SK015

CHEMISTRY
TUTORIAL TOPIC 4: CHEMICAL BONDING

6. Propose all possible structures for compounds below. Determine the formal charge for
each atom in the compounds and choose the most plausible structure.
a) CS2
b) NCO−

7. a) What is meant by resonance structures?
b) Draw all the resonance structures of the O3 and NO3−.

4.2: Molecular Shape and Polarity

1. a) State Valence Shell Electron Pair Repulsion (VSEPR) theory.
b) Explain the following statements:
i) The bond angle in the CH4, NH3 and H2O molecules are 109.5, 107.3
and 104.5 respectively.
ii) The shape of SiF4 is tetrahedral while that of SF4 is see-saw.

2. a) Define dipole moment.
b) State factors that affect the polarity of a molecule.
c) Can a molecule be non-polar although it has polar bonds? Explain using
tetrachloromethane, CCl4 as an example.

3. Based on section 4.1, question 4(a)-4(j), state the molecular shape and determine the
polarity of each molecule.

4.3: Orbital Overlap and Hybridisation

1. a) State the Valence Bond theory.
b) Show the formation of sigma and pi bonds.
c) Use the valence bond theory to explain the bonding in O2 and show how the
bond is formed by overlapping of atomic orbitals.

2. a) What is meant by hybridisation of atomic orbitals?
b) State the type of hybridisation on each of the carbon atoms in the molecule
below.

HH
CC

H C CN
HH

c) State the type of hybridisation of the underlined atoms in the following

compounds.

i. BF3 ii. CCl4

148

CHEMISTRY
TUTORIAL TOPIC 4: CHEMICAL BONDING

3. Describe the type of hybridisation of the central atom in the following species. Show

the orbital overlapping diagram.

a) BeH2 b) BH3 c) PCl5
e) SCl6 f) ICl2−
d) SeF4
g) ICl4+

4.4: Intermolecular Forces & 4.5: Metallic Bonding

1. a) How do polar molecules attract each other?

b) Explain how London dispersion forces are formed between non-polar

molecules.

c) Name the type of van der Waals forces that exist between molecules for each

of the following species:

i. Cl2 ii. PCl3 iii. H2S

iv. CCl4

2. a) Give the factors that influence the strength of van der Waals forces.
b) Which molecule has a higher boiling point? Explain.
i. Br2 or ICl
ii. I2 or Br2
iii. HBr or CF4

3. a) Describe and state the factors that influence the hydrogen bond.
b) Compare the boiling point between:
i. H2O and CH4
ii. H2O and NH3

4. Ethanol, C2H5OH, and dimethyl ether, CH3OCH3, have same molar mass. Which of
the compounds has a higher boiling point? Explain.

5. a) What is meant by energy band gap ?
b) How to differentiate between a conductor, an insulator and a semiconductor in
relation to energy gap ?

6. Aluminium atoms in aluminium metal are hold together by metallic bond.
a) Describe how metallic bond is formed in aluminium metal using electron-sea
model.
b) Explain the following statement:
i. Boiling point of Be and Mg are 2970oC and 1091oC.
ii. Boiling point of Mg and Al are 1091oC and 2470oC.

149

CHEMISTRY
MEKA TOPIC 4: CHEMICAL BONDING

1. a) Write the Lewis dot symbol for each of the following atoms;
i. H ii. O iii. Al

b) Using Lewis dot symbol, show the formation of;
i. ionic bond in Al2O3,
ii. covalent bond in H2O, and
iii. dative bond in H3O+

2. Draw the Lewis structure for each of the following species. Identify those that do not

obey the octet rule. State the type of exception for each case.

a) SeF4 b) AlBr3 c) NO3−
d) CCl2F2 e) NO2+ f) NO

3. For each of the following molecules, N3−
CO2 OCS

a) Draw all possible Lewis structure.
b) Calculate formal charge of each atoms in each Lewis structures.
c) Choose the most plausible structure and explain.

4. Draw resonance structure for O3 and NO2+.

5. a) Complete the sentence below,
i. Valence Shell Electron-pair Repulsion (VSEPR) theory state that electron
pairs around the ______________________ is located as ____________
as possible from the others in order to ____________________
repulsions.
ii. The strength of repulsion between _________________________>
____________________________>_____________________________.

150

CHEMISTRY
MEKA TOPIC 4: CHEMICAL BONDING

b) For each of the following molecules;

CO2 HCl BF3 SnCl2 SO2
H2O PCl5
CCl4 CH2Cl2 NF3 SF6 BrF5
SF4 ICl3 XeF2

XeF4

i. Draw the Lewis structure.
ii. State the electron pair arrangement.
iii. State the number of lone pair and bonding pair electrons.
iv. Compare type of bond repulsion exist.
v. State the molecular geometry and draw the shape of molecule.
vi. State the bond angle.
vii. Determine which atoms is more electronegative and show the direction of

the dipole moment on the molecular shape.
viii. State the polarity of the molecule and explain your answer.

6. a) Complete the sentence below,
i. Valence bond theory state that a _______________ bond is formed when
neighboring atomic orbitals ______________.
ii. ___________ bond is formed when atomic orbitals is overlap end-to-end
while pi bond is formed when atomic orbitals is overlap ______________.

b) Draw the shape of orbital for each of the following hybrid;

i. sp ii. sp2 iii. sp3 iv. sp3d v. sp3d2

c) For each of the following molecules;

BeCl2 C2H2 AlCl3 CH2O CCl4
SF4 ICl3
NF3 H2O PCl5 XeF4

XeF2 SF6 BrF5

i. Draw the Lewis structure.
ii. Calculate formal charge of each atoms in the Lewis structure.
iii. State the number of lone pair and bonding pair electrons.
iv. State the molecular geometry and draw the shape of molecule.
v. Write the valence orbital diagram of each atoms in each molecules.
vi. Draw orbital overlapping diagram to show the formation of all covalent

bonds in the molecules.

151

CHEMISTRY
MEKA TOPIC 4: CHEMICAL BONDING

7. a) State the type of intermolecular forces for each of the following statements.

Statements Intermolecular Forces

Electrostatic forces form between H that bonded to

highly electronegative atom (N, O, or F) in one

molecule and highly electronegative atom (N, O, or

F) in another molecule.

Electrostatic forces form between the permanent δ+

pole of a molecule and permanent δ- pole of

adjacent molecules.

Electrostatic forces from between positive charge

metallic ions and the sea of delocalized valence

electrons.

Electrostatic forces form between the instantaneous

δ+ pole of a molecules with instantaneous δ- pole of

adjacent molecules

b) In each pair, state the type of intermolecular forces present forces present for each
substance, and predict which substance has higher boiling point:
i. CH3Br or CH3F
ii. NH3 or CH4
iii. CCl4 or CH3Cl
iv. H2O or NH3
v. HF or H2O

c) For each of the following metals; Be
Mg

i. Using electron-sea model, show the formation of metallic bond in Mg and
Be.

ii. Compare the size of ion in Mg and Be.
iii. Determine which metals has higher melting bond.

152

CHEMISTRY
KUMBE TOPIC 4: CHEMICAL BONDING

1. a) Write two possible Lewis structures of HCOOH. Calculate the formal charge

on each atom. Which structure is more stable? Why?

b) By using Lewis structure, determine whether the following molecules stable or

unstable.

i. SF2 ii. SF3+ iii. SF4
iv. SF5+ v. SF6

2. a) The molecules of SiF4, SF4, and XeF4 have molecular formulas of the type
AB4, but the molecules have different molecular geometries. Predict the shape
of each molecule, and explain why the shapes differ.

b) The 3 species NH2−, NH3 and NH4+ have H-N-H bond angles of 105o, 107o
and 109o respectively. Explain this variation in bond angles.

3. Explain why PF3 molecule is polar, while BF3 is non-polar.

4. Consider the Lewis structure for glycine, the simplest amino acid:

HO
HNCCOH

HH

a) State the type of hybridization of C, O, and N atoms.
b) Draw and label the overlapping of orbitals to show all the σ and π bonds

formed.

5. Explain the following statements.
a) Boiling point of CH3CH2CH2OH is higher than CH3CH2OCH3.
b) Boiling point of calcium is higher than that of potassium.
c) Iceberg is floating in sea water.
d) Copper can easily be shaped into pipes and drawn into wires.

