Chemistry Form 6 Sem 1 04

CHEMISTRY FORM 6 PHYSICAL CHEMISTRY CHAPTER 4 : STATE OF MATTER SOLID, LIQUID AND GAS

2.1 THE KINETIC THEORY OF MATTER
The Gaseous State
A gas is composed of atoms or molecules that are separated from each









other by distances far greater than their own size. The gas particles can be considered as ‘point’ particles – they possess mass but have negligible volume There are no forces between the gas particles Gas particles can vibrate, rotate and move anywhere within the container where the gas is placed. A gas has no fixed shape or volume and can be easily compressed Particles of gas are in constant random motion, moving in straight line unless they collide with the wall of container or with other gas particles As the particles collide with the wall, they exert a pressure on container The collision are perfectly elastic. There’s no loss of kinetic energy during collision The average kinetic energy of the particles is directly proportional to the absolute temperature (Kelvin scale) of the gas

2.2 The Gas Laws 2.2.1 Boyle’s Law
Stated by Robert Boyle (1662) during his investigation of relationship between volume occupied by gas to the pressure of gas using the apparatus below.
The total pressure on the gas is the sum of the atmospheric pressure and the pressure exerted due to the difference in the height (h) of the mercury in the column


A plot of the result yields the following graph


Based on the results, he formulated Boyle’s Law which state that

“The volume occupied by a fixed mass of gas at constant temperature is inversely proportional to its pressure

P α 1/V

P x V = constant P1 x V1 = P2 x V2


Ideal gas obeys Boyle’s Law under all conditions of temperature and












pressure. Ideal laws does not exist as real gases obeys Boyle’s Law closely only at low pressures and high temperatures. Real gases do not obey Boyle’s Law closely at high pressures and low temperatures. Pressure = force per unit area, SI unit of pressure = Newton (N) per square metre 1 Nm-2 is called a pascal (Pa) 1 kPa = 1000 Pa or 1000 N m-2 Atmospheric pressure is expressed in units of milimetres of mercury (mm Hg). Standard atmospheric pressure is 760 mm Hg or 101325 Pa or 101325 N m-2. The SI Unit of volume is cubic metre, m3. 1 m3 = 103 dm3 = 106 cm3 ; 1dm3 = 1 litre The deviation of gases from Boyle’s Law is called non-ideal behaviour

(a)

(b) V

1/P against V 1/P

V against 1/P

1/P

V (c)

(d)

PV against P

V against P V

PV

P

P

2.2.2 CHARLES’ LAW
JAC Charles investigated the relationship between the volume occupied by a gas, at constant pressure with temperature which is describe by the apparatus setting below.

As T increases V increases


He found that the relationship is a linear, as shown in the graph

VαT V = constant x T V1/T1 = V2/T2

Temperature must Kelvin (absolute scale) T (K) = t (0C) + 273


When the temperature of a sample of gas is increased, the

kinetic energy of the gas particles will increase, thus increasing, the rate of collision between the gas particles and the wall of container. At the same time, the collision are more energetic and this lead to an increase in the pressure exerted by gas. Hence, in order to maintain the original pressure, the volume occupied by the gas must now increase.

Avogadro’s Law V α number of moles (n) V = constant x n

Constant temperature Constant pressure

V1/n1 = V2/n2

5.3

2.2.3 Avogadro’s Law
~ state that equal volumes of all gases at the same temperature and pressure contain equal number of atoms/molecules.
The volume occupied by 1 mole of any gas (molar volume) depends on the temperature and pressure of the gas. The molar volume of gas at standard temperature & pressure is 22.4 dm3.
The condition of s.t.p are :Temperature : 0 oC (273 K) Pressure : 101 kPa (1 atm)
Under room temperature and pressure, molar volume of gas is 24.4 dm3 / mol.
The condition of r.t.p are :Temperature : 25 oC (298 K) Pressure : 101 kPa (1 atm)

