Double bonds or not

1

Double bonds or not

• A saturated fat has no C=C double bonds (alkene functional groups) and is usually a solid fat like margarine or animal fat.

• An unsaturated fat has C=C double bonds and is usually an oil like vegetable oil.

2

Example question

3

Mark scheme

4

Making margarine

• To make margarine we have to saturate vegetable oil by bubbling hydrogen gas through the oil.

• This process is called hydrogenation

5

Is a fat or oil saturated or not?

• We can test for this by adding bromine water.

• If there are double bonds present the bromine water changes from orange/brown to colourless.

6

Example question

7

Mark scheme

8

Hydrolysis

• When an ester is hydrolysed it goes back to an acid and alcohol

• We can hydrolyse by adding acid or alkali (NaOH).

9

Example question

10

Mark scheme

11

Energy changes in chemistry

C7.2

12

Quiz

• When a chemical reaction takes place heat may be given out or taken in.

1. Can you remember the word we use when heat is given out?

2. Can you remember the word we use when heat is taken in?

13

What do I need to know?

1. Recall and use the terms ENDOTHERMIC and EXOTHERMIC

2. Describe examples of ENDOTHERMIC and EXOTHERMIC reactions.

3. Use simple energy level diagrams to represent ENDOTHERMIC and EXOTHERMIC reactions.

14

Change in energy

• Chemical reactants have a certain amount of chemical energy stored within them.

• When the reaction has taken place they have either more or less energy stored within them than before.

15

Definitions

When heat is given out (exothermic) then the products have less energy than they did before. They have lost it to the surroundings.

When heat is taken in (endothermic) then the products have more energy than they had before. They have taken it from the surroundings.

16

Energy level diagrams

Which diagram do you think is endothermic and which is exothermic?

Heat taken inHeat given out

17

Energy level diagrams

Endothermic Exothermic

Heat taken in Heat given out

Energy level of products is higher than reactants so heat taken in.

Energy level of products is lower than reactants so heat given out.

18

Example question

19

Mark scheme

20

Bond enthalpies

C7.2

21

Quick quiz1. Reactions where the products are at a lower energy than

the reactants are endothermic (TRUE/FALSE)2. Activation energy is the amount of energy given out when a

reaction takes place (TRUE/FALSE)3. A reaction which is exothermic transfers heat energy to the

surroundings (TRUE/FALSE)4. How can we tell if a reaction is exothermic or endothermic?5. Sketch the energy profile for an endothermic reaction.6. When methane (CH4) burns in oxygen (O2) bonds between

which atoms need to be broken?

22

Answers1. Reactions where the products are at

a lower energy than the reactants are endothermic (TRUE/FALSE)

2. Activation energy is the amount of energy given out when a reaction takes place (TRUE/FALSE)

3. A reaction which is exothermic transfers heat energy to the surroundings (TRUE/FALSE)

4. How can we tell if a reaction is exothermic or endothermic?

5. Sketch the energy profile for an endothermic reaction.

6. When methane (CH4) burns in oxygen (O2) bonds between which atoms need to be broken?

FALSE

FALSE

TRUE

Measure the temperature change

C—H bonds and O=O bonds

23


1. Recall that energy is needed to break chemical bonds and energy is given out when chemical bonds form2. Identify which bonds are broken and which are made when a chemical reaction takes place.3. Use data on the energy needed to break covalent bonds to estimate the overall energy change for a reaction.

24

Activation energy revisited

• What is the activation energy of a reaction?• The energy needed to start a reaction.• BUT what is that energy used for and why

does the reaction need it if energy is given out overall?

• The activation energy is used to break bonds so that the reaction can take place.

25

Burning methane

Consider the example of burning methane gas.

CH4 + 2O2 CO2 + 2H2O

This reaction is highly exothermic, it is the reaction that gives us the Bunsen flame. However mixing air (oxygen) with methane is not enough. I need to add energy (a flame).

26

What happens when the reaction gets the activation energy?

Bond Forming

BondBreaking

Progress of reaction

Ener

gy in

che

mic

als

OO

OO

H

CH

HH

O OOO

C H H H H

O C OO

OH H

H H

27

Using bond enthalpies

By using the energy that it takes to break/make a particular bond we can work out the overall enthalpy/energy change for the reaction.