153

CHAPTER 5: STATES OF
MATTER

LECTURE NOTE

lecture notes & questions

154

CHAPTER 5 2

STATES OF MATTER

5.1 Gas
5.2 Liquid
5.3 Solid
5.4 Phase Diagram

1

5.1 GAS LEARNING OUTCOMES

At the end of the lesson, student should be able to :
a) Explain qualitatively the basic assumptions of the

kinetic molecular theory of gases for an ideal gas.
b) Define gas laws

(i) Boyle’s law
(ii) Charles’ law

(iii) Avogadro’s law

c) Interpret the graphs of Boyle’s law and Charles’
law(*include sketch of graph).

d) Perform calculations involving gas laws and ideal

3 gas equation. 4

e) Determine molar mass of a volatile liquid using General properties of gases:
ideal gas equation. (Experiment 4 : Charles’ law
and ideal gas equation) 1. Occupy more space than their solids or liquids.
2. Have much lower densities then solids or liquids.
f) Define Dalton’s law. 3. Expand to fill their containers –assume the

g) Perform calculations using Dalton’s law. volume and shape of their containers.
4. Easily compressible compared to solids or
h) Explain qualitatively the ideal and non-ideal
behaviours of gases in terms of intermolecular liquids.
forces and molecular volume. (*use van der
Waals equation to explain the non-ideal 6
behaviour)

i) Explain the conditions at which real gases
approach the ideal behaviour.

5

155

General properties of gases: Kinetic Molecular Theory of Gases

5. Flows readily • Describes the behavior of an ideal gas.
6. Mix completely and evenly when put together in
• Ideal gas : gases which obey the ideal gas
the same container equation
7. 1 mole of gas occupies a volume of 24 L at room (PV = nRT)

conditions and 22.4 L at STP

78

Kinetic Molecular Theory of Gases 3. The volume of gas molecules are negligible
compared to the volume of its container.
Theory that explains the behavior of gases.
The theory is based on the following assumptions: 4. Attractive and repulsive forces between
molecules are negligible.
1. A gas consists of a very large number of
extremely small particles(molecules or atoms) in 5. The average kinetic energy of molecules is
constant, random, straight-line motion. proportional to the absolute temperature.

2. Collisions among molecules are perfectly *A gas corresponding to these assumptions is called
elastic. (energy can be transferred from one an ideal gas.
molecule to another during collision but the total
energy of all molecules in the system remain the 10
same)

9

The Gas Laws A molecular description of Boyle’s Law

a) Boyle’s Law : 12

Boyle’s law states that at constant temperature,
the volume of a fixed amount of gas is inversely
proportional to the gas pressure.

V 1 (at constant T & n)
P

where V = volume
P = pressure
T = temperature 11
n = number of moles

156

V = k1

Graphs based on Boyle’s Law

Graph of P versus V Graph of P versus 1
V
P
V = k1 (1/P) or P = k1 (1/V)
P

Therefore, PV = k1

P1V1 =P2V2

V 1
V
pressure is inversely
proportional to volume pprreospsourtrieonisadl tiorectvloyl1ume
14
13

Graph of PV versus P Example 1

PV A sample of chlorine gas occupies a volume of 2 L
PV = constant at a pressure of 1 atm. Calculate the pressure of the
gas if the volume is increased to 5 L at constant
P temperature.
15
(ans : 0.4 atm)

16

ANSWER Example 2

P1V1  P2V2 The pressure of a sample of hydrogen gas in a
50.0 mL container is 765 mmHg. The sample is
1atm2L  P25L then transferred into another container and the
measured pressure is 825 mmHg. What is the
P2  1atm 2L  volume of the second container?

5L (Ans :46.36 mL)

 0.4 atm 17 18

157

ANSWER b) Charles’ Law

P1V1  P2V2 Charles observed that at constant pressure, the
volume of a gas expands when heated and
V2  765 mmHg 50 mL contracts when cooled.

825 mmHg 19 20

 46.36 mL

Plot of Volume versus Temperature (C)

V

Plot of Volume versus Temperature (K)

21 22

Charles’ law states that at constant pressure, V  kT
the volume of a fixed amount of gas is directly 2
proportional to the absolute temperature of the
gas.

V  T (at constant P & n) Therefore, V  k
T 2

where V = volume V1  V2
T = absolute temperature (K) T1 T2
P = pressure
n = number of moles

23 24

158

Example 1 ANSWER

A sample of carbon monoxide gas occupies 3.2 L V1  V2
at 125 °C. The sample is then cooled at constant T1 T2
pressure until it contracts to 1.54 L. Calculate
the final temperature in degree celsius. 3.2 L K  1.54 L
(Ans : -81.54 °C) T2
125 273.15
25
T2 191.61K

  81.54 C

26

Example 2 ANSWER

A sample of gas trapped in a capillary tube by a V1  V2
plug of mercury at 22 oC has a volume of 4.5 mL. T1 T2
Calculate the volume of the gas when the capillary
tube is heated to 60 oC. 4.5 mL K  V2 K

(Ans : 5.08 mL) 22  273.15 60  273.15

V2  5.08 mL

27 28

The Combination of Boyle’s and Charles’ law Example 1

Boyle’s law: V  1 A sample of methane gas occupies 25.5 L at
Charles’ law: P 298.15 K and 153.3 kPa. Find its volume at STP.

V T (Ans : 35.35 L)

V  T 30
P

PV  k3
T

P1V1 = P2V2
T1 T2

29

159

ANSWER Example 2

P1V1  P2V2 A gas–filled weather balloon with a volume of 55.0
T1 T2 L is released at sea-level conditions of 759 mmHg
and 27 oC. Calculate the volume of the balloon
1.513 atm25.5 L  1x V2 when it rises to an altitude at which the
 298.15 K 273.15 temperature is -5 oC and the pressure is 0.066
 K atm.

V2 35.35 mL (Ans : 743.52 L)

31 32

ANSWER c) Avogadro’s Law

P1V1  P2V2 Avogadro’s law states that at constant
T1 T2 temperature and pressure, the volume of a
gas is directly proportional to the number of
0.9987 atm 55.0 L 0.066 atm x V2 moles of the gas.
 27  273.15 K
   5  273.15 K V  n (at constant T & P)

V  kn
4

V2 743.52 L V
n
 k V1 = V2
4 n1 n 2

33 34

A molecular description of Avogadro’s law Example 1

35 2 moles of chlorine gas kept in a cylinder with
piston occupies a volume of 49 L. When another
3 moles of chlorine gas is pumped into the
cylinder at constant temperature and pressure
the piston moves upwards to accommodate the
gas. Calculate the final volume of the gas.

(Ans :122.5L)

36

160

ANSWER (d) The Ideal Gas Equation

Combination of Boyle's law, Charles' law and

V1  V2 Avogadro's law : V  1
n1 n2 Boyle's Law: P

49 L  V2 Charles' Law: VT
2 5
Avogadro's Law: Vn nT
P
V2 122.5 L V

V=R nT

P

PV = nRT where :

37 Ideal gas equation R is the ideal gas constant 38

The value and unit of R depend on the unit of Further Application of The Ideal Gas Law
pressure and volume used in the equation.
Density and molar mass of a gas can be
unit of unit of value of R unit of R calculated by rearranging the
pressure volume 0.08206 ideal gas equation:
L atmmol1 K1
atm L PV  nRT
N m mol1 K1
N m2 or Pa m3 mol1 K1
or
Pa dm3 J mol1 K1
mmHg L mol1 K1
mmHg m3 8.314 d = PMr Mr = molar mass of a gas
or torr L mol1 K1 RT n = mol of gas
torr L m = mass of the gas
or 62.36 39
dm3 40

PV  nRT Example 1

PV=  m  RT A steel gas tank has a volume of 275 L and is
 Mr  filled with 0.485 kg of O2. Calculate the pressure
  of O2 if the temperature is 29 oC.

