Revision lecture 1.pptx

8/10/2019 Revision lecture 1.pptx

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Revision lecture Introduction and stoichiometry

Thermochemistry

Electrochemistry

corrosion


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Final Exam Paper Format There are seven (7) questions in the paper,

Question 1is compulsory(revise all topics)(40 marks).

Answer an additional four questions from questions 2 - 7

(15 marks each).

Q2: Introduction and stoichiometry

Q3: Thermochemistry

Q4: Electrochemistry and batteries

Q5: Water Chemistry

Q6: Corrosion

Q7: Acid rain, greenhouse effect and ozone layer

depletion (water and air chemistry)


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Atomic structure 3 subatomic particles

Atomic and mass numbers

Isotopes


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Periodic Table Periods

Groups

Halogens

Valency of each group

Metals and non-metals

Names and symbols of first 20 elements

Electron configuration Na(2,8,1) Ca(2,8,8,2)

Ionisation energy

Electron affinity

electronegativity


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Ionic and covalent

compounds Formation Properties-physical and chemical

Chemical formulas-oxides (based on groups)


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Moles,Mr and EF+MF m=mol x Mr

Calculation of EF

Balancing equations

Limiting reagents


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The percent composition by mass is the percent bymass of each element in a compound.

Percent composition of an element=n x molar mass of element x 100%

molar mass of compound

Example:

Cinnamic alcohol is used mainly in perfumery, particularly in soaps and

cosmetics. Its molecular formula is C9H10O. Calculate the percent composition

by mass of C, H and O in cinnamic alcohol. (Source: R. Chang (2007): Chemistry,

MaGraw-Hill, New York pg 108).

M =134.17 % C= (12.01 x 9)/134.17 x 100=80.56; %H=7.5 ; % O=11.93

Percent Composition ofcompounds


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Peroxyacylnitrate (PAN) is one of the componentsof smog. It is a compound of C, H, N and O.Determine the percent composition of oxygen andthe empirical formula from the following percent

composition by mass: 19.8 percent C, 2.50 percentH, 11.6 percent N. What is its molecular formulagiven that its molar mass is about 121g/mol?(Source: R. Chang (2007): Chemistry, MaGraw-Hill,

New York pg 108). The molar masses of C, H, N andO are 12.01, 1.008, 14.01 and 16.00 g/molrespectively.

E.F Example


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Mass ratio C: H: N: O=19.8: 2.5: 11.6: 66.1

Mole ratio C: H:N:O= 19.8: 2.5: 11.6: 66.1

12.01 1.008 14.01 16.00

=1.65:2.48:0.83:4.13Simplify by dividing each ration by the smallest ratio:

We get, 2:3:1:5

Hence, empirical formula is C2H3NO5

Steps to solution


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Empirical formula mass =121

Actual molar mass: empirical formula mass

=121:121=1:1

Therefore the empirical formula goes into themolecular formula once. Hence the molecularformula must be (C2H3NO5)

Cont


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Consider the reaction:

MnO2+ 4HCl MnCl2+ Cl2+ 2H2O (a)If 0.86 mole of MnO2and

48.2 g of HCl react, which reagent will be used first (limiting

reagent)? (b) How many grams of Cl2will be produced?[Source: R. Chang (2007): Chemistry, MaGraw-Hill,New York pg 111].

Limiting reagent-Example


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(a) Molar mass of HCl : 36.458g/mol

nHCl=m/M =48.2g/36.458g/mol=1.32 mol

nHCl :nMnO2 1:4 We have 0.86 molesMnO2 of and 1.32 moles of

HCl. Need 3.44 moles of HCl for the reaction. Insufficient HCl,

hence it is limiting.

(b)Mole ratio of HCl: Cl2 is 4:1. Thus, moles of Cl2 produced is

1.32 4=0.33mol. Mass of Cl2 produced=0.33molx 70.9g/mol=23g

Solution


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Disulfide dichloride (S2Cl2) is used in the vulcanization of

rubber, a process that prevents the slippage of rubber

molecules past one another when stretched. It is prepared by

heating sulphur in an atmosphere of chlorine:

S8(l)+ 4Cl2(g)4S2Cl2(l)

What is the theoretical yield of S2Cl2in grams when 4.06g of

S8 are heated with 6.24g of Cl2? If the actual yield of S2Cl2is

6.55g, what is the percent yield?Source: R. Chang

(2007): Chemistry, MaGraw-Hill, New York pg 112].

Percent yield-Example


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nS8 =4.06g/(8x32.07g/mol)=0.0158

nCl2=6.24g/(2x 32.07)=0.0973

S8 is the limiting reagent.

Mole ratio: S8 :S2Cl2 is 1:4.Moles of S2Cl2 produced will be 0.0158 x 4=0.0632

Mass of S2Cl2 produced is 0.0632mol x 135.04g/mol=8.53g

% yield =6.55/8.53 x 100=47.6%

solution


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The amount of solute present in a given quantity ofsolvent or solution.

Molarity (M)=moles of solute

litres of solution

Solution concentration


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Thermochemical equation Reactants and products

Enthalpy change with correct units must be shownand correct sign

Equation must be balanced.