Sum (bonds broken) – Sum (bonds made) = Energy change

28

BIN MIX

Breaking bonds is ENDOTHERMIC energy is TAKEN IN when bonds are broken

Making bonds is EXOTHERMIC energy is GIVEN OUT when bonds are made.

29

Bond enthalpiesBond Bond enthalpy (kJ) Bond Bond enthalpy (kJ)

C—H 435 Cl—Cl 243

C—C 348 C—Cl 346

H—H 436 H—Cl 452

H—O 463 O=O 498

C=O 804

C=C 614

30

Can you work out the energy change for this reaction?

CH4 + Cl2 CH3Cl + HCl

Tip: Draw the reactants and products and work out the bonds you are breaking and the ones you are making.

31

The answer is -120 kJ

32

Example question part 1

33

Question part 2

34

Question part 3

35

Mark scheme

36

Challenge question

• The true value for the energy change is often slightly different from the value calculated using bond enthalpies.

• Why do you think this is?

37

Example question

The calculated value is 120 kJ

38

Mark scheme

39

Definitions

Write each of these phrases in your book with a definition in your own words:

• Exothermic reaction• Endothermic reaction• Activation energy• Catalyst• Bond energy/enthalpy

40

How did you do?Exothermic reactionA reaction which gives energy out to the surroundings.Endothermic reactionA reaction which takes in energy from the surroundings.Activation energyThe energy required to start a reaction by breaking bonds in the reactantsCatalystA substance that increases the rate of a reaction by providing an alternative pathway with lower activation energy. It is not used up in the process of the reactionBond energy/enthalpyThe energy required to break a certain type of bond. The negative value is the energy given out when that bond is made.

41

Popular exam question

1. Explain why a reaction is either exothermic or endothermic?

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

42


1. Explain why a reaction is either exothermic or endothermic?

① In a chemical reactions some bonds are broken and some bonds are made.

② Breaking bonds takes in energy.③ Making bonds gives out energy.④ If the energy given out making bonds is higher than the

energy needed to break them the reaction is exothermic.⑤ If the energy needed to break bonds is higher than the

energy given out making them the reaction is endothermic.

43

Chemical Equilibria

C7.3 Reversible Reactions & Dynamic Equilibria

44


1. State that some chemical reactions are reversible

2. Describe how reversible reactions reach a state of equilibrium

3. Explain this using dynamic equilibrium model.

45

Reversible or not reversible

Until now, we were careful to say that most chemical reactions were not reversible –

They could not go back to the reactants once the products are formed.

46

Example

In the case of the vast majority of chemical reactions this is true, the reaction of methane and oxygen for example:

CH4(g) + O2(g) CO2(g) + 2H2O(l)

It is almost impossible to return the carbon dioxide and water to the original methane and oxygen.

47

Reversible

• Some chemical reactions, however, will go backwards and forwards depending on the conditions.

• CoCl2·6H2O(s) CoCl2(s) + 6H2O(l) pink blue

48

How do we write them down?

• This is the symbol for used for reversible reactions.

CoCl2·6H2O(s) CoCl2(s) + 6H2O(l)

49

What is equilibrium?

• Reversible reactions reach a balance point, where the amount of reactants and the amount of products formed remains constant.

• This is called a position of equilibrium.

50

Dynamic Equilibrium.

• In dynamic equilibrium the forward and backwards reactions continue at equal rates so the concentrations of reactants and products do not change.

• On a molecular scale there is continuous change.

• On the macroscopic scale nothing appears to be happening. The system needs to be closed – isolated from the outside world.

51

Example question

52

Mark scheme

53

Dynamic Equilibria

C7.3 Controlling equilibria

54


1. Recall that reversible reactions reach a state of dynamic equilibrium.2. Describe how dynamic equilibria can be affected by adding or removing products and reactants. 3. Explain the difference between a “strong” and “weak” acid in terms of equilibria

55

Position of the equilibrium

• Equilibrium can “lie” to the left or right.• This is “in favour of products” or “in favour of

reactants”• Meaning that once equilibrium has been

reached there could be more products or more reactants in the reaction vessel.

56

Le Chatelier’s principle

• If you remove product as it is made then equilibrium will move to the right to counteract the change

• If you add more reactant then equilibrium will move to the right to counteract the change.

• In industry we recycle reactants back in and remove product as it is made to push the equilibrium in favour of more product.