P = mRT (d = density of a gas) (Ans : 1.37 atm)
VMr
41 42
P = dRT
Mr

d = PMr
RT

161

ANSWER Example 2

n CO 2  0.485 x1000g A sample of chlorine gas is kept in a 5.0 L container
32 gmol1 at 228 torr and 27oC. How many moles of gas are
present in the sample?
15.156 mol
(Ans : 0.06 mol)
PV  nRT
44
 PO2
15.156 mol 0.08206 Latmmol1K1 (309.15K )
2.88 atm

1.37 atm

43

ANSWER Example 3

T 27  273.15 A chemist has synthesized a greenish-yellow
compound of chlorine and oxygen and finds that
300.15 K its density is 7.71 g L-1 at 36°C and 2.88 atm.
Calculate the molar mass of the compound.
PV  nRT
(Ans : 67.9 gmol-1)
n  PV
RT 45 46

0.3 atm x 5 L
0.08206 Latmmol1K1
 nCl (300.15K )

0.061 mol

ANSWER Example 4

T  36  273 .15 Calculate the density (in gL-1) of CO gas at:
 309 .15 K

PV  nRT a) room conditions (Ans : 1.14 gL-1)

PM b) STP (Ans : 1.25 gL-1)
RT
  r

M r  RT
P

 M r
 7.71gL1 x 0.08206 Latmmol 1K 1 309 .15 K
2.88 atm

M r  67 .9 gmol 1

47 48

162

ANSWER Example 5

a)   PM r Calculate the mass of KClO3 is required to
RT produce 2.40 L oxygen gas that measured at a
pressure of 1 atm and a temperature of 26oC.
1atm x 28 gmol 1
0.08206 Latmmol 1K 1 298 .15 K 2KClO3(s) → 2KCl(s) + 3O2(g)
  CO  (Ans : 7.99 g)

1.14 gL1 50

b)   PM r
RT

1atm x 28 gmol 1
0.08206 Latmmol 1K 1 273 .15 K
  CO 

1.25 gL1 49

ANSWER e) Dalton’s Law of Partial Pressure

PV  nRT Partial pressure is the pressure of an
individual gas component in a mixture.
n  PV
RT Dalton’s law of partial pressure states that the
1atm x 2.04 L total pressure of a mixture of non reacting
0.08206 Latmmol1K1 (299.15K ) gases is equal to the sum of the pressures
 nO2  that each gas would exert if it were present
alone.
0.0978 mol
52
3 mol O2  2 mol KClO3
0.0978 x 2
0.0978 mol O2  3

 0.0625 mol

mass KClO3  0.0625 mol x122.5 gmol1 51
 7.987 g

For a mixture of 3 gases, A,B and C :

PT = PA + PB + PC

53 54

163

According to ideal gas equation: Mole fraction and pressures
In the mixture of gas A and gas B:
Dalton’s theory allows us to form a relationship
between mole fraction, partial pressure and a total
pressure.

55 XA = mol fraction of gas A 56

Example 1 ANSWER

A gaseous mixture of 7.00 g N2 and 3.21 g CH4 is a) n N2  7g 1  0.25 mol n CH 4  3 .21 g 1  0.200 mol
placed in a 12.0 L cylinder at 25oC. 28 gmol 16 gmol
a) What is the partial pressure of each gas?
b) What is the total pressure in the cylinder? PN 2  n N 2 RT  n RT 

Ans: a) 0.51 atm , 0.409 atm V PCH 4  VCH 4
b) 0.92atm
 0.25 x 0.08206 x 298 .15  0.2006 x 0.08206 x 298 .15
57 12 .0 L 12 .0 L

 0.51atm  0.409 atm

b) PT  PN 2  PCH 4
 0.51  0.409
 0.92 atm

58

Example 2 ANSWER

A mixture of gases contains 4.53 moles of neon, n T  n Ne  n Ar  n Xe
0.82 moles of argon and 2.25 moles of xenon.  4.53  0.82  2.25
Calculate the partial pressure of the gases if the  7.6 mol
total pressure is 2.15 atm at a certain temperature.
PNe  n Ne PT  PXe  n Xe PT 
(Ans : P Ne = 1.28 atm,, P Ar = 0.23 atm, P Xe = 0.64 atm) nT nT
2.25
59  4.53 2 .15   7.6 2.15 
7.6
1.28 atm  0.64 atm

PAr  n Ar PT 
nT
0.82
 7.6 2.15 

 0.23 atm 60

164

Example 3 ANSWER

A sample of gas at 5.88 atm contains 1.2 g CH4, 0.4 nH2  0.4 0.2 mol n CH 4  1.2  0.075 mol
g H2 and 0.1 g He. Calculate : 2 1.6

(a) The partial pressure of CH4, H2 and He in the nHe  0.1  0.025mol n T  n CH 4  n H 2  n He
mixture. 4  0.075 mol  0.2 mol  0.025 mol
 0.3 mol
(b) What is the partial pressure of CH4 and H2 if He
is removed?  PCH4nCH4 PT PHe  n He PT  PH 2  nH2 PT 
nT nT nT
Ans : a) PCH4=1.47 atm,PH2=3.92 atm, PHe=0.49 atm 0 .025 0.2
b) PCH4= 1.47 atm,PH2= 3.92 atm  00.0.3755.88  0.3 5.88   0.3 5.88 

61 1.47atm  0.49 atm  3.92 atm

62

ANSWER Example 4

b) If He isremoved , n T  0.3  0.025 4.00 dm3 of oxygen at a pressure of 2 atm and 1.00
 0.275 mol dm3 nitrogen at a pressure of 1 atm are introduced
into a 2.00 dm3 vessel.
PT  5.88  0.49
 5.39 mol a) Calculate the total pressure in the vessel.
b) What is the mole fraction of oxygen in the
PH 2  nH2 PT   PCH 4n CH4PT vessel?
nT nT
0.2 Ans : (a) 4.5 atm (b) 0.89
 0 .275 5.39   0 .075 5.39 
0 .275 64
 3.92 atm 1.47 atm

63

ANSWER  One of the applications of Dalton’s Law is

a) P1V1  P2V2 b) PO2  XO2 (PT ) to calculate the pressure of a gas collected

 P2(O2) over water ( for gases that neither react nor
1atm 4dm3
 2 dm3 XO2  4 atm soluble in water). gas + water vapour
4.5 atm
gas

 4 atm  0.89

 P1V1  P2V2
P2 (N2 )  1atm 1dm3
2 dm3
The gas collected is actually a mixture of the
 0.5 atm gas and water vapour.

PT  PN2  PO2 PT  Pgas  PH2O
 4.5 atm
65 66

165

Example 1 ANSWER

Consider the reaction below : a) PT  PO2  PH2O
735.5  PO2  23.8
2KClO3 2KCl + 3O2 PO2 711.7 torr
 0.936 atm

A sample of 5.45 L of oxygen is collected over T  25  273.15
water at a total pressure of 735.5 torr at 25 °C.  298.15 K
How many grams of oxygen have been
collected? (at 25°C, Pwater = 23.8 torr) PO2 V  nO2 RT
PO2 V
nO2  RT

 0.936atm5.45 K
0.08206 Latmmol1K1 298.15 K
(Ans : 6.68 g)
0.2086 atm

67 massof O2  0.2086 x32 68
 6.68g

Example 2 ANSWER

Excess amount of hydrochloric acid is added to a) 2HCl  Zn  ZnCl2  H2
2.5 g of pure zinc. The gas produced is collected 2.5
over water in a gas cylinder at 25 oC and 100.0 n Zn  65.5
kNm-2. Calculate :
 0.0385 mol
a) the number of mole of gas produced in the
reaction. 1 mol Zn  1 mol H2
b) the volume of gas collected in the cylinder. moles of H2 produce 0.0382 mol

Ans : (a) 0.038 mol (b) 0.93 L b) PO2 V  nO2 RT
nO2 RT
69 V PH 2

 0.0382 mol x 0.08206 Latmmol1K1301.15 K

1.0132 atm

0.93 L

70

Ideal and non-ideal behaviors of gases Deviation from Ideal Behaviour

Ideal gas - any gas that obeys the ideal gas Real gases do not show ideal behaviour and
equation and has the properties as outlined by said as that they deviate from the ideal behaviour
the Kinetic Molecular Theory of gases
The deviation is due to gas molecules do have:
Real gas (non-ideal gas) - gases which do not i. its own volume and they occupy some
obey ideal gas properties
space.
71 ii. intermolecular forces acting between them .

72

166

Real gases do not behave ideally because: ii. There are attractive forces between molecules

Real gases do not behave ideally because:

i. Gas molecules have volumes

The free volume where the molecules move • Collision between molecules and the wall of
about is smaller than the volume of the container are less frequent
container.
• The actual pressure exerted by the gas is lower
73 than calculated by ideal gas equation

74

The deviation is more significant at high At low temperature
pressure and low temperature. • The kinetic energies of the molecules decreases
• The molecules move at low speed
At high pressure: The intermolecular forces between molecules

1. The volume of container decreases, thus the become significant.
molecules of gas are closer to each other and
begin to occupy a sizeable portion of the 76
container.

