Download - Presentation 1 - MATTER
-
8/6/2019 Presentation 1 - MATTER
1/131
-
8/6/2019 Presentation 1 - MATTER
2/131
THE STRUCTURE OF ATOM
j 3 fundamental particles of matter are the proton, neutron and
electron j Proton number and nucleon number
j a symbol below is used to designate a particular atom of anelement where X is the symbol of the element, A is the nucleonnumber (or mass number) and Z is the proton number (or atomic number)
j the nucleon number is the sum of the number of protons andneutrons in the nucleus of an atom. j the proton number is the number of protons present in thenucleus of an atom.
j in neutral atom, the number of protons is equal to its number of electrons.
X A
Z
-
8/6/2019 Presentation 1 - MATTER
3/131
j isotopes are atoms of the same element with the same proton
number but different nucleon numbers. The atoms have differentnumber of neutrons.
j for example: j hydrogen has 3 isotopes
I SOTOPES
Hyd rogen,
H
Deuterium,
D
Tritium,
T
1 proton 1 proton 1 proton
1 electron 1 electron 1 electron
0 neutron 1 neutron 2 neutrons
1
1
2
1
3
1
-
8/6/2019 Presentation 1 - MATTER
4/131
j isotopes of an element have the same :
j proton number
j number of electrons in a neutral atom j electronic configuration j chemical properties
j isotopes of an element have different: j number of neutrons in the nucleus
j mass
j density
j rate of diffusion
-
8/6/2019 Presentation 1 - MATTER
5/131
j Relative Isotopic Mass
j the relative isotopic mass of an isotope is the ratio of themass of one atom of the isotope the one-twelfth of the mass of an atom of the mass of an atom of the 12C isotope
j Since the relative masses of the proton ( 1 .0074 a.m.u) andthe neutron ( 1 .0089 a.m.u) are both very close to one another
and the electron has negligible mass, it follows that all relativeisotopes masses will be very close to whole numbers. j the relative isotopic mass of an isotope can be assumed to beidentical to its nucleon number.
RELAT IV E MASSES
Relative isotopic massMass of one atom of isotope
1/12 x mass of one atom 12 C=
-
8/6/2019 Presentation 1 - MATTER
6/131
j Relative Atomic Mass j the relative atomic mass (Ar) of an element isthe average mass of all the isotopes of theelement weight to take into account the neutral
relative abundances of the isotopes, comparedto one-twelfth of the mass of an atom of the12 C isotope.
Relative atomic massWeighted average mass of the atom
1/12 x mass of one atom 12 C=
-
8/6/2019 Presentation 1 - MATTER
7/131
j Relative Molecular Mass (Mr) and Relative FormulaMass
j the relative molecular mass (Mr) of a compound isthe ratio of the mass of one molecular of thecompound to one-twelfth of the mass of an atom of the 12C isotope.
j for ionic compounds, which do not consists of molecules, the term relative formula mass is usedinstead.
Relativemolecular mass
Mass of one molecule of the compound
1/12 x mass of one atom 12 C isotope=
-
8/6/2019 Presentation 1 - MATTER
8/131
-
8/6/2019 Presentation 1 - MATTER
9/131
MASS SPE C TROMETRYU The mass spectrometer
U the mass spectrometer is an instrument used:
U to determine the relative isotopic masses andabundances of isotopes.
U to determine the relative molecular masses andstructure of organic compounds.
U the mass spectrometer works on the principle thatwhen ions or charged particles pass through amagnetic field, they will be separated according totheir mass ( m) ratio
Charge ( e)
-
8/6/2019 Presentation 1 - MATTER
10/131
-
8/6/2019 Presentation 1 - MATTER
11/131
U the sample to be studied undergoes five separatestages once it has been injected into the mass
spectrometer:U Vaporisation = the sample is vaporisedU Ionisation = positive ions are produced from
the sampleU Acceleration = the positive ions produced areaccelerated by an electric fieldU Deflection = the positive ions are deflected bya magnetic fieldU Detection = the positive ions are detected anda record is made
-
8/6/2019 Presentation 1 - MATTER
12/131
U VaporisationU the sample must be in gaseous form, liquidsand solids are heated to vapourised them.
U for example:X(s) X(g)X(l) X(g)
U the vapour formed is then injected directly into the ionisation chamber
-
8/6/2019 Presentation 1 - MATTER
13/131
U I onisationU the ionisation chamber, the sample isbombarded with high-energy electronsproduced by the electron gun to causeionisation.U electrons from the sample are knocked out by the high-energy electrons and hencepositive ions are producedX(g) + e - X+ + 2e -
-
8/6/2019 Presentation 1 - MATTER
14/131
-
8/6/2019 Presentation 1 - MATTER
15/131
U D eflectionU the accelerated ions are passed through a magnetic field where they are deflected.U the amount of deflection depends on their
mass ratiocharge
U D etection
U ions that reach the detector produce a current.U this current is amplified and a record, the
mass spectrum of the sample, is obtained
-
8/6/2019 Presentation 1 - MATTER
16/131
SAMPLE GR APH P RODUC E DURINGDETE C T ION
-
8/6/2019 Presentation 1 - MATTER
17/131
MASS SPE C T RU M QU EST ION PATTE RN
T he mass spectrum shows that the element consists of three isotopes with relative masses m1, m2 and m3 inthe ration of b : a : c , the relative atomic mass (Ar) or the element can be calculated from the formula below:
m1 m2 m3
c
b
a
Ar = (m 1 x b) + (m 2 x a) + (m3 x c)
a + b + c
-
8/6/2019 Presentation 1 - MATTER
18/131
The mole concept and Avogadro C onstant
U a mole is the amount of substance that contains asmany particles (atoms, molecules or ions as specified
by the formula) as there are carbon atoms is 12 kg of
carbon-12.U the number of particles in 1 mol of any substance isa constant known as the Avogadro's constant (L) =
6.02 x 10 23 mol -1U the mass of 1 mol of any substance is the same asthe relative atomic mass, or relative molecular/formula
mass of that substance expressed in gram
-
8/6/2019 Presentation 1 - MATTER
19/131
U EX AMPL E : (1.9 PAG E 17)
U Determine the number of atoms in each of the following substances
(a) 1 mol of magnesium
(b) mol of oxygen gas
(c) 3 mol of carbon dioxide
-
8/6/2019 Presentation 1 - MATTER
20/131
U ANSW ER : (1.9 PAG E 17)(a)1 mol of Mg contains 6.02 x 10 23 atoms
(b) mol of oxygen gas = x 6.02 x 10 23
molecules of O 2= 2 x (1/2 x 6.02 x 10 23) atoms oxygen
= 6.02 x 1023
atoms(c) TRY Y OU R SE LF
-
8/6/2019 Presentation 1 - MATTER
21/131
Moles of gases
UIn reactions involving gases, the volume of the gasesthat take part in the reaction is usually more importantthan the mass of the gases involved.
