Chemistry Form Five

7/28/2019 Chemistry Form Five

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Chemistry Form Five: Chapter 1 - Rate of Reaction

Rate of reaction = change of quantity in reactant or product per unit time.

We usually use water displacement method to collect gas in school laboratory as shown below:
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The reaction is fastest at the start when the reactants are at a maximum

(steepest gradient) The gradient becomes progressively less as reactants are used up and the

reaction slows down.

Finally the graph levels out when one of the reactants is used up and thereaction stops.

The amount of product depends on the amount of reactants used.

The initial rate of reaction is obtained by measuring the gradient at the start

of the reaction. A tangent line is drawn to measure rate of reaction atinstataneous time

hemistry Form 5: Chapter 1 - Collision Theory

According to the collision theory, particles of reactant that achieve activation energy and collide with

correct orientation will result in reaction.

1. Correct Orientation


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Click on the diagram below to play!

2. Activation Energy

Activation energy is the minimum amount of energy that must be overcome by the colliding

particles so that the reaction can occur
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Chemistry Form 5: Chapter 1 - Effect of Concentration on Rate of

Reaction

Experiment to show the effect of concentration on reaction rate

Sodium thiosulphate solution react with dilute sulphuric acid to form a yellow precipitate

of sulphur. In this experiment, the time taken for the formation of sulphur to cover themark 'X' until it disappears from sight can be used to measure rate of reaction.

As the concentration of sulphuric acid is increased, the rate of reaction between sulphuric

acid and sodium thiosulphate increases.

Explanation using collision theory
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When the concentration of the solution of a reactant increases, the number ofparticles per unit volume of the solution also increases.

With more particles per unit volume of the solution of the reactant, the

frequency of collision increases. This causes the frequency of effective collision to increase. Hence, the rate of

reaction increases.

Chemistry Form 5: Chapter 1 - Catalyst Affects the Rate of Reaction

Catalyst is a chemical substance that change the rate of chemical reaction.

Characteristics of catalyst:

Catalyst remains chemically unchanged during reaction. Its chemical composition still

the same before and after reaction.

Catalyst only change the rate of reaction.

Catalyst does not change the quantity of the product formed.

Catalyst is specific in its action.

Only a small amount of catalyst is needed to achieve a big increase in rate of reaction.


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How catalyst increase the rate of reaction:

When a positive catalyst is used in a chemical reaction, it enables the reaction to occur

through an alternative path which requires lower activation energy.

As a result, more colliding particles are able to overcome the lower activation energy.

This causes the frequency of effective collision to increase.

Hence, the rate of reaction increases.

Decomposition of hydrogen peroxide by catalyst of manganese (IV) oxide

hemistry Form 5: Chapter 2 - Hydrogenation

Hydrogenation process is addition reaction to convert alkene becomes alkane. It converts

unsaturated compound to saturated compound.
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Ethene reacts with H2 at 180 C in the presence ofnickel or platinum catalyst to produce ethane


Application of hydrogenation: Making Margarine

Vegetable oils often contain high proportions of polyunsaturated and mono-unsaturatedfats (oils), and as a result are liquids at room temperature. That makes them messy tospread on your bread or toast, and inconvenient for some baking purposes.

You can "harden" (raise the melting point of) the oil by hydrogenating it in the presenceof a nickel catalyst. Conditions (like the precise temperature, or the length of time thehydrogen is passed through the oil) are carefully controlled so that some, but notnecessarily all, of the carbon-carbon double bonds are hydrogenated. This produces a"partially hydrogenated oil" or "partially hydrogenated fat".

hemistry Form 5: Chapter 2 - Manufacture of Ethanol (Hydration)

Ethanol is manufactured by reacting ethene with steam. The reaction is reversible, and

the formation of the ethanol is exothermic.
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Chemistry Form 5: Chapter 2 - Dehydration of Alcohol

ALCOHOL -------> ALKENE

In the dehydration of alcohols, a molecule of water is eliminated from each alcohol

molecule to produce alkene.

There are two methods of dehydration:

a) Ethanol vapour is passed over a heated unglazed porcelain chips, porous pot, pumice stone or

alumina (aluminium oxide).

b) Ethanol is heated under reflux at 170 C with excess concentrated sulphuric acid.
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School Laboratory Experiment

Alkene can be tested by decolourising brown bromine water or decolourising purple

acidified potassium manganate (VII) solution.

hemistry Form 5: Chapter 2 - Carboxylic Acid

Carboxylic acids are organic compounds which form an homologous series with

the general formula of CnH2n+1COOH. Carboxylic acids are compounds which contain a -COOH functional group.

Carboxylic acids are weak acid which ionize partially in water to produce lower

concentration of hydrogen ions compare to strong acid.