Ideal Gas Equation Boyle’s law: V α 1 (at constant n and T) P Charles’ law: V α T (at constant n and P) Avogadro’s law: V α n (at constant P and T) Vα

nT P

V = constant x

nT P

=R

nT P

R is the gas constant

PV = nRT

2.2.4 Ideal Gas Equation Pressure, P

Volume, V

Gas constant, R

Temperature, T

Pa or N m-2

m3

8.31 J K-1 mol-1

K

atm

dm3

0.0821 dm3 atm K-1 mol-1

K


For a fixed mass of gas
P V / T = constant

Density (d) Calculations d=

PM

m V

=

RT

P1 V1 P2 V2 = T1 T2 m is the mass of the gas in g M is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance dRT

M=

d is the density of the gas in g/L P

1. Calculate the volume of the following gases at s.t.p. (at 0oC and 1 atm = 101 kPa = 760 mm Hg) a) 200 cm3 hydrogen at 30°C and 2.00 x 104 Pa P1V1 / T1 = P2V2 / T2 (2.00 x 104)(200) / (273+30) = (101 x 103)(V2) / (0 + 273) V2 = 35.7 cm3

b) 4.25 dm3 hydrogen chloride at 50°C and 650 mm Hg P1V1 / T1 = P2V2 / T2 (650 mmHg)(4.25) / (273+50) = (760 mmHg)(V2) / (0 + 273) V2 = 3.07 dm3

c) 600 cm3 oxygen at 308 K and 2.20 atm P1V1 / T1 = P2V2 / T2 (2.20)(600) / (308) = (1.00)(V2) / (0 + 273) V2 = 1170 cm3

d) 1.70 dm3 neon at 95°C and 1.45 atm P1V1 / T1 = P2V2 / T2 (1.45)(1.70) / (273+95) = (1.00)(V2) / (0 + 273) V2 = 1.83 dm3

2.

An aerosol can containing helium gas at 30°C and 1.8 x 103 Pa is heated to 60°C. What is the pressure of helium in the can now? P1 / T1 = P2 / T2 (1.8 x 103) / (273+30) = P / (60 + 273) P2 = 2.0 x 103 Pa

3.

Determine the density of SO2 gas at 25°C and 101 kPa. PV = nRT @ P = mRT / MRV => P = dRT / MR d = (101 x 103) (32 + 2(16) / 8.31 (25 +273) d = 2610 g / m3 @ 2.61 x 10-3 g / cm3

4.

4. A 20.0 m3 steel tank was constructed to store liquefied natural gas (LNG) which contained mainly methane at 160°C and 101 kPa (a) How many grams of methane can be stored in the container if the density of the liquid is 416 kg m-3? Since density = 416 kg m-3 ; Mass = 20.0 x 416 = 8320 kg @ 8.32 x 106 g

(b) Calculate the volume of a storage tank capable of holding the same mass LNG as a gas at 28°C and 101 kPa. Since the temperature of gas decrease V1 / T1 = V2 / T2 20.0 / 160 + 273 = V2 / 28 + 273 V2 = 13.9 m3

5. An organic compound consists of 24.24% carbon, 4.04% hydrogen and 71.72% chlorine by mass. If a 1.803 g sample occupies 546 cm3 at 84.2°C and 745 mm Hg (Given 760 mm Hg = 101 kPa) (a) calculate its molar mass Pressure in kPa = 745 mm Hg x 101 kPa / 760 mm Hg Pressure in kPa = 99.0 kPa PV = nRT (99.0 x 103)(546 x 10-6) = (1.803 / MR) 8.31 (84.2 + 273) MR = 99.0

(b) determine its molecular formula Element

C

H

Cl

Mass

24.24

4.04

71.72

Mol

24.24 12 = 2.02

4.04 1 = 4.04

71.72 35.5 = 2.02

Ratio

2.02 / 2.02 =1

4.04 / 2.02 =2

2.02 / 2.02 =1

Empirical formula = CH2Cl (CH2Cl)n = 99.0 [(12(1) + 2(1) + 35.5(1)]n = 99 n=2 Molecular formula = (CH2Cl)2 = C2H4Cl2

2.3 Dalton’s Law Partial Pressure
Partial Pressure – In a mixture of gases which do not interact with one another, each gas in the mixture will exert its own pressure independent of the other gases.
Dalton’s Law of Partial Pressure –
In a mixture of gases which do not interact with one another, the total

pressure of the mixture is the sum of the partial pressure of the constituent gases

PT = Pa + Pb + Pc + ……..
The partial pressure of a gas is also given by the following expression:

Pa = mole fraction of A in mixture (Xa) X Total pressure Mole fraction of A (Xa) = moles of A Total moles

Dalton’s Law of Partial Pressures

V and T are constant

P1

P2

Ptotal = P1 + P2

Consider a case in which two gases, A and B, are in a container of volume V.

nART PA = V

nA is the number of moles of A

nBRT PB = V

nB is the number of moles of B

PT = PA + PB PA = XA PT

nA XA = nA + nB PB = XB PT Pi = Xi PT

nB XB = nA + nB

A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?