UT he relationship between the amount of gas (number of moles of gas) and the volume of gas is given inAvogadros law.
UAvogadros law = under the same conditions of temperature and pressure, equal volume of gasescontain equal number of moles
Volume number of moles
-
8/6/2019 Presentation 1 - MATTER
22/131
UE xample:
Under room conditions, 10 cm3
of CO 2 will containthe same number of moles or molecules as 10 cm 3
of N 2
UAt standard temperature and pressure (s.t.p; 273 K and 101 kPa), 1 mol of all gases will occupy a volumeof 22.4 dm 3. this volume is known as the molar
volume (V m).UUnder room conditions, 1 mol of all gases occupies24.0 dm 3
-
8/6/2019 Presentation 1 - MATTER
23/131
U when stating the volume of gas, thetemperature and pressure at which the volumewas measured must be stated.
U the conditions of standard temperature and pressure (s.t.p) are temperature of 0 oC (273
K) and 1.0 atmospheric pressure (1.01 x 105
Nm -2)
U at s.t.p one mole of gas occupies 22.4 dm 3
Reacting Volumes of Gases StandardTemperature and Pressure
-
8/6/2019 Presentation 1 - MATTER
24/131
U standard conditions should not be confusedwith standard temperature and pressure (s.t.p)
U under standard conditions, gases are at a
pressure of 1 atm ( 1 .0 1 x 1 05
Nm-2
), solutionsare at unit concentration ( 1 .0 mol dm -3) and thetemperature must be specified. If the temperatureis not stated, then it is assumed to be 2 98 K(2 5oC )
Standard conditions
-
8/6/2019 Presentation 1 - MATTER
25/131
UQUE ST IONS:Calculate the volume occupied by thefollowing gases at room conditions
(a) 40 g of NH 3(b) 1.5 g of CO 2
(c) 24 x 10 23 molecules of O 2
-
8/6/2019 Presentation 1 - MATTER
26/131
UANSW ER S:
(a) 40 g NH 3 = 40 mol of NH 317
Volume = 40/17 x 24 dm 3
Volume = 56.47 dm 3
for questions (b) and (c) try yourself
Answers:(b) 0.818 dm 3
(c) 96.0 dm 3
-
8/6/2019 Presentation 1 - MATTER
27/131
UEXER CIS E S:
1. Calculate the mass of methane gas, CH 4, whichwill occupy the same volume as 8.0 g of nitrogengas under the same conditions
2. T he equation for the complete combustion of methane is
CH 4(g) + 2O 2(g) CO 2 (g) + 2H 2O(g)
What volume of oxygen gas is required tocompletely burn 60.0 cm 3 of methane under thesame conditions?
Page 18 pre-u text physical chemistry
-
8/6/2019 Presentation 1 - MATTER
28/131
U the concentration of a solution is usuallyexpressed as the mass of solute per 1.0 dm 3 of solution (g dm -3) or expressed as the number of
moles of solute in 1 dm3
of solution (mol dm-3
)UT he concentration in the unit of mol dm -3 isalso known as the molarity of the solution
(symbol: M)
Moles An d Solutions
-
8/6/2019 Presentation 1 - MATTER
29/131
T he number of moles of solute present in a givenvolume of a solution (of molarity M) is
Number of moles of a solute = molarity x volume (indm 3)
Moles An d Solutions
Molarity, M (mol dm -3) = C oncentration (g dm -3)
Molar mass of the solute (g mol-1
)
-
8/6/2019 Presentation 1 - MATTER
30/131
(1) Calculate the molarity of a solution containing7.3 g HCl in 1 dm 3 of solution.
(2) Calculate the concentration (in g dm -3) in asolution of 0.24 M sodium nitrate.
