Introductory Chemistry B CH4751 Lecture Notes 7-10 Dr. Erzeng Xue

1
Introductory Chemistry B CH4751Lecture Notes
7-10Dr. Erzeng Xue
CH4751 Lecture Notes 7-10 (Erzeng Xue)
2
Matter - Molecule - Molecular Formulas
CH4751 Lecture Notes 7 (Erzeng Xue)

• Matter mixture (mixture of pure substances)
• pure substances single element compound (O2,
  Ar, Si)
• multi-element compounds (H2O, CaCO3)
• microscopic - molecule the smallest unit
  of a substance (usually consists of 2 or more
  atoms, e.g. O2, H2O) some single atom
  substances are called directly as xx atom, e.g.
  Ar atom, C atom, Si atom etc.)
• Written representation of molecules
• Molecular formulas - that indicates the actual
  number of atoms in a molecule,
• e.g. there are 2 hydrogen (H) atoms and 1 oxygen
  (O) atom in a water molecule H2O
• Structural formulas - that shows which atoms are
  attached to which in the molecule.
• e.g.

macroscopic
The subscript indicates the number of the same
atoms in the molecule
The short bar represents a chemical bond
water hydrogen peroxide methane H2O H2O2 CH4
Note Structure formula does not give geometry
information of the molecule, such as bond length
and angle
3
Counting the Atoms and Molecules
CH4751 Lecture Notes 7 (Erzeng Xue)
Matter - The Mole

• A molecule contains fixed number of atoms
• H2O, CO2, CaCO3, CH4, C4H10, etc.
• In a specific chemical reaction
• vAA vBB vCC vDD
• vA, vB, vC and vD are called reaction
  stoichiometry coefficients, which represent the
  ratio of number of molecules according to which
  the reaction proceeds.
• e.g. CH4 2O2 CO2 2H2O
• here, vA1, vB2, vC1 vD2, which means that
  ONE molecule of CH4 can react with TWO molecules
  of O2, producing ONE molecule of CO2 and TWO
  molecule of H2O.
• Exercise 2NO O2 2 NO2 what are the
  values of stoichiometry coefficient?
• In the microscopic scale, the composition of a
  substance or a chemical reaction are quantified
  by the number of atoms or molecules

4
The Mole
CH4751 Lecture Notes 7 (Erzeng Xue)
Matter - The Mole

• In industry/lab the quantities of matter are
  accounted by g, tons, litre, m3 etc
• e.g. we say 10m3 gas, 10 litres of water etc
• Questions
• 1) How to relate the a.m.u. of a H2O molecule
  to a daily mass scale, say, gram?
• 2) for the reaction CaCO3(s)CaOCO2
• - Can we say 1kg CaCO3 decompose to 1kg CaO
  plus 1kg CO2?
• 3) for the reactions CH42O2CO22H2O
• - Can we say 1m3 CH4 reacts with 2m3 O2 giving
  1m3 CO2 2m3 H2O?
• Why?
• - We need a quantitative conversion between
  the number of atoms/molecules and the everydays
  mass units (macroscopic).
• The conversion is done through the Mole, by
  definition
• I mole 6.023 x 1023 molecules/atoms
• Avocados number or Avocados constant

5
The Mole and Molar Mass
CH4751 Lecture Notes 7 (Erzeng Xue)
Matter - The Mole

• For any substances, the number of atoms,
  molecules (including isotopes) and ions in ONE
  mole is a fixed number
• I mole 6.023 x 1023 atoms/molecules/ions
• Conversion of mass in a.m.u. scale to that in
  gram scale
• numerically 1 a.m.u. x 6.023 x 1023 1 gram
• The mass of ONE mole of substance is called Molar
  Mass
• e.g. Mass of a carbon (C) atom12.0 a.m.u.
• number of C atoms in 1 mole6.023 x 1023
• MOLAR MASS of C 12 x 1 gram/(a.m.u.x6.023x1023)
  12 gram (per mole)
• Þ The molar mass has the same value as that of
  a.m.u for an atom/molecule, but it is in gram.
  Molar weight has the same value as that of molar
  mass on earth.