57

Complete

When a system is at__________ to make more product you can_________ product or add more __________ for example by recycling them back in.

To return to reactants you ______ product or remove_________.

[equilibrium, add, reactant, remove, product]

58

Strong and weak acids

A strong acid is one which is FULLY IONISED in water. It will have a high hydrogen ion concentration

HCl H+ + Cl-

A weak acid is one which is NOT fully ionised and is in equilibrium. It has a low hydrogen ion concentration

CH3COOH CH3COO- + H+

Caution – weak and strong are not the same as concentration.

60

Mark scheme

61

Example question

62

Mark scheme

63

Practicing definitions

Write each of these phrases in your book with a definition in your own words:

• Reversible reaction• Dynamic equilibrium• Position of equilibrium• Strong acid• Weak acid

64

How did you do?Reversible reactionA reaction that can proceed in the forward or reverse directions (represented by two arrows in an equation).Dynamic equilibriumThe point where the rate of the forward reaction = rate of the reverse reaction.Position of equilibriumThe point where there is no further change in the concentration of either reactants or products. The position can lie to the left (favouring reactants) or right (favouring products).Strong acidAn acid that is completely dissociated in waterWeak acidAn acid that is only partly dissociated in water because the reaction is in dynamic equilibrium and favours the reactants (LHS).

65


1. Ethanoic acid (CH3COOH) is a weak acid but hydrochloric acid is a strong acid. Use ideas about ion formation and dynamic equilibrium to explain this difference.

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

66

Popular exam questionEthanoic acid (CH3COOH) is a weak acid but hydrochloric acid is a strong acid. Use ideas about ion formation and dynamic equilibrium to explain this difference.

① Hydrochloric acid ionises completely② So hydrogen ion concentration is high③ Ethanoic acid only partly dissociates because the reaction

is reversible④ Equilibrium is mainly to the left⑤ So hydrogen ion concentration is low.

67

Analysis

C7.4 – Analytical Procedures

68


1. Recall the difference between qualitative and quantitative methods of analysis.

2. Describe how analysis must be carried out on a sample that represents the bulk of the material under test

3. Explain why we need standard procedures for the collection, storage and preparation of samples for analysis

69

Qualitative vs. Quantitative

• A qualitative test is usually very quick. It can give vital information without needing to wait too long for it.

• A quantitative test gives precise values, for example for a concentration in Moldm-3

• Examples of qualitative tests include universal indicator, silver nitrate for halide ions and bromine water for unsaturation.

• Examples of quantitative tests include titration, chromatography and spectroscopy.

70

Which sample should I test?

• It is important that the sample represents the bulk of the material under test.

• You may chose to test more than one sample from a range of points to ensure that you have results which represent the whole.

• For example is it well mixed? Are their pockets of higher concentration/different composition?

71

Chemical industry

• Analysis of samples is crucial to the chemical industry to ensure the quality of the chemicals they are manufacturing. Some are analysed numerous times a day or even within an hour.

• To maintain consistency it is essential that we use standard procedures to:o collect the sampleo store the sampleo prepare the sample for analysiso analyse the sample.

72

Chromatography

C7.4 – paper chromatography

73

Chromatography

74

Solvents

1. The mobile phase is the solvent – the part that moves

2. In paper chromatography it is water or ethanol

75

Paper/column

1. The stationary phase is the paper in paper chromatography or the column in gas chromatography.

2. In thin layer chromatography it is silica gel on a glass plate

3. The stationary phase does not move.

76

How does the technique work?

In chromatography, substances are separated by movement of a mobile phase through a stationary phase.

Each component in a mixture will prefer either the mobile phase OR the stationary phase.

The component will be in dynamic equilibrium between the stationary phase and the mobile phase.

77

Substance A

• This is substance A

• Substance A prefers the stationary phase and doesn’t move far up the paper/column.

• The equilibrium lies in favour of the stationary phase.

78

Substance B

• This is substance B

• Substance B prefers the mobile phase and moves a long way up the paper/column

• The equilibrium lies in favour of the mobile phase

79

Using a reference

• In chromatography we can sometimes use a known substance to measure other substances against.

• This will travel a known distance compared to the solvent and is known as a standard reference.

80

Advantages of TLC

TLC has a number of advantages over paper chromatography.

It is a more uniform surface chromotograms are neater and easier to interpret

Solvent can be used which is useful if a substance is insoluble in water.