Pi = Xi PT

PT = 1.37 atm

0.116 Xpropane = 8.24 + 0.421 + 0.116

= 0.0132

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

Example 2 A mixture of gases at 200 kPa contains the following composition of gases by volume. (20% CO2 ; 50% CO ; 30% O2) a) What is the partial pressure of each gas in the mixture? b) If CO2 is removed from the vessel, what would be the partial pressure of O2 and CO For CO2 PCO2 = xCO2 . Ptot PCO2 = (20) / (20 + 50 + 30) x 200 kPa PCO2 = 40 kPa For CO PCO = xCO . Ptot PCO = (50) / (20 + 50 + 30) x 200 kPa PCO = 100 kPa For O2 PO2 = xCO . Ptot PO2 = (30) / (20 + 50 + 30) x 200 kPa PO2 = 60 kPa Partial pressure of O2 and CO will remain as 60 kPa and 100 kPa respectively, since removing CO2 will not affect the partial pressure of each gas in the vessel

Example 3 : 5.00 dm3 of H2 at 200 kPa, 12.0 dm3 of N2 at 300 kPa and 1.50 dm3 of Cl2 at 120 kPa were forced into a vessel of capacity 10.0 dm3 at constant temperature. Calculate the pressure in the vessel as a result of the mixture of these gases.

For hydrogen P1V1 = P2V2 (200 kPa)(5.00) = P2(10.00) P2 = 100 kPa

For nitrogen P1V1 = P2V2 (300 kPa)(12.00) = P2(10.00) P2 = 360 kPa

For chlorine P1V1 = P2V2 (120 kPa)(1.50) = P2(10.00) P2 = 18 kPa Total pressure = PH2 + PN2 + PCl2 = 100 kPa + 360 kPa + 18 kPa = 478 kPa


Nitrogen monoxide and oxygen gas are mixed in the vessel as in the diagram below

Calculate its partial pressure of each gas and total pressure if the stopper is lifted up from the vessel.

For NO P1V1 = P2V2 (0.5 atm)(4) = P2(4 + 2)) P2 = 0.333 atm

For O2 P1V1 = P2V2 (1.0 atm)(2) = P2(4 + 2) P2 = 0.333 atm

Total pressure = PNO + PO2 = 0.333 + 0.333 = 0.666 atm

2.5 THE DISTRIBUTION OF MOLECULAR SPEEDS IN A GAS
Speed of various gas molecules differ widely and are constantly changing. This is due to their frequent collision and the resultant changes in energy.
At any time, the speed instantly increase and spread widely. This spread of molecular speeds is called Maxwell-Boltzmann distribution.
There are 2 factors which can only affect the distribution of the graph 1.Temperature.
At constant pressure, the Maxwell Boltzmann distribution graph can be varies with temperature
The graph below shows the distribution of the graph at 3 different temperature

Distribution of molecular speeds at three temperatures.

THE DISTRIBUTION OF MOLECULAR SPEEDS IN A GAS
Speed of various gas molecules differ widely and are constantly changing. This is due to their frequent collision and the resultant changes in energy.
At any time, the speed instantly increase and spread widely. This spread of molecular speeds is called Maxwell-Boltzmann distribution.
Important points to note about the distribution curves for molecular speed in a gas.
Area directly proportional to the total number of molecules in the gas
At any temp., a small proportion have high/low speed – mostly average
T ↑, distribution curve shift to right – peak become lower – number of gas molecules with high speed increase ; average speed decrease.
When heat is supplied, the energy is used to increase the average kinetic energy and to increase the motion of the gas molecules.
M-B apply not only to molecular speeds but also to molecular energy.

2.

