressure of Y = 48 – 39 = 9 kPa9 = P

Yo � 0.25; P

Yo = 36 kPa

3. B Let mass of butanoic acid in ether layer = x g∴ Mass of butanoic acid in water = (1.0 – x) g

x————— = 3.0

1.0 – x∴ x = 0.75 gNumber of moles of butanoic acid

0.75= ——— = 0.00852

88CH

3CH

2CH

2COOH + NaOH →

CH3CH

2CH

2COONa + H

2O

∴ Number of moles of alkali required= number of moles of acid present= 0.00852

MV 1 � V0.00852 = ———— = —————

1000 1000∴ V = 8.52 cm3

mA p

A � M

A 4. B ———— = ————————

mH2O

pH2O

� 18

0.188 (97.1 – 94.5) � MA———— = ————————————————

1 94.5 � 18M

A = 123

5. C MgCl2(aq) → Mg2+(aq) + 2Cl–(aq)

MgCl2 solution contains the most number of

ions per dm3 solution. 6. C The stronger the intermolecular forces of

attraction, the greater the energy liberated in forming a non-ideal solution. CHCl

3 is the

most polar because it has three chlorine atoms per molecule. Both CHCl

3 and CH

3Cl can

form hydrogen bonding with CH3COCH

3. CCl

4

molecule does not contain hydrogen atom. Hence, it cannot form hydrogen bonds with CH

3COCH

3.

7. C 8. B X is an azeotropic mixture. Its composition

and boiling point remains unchanged during distillation. (1 correct) Fractional distillation of Y produces pure B. (2 correct) The mixture shows negative deviation. (3 incorrect)


1. (a) Both water and methanol molecules form intermolecular hydrogen bonds. The intermolecular forces of attraction between

water molecules, between methanol molecules and between water-methanol molecules are almost the same. Hence, the mixture of methanol and water obeys Raoult’s law.The intermolecular forces of attraction between 1-propanol and water in the mixture is weaker than the intermolecular forces of attraction between 1-propanol-1-propanol molecules and between water-water molecules. Hence, a mixture of 1-propanol and water shows positive deviation from Raoult’s law and forms an azeotropic mixture with minimum boiling point.

(b) Two liquids are immiscible if the intermolecular forces in the liquids are vastly different. In water, the molecules are held strongly by hydrogen bonds. Hexanol is a much bigger molecule than water. The van der Waals forces in hexanol is very strong. Hence, it is immiscible with water.

2. (a) Sodium hydroxide solution; phenolphthalein (weak acid – strong base titration).

(b) More soluble in water because more alkali is required for neutralisation as shown in the table.

(c)

ANSWER

CHAPTER 12 Phase Equilibrium

ABGradient = ———

BC (2.42 – 1.60)

= —————————— = 0.13 (17.0 – 10.5)

(d) Let mass of butanedioic acid remaining in water = m g∴ Mass of butanedioic acid in ether

= (1.0 – m) g (1.0 – m)

——————— 40

0.13 = ——————— m

—— 20

m = 0.79 0.79

% = ———— � 100 = 79% 1



(e) First extractionLet mass of acid in water = x g∴ Mass of acid in ether = (1.0 – x) g

(1.0 – x)———————

200.13 = ——————— x

—— 20

x = 0.88 g

Second extractionLet mass of acid in water = y g∴ Mass of acid in ether = (0.88 – y) g

(0.88 – y)———————

20 0.13 = ——————— y

—— 20

y = 0.78 g

Essay Question

1. (a) (i) 60 °C; 8% H2O and 92% HF.

(ii) The intermolecular forces of attraction between HF --- HF in pure HF and betweenH

2O --- H

2O in pure H

2O are hydrogen

bonds. In aqueous solution, HF dissociates to form H+ and F- ions. The ion-solvent attraction between H+ and F–

ions with water is stronger than hydrogen bonds. Hence aqueous HF shows negative deviation from Raoult’s law.