6
The Mole and Molar Mass
CH4751 Lecture Notes 7 (Erzeng Xue)
Matter - The Mole

• I mole 6.023 x 1023 molecules/atoms
• Molar Mass - The mass of one mole substance

Element mass of one atom number of atoms in one mole The mass of one mole - molar mass
H 1.008 a.m.u 6.023 x 1023 1.008 gram
He 4.009 a.m.u 6.023 x 1023 4.009 gram
Li 6.941 a.m.u 6.023 x 1023 6.941 gram
molecule / ion mass of one molecule number of molecules /ions in one mole The mass of one mole - molar mass
CO2 44 a.m.u. 6.023 x 1023 44 gram
H2O 18 a.m.u. 6.023 x 1023 18 gram
SO4 96 a.m.u. 6.023 x 1023 96 gram
7
The Mole - Calculations
CH4751 Lecture Notes 7 (Erzeng Xue)
Matter - The Mole

• Questions
• 1). Calculate the number of molecules and the
  number of oxygen atoms in 0.25 moles of calcium
  nitrate Ca(NO3)2.
• 2). What is the molar mass of Ca(NO3)2? What is
  the mass of 0.25 mole of Ca(NO3)2?
• 3). What is the number of moles of 82 gram of
  Ca(NO3)2? How many nitrogen atoms in it?
• 1) Number of Ca(NO3)2 molecules
• 0.25mol x 6.022x1023molecules/mol 1.5x1023
  molecules
• Number of oxygen atoms
• 0.25mol x 6.022x1023molecules/mol x 6 O
  atoms/molecule 9.0x1023 atoms
• 2) Molar mass
• 1x40(Ca)2x14(N)6x16(O)164 gram/mole
• The mass of 0.25 mole of Ca(NO3)2
• 0.25 mol x 164 g/mole41 gram
• 3) Number of moles
• 82 gram / 164 gram/mole0.5 moles
• Number of N atoms 0.5 mole x
  6.022x1023molecules/mol x 2 N atoms/molecule
  6.0x1023 atoms

8
The Mole - Calculations
CH4751 Lecture Notes 7 (Erzeng Xue)
Matter - The Mole

• Questions
• 1). Calculate the mass of 2 moles of H2SO4 and
  the mass of of 2 moles of SO4.
• 2). Calculate the number of moles and the number
  of oxygen in 1 tea-spoon of sugar (2 g).
• 1) The mass of 2 moles of H2SO4
• 2 mole x molar mass of H2SO4 2 x
  2x1(H)1x32(S)4x16(O)2x98196 gram
• The mass of 2 moles of SO4
• 2 mole x molar mass of SO4 2 x
  1x32(S)4x16(O)2x96192 gram
• 2) Number of moles of 2 gram C12H22O11 molecules
• 2 g/ molar mass of C12H22O11
  2/12x12(C)22x1(H)11x16(O)2/3420.005848
  mole
• Number of oxygen atoms
• 0.005848 mol x 6.022x1023molecules/mol x 11 O
  atoms/molecule3.874x1022 oxygen atoms

9
The Mole - Calculations
CH4751 Lecture Notes 7 (Erzeng Xue)
Matter - The Mole

• Questions
• 1). The gas fired central heating system at home
  involves combustion of methane with oxygen. The
  reaction occurs being CH4 2O2 2H2O CO2
  Heat . It means that for every mole of methane
  burnt, two moles of oxygen are required and this
  gives, in addition to the heat, 2 moles of H2O
  and one mole of CO2 as the result of the
  reaction. Calculate how many grams of CO2 is
  generated by heating the house for 1 hour,
  assuming the rate of CH4 consumption is 5 m3
  (STP) per hour and the density of gas is 1 kg/m3.
• 1) The mass of gas consumed in 1 hour 1 kg/m3 x
  5 m3/hour5 kg/hour
• 2) The number of moles of gas consumed in one
  hour
• 5kg x 1000g/kg / molar mass of
  CH45000/1x12(C)4x1(H)312.5 moles/hour
• 3) The number of moles of CO2 generated
• 1 mole CO2 per mole of CH4 x moles of CH4
  consumed1 x 312.5312.5 moles CO2/hour
• 4) The mass (weight) of CO2 generated in one hour
• 312.5 mole x molar mass of CO2 312.5 x
  1x12(C)2x16(O)312.5x4413750 gram

10
Concentration of a Solution - The Molarity
CH4751 Lecture Notes 8 (Erzeng Xue)
Matter - Quantities

• When a solution (solutesolvent) is concerned,
  the concentration of solute is usually defined as
  the number of moles of solute in the total volume
  of solution.
• moles of solute
• Molarity unit mol/L or referred as molar
  or M
• litres of solution
• e.g. If a sulphuric acid solution is marked as
  4M. It means that one litre of solution
  (watersulphuric acid) contains 4 moles of H2SO4.
  