Molecular mass
At constant temperature and pressure, the molecular gas distribution can also be influenced by the mass of gas itself.
Distribution graph below shows the distribution of different gases molecules with different molar mass


Important points to note about the distribution curves for

molecular speed in a gas.
Area directly proportional to the total number of molecules in the gas
At constant temperature, a small proportion of molecules have high/low

speed – but mostly average
Smaller the molecular mass of the gaseous molecules, distribution curve shift to right & at the same time, peak become lower. This indicated lesser molecules travel at average speed while more molecules travel at high speed.
This is due to lighter molecules can move faster than a heavier molecule. More light molecules are able to travel at high speed, causing lesser molecules to travel at average speed (represent by lowered peak)

4.5 Real gases – Deviations from ideality
Real gases area gases that do not obey the ideal gas equation, PV = nRT low pressure high temperature
Real gases behave ideally at _____________ & ________________ liquefied
At high pressure and low temperature, gases would ______________
The deviation of real gases from ideal gas can be explained using PV/RT against P. The graph below shows how 3 real gases behave
At very low pressure, the gas molecules are far apart and attraction can be ignored forces between the molecules _____________________________
However, when pressure increased, the attraction forces become closer significant __________ since the molecules are now __________. Thus, when negative deviation, these attractions make the gas molecules are at _________ lower more compressible. Volume of the gas _________ more than < nRT and the gas exhibit _______________ expected. Thus, PV …… negative deviation deviation from ideality.

Graph of PV / RT against P

PV RT

1

NH3 H2 CH4

P


At higher pressure, the molecules are pushes very close together, and

now repulsive forces operate between them. These repulsive forces make the gas harder to compress. As a result, a higher pressure has to > nRT. Moreover, as high be used to compress the gas. Thus, PV …. pressure, volume of the gas become smaller and hence the actual volume of the gas molecules in comparison with the volume of the positive containing vessel cannot be ignored. Thus, the gas exhibit _________ deviation from ideality.
The same curve could also be given at different temperature. Example, for carbon dioxide gas at 100 K, 200 K and 300 K

Graph of PV / RT against P

PV RT

100K 700 K

400 K

1

P


At low temperatures, when a molecule is about to collide with the wall of

container, the intermolecular forces of attraction between molecules will the force exerted by the impact. As a result, the pressure exerted by the gas is < reduced. This will cause the value of PV …… RT (negative deviation)
At very high temperatures, the kinetic energy of the gas molecules is very high and there are no chances for molecules to interact with each other causing no intermolecular forces occur.
Negative deviation caused by polar bond hydrogen and …………… helium
Light gases such as ……………. have small molecular mass and are non-polar. Hence, they possessed very weak intermolecular forces, and behave almost to ideal (especially at room condition)
Gas such as ammonia, carbon monoxide and nitrogen dioxide are polar molecules so they have strong intermolecular forces. These gases will show deviation from ideal behavior

3.1

Properties of Liquid


Average content of kinetic energy.
Molecules arrangements are loose and a little apart.
Intermolecular forces among particles are intermediate.
Particles in liquid state can move around freely, rotate and vibrate freely.
Has fixed mass and volume.
Shape follow its container, and must be filled from beneath of the container.

Molar latent heat of fusion, ∆Hfusion :- amount of energies required to change 1 mole of solid to liquid under standard condition. Molar latent heat of vaporization, ∆Hvaporization :- amount of energies required to change 1 mole of liquid to gas under standard condition

3.1.1 : Vapour Pressure
Particles in liquid are in constant motion. Due to the presence of intermolecular forces, the particles can only move throughout the ‘body’ of liquid and not free from attraction by other liquid particles.
When some of these particles at surface of liquid have enough energy to overcome intermolecular forces, it ‘escape’ as vapour. This process is called evaporation.
In open container, evaporation on will continue until no liquid remain.
In a close container, vapour will trap in the container and collide with the walls of container to form vapour pressure.
Vapour pressure will increase with time. When the rate of vaporation = rate of condensation it will attain a maximum value of pressure ~ called as saturated vapour pressure.
Saturated vapour pressure – vapour pressure exerted in a close container, where rate of evaporation is at equilibrium with rate of condensation.