The volume of gas molecules is significant.

2. As the volume of container decreases at high
pressure, the molecules are so close to each
other.

The intermolecular forces between
molecules become important.

75

However, real gases behave almost ideally at a low (b) At high temperature:
pressure and high temperature. • The gas molecules have high kinetic energies

At low pressure and high temperature, the deviations and move at high speed
are small and ideal gas equation can be used without • The molecules are able to free themselves
generating serious error because:
from the intermolecular forces that act
(a) At low pressure: between them.
• The intermolecular forces can be neglected,
• To achieve a low pressure, the volume of a container thus they behave almost ideally.
is increased.
78
• When the volume of a container increases the

molecules will be far apart from one another, hence

the intermolecular forces can be neglected.

• At a low pressure the volume of the container is 77
extremely large compared to the size of the
molecules, thus the volume of molecules can be
neglected.

167

PV The van der Waals equation
Variation of RT vs P of 1 mole of N2 gas at different T

PV 203 K 293 K • Since real gases deviate from ideal behavior
RT especially at high pressure and low temperature

673 K • the ideal gas equation (PV=nRT) cannot be
used to accurately describe the gas variables of
1.0 1000 K real gases.
Ideal gas
The value approaches 1.0 at a • Ideal gas equation needs to be adjusted
very low pressure P/atm
• Adjusting the equation, two parameters need to

be reconsidered :

• the line converge to 1.0 mol when p≈ 0 • The lines approaching the ideal line • attractive forces between the gas
when T increases molecules
• gas behave ideally at a very low
pressure i.e high volume • gas behave ideally at a high • volume of the gas molecules
temperature
80
79

a) Attractive Forces Between Molecules b) Volume of Gas Molecules

Attractive forces which act between the gas Since the gas molecules occupy a sizeable portion of a container,
molecules will : the space in which the molecules are able to move are less than
the volume of the container.
● make the molecules move slower

● give less impact to the wall * Vreal = the space in which the molecules
able to move
● pressure exerted by the real gas is less compared Vcontainer
to the ideal gas
nb
Preal < Pideal Vreal < Vcontainer
● therefore, the term pressure need to be The correction factor to the volume is Vreal = Vcontainer – nb

corrected by adding coefficient n2 a where ;
V2

n = number of moles b = a constant representing the volume
occupied by the gas molecules
P ideal = P real + n2a a = a positive constant which depend on the
V2
strength of the attractive forces acting

between the molecules n = number of moles

V = volume of the container 81 82

 Thus, the pressure of gas equation becomes: Combining both factors into the ideal gas equation :
PV = nRT
nRT n 2a
P= V -n b - V2 Correction  Peq+uanVti2o2aneqVau-tenqbuat=ionnRT van der Waals
for molecular 
real gas attraction 
pressure
Correction for
volume of where P = pressure of real gas
molecules V = volume of container
n = mole of gas
* a and b = van der Waals constant

83 84

168

High value of a and b show great deviation
from ideal gas. (The gas tend to show the real
gas character)

Value of a
• Molecules with strong attractive forces has high

value of a

Value of b

• Large molecular size have high value of b due to

large volume of the gas molecules. 85

169

LEARNING OUTCOMES

At the end of this lesson, students should be able to :

i. Relate the properties of liquid to intermolecular

forces, molecular arrangement and molecular

motion in explaining shape, volume, surface

5.2 LIQUID tension, viscosity, compressibility and diffusion.

ii. Explain vaporization process and condensation

process based on kinetic molecular theory &

intermolecular forces.

iii. Define vapour pressure and boiling point.

iv. Explain boiling process.

v. Illustrate the relationship between:

- intermolecular forces and vapour pressure; &

86 - vapour pressure and temperature. 87

The Properties of Liquid 2. Surface tension

1. Volume and Shape • molecules within a liquid are pulled in all direction
• Liquid has a definite volume but not a definite by intermolecular forces (net attraction = 0)

shape. • molecules at the surface are only pulled
• The particles are arranged closely but not rigidly. downward and sideways (net attraction = into the
• Held together by a strong intermolecular forces liquid)

but not strong enough to hold the particles firmly a liquid surface tends to have the smallest
in place. possible area
• Particles are able to move freely.  liquid surface tighten like an elastic film
• Thus, a liquid flows to fit the shape of its
container and is confined to a certain volume. 89

88

• Surface tension is the energy required to Fig. 11.8
increase the surface area of a liquid by a unit
amount. molecule at
the surface
• e.g: surface tension of water = 7.29102 J m2

• The stronger the forces between particles in a
liquid, the greater the surface tension.

90 molecule within
the liquid
170
91

3. Viscosity c) The temperature of the liquid
- Increase in temperature, decrease the viscosity
• Viscosity is the resistance of a liquid to flow. of a liquid because :
- At higher temperature, molecules have higher
• The higher the viscosity, the more slowly the kinetic energy.
liquid flows. -The molecules move faster.
- Molecules can overcome the intermolecular
The factors affecting the viscosity : attractive forces more easily.
- Resistance to flow decrease.
a) The strength of intermolecular forces - Viscosity decrease.

- Liquids that have strong intermolecular forces 93
have higher viscosity

b) The size of the molecules

- Liquids with larger size (higher molar mass) is

more viscous because it has stronger
intermolecular forces

92

4. Compressibility Vaporisation Process

• In liquid, the particles are packed closely together. • Vaporization is a process of changing the state from
liquid to vapour.
• Thus, there is very little empty space between the
molecules. • Molecules in a liquid moves freely, collide with each
other and posses different magnitudes of kinetic
• Liquids are much more difficult to compress than energy.
gas.
• Some molecules have a relatively high kinetic
5. Diffusion energies.

• Diffusion is the movement of liquid from one fluid • Molecules at the surface possess sufficient enough
through another. kinetic energy to overcome the intermolecular
attractive forces that bind them.

• Diffusion can occur in a liquid because molecules • When the kinetic energy is sufficient enough to

are not tightly packed and can move randomly overcome the intermolecular forces acting on them,

around one another. 94 the molecules will escape as vapour. 95

Factors affecting the rate of vaporisation : ii) Temperature
• As temperature is increased, the total number of
i) Surface area
• The larger the surface area, the higher the chances molecules with high kinetic energy is increased.
• More molecules have enough energy to escape
for the molecules to escape from the surface.
• Surface area increases, evaporation rate increases from the surface of the liquid.
• Thus evaporation rate increases.
evaporation – the change state of a liquid into a
vapour at a temperature below bp of the liquid, iii) Intermolecular attractive forces
occurs at the surface of liquid. • The weaker the intermolecular attractive forces,

96 the faster the evaporation rate.

97

171

Condensation Process Vapour Pressure

• A process whereby vapour molecules turn to Definition:
liquid. Vapour pressure is the pressure exerted by a
vapour in equilibrium with its liquid phase.
• Some of the vapour molecules may lose their
kinetic energy during the collision. • Molecules can escape from the surface of liquid
at any temperature by evaporation.
• They do not have enough energy to remain as
vapour molecules. • In an open system, vapour molecules which
evaporate off will diffuse away.
• They reached the surface of the liquid and
trapped by the attractive forces. 99

• If they cannot overcome the attractive forces,
these vapour molecules return as liquid
molecules.

• The process is known as condensation. 98

•In a closed system, vapour molecules which leaves the Boiling – the process
surface are trapped in the close container.
•These vapour molecules are in constant random motion. • Increasing the temperature will increase in the
•The molecules strike the wall of container and exert vapour pressure.
some pressure.
•As the quantity of molecules in the vapour phase • As heat is applied, the vapour pressure of a
increase, some molecules may lose energy and system will increase until it reaches a point
condense. whereby the vapour pressure of the liquid
•Eventually, the rate of evaporation = the rate of system is equal to the atmospheric pressure.
condensation. (The system achieved dynamic
equilibrium) • Boiling occurs and the temperature taken at this
•At equilibrium, the number of vapour molecules above point is known as the boiling point.
liquid are constant.
•The pressure exerted by those molecules is called • At this point, the change of state from liquid to
vapour pressure. (or maximum vapour pressure) 100 gas occurs not only at the surface of the liquid
but also in the inner part of the liquid.