• Note Molarity refers to concentration. For a
  solution of the same molarity, the larger the
  volume is the more the number of moles of solute
  it contains.
• The volume used in calculating the molarity
  includes both solute and solvent.
• Question
• 1) How many moles of H2SO4 are there in 250ml of
  4M H2SO4 solution?
• 2) Are the molarities of the following
  solutions the same or different?
• 2 moles in 4 litres solution, 0.4 moles in
  800ml solution

11
Molarity Calculations
CH4751 Lecture Notes 8 (Erzeng Xue)
Matter - Quantities

• Examples
• 1). What is the molarity of a solution made by
  dissolving 2.355 g of sulphuric acid (H2SO4) in
  water and diluting to a final volume of 50 ml?
• Molar mass of H2SO4 2x1.0(H)1x32.1(S)4x16.0(O
  )98.1 gram/mol
• number of moles in 2.335 g H2SO4 2.335(g) x
  1/98.1(g/mol)0.024 mol of H2SO4
• Molarity moles of solute/ Litres of
  solution0.024 mol / 0.05 L0.48 M.
• 2). If the above solution is added to another 50
  ml water, what is the concentration of the
  resultant solution? Will the molarity of solution
  remain the same? Will the number of moles of
  H2SO4 in the solution remains the same?
• 3). If the we take 10 ml of the solution in 1)
  and add it into another 100 ml of water, what is
  the concentration of resultant solution?
• molarity(0.024mol/50ml)x10ml/(10ml100ml)0.0
  436 M (or mol/L)

12
Concentration of a Gas
CH4751 Lecture Notes 8 (Erzeng Xue)
Matter - Quantities

• When a gas is concerned, the concentration of a
  gas, A, in a gas mixture, AB, (usually in a
  confined space) can be expressed by means of mole
  fraction
• moles of A
• Mole fraction of A unit mol/mol or
  -/-
• total moles of AB
• moles of A
• or mole percentage of A x
  100
• total moles of AB
• Note Unlike a solution, the volume of a gas is
  very sensitive to the temperature and pressure.
  Therefore, when quoting the volume of a gas, T
  P must be given.
• e.g. Air contains 78 of nitrogen (N2) and
  21 of oxygen (O2 ), which means that in every
  100 moles of air there are 78 moles of N2 and 21
  moles of O2 (There are also trace amount of Ar,
  CO2 and other gases).
• N2 mole fraction78/1000.78 N2 mole
  (78/100)x10078
• O2 mole fraction21/1000.21 O2 mole
  (21/100)x10021

13
Concentration of a Gas
CH4751 Lecture Notes 8 (Erzeng Xue)
Matter - Quantities

• The pressure and concentration of a gas
• Pressure by definition is the force exerted
  collectively by gas molecules colliding on the
  surface (in all walls of a container) of unit
  area.
• Within a same space (volume) at the same
  temperature, the more gas molecules there are,
  the more intense the collision of these molecules
  on the wall, thus a higher pressure it will be.
• For most common gases, the following relation is
  observed
• (Note that the term n/V has meaning of volume
  concentration)
• The equation is called Ideal Gas Law (or Perfect
  Gas Law)
• A gas which obeys this relation is called ideal
  gas (or perfect gas)
• It is usually more convenient to measure the
  pressure of a gas.

