HSC extended responses to memorise

Question 1

Le Chatelier’s Principle

Answer 1

Le Chatelier’s principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to re-establish an equilibrium.

Question 2

Change in temperature

- Collision Theory

Answer 2

Increasing heat energy increases the kinetic energy of the particles. This increases their velocity. This also means they are more likely to have sufficient energy to overcome the activation energy needed for the reaction to occur. This increases the frequency of collisions and improves the chances of successful reactions when a collision occurs.

Question 3

Change in temperature

- LCP

Answer 3

• If the forward reaction is exothermic and the temperature is increased the reverse reaction will be favoured to absorb the extra heat.
• Similarly, if the temperature is decreased and the forward reaction is exothermic, the forward reaction will be favoured to produce more heat to bring the reaction back to equilibria.

Question 4

Change in concentration

- Collision theory

Answer 4

Increasing the concentration of a substance increases the numbers of particles and hence there is a higher likelihood of reactant particles colliding to form one or more products or vice versa.

Question 5

Change in concentration

-LCP

Answer 5

• If you increase the concentration of reactants in a reaction, the forward reaction will be favoured to produce more products.
• If you decrease the concentration of reactants in a reaction, the reverse reaction will be favoured to produce more reactants.

Question 6

Change in pressure

-Collision Theory

Answer 6

Increasing pressure for gases involves increasing the density or decreasing the volume. Pushing the same number of particles into a smaller space will increase the likelihood of collisions hence increasing the rate of reaction.

Question 7

Change in pressure

-LCP

Answer 7

• A change in pressure depends upon the amount of moles in the reaction. If you increase the pressure, and there are more moles of reactants than products, the forward reaction will be favoured in order to ‘decrease’ the pressure.
• If you decrease the pressure, and there are more moles of reactants than products then the reverse reaction will be favoured to ‘increase’ the pressure again.

Question 8

Explain, in terms of the collision theory, why an increase in temperature of an equilibrium system always favours the endothermic reaction.

Answer 8

collision theory states that particles must collide with sufficient energy and in the correct orientation to overcome the activation energy barrier. When the temperature is increased, the particles gain kinetic energy. For a chemical reaction, the activation energy barrier for the endothermic reaction is always greater than for the exothermic reaction. Hence, an increase in temperature will affect the percentage of particles able to overcome the endothermic activation energy barrier more than for the exothermic reaction. This will cause the equilibrium to shift in the endothermic direction when the temperature is increased.

Question 9

Cycad Toxins

Answer 9

Aboriginal and Torres strait islander peoples used a variety of processes to prepare plant materials for food. Some of these foods contain toxic substance’s such as cycasin in cycad seeds. Processes which reduced the concentration of the toxic substances include leeching and fermentation. Cycad seeds were cracked open to extract the kernel. The kernel was ground into a paste and placed in a dilly-bag. The bag was then secured between rocks in a flowing system. The soluble cycasin was leeched out of the paste over several day. Then the paste was dried as used as a flour.

Alternatively, the cycad seeds were dropped into a lined pit and covered with soil. Over a period of months, the seeds fermented in anerobic condition changing the chemical makeup of the kernels. The seeds are then dug-up and consumed.

Question 10

Buffers with a chemical equation

Answer 10

CH3COOH + H2O ⇌ CH3COO- + H3O+
A buffer is a mixture of a weak acid and its conjugate base. In this buffer, the acid is CH3COOH and its conjugate base is CH3COO- This buffer resists change to pH as follows.
If an acid is added to the buffer solution, the system will adjust itself to minimise the disturbance by driving the reaction to the left. It does this by the CH3COO- ions reacting with the acid added. As the solution is made up of acid and ions, there are lots of these in the solution, so it is able to absorb a lot of hydronium ions thus maintaining the pH

If a base is added to the buffer system, the OH- ions will react with the hydronium ions in the solution, the system will adjust itself to minimise the disturbance by driving the reaction to the right to produce more hydronium ions. As the CH3COOH is a weak acid, there are lots of molecules of this acid that are able to ionise to produce hydronium ions, thus maintaining the pH

Question 11

Explain why the colour did not change (buffers)

Answer 11

The colour of the universal indicator did not change because the pH did not change very much despite the addition of acid and/or base. This is due to the equilibrium position of the buffer equation shifting in response to the addition of acid and/or base in accordance with Le Chatelier’s Principle.