Saturated vapour pressure is directly proportional to the temperature. Higher the

temperature more liquid can escape as vapour and the collision of vapour with container will be more vigorous. P

T

3.1.2 : Boiling
Boiling point of liquid is the temperature where saturated vapour pressure is equal to the external pressure. VAPOUR PRESSURE/ ATM ETHER ETHANOL WATER 1

35

78

100

temperature


For example, the boiling point of water is 100 0C at 1 atm, while under the same

pressure, boiling point of ether and ethanol are 35 and 78oC respectively.
Bubbling at the surface of water is the sign of that the saturated vapour pressure is

equal to external pressure


Constant heat is supply during boiling process, so that molecules in water









receive enough energy to overcome the attraction forces among the liquid molecules. Volatility of liquid refers to the liquid with high saturated vapour pressure at any given temperature. Higher the vapour pressure, higher the volatility. Among the 3 liquids, volatility increased from Intermolecular forces holding the particles in the liquid state are weak, so the tendency of liquid to ‘escape’ as vapour, is high. More volatile a liquid, lower the boiling point, faster for it to achieve saturated vapour pressure at low temperature. Furthermore, when the pressure is lowered, the attraction forces between the particles become weaker. Hence, liquid are easier to vapourise and decrease its boiling point

3.2

LATTICE STRUCTURE & CRYSTAL SYSTEM


2 types of solids – a) crystalline and b) amorphous
In crystalline structure, atoms, molecules or ions are closely packed and have an

ordered 3-D arrangement.
Regular arrangement of atoms, molecules & ion is called as crystal lattice (space

lattice).
Lattice - pattern of point repeat
Unit cell ~ small repeating unit that made up crystal.
Crystal system (primitive unit cells) is a method of classifying crystalline substance

on the basis of their unit cell.
These are the 7 possible shapes of the unit cell

lattice point

Unit Cell

Unit cells in 3 dimensions

Shared by 8 unit cells

Shared by 2 unit cells


From the lattice formed, the number of particles involved can be calculated. In

each position, it only contributes a fraction of its volume or mass to its unit cell

Shared by 8 unit cells

Shared by 2 unit cells

11.4

1 atom/unit cell

2 atoms/unit cell

4 atoms/unit cell

(8 x 1/8 = 1)

(8 x 1/8 + 1 = 2)

(8 x 1/8 + 6 x 1/2 = 4)

EXAMPLE 1 : EXAMPLE 2 : CALCULATE THE NUMBER OF DIAGRAM SHOWS A UNIT CELL OF AN PARTICLES IN THE UNIT CELL OF THE OXIDE OF ELEMENT X. WHAT IS THE a) FACE-CENTRED CUBIC LATTICE FORMULA OF THE X OXIDE? b) BODY CENTRED CUBIC LATTICE a) FCC = 8 corner + 6 faces = 8 x 1/8 + 6 x ½ = 1+3 = 4 b) BCC = 8 corner + 1 body center = 8 x 1/8 + 1 2

X = 1 body + 2 face = 1 (1) + 2 (1/2) = 2 O = 8 corner = 8 (1/8) =1 Formula = X2O O = 12 middle edge = 12 (1/4) = 3 Rh = 8 corner = 8 (1/8) =1 b) Formula = RhO3

11.4

* Other than the regular 3 unit cell, there are other structure that can be found, such as a hexagonal close-pack (h.c.p.) and cubic close pack (c.c.p.).


3.3





Allotropes When an element exists in two or more physical forms in the same state, they exhibit allotropy Usually, allotropes have the different types of crystalline structure, due to different arrangement of atoms Examples of elements which exhibit allotropies are sulphur and carbon Allotropes can be changed under certain temperature and pressure. For example, sulphur allotropes change its crystalline structure at 95.6oC at 1 atm. Tm = 95.6oC

Rhombic Sulphur

Monoclinic Sulphur


This temperature is known as transition temperature, Tm
Both sulphur allotropes have 8 atoms, S8, which look like a crown shape

ALLOTROP ES

ALPHA-SULPHUR (RHOMBIC)

BETA-SULPHUR (MONOCLINIC)