(b) (i) Partial pressure of P = 120 � 0.6= 72 kPa

Partial pressure of Q = 40 � 0.4 = 16 kPa∴Total vapour pressure = 72 + 16

= 88 kPa 72 (ii) Mole fraction of P in the vapour = —— 88

= 0.82 16

Mole fraction of Q in the vapour = —— 88

= 0.18



Objective Questions

1. The standard electrode potential of some half-cells are given as follows:

13 Electrochemistry

Half-reaction

½ Cl2 + e– → Cl–

½ Br2 + e– → Br–

½ I2 + e– → I–

½ H2O

2 + H+ + e– → H

2O

O2 + 2H+ + 2e– → H

2O

2

E θ (V)

+1.36

+1.07

+ 0.54

+1.77

+ 0.68

Which of the following pair of reagents does not result in a chemical reaction under standard conditions?A H

2O

2 with Cl

2

B H2O

2 with Br

2

C H2O

2 with I

2

D H2O

2 with KI

2. The cell diagram of an electrochemical cell is given as

Pt(s) ⏐ Fe2+(aq), Fe3+(aq) � Cr2O

72–(aq), Cr3+(aq) ⏐ Pt(s)

Which of the following will increase the e.m.f. of the electrochemical cell?A Increase the concentration of Fe3+ ionsB Increase the concentration of Cr3+ ionsC Increase the concentration of Fe2+ ionsD Decrease the concentration of Cr

2O

72– ions

3. The electrode potential of two half-cells are given below:

Fe3+(aq) + e– → Fe2+(aq) ; E θ = + 0.76 V

Fe(CN)63– + e– → Fe(CN)

64– ; E θ = + 0.36 V

Which of the following statements is correct with regards to the cells above?A Fe(CN)

64– is more stable than Fe2+(aq).

B Fe(CN)63– is more stable than Fe3+(aq).

C Fe(CN)63– is a stronger oxidising agent than Fe3+(aq).

D Fe(CN)64– is a stronger reducing agent than Fe2+(aq).

4. What is the total volume of gas produced at 25.0 °C when 0.20 A of current is passed through

dilute sulphuric acid for 4825 seconds? (Faraday constant is 9.65 � 104 C mol–1, molar volume of gas at 25.0 °C is 24.4 dm3 mol–1) A 0.061 dm3

B 0.122 dm3

C 0.183 dm3

D 0.244 dm3

5. Electrolysis is used to remove ions in industrial waste. Which of the following ions is removed

in the treatment of effluent using electrolysis?A Al3+ ionsB Cr3+ ionsC PO

43– ions

D Cr2O

72– ions



A B C D




6. Chlorine gas is produced by electrolysis of brine in a diaphragm cell. Which of the following is also produced besides chlorine in the electrolysis process?1 Hydrogen gas2 Sodium hydroxide3 Sodium chlorate(I)

7. The electrode potential of two half-cells are given below:

Sn4+(aq) + 2e– → Sn2+(aq) ; E θ = + 0.15 V

Fe3+(aq) + e– → Fe2+(aq) ; E θ = + 0.77 V

Which of the following statement(s) is true of the electrochemical cell formed from these two half-cells?1 The cell diagram is represented by

Pt(s) ⏐ Fe2+(aq), Fe3+(aq) � Sn4+(aq), Sn2+(aq) ⏐ Pt(s)

2 The e.m.f. of the cell produced is 0.62 V.3 The cell e.m.f. will increase if the concentration of Sn4+ is decreased.


1. The half-equations and standard electrode potential at 25 °C for two half-cells are

Hg2Cl

2(l) + 2e– → 2Hg(l) + 2Cl–(aq) ; E θ = + 0.24 V

Fe3+(aq) + e– → Fe2+(aq) ; E θ = + 0.77 V

(a) Write a cell diagram for the electrochemical cell constructed using the two half-cells above.(b) Calculate the e.m.f. of the electrochemical cell at standard conditions.(c) Name a substance that can be used as salt bridge for the cell above and explain how it

works.(d) An oxidising substance Y is added to the Fe3+/Fe2+ half-cell and the e.m.f. of the cell is

observed. The e.m.f. of the cell can be calculated using the Nernst equation. (i) Explain how the addition of the oxidising agent Y causes the change of the e.m.f. of the

cell. (ii) Sketch a graph showing the change of e.m.f. versus the volume of oxidising agent Y

added into the Fe3+/Fe2+ half-cell. [Fe2+]

(iii) Calculate the cell e.m.f. if the ratio of ———— is 1.00 � 10–3 while the concentration of [Fe3+]

Cl– remains unchanged.