(3) Calculate the molarity of sodium hydroxidewhich contains 23.0 g NaOH in 250 cm 3 of solution
(4) Calculate the number of moles of copper sulphate in 400 cm 3 of 0.50 M CuSO4 solution
(5) Calculate the number of ions in 250.0 cm 3 of
0.3 M aqueous aluminium chloride
-
8/6/2019 Presentation 1 - MATTER
31/131
U the concentration of a solution is given byMoles An d Solutions
c = n /VWHERE :c = concentration (mol d m -3)n = amount of solute (mol)V = volume ( d m 3)
U the following expressions are commonly used:
Number of moles = MV/1 000WHERE :M = molarit y (mol d m -3)V = volume (cm 3)
-
8/6/2019 Presentation 1 - MATTER
32/131
-
8/6/2019 Presentation 1 - MATTER
33/131
ANSWERS:
(a)(i) OH-
(aq) + H+
(aq) H2
O(l)
MOH - VOH - = MH+ VH+MOH - x 2 5.0 = 0. 12 x 2 3. 1 0
MOH - = 0. 11 mol dm -3
-
8/6/2019 Presentation 1 - MATTER
34/131
ANSWERS:
(a) (ii) total number of moles of OH - in 500 cm 3 of C l
MV/1 000 = 0. 11 X 500 = 0.055 mol1 000
Number of moles of OH - from NaOH present in C l
MV/1 000 = 0. 1 0 X 500 = 0.050 mol1 000
Number of moles = total number of - number of moles of
of OH-
from A(OH) 2 moles of OH-
OH-
from NaOH
= 0.055 0.050= 0.005 mol
-
8/6/2019 Presentation 1 - MATTER
35/131
ANSWERS:(a) (iii) A(O H)2(aq) A
2+(aq) + 2O H-(aq)
2 mol of O H- is obtaine d from 1 mol of A(O H)2.
So 0.005 mol of O H- is obtaine d from:
0.005 x 1 = 0.0025 mol of A(O H)2
2
Number of moles = m/M0.0025 = 0.429/M
Molar mass of A(O H)2 = 0.429/0.0025
= 171.6 g mol -1
Molar mass of A(O H)2 = A r of A + 2(16 + 1)Ar of A = 171.6 - 2(17)
Ar of A 138
(b) A is Barium, Ba
-
8/6/2019 Presentation 1 - MATTER
36/131
Empirical Formula an d
Molecular FormulaU the empirical formula of the compoundshows the simples whole number ratio for
the atoms of all the different elementpresent in one molecule of the compound.
U the molecular formula of a compoundshows the actual number of atoms of different elements in one molecule of thecompound.
-
8/6/2019 Presentation 1 - MATTER
37/131
Compoun d Molecular formula
Empiricalformula
E thane C2H4 CH2Phosphorus (V) oxi d e P 4O 10 P 2O 5
Hyd rogen peroxi d e H2O 2 HO
E thanoic aci d C2H4O 2 CH2O
-
8/6/2019 Presentation 1 - MATTER
38/131
Example:U a hydrocarbon has the followingcomposition by mass: C = 92 .3%; H = 7.6%
(a) calculate its empirical formula
(b) given that the M r for the hydrocarbonis 78, determine its molecular formula
Question from the pre-u text stpm longman page 21
-
8/6/2019 Presentation 1 - MATTER
39/131
answers:
The empirical formula is C H
(b) Let the molecular formula be (C H)n
Mr of compoun d = (12 + 1)n = 13n13n = 78
n = 78/13 = 6
molecular formula is C 6H6
E lements present carbon h yd rogen
Mass per 100 g of compoun d 92.3 g 7.6 gNumber of moles 92.3/12 = 7.7 7.6/1 = 7.6
Simplest mole ratio 7.7/7.6 1 7.6/7.6 = 1
-
8/6/2019 Presentation 1 - MATTER
40/131
EXER C ISES: A compund Y has the following compositionby mass:
Na, 2 9. 1 % ; S, 40.5% ; O, 30.4%
C alculate the empirical formula of Y
Question from the pre-u text stpm longman page 21
-
8/6/2019 Presentation 1 - MATTER
41/131
-
8/6/2019 Presentation 1 - MATTER
42/131
-
8/6/2019 Presentation 1 - MATTER
43/131
Answer:
Mass of oxygen combined with1
0.8 g of Mg= 1 8.0 g 1 0.8 g = 7. 2 gC omposition by mass of oxide is 1 0.8 g Mg
: 7. 2 g OMole ratio of Mg : O = 1 0.8 /2 4 : 7. 2/1 6
= 0.45 : 0.45= 1 :1
Empirical formula is MgOQuestion from the pre-u text stpm longman page 22
-
8/6/2019 Presentation 1 - MATTER
44/131
Exercises:1
.1
00 cm3
of gaseous hydrocarbonrequires 450 cm 3 of oxygen for completecombustion to give 300 cm 3 of carbon
dioxide. All volumes are measuredunder the same conditions. C alculatethe molecular formula of the
hydrocarbon.
Question from the pre-u text stpm longman page 22
-
8/6/2019 Presentation 1 - MATTER
45/131
G as Law
-
8/6/2019 Presentation 1 - MATTER
46/131
-
8/6/2019 Presentation 1 - MATTER
47/131
* The Liquid State* Forces holding the particles together are weaker than
those holding the particles together in the solid state* Packed closely together in clusters* Not in an orderly arrangement
* Particles can vibrate, rotate and move freely throughout the liquid* Not easily compressed*
Have no fixed shape (take the shape of the container)but have no fixed volume* The particles in liquids have more energy compared toparticles in solid but have less energy that those in gases.