DIAGRAM

SHAPE

orthorhombic

monoclinic

COLOUR

Light yellow

Deep yellow

DENSITY

Greater

Lower

MELTING POINT

Lower

Higher

ALLOTRO PES

DIAMOND

GRAPHITE

DIAGRAM

- Each carbon is surrounded by 4 other carbon atoms in tetrahedral shape, building a strong giant covalent network among C–C. Because of this, diamond is EXPLANA extremely hard and strong. TION - Since all 4 valence electron in carbon are strongly bonded, so there is no delocalized electron around the structure, making diamond an insulator

- Each carbon is surrounded by 3 other carbon atoms in a planar, building a sheet like honeycomb hexagon structure. Each layer of graphite is hold by mere weak Van Der Waals forces. - Since there are only 3 electrons used for bonding, so there’s 1 unbonded ‘delocalised’ electron which is locate among/within the layer

ALLOTRO PES

DIAMOND

GRAPHITE

DIAGRAM

TYPE OF HYBRIDIS ATION ELECTRIC AL CONDUC TIVITY DENSITY APPLICAT ION / USES

sp3

sp2

insulator

conductor

higher

lower

Diamond cutter, jewelry

Lubricant, electrode


Other allotrope of carbon is fullerene, which is with the formula C60. It is also known

as buckminsterfullerene, where it consists of 12 pentagonal faces and 20 hexagonal faces (similar to a soccer ball shape).

Each carbon atom is _____________ hybridized.
It is used to make superconductor, lubricant, micro-ball bearing in nano machine and also abrasives

Phase Diagram of Water

Water is a Unique Substance

Maximum Density 40C Density of Water

Ice is less dense than water

11.3

11.8

Uses of Solid Carbon Dioxide (Dry Ice) • Solid carbon dioxide is used as a refrigerant for foodstuffs such as ice cream since it does not melt on warming. • it is also used in stage performance to give a smoky effect. This is due to the effect of condensation of water vapour around the atmosphere by the cold air due to sublimation of carbon dioxide. • It is also used as fire extinguisher to put off fire. Fire extinguisher is stored in liquid phase under high pressure. Once the gauge is released, the pressure is reduced and CO2 gas is given off. • It is also used in ‘cloud seeding’ to encourage the formation of ice crystals in clouds. As the dry ice sublimes, it absorbs heat in the clouds, thereby lowering the Temperature and causing the water vapour to condense and form water. Cloud seeding is carried out to induce rainfall especially after long periods without rain

4.10 Colligative properties - effect of a non volatile solute on the vapour pressure of a solvent • Colligative properties are those properties of solutions that depend on the number of dissolved particles in solution, but not on the identities of the solutes. • For example, the freezing point of salt water is lower than that of pure water, due to the presence of the salt dissolved in the water. • To a good approximation, it does not matter whether the salt dissolved in water is sodium chloride or potassium nitrate; if the molar amounts of solute are the same and the number of ions are the same, the freezing points will be the same • The four commonly studied colligative properties are freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure. In this Chapter, we shall study the first two while at the end of the Chapter we shall discuss the next two.

4.10.1 The effect of a solute on the Vapour Pressure of a solvent
The addition of a non-volatile solute to a pure liquid caused the vapour pressure of liquid Y to decrease. This occurs because the presence of solute molecules decrease the number of molecules of solvent that can reach the surface of the liquid. As a result, the number of solvent that can 'escape' as vapour decrease

Freezing point depression Boiling point elevation ↓ ∆Tf is defined as the freezing point ↑ ∆Tb is defined as the boiling point of the pure solvent (Tof) minus the of the solution (Tb) minus the freezing point of the solution(Tf) boiling point of the pure solvent (T°b) ↓ the freezing point is lower compared to the pure solvent. ↑ the boiling point is higher compared to the pure solvent ↓ Freezing involves a transition from the disordered state to the ordered ↑ Boiling involved a transition from state. For this to happen, energy liquid to gaseous state. During the must be removed from the system. process occur, energy is absorbed to Because a solution has greater the system. Due to the presence of disorder than the solvent, more solute in solvent (which cause energy needs to be removed from it interference of other intermolecular to create order than in the case of a forces) , more energy is required to pure solvent. Therefore, the solution vaporise the solvent, hence increase has a lower freezing point than its the boiling point of the solvent. solvent. Note that when a solution freezes, the solid that separates is the pure solvent component