2. In an electrolysis experiment, two electrolytic cells cell A and cell B are connected in series as show below. Cell A consists of two copper electrodes immersed in aqueous copper(II) sulphate solution. Cell B consists of silver electrodes immersed in silver nitrate solution.

cell A cell B

copperelectrodes

silverelectrodes

silver nitratesolution

copper(II)sulphatesolution



When a current of 0.50 A is allowed to pass through the two cells for 20.0 min, the mass of copper and silver deposited in cell A and cell B are 0.197 g and 0.672 g respectively. [Relative atomic mass: Cu, 63.5, Ag, 108](a) Calculate the number of coulomb that passed through the two cells.(b) Calculate the number of coulomb that is required to deposit 1.0 mol of

(i) copper atoms, (ii) silver atoms.

(c) Comment on the results obtained in (b) in relation to the charge of the ions.(d) If the charge of an electron is 1.60 � 10–19 C, calculate Avogadro’s constant using the result

obtained in (b)(i).(e) Predict the number of coulomb that is required to deposit 1.0 mol of gold when AuCl

3 is

electrolysed with gold electrodes using the result obtained in (b)(i).

Essay Question

1. (a) The overall chemical change taking place during the industrial electrolytic manufacture of aluminium metal is represented by

2Al2O

3 + 3C → 4Al + 3CO

2

(i) Explain all the chemical changes that take place in the process that result in the overall equation in the electrolyic process.

(ii) Using the electrode potential values, explain why aluminium metal cannot be produced from the electrolysis of aluminium salt solutions using graphite electrodes.

(b) Explain what is anodisation. An aluminium object with a surface area of 100 cm3 is to be anodised. Calculate the quantity of electricity (in coulomb) required to increase its oxide layer by 1.00 � 10–4 cm in thickness. [Density of Al

2O

3 is 4.00 g cm–3]



Objective Questions

1. C Cl2 + H

2O

2 → 2Cl– + O

2 + 2H+;

∆Eθ = (+1.36 – 0.68) V= + 0.68 V

Br2 + H

2O

2 → 2Br– + O

2 + 2H+ ;

∆Eθ = (+1.07 – 0.68) V= + 0.39 V

The difference of the standard redox potential, ∆Eθ of the overall reactions are positive. Hence H

2O

2 will react with Cl

2 and Br

2.

I2 + H

2O

2 → 2I– + O

2 + 2H+;∆Eθ = (+ 0.54 – 0.68) V

= – 0.14 V∆Eθ of overall reaction is negative. Hence H

2O

2 does not react with I

2.

2I– + H2O

2 + 2H+ → I

2 +

2H

2O;

∆Eθ = (– 0.54 + 1.77) V= + 1.23 V

∆Eθ of overall reaction is positive. Hence H2O

2

will react with KI. 2. C The reaction that takes place in the

electrochemical cell is6Fe2+(aq) + Cr

2O

72–(aq) + 14H+(aq) →

6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)

According to the Nernst equation: 0.059 [Fe2+]6 [Cr

2O

72–]

E = Eθ + ———— log ——————— —————— 6 [Fe3+]6[Cr3+]2

0.059 [reactants]x

Or: E = Eθ + ———— log ——————— —— 6 [products]y

Hence the increase in the concentration of the reactants or the decrease in the concentration of the products will increase the E value.

3. B A more positive E value indicates that Fe3+(aq) has a higher tendency to be reduced to Fe2+ (aq) than Fe(CN)

63– to Fe(CN)

64– (C and D

incorrect). Hence Fe(CN)63– is more stable

than Fe3+ (aq) and Fe2+ (aq) is more stable than Fe(CN)

64– (A incorrect, B correct).

4. C Q = 0.20 A � 4825 s= 965 C= 0.01 FAt the cathode: 2H+ + 2e– → H

2

0.01 F electricity will produce½ � 0.01 � 24.4 = 0.122 dm3 of H

2 gas.

At the anode: 4OH– → O2+ 2H

2O + 4e–

0.01 F electricity will produce¼ � 0.01 � 24.4 = 0.061 dm3 of O

2 gas.