81

Past Paper Questions

82

Past paper question

83

Mark scheme

84

Describing how chromatography works – exam definition

• stationary phase is paper and mobile phase is solvent / mobile phase moves up through stationary phase (1)

• for each compound there is a dynamic equilibrium between the two phases (1)

• how far each compound moves depends on its distribution between the two phases (1)

85

Using an Rf value

• In order to be more precise we can use measurements on the TLC plate to compare the distance travelled by our substance (the solute) with the distance travelled by the solvent.

• The Rf value is constant for a particular compound.

• The distance travelled however could be different on different chromatograms.

• The Rf value is always less than 1.

86

Rf value

87

Example questionThis question relates to the chromatogram shown in the earlier question. Refer back…

88

Mark scheme

89

Example question

90

Mark scheme

91

Past paper question

95

Mark scheme

96

Gas-liquid chromatography

C7.4 GLC

97


1. recall in outline the procedure for separating a mixture by gas chromatography (gc);

2. understand the term retention time as applied to gc;

3. interpret print-outs from gc analyses.

98

Gas chromatography

• The mobile phase is an unreactive gas known as the carrier gas this is usually nitrogen

• The stationary phase is held inside a long column and is lots of pieces of inert solid coated in high bp liquid.

• The column is coiled in an oven

• The sample to be analysed is injected into the carrier gas stream at the start of the column.

99

GC

100

GC analysis• Each component of the sample mixture has a different

affinity for the stationary phase compared with the mobile phase

• Therefore each component travels through the column in a different time.

• Compounds favouring the mobile phase (usually more volatile) emerge first.

• A detector monitors the compounds coming out of the column and a recorder plots the signal as a chromatogram

101

GLC Chromatograph

102

Interpretation

• The time in the column is called the retention time

• Retention times are characteristic so can identify a compound

• Area under peak or relative heights can be used to work out relative amounts of substances

103

The key points – revise this!

• the mobile phase carries the sample (1) • components are differently attracted to the

stationary and mobile phases (1) • the components that are more strongly attracted

to the stationary phase move more slowly (1) • the amount of each component in the stationary

phase and in the mobile phase is determined by a dynamic equilibrium (1)

104

Past paper question

108

Mark scheme

109

Titration

C7.4

110


1. Calculate the concentration of a given volume of solution given the mass of solvent;

2. Calculate the mass of solute in a given volume of solution with a specified concentration;

3. Use the balanced equation and relative formula-masses to interpret the results of a titration;

111

Concentration• We can measure the concentration of solution in grams/litre. This

is the same as g/dm3

• 1dm3 = 1000cm3

• If I want to make a solution of 17 g/dm3 how much will I dissolve in 1dm3.

• 17 g

• If I want to make a solution of 17g/dm3 but I only want to make 100cm3 of it how much will I dissolve?

• 1.7g

112

Making standard solutions

• For a solution of 17g/dm3

• First I will measure 17g of solid on an electronic balance

113

Making standard solutions

• Now I must dissolve it in a known 1dm3 of water.

• I transfer it to a volumetric flask and fill up with distilled water to about half the flask.

• I then swirl to dissolve• Top up with a dropping pipette so that

the meniscus is on the line.

114

How much to dissolve?

• Worked example:

• I want to make 250cm3 of a solution of 100g/dm3.

• How much solid do I transfer to my 250cm3 volumetric flask?

115

How much to dissolve?

Worked example:I want to make 250cm3 of a solution of 100g/dm3.

1. Work out the ratio of 250cm3 to 1000cm3

250/1000 = 0.25

2. I therefore need 0.25 of 100g in 250cm3 which is 0.25x100=25g

116

General rule

117

Practie - how much to dissolve?

• I want to make 250cm3 of a solution of 63.5g/dm3.


• 250/1000 x 63.5 = 15.9 g

118

Practice - how much to dissolve?

• I want to make 100cm3 of a solution of 63.5g/dm3.


• 100/1000 x 63.5 = 6.35 g

119

Concentration from mass and volume

We need to rearrange this:

To give

120

What is the concentration of?

1. 12g dissolved in 50cm3




121

What is the concentration of?