Where P - absolute pressure V - volume of gas
n - number of mole of gas R - gas constant T
- absolute temperature (Kelvin scale)
14
Concentration of a Gas
CH4751 Lecture Notes 8 (Erzeng Xue)
Matter - Quantities

• Concept of partial pressure
• If a gas mixture contains n1 moles of A, n2 moles
  of B, n3 moles of C,, the mixture is held in a
  container of vol. V and kept at temperature T,
  for each gas, we can write
• where PtotalP1P2P3
• ntotaln1n2n3
• Pi is called partial pressure of a component i,
  which can be regarded as a fraction of pressure
  contributed by each component to the overall
  pressure.
• The pressure/volume ratio of gases in a mixture
  is directly proportional to their mole ratio
• The pressure/volume fraction of a gas in a
  mixture is proportional to its mole fraction

pressure ratio
volume ratio
mole ratio
volume fraction
mole fraction
pressure fraction
15
Concentration of a Gas
CH4751 Lecture Notes 8 (Erzeng Xue)
Matter - Quantities

• Some important concepts about gas
• Most common gases show behaviour close to ideal
  gas, especially at low pressure and high
  temperature (when gas molecules are far apart)
• For ideal gas its volume, V, is proportional to
  the number of moles of matter, n, and is
  independent of the type of substance - which
  means that ONE mole of any gases will have the
  same volume (molar volume) at the same pressure
  temperature.
• From ideal gas law, any 3 of 4 variables
  (P-V-n-T) will determine the 4th variable.
• Some usage of partial pressure concept
• Quantifying each of gas components in a gas
  mixture
• Characterising systems that contain multi-phase
  such as gas-liquid, gas solid etc.
• Analysing systems that involve phase change
• Describing reactions that have reactants and/or
  products present in both gl phases
• Relating certain solution properties with gas
  phase pressure (e.g. solubility)
• etc.

16
Concentration of a Gas
CH4751 Lecture Notes 8 (Erzeng Xue)
Matter - Quantities

• Examples
• 1). Air contains 78 of N2 and 21 of O2
• Þ these can be considered as mole fractions as
  well as volume fractions
• (Note When quoting a percentage of a gas in a
  mixture, it is better to label the percentage
  sign with a proper abbreviation, e.g. vol. or
  wt. to avoid confusion)
• if air pressure is 1 atm., the N2 partial
  pressure PN20.78 atm
• the O2 partial pressure PO20.21 atm
• 2). A gas cylinder at 200 bar contains 5
  hydrogen (H2) in Argon (Ar). What is the mole
  fraction of each gas? What is the partial
  pressure of each gas?
• Þ When we say 5 H2 in Ar, usually we mean the
  vol. as for gas the vol. has the same value as
  that of mol, we dont normally put vol. before
  the sign. The here however, is definitely not
  wt. (wt. is rarely used to describe a gas
  mixture).
• mole fraction for H2 50.05, and for
  Ar (100-5)950.95
• partial pressure since
• for H2 0.05x20010 bar, for Ar
  0.95x200190 bar

17
Concentration of a Gas
CH4751 Lecture Notes 8 (Erzeng Xue)
Matter - Quantities

• Examples
• 1). Calculate from the ideal gas law the volume
  of 1 mole gas at 0C and 20C and 1 atm.
• at 0C
• at 20C
• 2). Reaction 2NO O2 2 NO2 occurs at 2 bar
  and 300C. Initially there are 200 moles of NO
  and 200 moles of O2. After time t, 120 moles of
  NO is converted. Calculate the partial pressure
  of each component in the gas phase at t, assuming
  the pressure remains constant.
• Ans. 2NO O2 2 NO2
• initial moles 200 200 0
• change at t -120 -60 120 total moles at t
• gas mixture at t 200-12080 200-60140 120
  ntotal80140120340
• mole fraction(ni /ntotal) 80/3400.235 140/3400.
  411 120/3400.353
• pressure fraction (Pi /Ptotal) 0.235 0.412 0.353
• since Ptotal2 bar
• the partial pressure at t, PNO0.470 PO20.824 P
  NO20.706 bar

18
Change of Matter - Chemical Reaction
CH4751 Lecture Notes 9 (Erzeng Xue)

• What is a chemical reaction?
• A chemical reaction is a process by which one or
  more substances are changed into other substances
  by rearrangement, combination or separation of
  atoms
• Process of a chemical reaction
• 1. meet 2. react 3. to leave