Total volume of gas produced = 0.122 + 0.061= 0.183 dm3

5. B Heavy metal ions are removed from the effl uent using electrolysis under certain conditions. Al3+ ions are not discharged because of its very

positive Eθ value. Anions are not removed in electrolysis.

6. B The ions present in brine solution are H+, Na+, OH– and Cl–.H+ ions are discharged at the cathode forming H

2 gas. (1 correct)

Cl– ions are discharged forming Cl2 gas, leaving

Na+ and OH– ions in the electrolyte forming NaOH. (2 correct)Cl

2 and NaOH are prevented from reacting by

the diaphragm, hence NaClO is not formed.(3 incorrect)

7. C Sn4+/ Sn2+ will be the negative terminal of the cell since it has a less positive Eθ value.The cell diagram is Pt(s) | Sn2+(aq), Sn4+(aq) || Fe3+(aq), Fe2+(aq) | Pt(s). (1 incorrect)E.m.f. of cell = (0.77 – 0.15) V

= 0.62 V (2 correct)According to the Nernst equation:

0.059 [Sn2+] [Fe3+ ]2

E = Eθ + ———— log —————————— 2 [Sn4+][Fe2+]2

The cell e.m.f. will increase when [Sn4+] is decreased. (3 correct)


1. (a) Pt(s) | Hg(l),Hg2Cl

2(l) || Fe3+(aq),Fe2+(aq) | Pt(s)

(b) Eθcell

= Eθ Fe3+ / Fe2+ – Eθ

Hg+ / Hg

= + 0.77 – 0.24

= + 0.53 V(c) Ammonium chloride / potassium chloride /

any suitable electrolyte.The salt allows the movement of ions so as to maintain electrical neutrality of the two half-cells and also prevent the electrolytes of the two half-cells from mixing.

(d) (i) The reaction of the cell is:2Hg(l) + 2Cl–(aq) + 2Fe3+(aq) → Hg

2Cl

2(l) + 2Fe2+(aq)

According to the Nernst equation: 0.059 [Cl–]2 [Fe3+]2

E = Eθ + ———— log ————————— 2 [Fe2+]2

The addition of the oxidising agent Y will increase the concentration of Fe3+ ions and hence the e.m.f. of the cell is increased.

(ii)

ANSWER

CHAPTER 13 Electrochemistry

0.53

e.m.f of cell (V)

volume of Y (cm3)

0.059 [Cl–]2 [Fe3+]2

(iii) Ecell

= Eθcell

+ ———— log ————————— 2 [ Fe2+ ]2



0.059 1= 0.53 + ———— log ———————————

2 (1.00 � 10–3)2

= + 0.707 V 2. (a) Number of coulomb = I � t

= 0.50 � 20.0 � 60= 600

(b) (i) 0.197 g of Cu is deposited by 600 C.l.0 mol of Cu is deposited by

63.5———— � 600 C = 193 000 C

0.197 (ii) 0.672 g of Ag is deposited by 600 C.

l.0 mol of Ag is deposited by 108.0

———— � 600 C = 96 400 C 0.672

(c) 193 000 C is approximately equals to 2 �96 400 C.The number of coulomb required to deposit 1.0 mol of Cu is twice that required to deposit 1.0 mol of Ag. This is because the charge of Cu2+ ions is twice of that of Ag+ ions.

193 000(d) —————— = NA � e–

296 500 = N

A � 1.60 � 10–19

NA = 6.03 � 1023 mol–1

(e) Since Au3+ has a charge of 3/2 times of Cu2+, the number of coulomb to deposit 1.0 mol of Au is 3/2 � 193 000 C = 289 500 C

Essay Question

1. (a) (i) Aluminium is manufactured by the electrolysis of molten aluminium oxide using graphite electrodes.Molten Al

2O

3 dissociates into Al3+ ions

and O2– ions.The reactions that take place at the electrodes are

anode: 2O2– → O2 + 4e– …… (1)

cathode: Al3+ + 3e– → Al …… (2)The O

2 gas evolved at the anode reacts

with the carbon electrode to produce carbon dioxide gas: C + O

2 → CO

2

…… (3)The overall equation = (1) � 3 + (2) �

4 + (3) � 34Al3+ + 6O2– + 3C → 4Al + 3CO

2

Or: 2Al2O

3 + 3C → 4Al + 3CO

2

(ii) Al3+ + 3e– → Al ; Eθ = –1.66 V2H+ + 2e– → Η

2 ; Eθ = 0 V

The electrode potential of Al3+/Al is much more negative than that of H+/H

2,

resulting in a much higher discharge potential for Al3+ ions to be discharged.Electrolysis of aluminium salt solutions using graphite electrodes will result in H+ ions being discharged to produce hydrogen gas instead of aluminium metal.