-
8/6/2019 Presentation 1 - MATTER
48/131
* The Gaseous State* Composed of atoms or molecules that are separated
from each other by distances far greater than their ownsize* Can be considered as point particles, that is they possess mass but have no volume* No forces between the gas particles* The particles can vibrate, rotate and move anywhere
within the container where the gas is placed* Easily compressed* Gas are in constant random motion, moving in straight lines unless they collide with the walls of container or
with other gas particles
-
8/6/2019 Presentation 1 - MATTER
49/131
* W hen the particles collide with the walls of thecontainer, they exert a pressure on the container
* Collision are perfectly elastic, no loss of kinetic energy during collisions* A verage kinetic energy of the particles is directly
proportional to the absolute temperature of the gas(Kelvin scale)
-
8/6/2019 Presentation 1 - MATTER
50/131
Bo yles Law: The relationship of the volume of a gas
* B oyles law states that the pressure of a fixed mass of gas at a constant temperature is inversely proportional to its volume.* Mathematically, the law may be expressed as:
P 1
V (At the constant mass and temperature)
* Putting it in equation:
P =k
VWhere k is the proportionality constant
* From this relationship, we can deduce that:
PV=
constant or PV=
k(At the constant mass and temperature)
P 1V1 = P 2V2
-
8/6/2019 Presentation 1 - MATTER
51/131
-
8/6/2019 Presentation 1 - MATTER
52/131
* according to B oyles Law, when the pressure acting onthe gas is increased, its volume will decrease. A s
pressure is increased to infinity, 1/p approaches zeroand the volume of the gas is reduced to zero.
* A gas which obeys B oyles Law perfectly under any condition is known as an ideal gas or a perfect gas, a gas
which exists only in theory *
* However, all real gases (eg: Nitrogen, carbon dioxide,chlorine, oxygen, ect) do not obey B oyles Law perfectly, especially under high pressure and low temperature.
-
8/6/2019 Presentation 1 - MATTER
53/131
* This is because at high pressure, the gas particlesare very close to one another, and the
intermolecular attractive forces are strong enough tocause the gas to condense into liquid.
* For a gas to obey B oyles Law perfectly under any condition, the gas particles must have zero volume(so that at infinite pressure, the volume occupied by the gas will become zero), and there must not be any
intermolecular forces between the gas particles(sothat the gas does not condense due to any intermolecular attractive force)
-
8/6/2019 Presentation 1 - MATTER
54/131
EXE RCI SES:
-
8/6/2019 Presentation 1 - MATTER
55/131
EXE RCI SES:A 1.00 L sample of gas at atmospheric pressure is compressed to 0.700 L
at constant temperature. Calculate the final pressure of the gas
P2V2 = P 1V1P2 = P 1V1
V2P2 = (1 atm)(1.00 L)
0.700 LP2 = 1.43 atm.
-
8/6/2019 Presentation 1 - MATTER
56/131
EXE RCI SES:A sample of oxygen gas occupied a volume of 42.5 cm 3 at 25 oC and 1
atm pressure. What will be its volume when the pressure is increasedto 1.8 atm, at constant temperature?
P1V1 = P 2V21.0 x 42.5 = 1.8 x V 2V2 = 23.6 cm 3
-
8/6/2019 Presentation 1 - MATTER
57/131
EXE RCI SES:250 cm 3 of chlorine gas at 200 kPa is compressed, at constant
temperature, until its pressure is 400 kPa. What is the final volumeoccupied by the sample of chlorine gas?
P1V1 = P2V2200 x 250 = 400 x V 2V2 = 125 cm 3
-
8/6/2019 Presentation 1 - MATTER
58/131
EXE RCI SES:9.97 dm 3 of air at 119.0 kPa is allowed to expand, at constant
temperature, to 12.0 dm 3. calculate the final pressure.
P1V1 = P2V2119.0 X 9.97 = P 2 x 12.0P2 = 98.9 kPa
1 atm = 760 torr = 101 kPa = 101000 Pa = 101000 Nm -2
-
8/6/2019 Presentation 1 - MATTER
59/131
1 atm = 760 torr = 101 kPa = 101000 Pa = 101000 Nm 2
1 oC = 273 K = 33.8 oF1 cm3 = 1 x 10-6 m31 dm3 = 1 x 10-3 m31 dm3 = 1000 cm3
-
8/6/2019 Presentation 1 - MATTER
60/131
V
T
Charles law: the effect of temperature on the volume of a gas
* Charles law state that at constant pressure, the volume of a fixed massof gas is directly proportional to its absolute temperature* Mathematically expressed as:
V T(At the constant mass and pressure)
* or it can be express in equation as:
= k
V 1T 1
V 2T 2
=
-
8/6/2019 Presentation 1 - MATTER
61/131
* The volume-temperature relationship can be shown using the graphbelow:
-
8/6/2019 Presentation 1 - MATTER
62/131
EXER CIS E S:
-
8/6/2019 Presentation 1 - MATTER
63/131
EXER CIS E S:
A sample of gas occupied a volume of 10.0 L at 50 oC. Assuming that the pressure remains constant, what temperature in oC is needed to reducethe volume to half?
V1 = V2
T 1 T 210.0 L = 5.0 L(273 + 50) T 2T 2 = 5.0 X 323
10
-
8/6/2019 Presentation 1 - MATTER
64/131
-
8/6/2019 Presentation 1 - MATTER
65/131
P RE SSU RE LAW
-
8/6/2019 Presentation 1 - MATTER
66/131
P RE SSU RE LAW
* Pressure law states that the pressure of a fixed mass of gas at constant volume varies directly with its absolute temperature.
* This expressed mathematically as
P T* or it can be express in equation as:
PT
= k
P1T 1
P2T 2
=
* EX A MPLE
-
8/6/2019 Presentation 1 - MATTER
67/131
* EX A MPLE:* A fixed volume of gas at 30 oC has a pressure of 101 kPa. I f thepressure is increase to 113 kPa, calculate its temperature.