• Bubbles form within the liquid. 101

Definition: The Relationship Between Intermolecular
Forces and Vapour Pressure
Boiling Point : the temperature at which the
vapour pressure of a liquid is equal to the • Molecules with weak intermolecular forces need to
external atmospheric pressure. have less kinetic energy to escape from the liquid
surface.
Normal Boiling Point: the temperature at which
liquid boils when the external pressure is 1 atm. • Less energy is needed to overcome the
(that is the vapour pressure is 760 mmHg) intermolecular forces between liquid molecules.

102 • Thus, a liquid can easily vapourise.

• More vapour molecules will be present and exert
higher pressure.

• Vapour pressure increases.

• The weaker the intermolecular forces, the higher the

vapour pressure. 103

172

The Relationship Between Evaporation Boiling
Vapour Pressure and Boiling Point
 The changes  The changes of liquid into
• A liquid boils when its vapour pressure equals to of a liquid into a vapour state, occurs at the inner
atmospheric pressure. vapour state part (body) of the liquid at a
occuring at the certain temperature. (when the
• The temperature in which the vapour pressure of its surface of a vapour pressure is equal to the
liquid equal to atmospheric pressure is called boiling liquid at any atmospheric pressure)
point. temperatures
and pressures.  The temperature is called the
• Liquids with weaker intermolecular forces has higher boiling point.
vapour pressure.
105
• Thus, a molecule with weak intermolecular forces can
easily vapourise and the system requires less heat to
achieve atmospheric pressure, therefore it boils at a
lower temperature.

Intermolecular forces  , Vapour pressure, Boiling point 

104

173

LEARNING OUTCOMES

At the end of the lesson, students should be able to :

i. Explain the following process using kinetic

molecular theory:

5.3 SOLID (i) freezing (solidification);

(ii) melting (fusion);

(iii) sublimation; &

(iv) deposition

ii. Differentiate between amorphous and crystalline

solids.

iii. State the following types of crystalline solids with

appropriate examples.

i. metallic; iii. molecular covalent;

ii. ionic; iv. giant covalent. 107

Fixed shape of a solid In principle, solid, liquid and gas states are
interconvertible
• Intermolecular forces between solid molecules
are strong enough to hold molecules together solid liquid gas
and lock them in place sublimation
deposition
• There is very little free space between the
molecules 5

• Particles are closely arranged and regularly in 109
order

• Rigid arrangement- particles can vibrate, rotate
about fixed position and cannot move freely
without disrupting the whole structure.

108

Freezing (solidification) Melting (fusion)

• Occurs when a liquid changes to solid • A process at which a solid changes into a liquid

• Freezing process: when the temperature of a • The melting process:
liquid is lowered:
- When a solid substance is heated, kinetic energy
- the kinetic energy of the liquid molecules decrease of the molecules increase.

- the molecules move slowly - The molecules vibrate and rotate more rapidly.

- the intermolecular attractions overcome the - At certain temperature, the kinetic energy is
motion of the molecules higher enough to overcome the intermolecular
forces of attraction between solid particles.
- when the intermolecular attraction are strong
enough to hold the particles together in a fixed and - The particles are free to move and the solid starts
orderly arrangement, the liquid freezes. to melt.

Freezing point is a temperature at which the liquid Melting point is the temperature at which solid

and solid phases of a substances coexist at 110 and liquid coexist in equilibrium. 111

equilibrium.

174

Sublimation TYPES OF SOLID
• The process by which a substance changes
CRYSTALLINE SOLID AMORPHOUS SOLID
directly from solid to the gaseous state
without passing through the liquid state. • A solid that has highly • Solid that does not have
• Occurs on solid with weak intermolecular forces ordered structure where a regular three
of attraction. atoms, ions or molecules dimensional
show a regular repetition arrangement of atoms or
Deposition in 3D-arrangement molecules
• The process where molecules from gaseous • Formed when a • Formed when a
saturated liquid is cooled saturated liquid is cooled
state change to the solid state. slowly. rapidly
• The opposite process of sublimation. • Its atoms, molecules or • Example : glass,
ions occupy specific
112 position. plastic material
• Example : ice, sugar, salt
113

Types of crystalline solid ii. Ionic crystal

i. Metallic crystal • Ionic crystals are composed of positive and
negative ions which are held together by ionic
• Metallic crystals are formed by metal bonds

• These crystals are made up of atoms of the same • Physical properties of ionic crystal:
element, thus they can pack very closely in 3-
dimensional structure - high melting point

• Composed of atoms of the same metal linked - hard but brittle
together by metallic bonds
- does not conduct electricity in the solid state but
• The physical properties of metal: does

- high electrical and thermal conductivity conduct electricity in molten or in aqueous state.

- lustre • Example : NaCl 115
- ductile and malleable

• Examples: all metallic elements – Na, Mg, Fe 114

iii. Molecular covalent crystals Diamond
• Composed of molecules held together by • In diamond, each carbon atom bonded to four

intermolecular forces(van der Waals or other carbon atoms through strong covalent
hydrogen bonds) bonds in a tetrahedral arrangement.
• Example : Iodine, I2 • This arrangement is repeated to give a three-
dimensional giant structure
iv. Giant covalent solids • The physical properties of diamond:
• Very large molecules / gigantic structure - transparency
• Composed of atoms linked together in large - great hardness
- non-conductor of electricity
networks by covalent bonds - high melting point (about 3500oC)
• Example : Diamond, graphite, SiO2
117
116

175

Graphite

• Graphite is an example of layered structure in
the hexagonal crystalline system.

• In the layers, each carbon atom is covalently
bonded to three other carbon atoms to form a
hexagonal ring.

• The layers are held together by weak van der
Waals forces.

• The physical properties of graphite:

- grey in colour

- metallic lustre

- soft

118 - conductor of electricity and heat 119

120

176

LEARNING OUTCOMES

At the end of this lesson students should be able to:

a) Define phase, triple point and critical point

5.4 PHASE DIAGRAM b) Identify triple point and critical point on the phase
diagram.
121
c) Sketch the phase diagram of H2O and CO2.
d) Compare the phase diagram of H2O and CO2.
e) Explain the anomalous behaviour of H2O.
(e) Describe the changes in phase with respect to

i. temperature (at constant pressure); and

ii. pressure (at constant temperature). 122

• Phase is a homogeneous part of a system in Phase diagram of carbon dioxide, CO2
contact with other parts of the system but
separated from them by a well-defined boundary BC

• Phase Diagram is a diagram that describes the T
stable phases and phase changes of a A
substance as a function of temperature and
pressure 124

• Phase diagram is used to predict the phase that
exist under a certain temperature and pressure

123

Regions of the diagram TA curve:
• at the temperatures and pressures along TA curve,
- The diagram has three main regions :
solid and vapor are at equilibrium. (sublimation or
Solid, liquid and gas deposition)

- Each region corresponds to one phase of the CO2(s) ⇌ CO2(g)
substance. • TA curve represents the vapor pressure of solid CO2

- CO2 is a gas under normal conditions of as it sublimes at different temperature
temperature and pressure ( 25oC and 1 • sublimation is a process of a solid changing directly
atm)
into a gas
Lines between regions • Solid CO2 does not melt, but sublimes when heated

Any point along a line shows the pressure and below 5.2 atm
• At 1 atm, CO2 sublimes at -78oC
temperature at which the two phases exist in • CO2 never exists as liquid under normal atmospheric

equilibrium. 125 pressure
• reverse process of sublimation is deposition 126

177

TB curve: TC Curve:
• At the temperatures and pressures along TC
• At the temperatures and pressures along TB curve,
solid and liquid are at equilibrium. (melting or curve, liquid and vapour are at equilibrium.
freezing) CO2(l) ⇌ CO2(g)

CO2(s) ⇌ CO2(l) • Known as the boiling point curve.
• Known as the melting point or freezing point curve. • TC curve shows that the boiling point of this

• TB curve represents the change in melting point of liquid increases with increase in pressure.
CO2 with increasing pressure.
128
• TB curve slant to the right because solid is denser
than liquid. This is normal characteristic of most
substances

• Melting point of CO2 increases with pressure.
• Solid is harder to melt at higher pressure.