(b) Anodisation is a process in which the thickness of aluminium oxide of an aluminium object is thickened by electrolysis. In this process, the aluminium object is made the anode, using dilute sulphuric acid or chromic acid as the electrolyte.Volume of Al

2O

3 to be deposited on the

aluminium object = 100 cm2 � 1.0 �10–4 cm= 0.01 cm3

Mass of Al2O

3 to be deposited

= density � volume = 4.0 � 0.01 = 0.04 gDuring anodising at the anode:4OH– → O

2 + 2H

2O + 4e–

1 mol of O2 to be formed requires 4 Faradays.

O2 reacts with the aluminium anode object:

3O2 + 4Al → 2Al

2O

3

1 mol (102 g) of Al2O

3 requires 4 � 3/2

Faradays to be formed.0.04 g Al

2O

3 requires

4 � 3/2 � 96 500—————————————— � 0.04 = 227 C

102



Objective Questions

1. Which of the following chemical changes releases heat energy?A O

2(g) → 2O(g)

B O(g) + e– → O–(g)C O–(g) + e– → O2–(g)D Mg+(g) → Mg2+(g) + e–

2. The standard enthalpy change of neutralisation for dilute sulphuric acid and sodium hydroxide is the heat liberated whenA 1.0 g of aqueous H

2SO

4 reacts with 1.0 g of aqueous NaOH.

B ½ mol of aqueous H2SO

4 reacts with 1.0 mol of aqueous NaOH.

C 1.0 mol of aqueous H2SO

4 reacts with ½ mol of aqueous NaOH.

D 1.0 mol of aqueous H2SO

4 reacts with 2.0 g of aqueous NaOH.

3. The standard enthalpy change of formation of carbon monoxide is the standard enthalpy change for the reactionA C(s) + O(g) → CO(g)B C(s) + ½O

2(g) → CO(g)

C C(g) + O(g) → CO(g)D C(g) + ½ O

2(g) → CO(g)

4. Which of the following represents the correct ascending order of the lattice energies of ionic compounds? A NaBr < NaCl < NaIB KI < LiF < NaClC KI < NaBr < LiClD KCl < KBr < KI

5. The second ionisation energy of magnesium is +1450 kJ mol–1.Which of the following thermochemical equations represents the second ionisation energy of magnesium?A Mg+(g) → Mg2+(g) + e– ; ∆H = +1450 kJ mol–1 B Mg+(s) → Mg2+(g) + e– ; ∆H = +1450 kJ mol–1 C Mg(s) → Mg2+(s) + 2e– ; ∆H = +1450 kJ mol–1 D Mg(g) → Mg2+(s) + 2e– ; ∆H = +1450 kJ mol–1

6 Which of the following ions has the highest (most exothermic) enthalpy of hydration?

Ion A B C DCharge +1 –2 +2 +2Ionic radius (nm) 0.095 0.14 0.065 0.11

7. By using the following information, calculate the standard enthalpy of formation of ethanol.

Standard enthalpy of combustion of graphite = –393 kJ mol–1

Standard enthalpy of combustion of hydrogen = –286 kJ mol–1

Standard enthalpy of combustion of ethanol = –1367 kJ mol–1

A –191 kJ mol–1

B +191 kJ mol–1

C –277 kJ mol–1

D +277 kJ mol–1

14 Thermochemistry



A B C D




8. Which of the following processes is/are always exothermic?1 Na+(g) + water → Na+(aq)2 Cl(g) + e– → Cl–(g)3 Na+(g) + Cl–(g) → NaCl(s)

9. Na(s) + ½Cl2(g) → Na(g) + ½Cl

2(g) ; ∆H

1 = +108 kJ mol–1

Na(g) + ½Cl2(g) → Na+(g) + Cl–(g) ; ∆H

2 = +257 kJ mol–1

Na+(g) + Cl–(g) → NaCl(s) ; ∆H3 = -766 kJ mol–1

Based on the information given above, which of the following statements is/are true?1 ∆H

1 is the enthalpy of atomisation of sodium.