SOLUT I ON:
P1 = P 2T 1 T 2
101 = 113
(273 + 30) (273 + T 2)
T 2 = 339 K
AVO G ADR OS LAW
-
8/6/2019 Presentation 1 - MATTER
68/131
AVO G ADR OS LAW
* A vogadros law states that all gases at the same temperature, pressureand volume contain the same number of particles.* This expressed mathematically as
V n (At the constant pressure and temperature)Where n = number of moles
* the molar volume of a gas is the volume occupied by 1 mole of gas at standard temperature and pressure (s.t.p). This volume is 22.4 dm 3 or22400 cm 3* under room conditions, the volume is usually stated as 24.0 dm 3
Th i d l g l
-
8/6/2019 Presentation 1 - MATTER
69/131
The i d eal gas law
* Using B oyles law and Charles law, it is possible to derive an equationrelating all three functions of volume, temperature and pressure as
followsP 1 V1 P 2 V2
T1 T2=
PVT
* This means that for a given mass of gas, = a constant
* For 1 mole of a gas, this becomes = R where R is the gas
constant. A nd for n moles of gas,
PV
T
P V n RT=
* W h
-
8/6/2019 Presentation 1 - MATTER
70/131
W here:
* R has a value of 8.314 J K -1 mol -1 in S I units if:
*
the pressure (P) is measured in pascals (Pa)* the volume (V) is measured in cubic metres (m 3)
* the temperature (T) is measured in Kelvin's (K)
*
R has a value of 0.082 dm3
atm k -1
if:* the pressure (P) is measured in atmospheres (atm) which is 1atm = 101 kilopascals (kPa), while 101 x 10 3 Pa = 760 torr or760 mm Hg
* the volume (V) is measured in dm 3
* the temperature (T) is measured in Kelvin's (K)
* and n in moles (mol)
DALTONS LAW OF PA RTI AL P RE SSU RE
-
8/6/2019 Presentation 1 - MATTER
71/131
DALTONS LAW OF PA RTI AL P RE SSU RE
* D altons law state that the total pressure of a mixture of gases equalsthe sum of the partial pressures of the gas present in the mixture.
* P TOT A L = P A + P B + P C +
* W here P TOT A L = Total pressure of the gas mixture P A , P B and P Crespectively
* W hile the partial pressure of gas is the pressure that a gas in a mixture would exert if it alone occupied the whole volume of the gasmixture at the same temperature.
The partialpressure (P A) of
gas A
=
Number of moles
of A
Total number of moles of gases in
the mixture
x P TOTAL
EXAMPL E
-
8/6/2019 Presentation 1 - MATTER
72/131
EXAMPL E
* A container of volume 2 dm3 contains 0.4 mol of oxygen and
1.2 mol of carbon dioxide under a total pressure of 100 kPa.Calculate the partial pressure of each gas in the mixture.
* A nswer
Total number of moles of gas = 0.4 + 1.2 = 2.6 mol
Mole fraction of oxygen = 0.4/1.6 = 0.25
Mole fraction of CO2 = 1.2/1.6 = 0.75
Partial pressure of oxygen = 0.25 x 100 kPa = 25 kPa
Partial pressure of CO2 = 0.75 x 100 kPa = 75 kPa
Brain test
-
8/6/2019 Presentation 1 - MATTER
73/131
Brain test
-
8/6/2019 Presentation 1 - MATTER
74/131
TRY THIS QU ES TIONS
-
8/6/2019 Presentation 1 - MATTER
75/131
TRY THIS QU ES TIONS
* 2.5 dm 3 container contains 0.1 mol nitrogen and 0.25 mol
oxygen at 25 oC. Calculate the partial pressure of each gas andthe total pressure of the mixture.
* A nswer
Using pV = nRTFor nitrogen:p X (2.5 X 10 -3) = 0.1 x 8.31 x(25 + 273)
p = 9.91 x 10 4 Pa
For oxygen:p X (2.5 X 10 -3) = 0.25 x 8.31 x(25 + 273)p = 2.48 x 10 5 Pa
Total pressure of the mixture = (9.91 x 10 4) + (2.48 x 10 5)Pa =3.47 x 10 5 Pa
-
8/6/2019 Presentation 1 - MATTER
76/131
TRY THIS QU ES TIONS
-
8/6/2019 Presentation 1 - MATTER
77/131
TRY THIS QU ES TIONS* 1.0 dm 3 of gas A at a pressure of 505 kPa and 2.5 dm 3 of gasB at a pressure of 232.3 kPa were forced into a container of 0.7dm 3 capacity. Calculate the total pressure in the 0.7 dm 3
container. The temperature remain constant throughout.* A nswer
For gas A The volume was reduced from 1.0 dm 3 to 0.7 dm 3 underconstant pressure.
Using B
oyles Law: p 1 V 1 = p 2 v 2505 x 1 = p 2 x 0.7p2 = 721.4 kPa
TRY THIS QU ES TIONS
-
8/6/2019 Presentation 1 - MATTER
78/131
TRY THIS QU ES TIONS
* A nswer
For gas BThe volume was reduced from 2.5 dm 3 to 0.7 dm 3
Using B oyles Law 232.2 x 2.5 = P B x 0.7P B = 829.3 kPa
The total pressure = 721.4 + 829.3The total pressure = 1550. 7 kPa
DETER MINING RE LATIVE MOL ECULA R MASS
-
8/6/2019 Presentation 1 - MATTER
79/131
* The ideal gas equation can be used to determine the relativemolecular mass of a substance.
* PV = nRT and n = m/M r* W here n = number of moles, m = mass of the sample, M r =relative molecular mass
* Thus, after making the modification, the equations below was produced:
P V RT= mMr
P V
mRT=Mr
Explaining d eviations from i d ealit y
-
8/6/2019 Presentation 1 - MATTER
80/131
Explaining d eviations from i d ealit y
* I deal gas is the gas molecules that have no volume, and thereare no intermolecular forces between the molecules.
* However for real gases, the molecules have a volume andthere are intermolecular forces between them
* A t high pressures, the volume occupied by the gas is smalland the volume of the gas molecules cannot be ignored
* The molecules are pushed so close together that a repulsiveforce operate between them making the gas less compressible
* A t low temperatures, the kinetic energy of the molecules islow. The intermolecular forces between the moleculesbecome more apparent.