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Hesss Law-ExampleFrom the following dataC graphite(s) + O2(g) CO2(g) H

0=-394KJCdiamond (s) + O2(g) CO2(g) H

0=-396 KJCalculate the H0for the reaction

Cgraphite(s) Cdiamond(s)SolutionIf we reverse the second equation and add tothe first equation, we get the desired reactionCgraphite(s) + O2(g) CO2(g) H

0= -394KJCO2(g) C diamond(s) + O2(g) H

0=+396KJ

_________________________ ______Cgraphite(s) Cdiamond(s) H

0=+2KJ


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Standard enthalpy offormation

This is the enthalpy for the formation of one mole ofa compound in its standard state, from its elementsin their normal (reference) forms in their standardstates.

Denoted DHfo

E.g., C (graphite) + 2 Cl2(g) CCl4(l)Hf

o = 135.4 kJ mol-1


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Standard enthalpies offormation (cond)

DHfo values have been tabulated and can be used to

calculate other DHovalues, using Hesss Law.

By definition, DHfo values of elements in their standard

(reference) states are zero. Standard state of anelement is its most stable physical state. More examples:

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

DHf

o=45.9 kJ mol-1

2 C (graphite) + 3 H2(g) + O2(g) C2H6O (l)(ethanol)DHf

o=277.7 kJ mol-1


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Heat capacity Heat capacity (quantitative measure of overall

effect of heat on temperature of a givenobject)depends ono The material; water has a high heat capacity, while metals are lower.

o The amount of material

It can be expressed per gram of substance (specificheat, s)

q = m c DT

mass (g) specific heat (J g-1 K-1)


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CrO42-

O.S. (Cr) + 4 x O.S.(O) = -2

O.S. (Cr) + 4 x (-2) = -2

O.S. (Cr) = +6

Oxidation state:Example:chromate


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MnO4- + 8 H+ + 5 Fe2+ Mn2+ + 4 H2O + 5 Fe

3+

(in Basic medium)

MnO4- + 8 H+ + 5 Fe2+ +8OH- Mn2+ + 4 H2O + 5 Fe

3+

+8OH-

MnO4- + 8 H2O+ 5 Fe

2+ Mn2+ + 4 H2O + 5 Fe3+ + 8OH-

MnO4-

+ 4 H2O+ 5 Fe2+

Mn2+

+ 5 Fe3+

+8OH-

Example

El

h i l k


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Fis known as Faradays constant.

F= 96,480 C mol-1(J V-1mol-1).

The work that can be done is referred to as freeenergy, DG.

Therefore

DG = - n F Ecell

Electrochemical work(contd.)

N

i hi


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Ecell= E0

cell- (RT/nF)lnQEcell= cell potential under nonstandard conditions

(V)E0

cell

= cell potential under standard conditionsR = gas constant, which is 8.31 (volt-

coulomb)/(mol-K)T = temperature (K)

n = number of moles of electrons exchanged inthe electrochemical reaction (mol)

F = Faraday's constant, 96500 coulombs/molQ = reaction quotient, which is the equilibriumexpression with initial concentrations rather thanequilibrium concentrations

Nernst equation-use thisin exam

F i


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Factors promotingcorrosion

Moisture O2, other oxidizing agents.

Acidity (low pH)

The presence of aggressive anions, such as

chloride ion.

Corrosion can take place in the absence ofmoisture, in dry atmosphere, but only at hightemperatures (dry corrosion.)

Wet corrosion can take place in air, immersedin water, or in the soil.


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Rusting of iron

El t h i t f


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Electrochemistry ofrusting

Certain parts of the iron surface act as anodes.o Dents, imperfections, areas of stress.

o Areas covered in water droplets, in which the cations can dissolve.

Anodic half-reaction:

Fe (s) Fe2+

(aq) + 2 e-

The electrons travel through the metals toregions acting as cathodes, in contact withhigher levels of O2(e.g., edge of waterdroplets).


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Inhibition of corrosion Passivation: protective coating of metal oxide. Use of corrosion inhibitors.

Inert protective coatings: paints and varnishes.These act as a barrier between the metal and

O2. Sacrificial anodes.

Coatings of other metals, e.g., galvanised iron.


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Sacrificial anodes If the metal, e.g., Fe, is electrically connected

to a piece of another metal which is morereactive in the electrochemical series, e.g., Znor Mg, the latter functions as the anode,tending to oxidise; the Fe then becomes a

negatively charged cathode, and isprevented from oxidising until the Zn or Mg isused up.

The piece of Zn or Mg is called a sacrificialanode, since it is consumed while preventingthe Fe from rusting.

If the Fe is connected to a less reactive metal,e.g., Cu or Sn, it becomes an anode, and ismore reactive towards corrosion. Especially a

problem with Al.

G l i i d t l


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Galvanising and metalcoatings

Coating a metal such as Fe with another metalsuch as Zn (e.g., galvanised iron) or Cr (chromeplating) provides two types of protection.o Barrier to prevent contact with O2, including passivation.

o In the event of the coating being scratched, sacrificial anode protection.

Achieved by coating with molten metal(galvanising with Zn) or electroplating.

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Revision lecture 1.pptx