Formula Common Name Source IUPAC Name


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HCO2H formic acid ants (L. formica) methanoic acid

CH3CO2H acetic acid vinegar (L. acetum) ethanoic acid

CH3CH2CO2H propionic acid milk (Gk. protus prion) propanoic acid

CH3(CH2)2CO2H butyric acid butter (L. butyrum) butanoic acid

CH3(CH2)3CO2H valeric acid valerian root pentanoic acid

CH3(CH2)4CO2H caproic acid goats (L. caper) hexanoic acid

CH3(CH2)5CO2H enanthic acid vines (Gk. oenanthe) heptanoic acid

CH3(CH2)6CO2H caprylic acid goats (L. caper) octanoic acid

CH3(CH2)7CO2H pelargonic acid pelargonium (an herb) nonanoic acid

CH3(CH2)8CO2H capric acid goats (L. caper) decanoic acid

hemistry Form 5: Chapter 2 - Esterification

Esterification is a chemical reaction between carboxylic acid combines with an

alcohol in the presence of a catalyst (commonly concentrated sulphuric acid) to

form an ester.
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Examples of Esters methyl butanoate (apple) :

methan

ol

+

butanoic

acid

methylbutanoate

(ester)

+

wate

r

CH3OH+

C3H7COOH C3H7COOCH3 (ester)

+

H2O

ethyl methanoate (rum essence) :

ethan

ol

+

methanoic

acid

ethylmethanoate

(ester)

+

wate

r

C2H5O

H

+

HCOOH HCOOC2H5 (ester)

+

H2O

ethyl butanoate (pineapple) :

ethan

ol

+

butanoic

acid

ethylbutanoate

(ester)

+

wate

r

C2H5O

H

+

C3H7COOH

C3H7COOC2H5

(ester)

+

H2O

pentyl ethanoate (banana) :
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pentan

ol

+

ethanoic

acid

pentylethanoate

(ester)

+

wate

r

C5H11O

H

+

CH3COOH

CH3COOC5H11

(ester)

+

H2O

pentyl butanoate (apricot) :

pentan

ol

+

butanoic

acid

pentylbutanoate

(ester)

+

wate

r

C5H11O

H

+

C3H7COOH

C3H7COOC5H11

(ester)

+

H2O

octyl butanoate (orange) :

octano

l

+

butanoic

acid

octylbutanoate

(ester)

+

wate

r

C8H17O

H

+

C3H7COOH

C3H7COOC8H17

(ester)

+

H2O

methyl ethanoate (solvent) :

methan

ol

+

ethanoic

acid

methylethanoate

(ester)

+

wate

r

CH3OH+

CH3COOH CH3COOCH3 (ester)

+

H2O

ethyl ethanoate (solvent) :

ethan

ol

+

ethanoic

acid

ethylethanoate

(ester)

+

wate

r

C2H5O

H

+

CH3COOH

CH3COOC2H5

(ester)

+

H2O

Esterification by refluxing


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Chemistry Form 5: Chapter 3 - Redox Reaction
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Redox reactions are chemical reactions involving oxidation and reduction

occurring simultaneously. Oxidising agent is the substance that causes oxidation.

Reducing agent is the substance that causes reduction.

Oxidation involves loss of electrons and increase in oxidation number.Reduction involves gain of electrons and decrease in oxidation number.

Example:

The magnesium's oxidation state has increased from 0 to +2 , it has been oxidised.

Magnesium acts as reducing agent. The hydrogen's oxidation state has decreased from+1 to 0 , it has been reduced. The hydrogen ion acts as a oxidising agent.

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There is no change in oxidation number. Therefore, this is not a redox reaction.

Chemistry Form 5: Chapter 3 - Rules of Oxidation Number

There are several rules for assigning the oxidation number to an element. Learning these rules

will simplify the task of determining the oxidation state of an element, and thus, whether it has

undergone oxidation or reduction.

1. The oxidation number of an atom in the elemental state is zero.Example: Cl2 and Al both are 0

2. The oxidation number of a monatomic ion is equal to its charge.

Example: In the compound NaCl, the sodium has an oxidation number of 1+ and

the chlorine is 1-.

3. The algebraic sum of the oxidation numbers in the formula of a compound is zero.

Example: the oxidation numbers in the NaCl above add up to 0

4. The oxidation number of hydrogen in a compound is 1+, except when hydrogen forms

compounds called hydrides with active metals, and then it is 1-.

Examples: H is 1+ in H2O, but 1- in NaH (sodium hydride).

5. The oxidation number of oxygen in a compound is 2-, except in peroxides when it is 1-,

and when combined with fluorine. Then it is 2+.

Example: In H2O the oxygen is 2-, in H2O2 it is 1-.

6. The algebraic sum of the oxidation numbers in the formula for a polyatomic ion is equal

to the charge on that ion.

Example: in the sulfate ion, SO42-, the oxidation numbers of the sulfur and the

oxygens add up to 2-. The oxygens are 2- each, and the sulfur is 6+.

Chemistry Form 5: Chapter 3 - Conversion of Iron (II) to Iron (III) and Iron

(III) to Iron (II)
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Conversion ofFe2+ to Fe3+

Oxidising agent: bromine water

Reducing agent: Fe2+ ions

Oxidation half equation:

Fe2+ ions lose electrons and are oxidized to Fe3+.