E l i i d i i f i d li
-
8/6/2019 Presentation 1 - MATTER
81/131
Explaining d eviations from i d ealit y
* W hile real gases show a marked deviation from ideal
behavior under high pressure and low temperature* A ll real gases behave almost ideally under very low pressureand very high temperature.
*
Low pressure = the volume occupied by the gas is very large, the volume of the gas molecules by comparison isextremely small and can be ignored
* High temperatures = the kinetic energy of the moleculesis so high that the intermolecular forces operating betweenthem can be ignored
Explaining d eviations from i d ealit y
-
8/6/2019 Presentation 1 - MATTER
82/131
Explaining d eviations from i d ealit y
* A t room conditions (25 oC and 1 atm), the deviation of realgases from ideal behavior is about 3%
* Small non-polar molecules such as hydrogen and heliumshow the least deviation, while big polar molecules such asammonia and sulphur dioxide show a marked deviation from
ideal behavior.
K inetic theor y of liqui d s
-
8/6/2019 Presentation 1 - MATTER
83/131
y q
* Liquid are intermediate in character between solids and gases.
* Properties of solids, liquid and gases:
Soli d Liqui d G as
Particles arranged fixedand very close
Particles arrangedrandom and close
Particles arrangedrandom and far apart
Forces very strong Forces strong Forces very weak
Space between particlescan be negligible
Space between particlesis very small
Space between particlesis very big
-
8/6/2019 Presentation 1 - MATTER
84/131
Evaporation an d con d ensation process
-
8/6/2019 Presentation 1 - MATTER
85/131
* The particles in a liquid are in constant motion, however, theseparticles do not possess the same kinetic energy. A s a result of random collisions, some particles at the surfaces holding them
within the liquid. The process in which liquid particles withsufficient energy escape from the liquid surface and enter the gasphase is called evaporation.
* Evaporation occurs at the surface of a liquid at any temperaturebelow the boiling point of the liquid. Since the particles whichescape from the liquid during evaporation are the ones with thehighest kinetic energies, the average kinetic energy of its remaining particles will fall. Consequently, the temperature of the liquid willfall as the liquid evaporates
* The evaporation process occurs continuously.
p p
Evaporation process in the close d container
-
8/6/2019 Presentation 1 - MATTER
86/131
Vaporisation condensation
Vapour
* W hen, the evaporation occurs, the particles in the vapour state willcollide with the walls of the container.
* Some of the particles while bouncing within the enclosed space, willhit the liquid surface and re enter the liquid phase. The change of statein which a vapour or a gas is changed into a liquid is calledcondensation.
Boiling of liqui d
-
8/6/2019 Presentation 1 - MATTER
87/131
* I f a liquid is heated in an open container, its saturated vapour pressure increases.
* W hen the saturated vapour pressure becomes equal toatmospheric pressure, bubbles of vapour will form in theliquid and escape into the atmosphere because the vapourpressure is high enough to overcome the external pressure.* The saturated vapour pressure of water is equal to theexternal pressure (1 atm) at 100 oC. Hence water boils at 100oC* Heat must be supplied continuously to sustain boiling. W hen a liquid boils, the heat supplied is used to break theintermolecular forces between the particles in the liquid andnot used to increase their kinetic energy. The temperature
will not change as long as there is liquid left
-
8/6/2019 Presentation 1 - MATTER
88/131
FREE ZING
-
8/6/2019 Presentation 1 - MATTER
89/131
* W hen the temperature of a liquid is lowered, the kinetic
energy of the particles decreases* A point is reached when the intermolecular forces are strong enough to hold the particles together in a fixed, orderly arrangement.* The freezing point is the temperature where a solid is inequilibrium with its liquid under a pressure of 1 atm
Soli d : lattice structure
-
8/6/2019 Presentation 1 - MATTER
90/131
* most solid are crystalline. I n crystalline solids, theparticles are arranged in an orderly manner.* I n solid, the particles are held rigidly together by strong attractive forces in a three dimensional structurecalled the lattice structure.* Solids are more dense than liquids* I n amorphous solids, the particles are not arranged inan orderly manner* The basic repeating unit of a crystalline solid is called a unit cell
-
8/6/2019 Presentation 1 - MATTER
91/131
Unit cell
CRYS TAL S TR UC TURE
-
8/6/2019 Presentation 1 - MATTER
92/131
* Crystal lattice regular arrangement of particles solid.* The unit cell is the smallest block of particles which whenrepeated throughout the crystal reproduce the crystalstructure* There are 7 types of unit cells:
*
Cubic* Hexagonal* Tetragonal* Orthorhombic* Rhombohedral* Monoclinic* Triclinic
Cubic
-
8/6/2019 Presentation 1 - MATTER
93/131
* a = b = c
* xo = y o = zo = 90 o* Example : Sodium chloride, barium flouride,potassium iodide, iron, copper
Hexagonal
-
8/6/2019 Presentation 1 - MATTER
94/131
* a = b c* xo = y o = 90 o, zo = 120 o* Example : scandium, aluminium chloride,iron ( II ) sulphide, phosphorus (V) oxide
Tetragonal
-
8/6/2019 Presentation 1 - MATTER
95/131
et ago a
* a = b c* xo = y o = zo = 90 o,* Example : tin, calcium chromate (V I ), barium
peroxide
-
8/6/2019 Presentation 1 - MATTER
96/131
-
8/6/2019 Presentation 1 - MATTER
97/131
Monoclinic
-
8/6/2019 Presentation 1 - MATTER
98/131
* a b c* xo = zo = 90 o y o 90 o* Example : sulphur, magnesium chloride,calcium iodide (V), copper ( II ) fluoride
Triclinic
-
8/6/2019 Presentation 1 - MATTER
99/131
* a b c* xo z o y o 90 o* Example : copper ( II ) sulphate (V I ),
pentahydrate, bismuth nitrate, calcium carbide
Cubic LA TTICE S YS TE M
-
8/6/2019 Presentation 1 - MATTER
100/131
* There are 3 types of cubic lattices* Simple cubic* B ody-centred cell* Face-centred cell
Cubic LA TTICE S YS TE M* l b
-
8/6/2019 Presentation 1 - MATTER
101/131
* Simple cubic* I n a simple cubic cell, the particles occupy the eight corners
of the cube.* Example of solid with simple cubic is potassium chloride
Simple cubic
Cubic LA TTICE S YS TE M* B d C d C bi
-
8/6/2019 Presentation 1 - MATTER
102/131
* B ody-Centred Cubic* I n the body-centred cubic, the particles occupy the eight
corners of the cube as well as the centre of the cube.* Example of body-centred cubic are sodium and caesiumchloride.