2 The enthalpy of formation of sodium chloride is -401 kJ mol–1.3 ∆H

2 is the sum of electron affinity for chlorine and the first ionisation energy for sodium.


1. (a) Each cyclohexene molecule, C6H

10, has one C = C bond. 1 mol of cyclohexene reacts with 1

mol of hydrogen to form cyclohexane, C6H

12.

C6H

10 + H

2 → C

6H

12 ; ∆H

1

Calculate the value of ∆H1 from the following data:

Enthalpy of formation of cyclohexene = –36 kJ mol–1

Enthalpy of formation of cyclohexane = –156 kJ mol–1

(b) Benzene, C6H

6 undergoes similar reaction with hydrogen to form cyclohexane.

C6H

6 + 3H

2 → C

6H

12 ; ∆H

2

Predict the value of ∆H2 assuming that each benzene molecule contains three C = C bonds.

(c) The value of ∆H2 as determined by experiment is –207 kJ mol–1. What inference can you

make with regard to the stability of the benzene ring.

2. The enthalpies of hydration of some ions are shown below.

Ion

∆H (kJ mol–1)

Na+

–390

Mg2+

–1891

Al3+

– 4613

Cl–

–381

Br–

–350

I–

–307

(a) Why are the hydration energies of both anions and cations negative?(b) What is the hydration energy of MgI

2?

(c) Why does the hydration energy get progressively less exothermic for the series (i) Cl–, Br –, I–

(ii) Al3+, Mg2+, Na+

(d) Given that,

NaCl(s) → Na+(g) + Cl–(g) ; ∆H = +776 kJ mol–1

Calculate the heat of solution of sodium chloride.



Essay Question

1. (a) When 0.20 g of magnesium powder is dissolved in dilute hydrochloric acid in a polystyrene cup, the temperature of the solution increases by 8.6 °C. In another experiment, the plastic cup and its contents are found to require 500 J of heat to raise the temperature by 1 °C.

(i) Write a chemical equation for the reaction between magnesium and hydrochloric acid. (ii) Calculate the heat liberated in the experiment. (iii) Hence, calculate the enthalpy change of reaction for 1 mol of magnesium.(b) In a similar experiment, it was found that when magnesium carbonate is dissolved in excess

hydrochloric acid, the enthalpy change is –90 kJ per mole of magnesium carbonate.Use your results in (a)(iii) to calculate the enthalpy change of formation of magnesium carbonate.(Standard enthalpy changes of formation of water and carbon dioxide are –285 kJ mol–1 and –393 kJ mol–1 respectively.)

(c) Calcium carbonate and magnesium carbonate decompose on heating. Which would decompose at a higher temperature, calcium carbonate or magnesium carbonate? Explain your answer.



Objective Questions

1. B 2. B ½ mol of H

2SO

4 contains 1 mol of H+ ions.

3. B 4. C The ions in all these ionic compounds carry

the same charge.∴ Lattice energy is inversely proportional to

interionic distance.Interionic distance decreases in the order:

KI > NaBr > LiCl∴ Lattice energy increases in the order:

KI < NaBr < LiCl 5. A 6. C The higher the charge, the higher the enthalpy

of hydration.The smaller the ionic radius, the higher the enthalpy of hydration.