19
Topics To Study
CH4751 Lecture Notes 9 (Erzeng Xue)
Chemical Reactions

• What we are going to study in this module
• Quantitative representation of a chemical
  reaction
• Reaction stoichiometry / reaction equation
• Factors deciding if a chemical reaction can
  proceed in the direction specified
• Reaction thermodynamics
• If a chemical reaction can proceed, what
  conversion level can be achieved?
• Chemical equilibrium
• If a chemical reaction can proceed, how fast it
  will go?
• Reaction kinetics
• Catalysis
• Types of chemical reaction commonly encountered
• What are the objectives we want to achieve after
  study
• When you encounter a chemical reaction
• You should be able to make judgement if a
  reaction is feasible
• If reaction is feasible, you should be able
  predict the level of conversion
• You should be able to understand some basic
  concepts of reaction kinetics

20
Reaction Equation
CH4751 Lecture Notes 9 (Erzeng Xue)
Chemical Reactions

• Chemical reaction equation
• vAA vBB vCC vDD (DHT)
• It shows the substances participating in reaction
  (reactants) and the substances that are formed as
  the consequence of the reaction (products).
• It shows the direction of reaction (reactants
  on the left products on the right).
• It shows the ratio of number of molecules or
  moles of reactants as well as products, the
  ratios are given by vA, vB, vC vD which are
  reaction stoichiometry coefficients.
• stoichiometry coefficient
• e.g. vA vB vC vD
• CH4 2 O2 CO2 2 H2O 1 2 1 2
• CaCO3 CaO CO2 1 - 1 1
• C6H12O6 6 O2 6 CO2 6 H2O 1 6 6 6
• Sometime additional information, e.g. the amount
  reaction heat involved is indicated.

21
Reaction Equation
CH4751 Lecture Notes 9 (Erzeng Xue)
Chemical Reactions

• More on reaction equation
• Sometime a chemical equation may appear in
  different forms
• The sign between reactants and products
• If an equal sign is used, it indicates the
  exact number of reactant molecules required to
  react each other and the number of product
  molecules to form
• e.g 2 NOO22NO2 (2 moles of NO react with
  1 mole of O2 forming 2 moles of NO2)
• If an arrow sign is used, it usually
  indicates the substances to react and the
  substances formed without emphasising the no. of
  molecules of each substances
• e.g NOO2NO2 (NO will react with O2
  giving NO2)
• If a sign is used, it usually indicates
  that the reactants cannot convert completely to
  products because the presence of reverse reaction
  (equilibrium controlled)
• e.g 2NOO2 2NO2 (Reaction of NO with O2
  forming NO2 is equilibrium controlled)

22
Reaction Stoichiometry
CH4751 Lecture Notes 9 (Erzeng Xue)
Chemical Reactions

• More on reaction equation vAA vBB vCC
  vDD
• The reaction stoichiometry coefficients, vA, vB,
  vC vD
• These numbers are usually integral, Sometime
  fraction numbers are seen in literature.
  Generally integral number is preferred.
• When the value of a coefficient is 1, it is
  commonly omitted.
• The coefficients are determined in such way that
  the sum of each atom in all reactants should be
  equal to the sum of the same atom in all
  products.
• The process to determine these coefficients is
  called balancing reaction equation
• General methods of balancing reaction equation
• Simple reaction equation - by inspection
• When no. of coefficients - 1 no. of atom types
  Þ solve coefficient equations
• From oxidation state changes.

23
Balancing Reaction Equation
CH4751 Lecture Notes 9 (Erzeng Xue)
Chemical Reactions

• Balancing equation by inspection
• Start with most complex formula
• Þ balance by changing adjustable coefficients
• Þ leave H and O until last
• Þ check your answer by the sum of no. atoms
  on each side
• e.g. Al HCl AlCl3 H2
• AlCl3 most complex Þ
• 3 Cl on r.h.s., 1 Cl on l.h.s. Þ set 3 for HCl
• Þ Al 3 HCl AlCl3 H2
• 3 H on l.h.s., 2 H on r.h.s. Þ set 6 for HCl
  and 3 for H2 and 2 for AlCl3
• Þ Al 6 HCl 2 AlCl3 3 H2,
• 1 Al on l.h.s., 2 Al on r.h.s. Þ set 2 for
  Al,
• Þ we now have 2 Al 6 HCl 2 AlCl3 3
  H2
• Check Al 2 on l.h.s. º 2 on r.h.s.
• Cl 6 on l.h.s. º 6 on r.h.s.
• H 6 on l.h.s. º 6 on r.h.s.6. Þ Balancing
  completed.