• Higher melting point at higher pressure. 127

• The vapor pressure curve ends at critical point, C. • Point T is known as the triple point.
• Critical point is the temperature and pressure at
• Triple point is the temperature and pressure
which the liquid and its vapor become identical at which the vapour, liquid and solid states
and indistinguishable. of a substance are in equilibrium.
• Critical point for CO2 is at 73 atm pressure and CO2(s) ⇌ CO2(l) ⇌ CO2(g)
temperature 31oC.
• At the temperature above 31oC the CO2 gas cannot • Triple point for carbon dioxide is -57°C and 5.2
be condensed no matter how high the applied atm.
pressure is.
130
129

Phase diagram of water • Water is a liquid under normal conditions of
temperature and pressure. ( 25oC and 1 atm )
BC
• At point T, the triple point; ice, water and steam
T are in equilibrium.
A H2O(s) ⇌ H2O(l) ⇌ H2O(g)

131 • Triple point for H2O is at 0.006 atm pressure and
temperature 0.01oC.

• At the temperatures and pressures along the
curves, 2 phases are in equilibrium.

132

178

TA curve Anomalous behaviour of water

• at the temperatures and pressures along TA curve, • For H2O, the TB curve (melting point or freezing
point curve) slant to the left, that is, the melting
ice and steam are at equilibrium point decreases with pressure.

H2O(s) ⇌ H2O(g) • This is because, water is denser than ice
TB curve • Ice is easier to melt at higher pressure

• at the temperatures and pressures along TB curve, - lower melting point at higher pressure

ice and water are at equilibrium 134

H2O(s) ⇌ H2O(l)
• TB curve represents the change in melting point or

freezing point of water with increasing pressure.

• Melting point of a substance is identical to its

freezing point.

• Melting point of a liquid at 1 atm is the normal

melting point.

• Normal melting point for water is 0oC. 133

TC Curve • Critical point for water is at 218 atm and
• At the temperatures and pressures along TC temperature 374oC

curve, water and steam are at equilibrium. • beyond the critical point, water and steam are
H2O(l) ⇌ H2O(g) indistinguishable

• TC curve also represent the boiling point of • At the temperature above 374oC steam cannot
water at different pressure. be condensed no matter how high the applied
pressure
• Boiling point of a liquid when its vapour pressure
is 1 atm is the normal boiling point. 136

• Normal boiling point for water is 100oC.

135

Exercise:
1. Describe the phase change when carbon

dioxide undergoes isobaric heating at 5.2 atm
pressure.
2. Describe the phase change when pressure is
applied to water isothermally at 0.01oC.

137

179

CHAPTER 5: STATES OF
MATTER

QUESTIONS

TUTORIAL.MEKA.KUMBE

lecture notes & questions

180

CHEMISTRY SK015
TUTORIAL CHAPTER 5: STATES OF MATTER

5.1: Gas

1. a) State the basic assumptions of the Kinetic Molecular Theory of Gases for an
ideal gas. Using this theory, explain Boyle’s law qualitatively.

b) A gas occupying a volume of 735 mL at a pressure of 0.975 atm is allowed to
expand at constant temperature until its pressure reaches 0.555 atm. What is its
final volume?

c) A sample of air occupies a 3.94 L container when the pressure is 1.35 atm. What
pressure in mmHg is required to compress it to 0.175 L?

2. a) Define Charles’ law.

b) Refer to the diagram below.

Explain how temperature affects the volume.
c) A sample of 95.0 L dry air is cooled from 130.0oC to 20.0 oC while the pressure

is maintained at 2.95 atm. Calculate the final volume.
d) Under constant-pressure condition, a sample of 9.75 L of hydrogen gas at 92 oC

is initially cooled until its final volume is 3.55 L. What is the final temperature?

3. a) Chlorine is widely used to purify municipal water supplies and to treat swimming
pool water. Suppose that the volume of a particular sample of Cl2 gas is 8.50 L at
885 torr and 25.0 oC, at what temperature will the volume be 15.50 L if the
pressure is 872 torr?

b) Ammonia is kept in a cylinder with a movable piston at 25.0 oC and 2.30 atm.
Calculate the pressure in the cylinder if the temperature is increased to 48.0 oC
and the piston is raised so that the volume occupied by the gas is doubled.

4. a) A sample of chlorine gas is confined in a 6.0-L container at 230 torr and 25 oC.
How many moles of chlorine gas are there in the sample?

b) You have a sample of 215 mL of chlorine trifluoride, ClF3 gas at 725 mmHg and
45.0 oC. What is the mass (in gram) of the sample?

c) What is the volume of 250 g of steam at 125.0°C and 5.35 atm?

181

CHEMISTRY SK015
TUTORIAL CHAPTER 5: STATES OF MATTER

5. a) Using ideal gas equation, derive an equation that relates the density of gas, , to
its pressure, P.

b) The density of gas X is 2.60 g L–1 at 25.0C and 101 kPa. What is the molar mass
of X?

c) The density of an unknown vapor at 446.0oC and 100.8 kPa is 4.33 g L−1.
Calculate the molar mass of the vapor.

6. The diagram below represents a mixture of four different gases in a fixed volume
container.

a) Which gas exerts the highest partial pressure?
b) Which gas exerts the lowest partial pressure?
c) Using Dalton’s law, what is the partial pressure of gas D if the total pressure is

0.86 atm?
7. A total pressure of a gas mixture that contains 6.50 g methane, CH4, 1.80 g hydrogen,

H2 and 8.40 g nitrogen, N2 at 25.0 oC is 6.50 atm. Calculate:
a) partial pressure of each gas.
b) the volume of the container used.
8.

Two tanks are connected by a closed valve. Each tank is filled with gas as shown, and
both tanks are held at the same temperature. When the valve is opened, the gases are
allowed to mix. Calculate:
a) the partial pressure of each gas.
b) the total pressure.
c) the mole fraction of each gas in the mixture.

182

CHEMISTRY SK015
TUTORIAL CHAPTER 5: STATES OF MATTER

9. The decomposition of potassium chlorate, KClO3 at 27.0 C produces potassium
chloride, KCl and oxygen gas. The gas was then collected over water and 135 mL of
oxygen gas was obtained at the pressure of 762.0 mmHg. Calculate the mass of the
oxygen gas collected if the vapour pressure of water is 23.4 mmHg at 24.0C.

10. How many milligrams of magnesium, Mg, need to react with excess hydrochloric acid
solution, HCl, to produce 30.0 mL of hydrogen gas measured at 25.0oC and 762 torr?

11. The van der Waals equation for 1 mole of real gas is as follows:

P + a (V − b) = RT
 V2


a) Explain the corrective terms for pressure and volume.
b) H2S gas has a larger a value than H2 gas. Explain.

5.2: Liquid

1. Explain the phenomenon below;
a) Aqueous solution of ethanol, CH3CH2OH has a lower surface tension than water.
b) As temperature increases, oil flows faster through a narrow tube.

2. Explain the phase transition that occurs in each of the following situation in terms of
molecular arrangement and motion.
a) When ice is heated, it turns to water.
b) Wet clothes dry on a hot sunny day.
c) Water droplets appear on a glass of cold water at room conditions.

3. Tetrabromomethane, CBr4, is also known as carbon tetrabromide. Determine whether
the statements about CBr4 below is TRUE or FALSE;
a) CBr4 has weaker intermolecular forces than CCl4.
b) CBr4 is more volatile than CCl4.
c) CBr4 has a higher boiling point than CCl4.
d) CBr4 has a higher vapour pressure than CCl4 at the same temperature.

4. Boiling point of two different liquids are given below:

Liquid Boiling Point (oC)

ethanol, 78
CH3CH2OH

acetone, 56
CH3COCH3

Determine which liquid will have a lower vapour pressure at 25oC.

183

CHEMISTRY SK015
TUTORIAL CHAPTER 5: STATES OF MATTER

5.3: Solid

1. Diagram below illustrates what happen to dry ice when exposed at room temperature.

Describe phase changes that occur in process A.

2. a) Differentiate between amorphous and crystalline solids on macroscopic and
molecular levels.

b) State the interparticle forces involved in each of the following crystalline solids:
i. metallic
ii. ionic

iii. molecular covalent
iv. giant covalent

5.4: Phase Diagram

1. The diagram below shows a portion of phase diagram for phosphorus.

Pressure

A B
C
43 atm

o Temperature

590 C

a) Indicate the phases presents in the region labeled A, B and C.

b) A sample of solid phosphorus cannot be melted by heating in an open container

at atmospheric pressure. Explain.