7. C C2H

5OH + 3O

2 → 2CO

2 + 3H

2O;

x 0 2 � (–393) 3 � (–286)∆H = –1367 kJ mol–1

–1367 = [2 � (–393) + 3 � (–286)] – xx = –277

8. D Na+(g) + water → Na+(aq) ;Hydration energy

Cl(g) + e– → Cl–(g) ;First electron affi nity

Na+(g) + Cl–(g) → NaCl(s) ;Lattice energy

9. B In equation (1), Na(s) → Na(g). Hence, ∆H1 is

enthalpy of atomisation for sodium. (1 correct)On adding equations (1), (2) and (3), we haveNa(s) + ½Cl

2(g) → NaCl(s) ; ∆Hθ

f

Thus, ∆Hθf

= (+108) + (+257) + (–766)= – 401 kJ mol–1 (2 correct)

½Cl2(g) → Cl–(g) is not electron affi nity for

chlorine. (3 incorrect)


1. (a) C6H

10 + H

2 → C

6H

12 ; ∆H

1

–36 0 –156∆H

1 = (–156) – [(–36) + 0]= –120 kJ mol–1

(b) ∆H2

= 3 � (–120)= –360 kJ mol–1

(c) The actual heat energy released is less than the theoretical value. This implies that benzene has a more stable structure than the predicted structure with three C = C bonds.The structure of a benzene molecule is a resonance hybrid of two Kekule structures. The delocalised π electrons confer stability on the benzene ring.

2. (a) In the hydration process, the H atoms (positive polarity) of H

2O are attracted to the negative

ions while the O atoms (negative polarity) of H

2O are attracted to the positive ions.

The formation of ion-solvent bonds results in a release of energy, known as hydration energy.

(b) –1891 + 2 � (–307) = –2505 kJ(c) The smaller the ionic size, the stronger the

ion-solvent bond and the greater the energy released. (i) The ionic size decreases from I– to CI–.

Hence, hydration energy decreases from Cl– to I–.

(ii) The ionic size decreases from Na+ to Al3+.Hence hydration energy decreases from Al3+ to Na+.

(d) NaCl(s) → Na+(g) + Cl–(g) ;∆H = +776 kJ mol–1

Na+(g) + water → Na+(aq) ;∆H = –390 kJ mol–1

Cl–(g) + water → Cl–(aq) ;∆H = –381 kJ mol–1

ANSWER

CHAPTER 14 Thermochemistry

∆H = (+776) + (–390) + (–381)= +5 kJ mol–1

Essay Question

1. (a) (i) Mg + 2HCl → MgCl2 + H

2

(ii) Heat liberated= 500 � 8.6= 4300 J = 4.3 kJ

(iii) Number of moles of Mg 0.2

= ——— 24.3

= 8.23 � 10–3

∴ Enthalpy change of reaction per mole of Mg

4.3= —————————

8.23 � 10–3

= 522.5 kJ mol–1

That is,

Mg(s) + 2HCl(aq) → MgCl2(aq) +

H2(g)

∆H1θ = –522.5 kJ mol–1 …… (1)

(b) Let the enthalpy change of formation of MgCl

2, HCl and MgCO

3 be x, y and z kJ

mol–1 respectively.

MgCO3 + 2HCl → MgCl

2 + H

2O + CO

2 … (2)

x 2y z –285 –393

Na+(g) + CI–(g)

–381 kJ–390 kJ+776 kJ

NaCl(s) + water → Na+(aq) + Cl–(aq) ; ∆H



∆H2θ = –90 kJ mol–1

∆Hθ of reaction = Σ∆H

fθ (products) – Σ∆H

fθ (reactants)

∴ –90 = [z + (–285) + (–393)] – [x + 2y]That is x + 2y – z = –588 … (3)From equation (1), we have

Mg(s) + 2HCl(aq) → MgCl2(aq) + H

2(g) … (1)

0 2y z 0∆Hθ of reaction = Σ∆H

fθ (products) – Σ∆H

fθ (reactants)

–522.5 = z – 2y … (4)Substituting equation (4) into equation (3),x + 522.5 = –588

x = –1110.5 kJ mol–1

That is, the enthalpy change of formation of magnesium carbonate is –1110.5 kJ mol–1

(c) MgCO3 → MgO + CO

2

CaCO3 → CaO + CO

2

Ca2+ ion has a bigger ionic radius than Mg2+ ion. Therefore Ca2+ ion has a weaker polarising power. As a result, the polarisation of CO

32– ion by Ca2+ ion is weaker. Hence, it

is diffi cult to decompose the CO32– ion to give

CO2 and O2– ions. This means that CaCO

3 is

stable towards thermal decomposition and decomposes at a higher temperature.


Question Bank (Pg 1-58)

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