Body-centered cubic
Cubic LA TTICE S YS TE M* F C d C bi
-
8/6/2019 Presentation 1 - MATTER
103/131
* Face-Centred Cubic* I n the face-centred cubic cell, the particles occupy the
centre of each of the six faces in addition to the eight corners of the cube* Example of the face-centered cubic structure are sodiumchloride and iodine
Face-centered cubic
-
8/6/2019 Presentation 1 - MATTER
104/131
-
8/6/2019 Presentation 1 - MATTER
105/131
allotrop y
-
8/6/2019 Presentation 1 - MATTER
106/131
* I s the existence of different forms of the same element in thesame physical state* A llotropes have different physical and chemical properties* The allotropes of carbon:
* There are three allotropes of carbon
* Graphite* D iamond* Fullerene
graphite Each carbon atom issp 2 hybridised, bond
-
8/6/2019 Presentation 1 - MATTER
107/131
Uses:Lubricants andas electrodes
inelectrochemica
l andelectrolytic
cells
Each carbonatom is
covalentlybonded tothree other
carbon atoms
to form flatsheets of interconnected hexagons;weak non-directional
van der Waals forces
attract thesheets or layers to
each other
angle is 12 0o
The unhybridised p electronsform a delocalised cloud of electrons enabling graphite
to conduct electricity andgiving graphite its shiny
appearance
The weak van
der Waalsforces between
the layersenable them toslide over each
other.Therefore,
graphite is usedas a lubricant
Density : 2 .3 gcm -3
-
8/6/2019 Presentation 1 - MATTER
108/131
fullerene
-
8/6/2019 Presentation 1 - MATTER
109/131
Also known as buckminsterfullerene with the formula C 60
Other forms of fullerenes with the formulae C 70 , C 76 , C 82and C 90 are also known
The C 60 fullerene, commonly known as Buckyball isspherical and consists of 60 carbon atoms
It is made up 3 2 faces that is 12 pentagonal faces and 2 0hexagonal faces
Each carbon atom in the fullerene is sp 2 hybridised
Fullerene are used to make superconductors, lubricants,micro-ball bearings in nanomachines such as micro-motorsand also as abrasives
The allotropes of sulphur
Th 2 ll f l h
-
8/6/2019 Presentation 1 - MATTER
110/131
There are 2 allotropes of sulphur
Rhombic sulphur ( -sulphur )
Monoclinic sulphur ( -sulphur )
Rhombic sulphur
monoclinic sulphur
THE C HARAC TERISTI C OF THE 2 ALLOTR
-
8/6/2019 Presentation 1 - MATTER
111/131
Formula S 8 Formula S 8
Yellow transparent crystal
Needle shapedOctahedral shape
Yellow translucent crystalDensity = 2 .06 g cm -3 Density = 1 .96 g cm -3
Melting point = 11 3oC Melting point = 11 9oCStability : stable at temperature
< 95.6 oC at 1 atmStability : stable at temperature
> 95.6 oC at 1 atmSoluble in carbon disulphide,
CS 2
Insoluble in carbon disulphide,C
S 2
Phase diagrams
-
8/6/2019 Presentation 1 - MATTER
112/131
A phase diagram summarises the relationship between the
solid, liquid and gaseous states of a given substance as a functionof the pressure and temperature on a single graph
E very substance has its own individual phase diagram. T he phase diagram is a graph plotted from experimentally obtained
results
Phase d iagram of water
Represents the vapour pressure of water below its freezing
-
8/6/2019 Presentation 1 - MATTER
113/131
Represents the sublimation vapour pressure-temperature curve for iceIt gives the conditions of pressureand temperature at which ice andwater vapour are equilibrium
Represents the vapour pressure-temperature curve for water Along this curve, liquid water and water
vapour are equilibriumIt shows that as the pressure (at which
water is held) is increased, its boilingpoint also increases
This line is the melting temperature curve. It showsthe effect of pressure on the melting point of ice
The curve slopes to the left with increasing
pressure showing that as the pressure is increased,the melting point of ice decreases slightly.