184

CHEMISTRY SK015
TUTORIAL CHAPTER 5: STATES OF MATTER

2. The diagram below shows the phase diagram of water.

Pressure

C

A

EF

1 atm

B

Do
Temperature, C

a) Name point A and B.
b) Explain the physical properties of water at point A and B.
c) Which curve represents the equilibrium between ice and water vapour?
d) Describe the phase changes when a sample at point E is heated at constant

pressure until point F.
e) Name the point at which the BC line intersects 1 atm line.
f) The BC line has a negative slope. Explain.

3. a) Define critical point and triple point.
b) Substance X has its triple point at 18C and 0.5 atm. The normal melting point
and boiling point are 20C and 300C respectively. Sketch the phase diagram of
X.

185

CHEMISTRY
KUMBE TOPIC 5: STATES OF MATTER

1. When an empty glass vessel with a mass of 134.74 g was filled with an unknown gas
Y, the pressure was found to be 99.3 k Nm−2 at 31.0C, while the mass was 137.28 g.
The glass vessel was then filled completely with water and the mass recorded was
1067.90 g. By using the ideal gas equation, determine the relative molecular mass of
Y. [69.3]

2. The combustion of ammonia gas, NH3 in oxygen produces nitrogen monoxide, NO
and water vapour.
a) Write a balanced equation of the above reaction.
b) In an experiment conducted at 100C, 2.0 moles of ammonia was combusted
with
6.0 moles of oxygen gas in a 5.0 L tank.
i. Determine the number of moles of each product formed.
ii. What would the total pressure in the tank be when the reaction
completed?
[51 atm]
iii. Calculate the partial pressure of each gas remained in the tank.
[NO=12 atm, H2O=18 atm, O2=21 atm]

3. One useful equation developed to predict the behaviour of a real gas was proposed by

Johannes Van der Waals. The real gas equation was:

P + n2a  (V − nb) = nRT
V2

a) What do the magnitude of a and b reflect?

b) Predict what would happen to the values of a and b when large gas molecules

such as propane, CH3CH2CH3 are used in an experiment.
c) Explain the differences between a real gas and an ideal gas.

d) Explain the conditions at which real gases approach the ideal gas behaviour.

4. a) State three properties of a liquid.

b) The table below shows the boiling points of several liquids.

Liquid Ethanal, Ethanol, Methanol,
CH3CHO CH3CH2OH CH3OH

Boiling 20 78 65

point(C)

Arrange the liquids in the order of ascending vapour pressure. Explain your
answer.

186

CHEMISTRY
KUMBE TOPIC 5: STATES OF MATTER

5. a) State the types of crystalline solid and interparticles forces involved in each
molecules/ atoms.

Molecules/ atom Types of crystalline Interparticles forces
solid
NaCl
Mg
SiO2
S8

b) The table below shows data for the phase diagram of carbon dioxide.

Pressure (atm) Temperature (C)

Triple point 5.2 −57

Critical point 73 31

i. Sketch the phase diagram of carbon dioxide.
ii. By referring to the phase diagram, explain why solid carbon dioxide

does not melt at ordinary temperature and pressure, instead it sublimes.
iii. Explain the phase changes that occur when a sample of carbon dioxide

in a closed vessel under 1 atm pressure and a temperature of −78C is
pressurised isothermally to 10 atm, followed by an isobaric heating to
20C.

187

CHEMISTRY
MEKA TOPIC 5: STATES OF MATTER

1. a) State Boyle's Law.
b) A quantity of gas at 23.0oC exerts a pressure of 748 torr and occupies a
volume of
10.3 L. Calculate the volume the gas will occupy at 23.0oC if the pressure is
increased to 1.88 atm.

2. a) Derive an equation that relates density, ρ and pressure,p from ideal gas
equation.
b) Calculate the temperature (in oC) of benzene vapour, C6H6, if the density is
1.50 g L−1 and the pressure is 800 torr.

3. a) State the value of temperature and pressure at STP.
b) What would be the volume of a gas at STP if it occupies a volume of 255 mL
at 25.0oC and 0.85 atm?

4. a) State Dalton's law of partial pressure.
b) A 4.0-L nitrogen at a pressure of 400 k Nm−2 and 1.0 L argon at a pressure of
200 k Nm−2 are transferred into a 2.0 L container. Calculate;
i. the partial pressure of nitrogen.
ii. the partial pressure of argon.
iii. the total pressure of the mixture.

5. A sample of 1.650 g of Ag2O decomposed on heating and the oxygen produced was
collected by displacement of water at 25.0C and 750 mmHg. The reaction equation
is,
2Ag2O(s) ⎯→ O2(g) + 4Ag(s)
a) Calculate the number of mole of Ag2O decomposed.
b) Calculate the number of mole of O2 gas produced.
c) Determine the partial pressure of O2 gas produced.
d) Calculate the volume of oxygen in L.

6. Explain the phase change that occurs in the following process in terms of molecular
motion and interparticle forces.
a) Melting
b) Condensation
c) Sublimation

7. Explain how each of the following affects the vapour pressure of a liquid.
a) Surface area.
b) Intermolecular forces.
c) Temperature.

188

CHEMISTRY
MEKA TOPIC 5: STATES OF MATTER

8. The diagram below represents the structure of diamond. State the interparticle forces
that exist in diamond. Explain why diamond is extremely hard.

9. Refer to the diagram below:

P, atm

1.00
0.006

0.01 T,oC

Describe all the phase changes that occur in each of the following cases:
a) Water vapour originally at 1.0×10−3 atm and −0.10oC is slowly compressed at

constant temperature until the final pressure is 10 atm.
b) Water originally at 100.00oC and 0.50 atm is cooled at constant pressure until

the temperature is −10oC.

189

CHAPTER 6: CHEMICAL
EQUILIBRIUM

LECTURE NOTE

lecture notes & questions

190

CHAPTER 6 6.1
Dynamic Equilibrium
CHEMICAL EQUILIBRIUM
2
6.1 DYNAMIC EQUILIBRIUM
6.2 EQUILIBRIUM CONSTANT
6.3 LE CHATELIER’S PRINCIPLE

1

LEARNING OUTCOMES Reversible and Non-reversible Reactions
At the end of this topic, students should be able to :
(a) Non-reversible reaction
(a) Explain the following terms
i. reversible reaction • The reaction that proceed in only one direction that is
ii. dynamic equilibrium towards the formation of product.
iii. law of mass action
A+B C+D
(b) State the characteristics of a system in
equilibrium. • The sign indicates non-reversible reaction.

(c) Interpret the curve of concentration reactants  A and B react to form C and D (forward reaction)
and products against time for a reversible
reaction.  C and D do not react to form A and B.

3 • Example : Mg (s) + 2HCl (aq) MgCl2 (aq) + H2 (g)

4

( b ) Reversible reaction A+B C+D

A+B C+D

• Consider the above reversible reaction which • The sign indicates reversible reaction.
occurs in both forward and reverse directions.
 Forward reaction : reaction proceeds from left to
• When molecules of reactants, A and B react, right (formation of C and D)
molecules of products C and D are formed.
 Reverse reaction : reaction proceeds from right to left
( formation A and B )

• As soon as some product molecules (C and D )are • The net concentrations of A, B, C and D do not change.
formed, the reverse process (reaction) begins to
take place in which reactants, (A and B) are formed
from the products.

5 6

191

Some examples of reversible reactions are State of Equilibrium
• Equilibrium is a state in which there are no
i. The Haber Process :
N2 (g) + 3H2 (g) observable changes as time goes by.

2NH3 (g)

ii. Decomposition of PCl5 : • Chemical equilibrium – chemical composition
of the system does not change at dynamic
PCl5 (g) PCl3 (g) + Cl2 (g) equilibrium.

e.g : N2O4 (g) 2NO2 (g)

iii. Stage two in the Contact Process : • For the above system, at equilibrium state, the

2SO2 (g) + O2 (g) 2SO3 (g) concentrations of trheemraeinacctaonntsst,anNt2Oo4v(egr) and
products, NO2(g) time
and there are no visible changes in the system.