24
Balancing Reaction Equation
CH4751 Lecture Notes 9 (Erzeng Xue)
Chemical Reactions

• Balancing equation by solving coefficient
  equations
• e.g. a C2H4 b O2 m CO2 n H2O
• l.h.s CL 2a, HL 4a, OL 2b
• r.h.s CR m, HR 2n, OR 2mn
• For CL CR Þ 2a m
• HL HR Þ 4a 2n
• OL OR Þ 2b 2mn
• There are 4 coefficients and 3 equations. Let a
  1, the rest coefficients can be solved.
• We have m 2, n 2, b 3
• C2H4 3 O2 2 CO2 2 H2O
• Balancing completed.

coefficient equations
25
Balancing Reaction Equation
CH4751 Lecture Notes 9 (Erzeng Xue)
Chemical Reactions

• Balancing equation by oxidation state changes
• a MnO4- b C2O42- c H d Mn2 e H2O f
  CO2
• Lose 1e- / C or 2e- / C2O42-
• Gain 5e- / from MnO4- to Mn2
• Lost e- by C2O42- gain e- by MnO4-
• Þ ratio C2O42- MnO4- is 5 2
• Þ 2 MnO4- 5C2O42- H Mn2 H2O CO2
• By inspection
• 2MnO4- 5C2O42- 16H 2Mn2 8H2O
  10CO2
• Check l.h.s. 2 Mn, 28 O, 10 C and 16 H
  r.h.s. 2 Mn, 28 O, 10 C and 16 H
• l.h.s. º r.h.s. balancing is complete

3
4
7
2
Coefficient equation method fails to produce the
coefficient.
reduction
oxidation
I.h.s. r.h.s. coeff. eqns
Mn a d ad
O 4a4b e2f 4a4be2f
C 2b f 2bf
H c 2e c2e
26
Types of Chemical Reactions
CH4751 Lecture Notes 10 (Erzeng Xue)
Chemical Reactions

• Based on the ways substances change in a chemical
  reaction, we can classify chemical reactions into
  5 basic types
• Combination reactions
• Decomposition reactions
• Double displacement or exchange reactions
• Neutralisation reactions
• Oxidation - Reduction - (redox) reactions

27
Combination Reactions
CH4751 Lecture Notes 10 (Erzeng Xue)
Chemical Reactions

• Two or more substances react and forms one
  product
• Combination reaction A B C
• e.g. 2 H2 O2 2 H2O
• CaO H2O Ca(OH)2
• Many elements react with each other in
  combination reactions to form compounds
• e.g. N2 3 H2 2 NH3
• A combination reaction betw. a metal a nonmetal
  produces an ionic compound
• e.g. 2 Mg O2 2 MgO

28
Decomposition Reactions
CH4751 Lecture Notes 10 (Erzeng Xue)
Chemical Reactions

• One substance undergoes a reaction producing more
  than 1 products
• Decomposition reaction A B C
• e.g. 2 H2O 2 H2 O2
• 2 NaN3 2 Na 3 N2
• A decomposition reaction used to inflate the air
  bag in cars. On impact reaction is ignited and
  produce quickly a large volume gas (10g NaN3
  gives 50 litres N2)
• Many compounds undergo decomposition when heated
• e.g. CaCO3 CaO CO2
• A commercial process to produce lime from
  limestone decomposition.

29
Double Displacement or Exchange Reactions
CH4751 Lecture Notes 10 (Erzeng Xue)
Chemical Reactions

• Exchange ionic partners between two ionic
  compounds
• Double displacement reaction A X B Y A
  Y B X
• where A, B are cations (positive ions)
• X, Y are anions (negative ions)
• e.g. 2NaOH (aq) Mg(NO3)2(aq) Mg(OH)2(s)
  2NaNO3 (aq)
• KCl (aq) AgNO3(aq) AgCl(s) KNO3(aq)
• Precipitates are often formed in this type of
  reaction.