The presence of Fe3+

ions is confirmed by the formation ofbrown precipitate with excess of NaOH solution.

Fe2+ --------> Fe3+ + e

Reduction half equation:

Bromine molecules which give bromine water its brown

colour gain electrons and are reduced to colourless bromide

ions.

Br2 + 2 e -------> 2 Br-

Overall ionic equation:

2 Fe2+ + Br2 2 Fe3+ + 2 Br-

Observation:

Brown bromine water decolourises. The solution changes

colour from pale green (Fe2+) to yellow (Fe3+).

Conversion ofFe3+ to Fe2+
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Oxidising agent: Fe2+ ions

Reducing agent: zinc

Oxidation half equation:

Zinc atoms lose their electrons and are oxidized to zinc ions,

Zn2+.

Zinc powder dissolves in iron (III) chloride.

Zn -------> Zn2+ + 2 e

Reduction half equation:

Fe3+ ions accept electrons and are reduced to Fe2+.

The presence of Fe2+ ions is confirmed by the formation of

green precipitate with excess of NaOH solution.

Fe3+ + e Fe2+

Overall ionic equation:

2 Fe3+ + Zn ---------> 2 Fe2+ + Zn2+

Observation:

Zinc powder dissolves into solution. The solution changes

colour from brown (Fe3+) to pale green (Fe2+).

Chemistry Form 5: Chapter 3 - Redox Reaction in Displacement of Metal

Reactivity Series of metals

Displacement of metals from solution is a redoxreaction whereby a less reactive metal ion is

displaced from its salt solution by a more

reactive metal. As a result, the less reactivemetal ion is deposited as a solid metal while the

more reactive metal dissolves in the solution.

The general formula for a displacement reaction

is:

M (s) + Xn+ (aq) -----> Mn+ (aq) + X (s)

where metal M is the more reactive than metal

X.

Most reactive

K

Na

Ca

Mg

Al

Zn

Fe

Pb

[H]Cu

Ag

Least reactive

Example of displacement reaction:


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Fe (s) + CuSO4 (aq) -----> FeSO4 (aq) + Cu (s)

Iron displaces copper from the solution because it is more reactive than copper metal.Iron, being more reactive, loses its electrons readily. The electrons are transferred fromthe iron atoms to the copper(II) ions in the solution. Copper(II) ions are reduced tocopper metal and iron atoms become oxidised to iron (II) ions. Iron acts as reducingagent whereas copper (II) ions act as a oxidising agent.

Oxidation reaction: Fe (s) -----> Fe2+ (aq) + 2e-

Reduction reaction: Cu2+ (aq) + 2e- -----> Cu (s)Overall Redox reaction: Fe (s) + Cu2+ (aq) -----> Fe2+ (aq) + Cu (s)

hemistry Form 5: Chapter 3 - Rusting of Iron
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Rusting is a corrosion of iron.

For iron to rust, oxygen and water must be present.

In the presence of acids and salts, rusting occurs faster because thesesubstances increase the electrical conductivity of water, making water a betterelectrolyte.

Oxygen acts as the oxidizing agent and iron acts as the reducing agent .

The surface of iron at the middle of the water droplet serves as the anode at which oxidation occurs. The

iron atoms lose electrons to form iron (II) ions.

The electrons flow to the edge of the water droplet where there is plenty of dissolved oxygen. The iron

surface there serves as cathode at which reduction occurs. Oxygen gains the electrons and is reduced to

hydroxide ions.
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The iron (II) ions produced combine with the hydroxide ions to iron (II) hydroxide. The Fe(OH)2 is then

further oxidized by oxygen to form iron (III) oxide, Fe2O3 known as rust

Chemistry Form 5: Chapter 3 - Redox Reaction in Electrochemistry Reaction

More electropositive metal undergoes oxidation reaction by releasing electronsand act as a reducing agent.

Less electropositive metal undergoes reduction reaction by gaining electrons and

act as a oxidising agent.

Electrons flow from more electropositive metal to less electropositive metal.

Chemistry Form 5: Chapter 4 - Exothermic and Endothermic Reaction

Chemical energy is needed to transform a chemical substance into a new product through

chemical reaction. Therefore, breaking or formation of chemical bond involves energy, whichmay be either absorbed or released from a chemical reaction.

To break the chemical bond, energy from surrounding is absorbed resulting decrease of

temperature of surrounding.

To form the chemical bond, energy from reaction is released to the surrounding resulting

increase of temperature of surrounding.
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Soaps are sodium or potassium salts of fatty acids. Soaps are prepared by hydrolyzingfats or oils under alkaline condition. This reaction is called saponification.

The fats or oils are hydrolysed first to form glycerol and fatty acids. The acids then react with an

alkali to form the corresponding sodium or potassium salts.

The soap formed can be precipitated by adding sodium chloride. This is because sodium

chloride lowers the solubility of soap in water.

The glycerol and excess sodium hydroxide solution are removed by rinsing the soap formed

with water.
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Chemistry Form Five