7 8

• Physical Equilibrium : a system whose physical Characteristic of a system in equilibrium
state does not change • A system in chemical equilibrium has the following
when dynamic equilibrium
is reached. characteristic :
i. occurs only in a closed system
e.g : ii. forward and a reverse reaction proceed at

H2O (l) H2O (g) equal rates
iii. the reaction quotient, Q = the equilibrium
• Both water liquid and water vapour exist at dynamic
equilibrium constant, K.
iv. the properties (concentrations, pressure and
9
colour) of the reactants and products are
constant over time.

• The point at which the rate of forward reaction
equals the rate of reverse reaction is known as
dynamic equilibrium.

10

• For the following reversible reaction : • Refer to Figure 1

A+B C+D • As the reaction progresses :
 [C] and [D] increases with time
Concentration Concentrations of  [A] and [B] decreases with time
reactants and  After time, ta, there is no net change in the
products are concentration of products or reactants.
constant  The system is in the state of equilibrium
 The equilibrium is dynamic equilibrium where
C and D (products) after ta, the reaction did not stop.
 But, the rate of forward reaction equals to the
A and B (reactants) rate of reverse reaction.

ta Time 12
Figure 1 : Concentration versus Time Curve

11

192

• Figure 2 shows the rate changes of both forward • Figure 2 shows:
and the reverse reactions over time. • The rate of forward reaction decrease in time

rate of foward Forward reaction rate because the amount of reactant decrease and
reaction the rate of reverse reaction increase since the
decreases concentration of product increase.
• At first, the rate of reverse reaction zero because
Rate A+B C+D no products are formed.
• As the reaction proceeds :
Reverse reaction rate
 [C] and [D] increase with time, hence the
rate of reverse C+D A+B rate of reaction also increase.
reaction
increases 14

ta Time Law of Chemical Equilibrium

Figure 2 : Rate versus Time Curve The equilibrium concentration of reactants and
products are related by law of mass action / law of
13 chemical equilibrium.
Law of Mass Action or Equilibrium Law
Example: The rate at which a chemical reaction takes
place at a given temperature is proportional to
N2O4(g) 2NO2(g) the product of the active masses of the
reactants
At equilibrium,
16
• Rate of forward reaction = Rate of reverse reaction

N2O4(g)  2NO2(g) 2NO2(g)  N2O4(g)

 [N2O4(g)] and [NO2(g)] remain constant
But!

[N2O4(g)] ≠ [NO2(g)]

15

When a reversible reaction has achieved dynamic • At equilibrium ; c d  
equilibrium at constant temperature, the ratio of where : C D  
reactant to product concentrations has a constant 
value, K.  
K  a b   
A B   
  

Consider general reaction : K = equilibrium constant
[ ] = concentration in the unit of mol L-1 @ M
aA + bB cC + dD a, b, c, d = stoichiometric coefficient for the

reacting species A, B, C, and D in
a balanced equation.

• Note : magnitude of K indicates how far the reaction
proceeds toward product at a given
temperature.

17 18

193

• For a reverse reaction :

K

cC + dD K’ aA + bB

• At equilibrium ; or K  1
a b    K
A B   

  

K  c d   
C D   

  

where ;

K’ - equilibrium constant for reverse reaction.

19

194

6.2 LEARNING OUTCOMES 6.2
The Equilibrium Constant • At the end of this topic, students should be able to :

(KC and KP) (a) Define homogeneous and heterogeneous
equilibria.
20
(b) Write expression for equilibrium constant in
terms of concentration, Kc and partial
pressure, Kp for homogeneous and
heterogeneous systems.

21

Homogeneous Equilibrium C c D d   

K c    

• In homogeneous equilibrium, all products and A a B b   
reactants are in the same phase.
  
  
  

(a) Reaction in a Aqueous Phase where :

• The quantity of reactants and products are [ ] = concentration in mol L-1 @ M
mention in concentration unit, mol L-1 or M.
Kc = equilibrium constant expression in terms
of concentration
• So, the equilibrium constant expression symbol is
Kc.
aA (aq) + bB (aq) cC (aq) + dD (aq) a, b, c & d = stoichiometric coefficients

22 23

(b). Reaction in a Gaseous Phase Exercise 1 :
• The concentration of reactant and product can be
Write the equilibrium constant expression in terms
mention in concentration unit or in partial pressure of concentration and pressure the following
reactions.
• So, the equilibrium constant can be expressed as
Kc or Kp.
aA (g) + bB (g) cC (g) + dD (g) i. 2SO2 (g) + O2 (g) 2SO3 (g)

Cc Dd  P C  c  P D d ii. 4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g)
AaBb  P A   P B
Kc  or K p  a 

    b iii. CH3COOH (aq) CH3COO- (aq) + H+ (aq)



where : iv. C3H8 (g) + O2 (g) CO2 (g) + H2O (g)

PA = partial pressure of gas A and etc. in terms
Kp = equilibrium constant expression of
pressures

24 25

195

Answer • The equilibrium constant ( Kc and Kp ) is a
dimensionless (no unit) quantity.

i. Kc  SO3 2     Kp P2 • Example :
SO2 2 O2  SO3 PCl3 (g) + Cl2 (g)

P P2 O2
SO2

    Kp  6 PCl5 (g)
NO4 H2O6 P P4 H2O
ii. Kc  NH3 4 O2 5 NO

P P4 5 Kc = 1.67 (at 500 K)
NH3 O2

   iii.
Kc  CH3COO H Kp   Kp = 4.07 x 10-2 (at 500 K)

CH3COOH

iv. Kc CO2  H2O   Kp PCO2PH2O • NOTE:
C3H8  O2  PC3H8 PO2
 The values of Kp and Kc will change when
temperature changes

26 27

Exercise 2 : Answer
Mtheethcaantaollys(CedH3rOeHac)tiiosnproofdcuacrebdoncommomneorxcidiaellyanbdy
hydrogen : KC  CH3OH
COH2 2

CO (g) + 2H2 (g) CH3OH (g)   0.0406 
 2 

An equilibrium mixture in a 2.00 L vessel is found  0.17  0.302 2
to contain 0.0406 mol CH3OH, 0.170 mol CO and  2  2 
0.302 mol H2 at 500 K. What is the value of Kc at
this temperature?  10.47

28 29

Exercise 3 : Answer
Equilibrium is established at 25°C in the reaction
KC  NO2 2
N2O4 

N2O4 (g) 2NO2 (g) Kc = 4.61 x 10-3 4.61 x 103  0.0236 2
N2O4 

Igfra[NmOs2]of= 0.0236 M in a 2.26 L flask, how many N2O4   0.1208 M
N2O4 are also present?

n  MV

 0.1208 molL1 2.26 L

 0.237 mol

mass N2O4  0.273 x 92
 25.12 g

30 31

196

Exercise 4 : Answer

   KP 
The equilibrium constant Kp for the reaction PNO 2 PO2
PNO 2 2

2NO2 (g) 2NO (g) + O2 (g)  158
0.27 2 PO2
is 158 at 1000 K. What is the equilibrium pressure  0.4 2
of O2 if the PNO2 = 0.400 atm and PNO = 0.270 atm?

PO2  346 .78 atm

32 33

Exercise 5 : Answer
The following equilibrium process has been
studied at 50°C. KC NH 3 2
N 2 H 2 3


N2 (g) + 3H2 (g) 2NH3 (g)

In one experiment the concentration of the reacting 6.0 x102  0.050 2
species at equilibrium are found to be [H2] = 0.250 N 2 0.253
M, [NH3] = 0.050 M. calculate the [N2] if the value
of Kc at this temperature is 6.0 x 102. N 2   2.67 x104 M

34 35

• wThheichvathlueebsaolaf nKccedorchKepmaircealdeeqpueantdiononistwherittweany. in Exercise 6 :

• Example : The value of Kc for the following reaction is
1.0 x 10-3 at 200 oC.
1. N2O4 (g) 2NO2 (g) (at 25oC)

Kc = [NO2]2 = 4.63 x 10-3 2NOBr (g) 2NO (g) + Br2 (g)
[N2O4]

2. 2NO2 (g) N2O4 (g) (at 25oC)

K’c = [N2O4] ; K’c = 1 Calculate Kc for the following reaction at the
[NO2]2 Kc = 216 same temperature.

NO (g) + ½Br2 (g) NOBr (g)

3. ½N2O4 (g) NO2 (g (at 25oC)

K’’c = [NO2] ; K’’c = 1 OR K’’c = √Kc = 0.0680
[N2O4]1/2 √ K’c

36 37

197