= 1000/50 x 12= 240g/dm3


=1000/100 x 50 = 500g/dm3


= 1000/1000 x 47= 47g/dm3


1000/250 x 200= 800g/dm3

122

Solutions from stock solutions

Stock solution•highest concentration

•use to make other solutions

Extract a portion of stock solution•as calculated

Dilute with distilled water•Making a known volume of a lower concentration

123

Making solutions from stock solutions

If I have a solution containing 63g/dm3, how do I make up 250cm3 of a solution of concentration 6.3g/dm3?

To make 1dm3 of 6.3g/dm3 I would need 100cm3

To make 250cm3 of 6.3g/dm3 I would therefore need 25cm3 and make it up to 250cm3 with distilled water

124

Working out masses

• We can use the useful relationship

• Where Mr is the molecular mass• eg Mr of NaOH is (23 + 16 + 1) = 40• This can help us to calculate an unknown mass

125

Titration calculations

• In a titration we have added a known amount of one substance usually an acid (in the burette) to a known amount of another substance usually an alkali (in the conical flask).

• The amount added allows us to determine the concentration of the unknown.

126

Titration equipment

127

Using a table

• It can be helpful to sketch a table to keep track of information you know…

Value Acid Alkali

Volume (cm3)

Mass (g)

Concentration (g/dm3)

Molecular weight (Mr)

129

Mark scheme

131

Mark scheme

132

Uncertainty

• Uncertainty is a quantification of the doubt about the measurement result.

• In a titration the uncertainty is the range of the results.

• If results are reliable then it will be within 0.2cm3

• NOTE THAT THIS IS RELIABLE NOT NECESSARILY ACCURATE

134

Mark scheme

136

Mark scheme

137

C7.5 Green Chemistry

The Chemical Industry

138


1. Recall and use the terms 'bulk' (made on a large scale) and 'fine' (made on a small scale) in terms of the chemical industry with examples;

2. Describe how new chemical products or processes are the result of an extensive programme of research and development;

3. Explain the need for strict regulations that control chemical processes, storage and transport.

139

Bulk processes

• A bulk process manufactures large quantities of relatively simple chemicals often used as feedstocks (ingredients) for other processes.

• Examples include ammonia, sulfuric acid, sodium hydroxide and phosphoric acid.

• 40 million tonnes of H2SO4 are made in the US every year.

140

Fine processes

• Fine processes manufacture smaller quantities of much more complex chemicals including pharmaceuticals, dyes and agrochemicals.

• Examples include drugs, food additives and fragrances

• 35 thousand tonnes of paracetamol are made in the US every year.

141

Example question

142

Mark scheme

143

Research and Development

• All chemicals are produced following an extensive period of research and development.

• Chemicals made in the laboratory need to be “scaled up” to be manufactured on the plant.

144

Research in the lab

145

Examples of making a process viable

• Trying to find suitable conditions – compromise between rate and equilibrium

• Trying to find a suitable catalyst – increases rate and cost effective as not used up in the process.

146

Catalysts

• Can you give a definition of a catalyst?• A substance which speeds up the rate of a

chemical reaction by providing an alternative reaction pathway.

• The catalyst is not used up in the process• Catalysts can control the substance formed eg

Ziegler Natta catalysts.

147

Regulation of the chemical industry

• Governments have strict regulations to control chemical processes

• Storage and transport of chemicals requires licenses and strict protocol.

• Why?

• To protect people and the environment.

148

Example question

149

Mark scheme

150

Process development

preparation of

feedstockssynthesis

separation of

products

handling of by-

products and wastes

monitoring of purity

151

Example question – part 1

152

Example question part - 2

153

Mark scheme

154


155

Mark scheme

156

Factors affecting the sustainability of a process

Sustainability

renewable feedstock

atom economy

type of waste and disposal

energy inputs and outputsenvironmental

impacthealth and safety risks

social and economic benefits

157

Example question

158

Mark scheme

159

Atom economy

160

Atom economy calculationFor example, what is the atom economy for making hydrogen by reacting coal with steam?

Write the balanced equation:

C(s) + 2H2O(g) → CO2(g) + 2H2(g)

Write out the Mr values underneath:

C(s) + 2H2O(g) → CO2(g) + 2H2(g)

12 2 × 18 44 2 × 2

Total mass of reactants 12 + 36 = 48gMass of desired product (H2) = 4g

% atom economy = 4⁄48 × 100 = 8.3%

161

Exam

ple

ques

tion

162


163

Mark scheme

Double bonds or not

Documents

Transcript of Double bonds or not