This is unusual because an increase in pressureusually favour the formation of solid. However, thisbehaviour is due to ice having an open structure. As
the pressure is exerted is increased, the hydrogenbond between the H 2 O molecules in ice are brokendown, changing the ice into a denser liquid phasewhich occupies a smaller volume
Represents the vapour pressure of water below its freezingpoint
At any point on this curve, water exists in a metastable
condition since water dies not normally exist as a liquid belowits freezing pointThe phenomenon represented by this curve is termedsupercooling
This is the melting point of iceof the freezing point of water at
This is boiling point of water at
-
8/6/2019 Presentation 1 - MATTER
114/131
This is known as the triple point where
all the phases, ice (solid), water (liquid)and water vapour (gas) are equilibrium.This occurs at 0.0 1 oC and 0.006 atm.Under these conditions, the vapour pressure of ice and water are the same
of the freezing point of water at1 atm At this point (0 oC and 1 atm),water and ice are in equilibriumwith each other
This is called critical point of
water The critical temperature (T c) is
32 7oC and critical pressure (P c) is22 0 atm
Beyond the critical point, theliquid form water cannot be
distinguished from its vapour form, that is there is no meniscusseparating liquid water from water vapour. Also, beyond this pointwater vapour cannot be changedinto liquid water no matter how
much pressure is applied
This is boiling point of water at1 atm. The temperature is1 00 oC
At this point ( 1 00oC and 1 atm),both water and water vapour
are in equilibrium with eachother
Example 1
-
8/6/2019 Presentation 1 - MATTER
115/131
ANSW ER
-
8/6/2019 Presentation 1 - MATTER
116/131
Phase diagram of carbon dioxide
-
8/6/2019 Presentation 1 - MATTER
117/131
x
T he solid-liquid equilibrium line, OC slope to the right, as is typicalf b
-
8/6/2019 Presentation 1 - MATTER
118/131
of most substances.
T his indicates that the melting point of carbon dioxide increase in pressure. T his is because solid carbon dioxide is denser than liquidcarbon dioxide since the molecules in the solid state are more closely
packed.
According to Le Chateliers Principle, increasing pressure will shift
the position of equilibrium towards the left hand side. Hence, it is moredifficult to melt solid carbon dioxide at high pressure.
T he triple point occurs at 5.1 atm that is above 1.0 atm.
T herefore, at all pressures below 5.1 atm, no liquid carbon dioxide canexist.
-
8/6/2019 Presentation 1 - MATTER
119/131
Its also indicates that the solid carbon dioxide will not melt butsublime at room condition
If solid carbon dioxide is warmed at 1.0 atm, it sublimes at -78 oC, passing directly from the solid phase into the gaseous phase withoutthrough the liquid phase.
This phenomenon is used in industry to freeze foods like ice cream.
Solid carbon dioxide or dry ice is used because it will not melt andwill not wet the food
T o obtain liquid carbon dioxide, a pressure greater that 5.1 atm would
have to be obtain
Uses of solid carbon dioxide (dry ice)
-
8/6/2019 Presentation 1 - MATTER
120/131
solid carbon dioxide is used as a refrigerantfor foodstuffs such as ice cream since it does notmelt on warming
It is also used in cloud seeding toencourage the formation of ice crystals inclouds. As the dry ice sublimes, it absorbs heatin the clouds, thereby lowering the temperature
and causing the water vapour to condense andform water. C loud seeding is carried out toinduce rainfall especially after long periodwithout rain
Effect of non-volatile solute on water vapour pre
-
8/6/2019 Presentation 1 - MATTER
121/131
Lowering of vapour pressureT he vapour pressure exerted by a pure liquid is dependant
only on the temperature and not the amount of liquid presentas long as the liquid is in equilibrium with water
When a non volatile solute is added to a pure liquid, thevapour pressure exerted by the solution is less than the vapour
pressure exerted by the pure liquid at the same temperature.
-
8/6/2019 Presentation 1 - MATTER
122/131
-
8/6/2019 Presentation 1 - MATTER
123/131
Effect of non-volatile solute on water vapour pre
-
8/6/2019 Presentation 1 - MATTER
124/131
C olligative properties
is a properties which depends on the number of dissolve particles and not on the nature of the solute
particles in the solution.
T he 4 common colligative properties are:
E levation of the boiling point of a solution
Depression of the freezing point of a solution
Lowering of the vapour pressure of a solution
Lowering of the osmotic pressure of a solution
-
8/6/2019 Presentation 1 - MATTER
125/131
when a non volatile solute (for example, sodium chloride orglucose) is dissolved in a solvent the vapour pressure of the
-
8/6/2019 Presentation 1 - MATTER
126/131
glucose) is dissolved in a solvent, the vapour pressure of thesolvent is lowered. This is shown by line Q in the previous
figure.As more solute is added, the solution becomes more
concentrated and the vapour pressure is lowered evenfurther (line R)
This happen because solute molecules at the surface of thesolution hinders the escape of the solvent molecules.However, the return of the solvent molecules from the vapourphase to the liquid phase is unaffected.
The result is a reduction in the vapour pressure of thesolvent. This can be se seen in the previous figure where Frand Fq represent the freezing points of the dilute andconcentrated solutions and Br and Bq represent the boilingpoints of the dilute and concentrated solution
Depression of freezing point and theDepression of freezing point and thecryoscopy constant Kcryoscopy constant K
-
8/6/2019 Presentation 1 - MATTER
127/131
cryoscopy constant, Kcryoscopy constant, K f f When a solute is dissolved in a solvent, it will cause a decrease
in the freezing point of the solvent
T he equation relating this decrease in the freezing point to theconcentration of the solution is given by:
-
8/6/2019 Presentation 1 - MATTER
128/131
Elevation of boiling point and the ebullioscopyElevation of boiling point and the ebullioscopyconstant Kconstant K
-
8/6/2019 Presentation 1 - MATTER
129/131
constant Kconstant K bbthe elevation in boiling point caused by a solute dissolving in a
solvent is given by:
Example
-
8/6/2019 Presentation 1 - MATTER
130/131
-
8/6/2019 Presentation 1 - MATTER
131/131
Finish Daa