30
Neutralisation Reactions
CH4751 Lecture Notes 10 (Erzeng Xue)
Chemical Reactions

• Reactions that involve an acid and a base produce
  a salt and water
• Neutralisation reaction HX BOH BX H2O
• where HX is an acid, usually
  containing H (hydrogen)
• BOH is a base, usually containing OH group
• e.g. HCl(aq) Mg(OH)2(s) MgCl2(aq)
  2H2O (l)
• Both an acid and base can be formed from oxides
• Acid oxides SO3 H2SO4
• Basic oxides (metal oxides) CaO
  Ca(OH)2
• Note There will be more detailed discussion
  about acid and base later in this module.

31
Oxidation-Reduction Reactions
CH4751 Lecture Notes 10 (Erzeng Xue)
Chemical Reactions

• Reactions that involve oxidation of one reactant
  and reduction of another
• Oxidation-Reduction reaction A BZ AZ
  B
• Oxidation - a chemical process by which an
  entity loses electrons.
• Reduction - a chemical process by which an
  entity gains electrons.
• (Remember as OIL RIG - Oxidation Is Loss,
  Reduction Is Gain)
• e.g. 2Mg0 O20 2Mg2O2- (2MgO) Mg (loses
  2e-s) is oxidised, O (gains 2e-s) is reduced
• H20 F20 2 HF- (2HF) H (loses 1 e-) is
  oxidised, F (gains 1e-) is reduced
• Each electron that is gained by one species must
  be lost by another, which means the oxidation and
  reduction always occurs in pair.
• This type reaction is often referred as Redox
  reaction (Reduction Oxidation)
• (Note The Redox reaction does not necessitate
  the participation of oxygen.)

32
More about Redox Reactions
CH4751 Lecture Notes 10 (Erzeng Xue)
Chemical Reactions

• Redox reaction A BZ AZ B
• In the rxn reactant A is oxidised by reactant BZ
  in the meantime BZ is reduced by A.
• The reactant that oxidises another is called
  oxidant (e.g. BZ in the above reaction), the
  oxidant itself is reduced as the result of
  oxidising the other reactant (A).
• The reactant that reduces another is called
  reductant (e.g. A in the above rxn), the
  reductant itself is oxidised as the result of
  reducing the other reactant (BZ).
• A species carries electronic charge if gained or
  lost electrons. The sign and magnitude of the
  charge is called Oxidation State or Oxidation
  Number.
• e.g. 2Mg0 O20 2Mg2O2- (2MgO) In MgO the
  O.S. of Mg is 2 O is 2-
• H20 F20 2 HF- (2HF) In HF the
  O.S. of H is 1 F is 1-
• Oxidation state of some elements in compounds
• - Usually H 1, O 2-, halogens (F, Cl, Br, I)
  1-.
• - Metal ions 1A group 1, 2A group 2,
  transition groups variable, but all metal ions
  are invariably positive.
• - Some elements have variable o.s., depending on
  the compounds they are in, e.g. Cu, Cu2, S2,
  S6, S7.

33
Examples
CH4751 Lecture Notes 10 (Erzeng Xue)
Chemical Reactions

• Classify the following reactions
• HNO3 (aq) NaOH (aq) NaNO3 (aq)
  H2O Neutralisation
• Cu(OH)2(s) CuO(s) H2O(l) Decomposition
  
• BaCl2(aq) K2SO4(aq) BaSO4(s) 2KCl(aq)
  Exchange
• 2Ca(s) O2(g) 2CaO(s) Redox / combination
• Cl2(g) 2NaBr(aq) 2NaCl(aq)
  Br2(aq) Redox
• Calculate the oxidation state of S in 1) SCl2, 2)
  SO42-, 3) S2O82- Mn in KMnO4
• 1). In a molecule, the sum of O.S. (charges) of
  all atoms in the molecule is zero
• (O.S.S) 2x(O.S.Cl)(0) Þ (O.S.S)
  2x(1-)(0) Þ (O.S.S)(2)
• 2). In an ion, the sum of O.S. of all atoms in
  the ions equal to overall charge
• (O.S.S) 4x(O.S.O)(2-) Þ (O.S.S)
  4x(2-)(2-) Þ (O.S.S)(6)
• 3) 2x(O.S.S) 8x(O.S.O)(2-) Þ 2x(O.S.S)
  8x(2-)(2-) Þ (O.S.S)(7)
• 4) (O.S.K)(O.S.Mn)4x(O.S.O)(0) Þ
  (1)(O.S.Mn)4x(2-)(0) Þ (O.S.Mn)(7)