If acid is added, the concentration of hydronium ions increases. The equilibrium position will shift left (according to Le Chatelier’s Principle) to reduce the concentration, and the pH does not change much.

If a base is added, the hydroxide ions react with the hydronium ions, lowering the hydronium ion concentration. The reaction shifts right (according to Le Chatelier’s Principle) to increase the concentration of hydronium ions, and the pH remains nearly constant.

Question 12

Lavoisier (1766) theory of acids and bases

Answer 12

After experimenting on oxides of non-metals, Lavoisier concluded that acids were substances that contained oxygen
Many acids, such as hydrochloric acid, do not contain oxygen

Question 13

Davy (1810) theory of acids and bases

Answer 13

Nothing that hydrochloric acid did not contain oxygen, whilst still acting as an acid. Davy proposed that acids contained replaceable hydrogen atoms

This theory did not really have an explanation as to when or how the molecules interacted

Question 14

Arrhenius (1884) theory of acids and bases

Answer 14

Rather than a substance with replaceable hydrogen atoms, Arrhenius suggested that acids would ionise in water to produce hydrogen ions (H+) and bases would ionise in water to produce hydroxide ions (OH-)

• Only applied for aqueous solutions
• Does not account for the behaviour of amphiprotic species that can donate and accept a proton

Question 15

Bronsted-Lowry (1923) theory of acids and bases

Answer 15

This theory removed the need for ionisation at all, and thus can explain acids which are acidic without being introduced to water. This definition is that acids are proton donors and bases are proton acceptors.

This theory still requires hydrogen atoms to be present within a molecule (e.g: BF3, AICI3) act as an acid without containing any hydrogen.
• A limitation of B-L is that it cannot account for the behaviour of acidic oxides such as SO2 or SO3 and their reaction with basic oxides like CaO.
• As there is no proton transfer, the B-L theory cannot explain it
could not account for ammonia being a base, as it does not dissociate in water to form hydroxide

Question 16

The boiling point of organic compounds

Answer 16

• Amides have the highest boiling point for the same reason as above as well as having dipole-dipole forces from their oxygen
• Amines have a higher boiling point than carboxylic acids due to their being 2 hydrogen bonds attached to nitrogen rather than only in one in carboxylic acids
  • Carboxylic acids can form both hydrogen bonds from its hydroxyl group, dipole-dipole forces from its oxygen, and dispersion forces from its non-polar region.
  • Alcohols can form hydrogen bonds due to their hydroxyl group, and dispersion forces from its non-polar region
  • Esters have less polarity than alcohols so have weaker dipole-dipole forces than both carboxylic acids and alcohols.
  • Assuming equal molar mass, esters have the lowest boiling point, then alcohols, then carboxylic acids, as more energy is needed to separate stronger intermolecular forces

Question 17

a procedure that can be used to produce the ester, ethyl ethanoate

Answer 17

1. Set up a mantle or hot plate and reflux equipment.
2. Pour 10 mL ethanol into a round bottom flask and add 20 mL of glacial acetic acid.
3. Add 1 mL of concentrated sulfuric acid to the mixture and boiling chips and heat under reflux for 45 minutes.
4. Allow to cool and then transfer contents of the round bottom flask to a separating funnel.
5. Wash with plenty of distilled water to remove any excess alcohol and acetic acid. The ester layer will float on top of the aqueous layer so expel the lower aqueous layer in the separating funnel.
6. Finally, add a solution of 1 mol L–1 sodium carbonate to remove any final traces of acid. Once again, the ester will float on top of the aqueous layer which can be let out of the bottom of the separating funnel.
   Safety: Ethanol is flammable. Use a heating mantle/hot plate rather than a naked flame to heat the solution to prevent fires.

Question 18

Difference between soaps and detergents

Answer 18

Soap and detergent molecules all consist of two parts: a hydrophobic ‘tail’ consisting of fatty acids; and a hydrophilic, charged ‘head’. The head groups vary between soap and detergents.

soaps do not have a polar head – the head of soaps and detergents (except the non-ionic ones) are IONIC
Na+ is not the end of the soap or detergent molecule. As the soap dissolves in water, the sodium ion dissociates from the soap molecule.
soap does not have a glycerol “backbone” – Soap is made from tricycerides and the triglyceride does but this is converted into two seperate species, gycerol and a soap.

Question 19

How soaps and detergents clean

Answer 19

The hydrophobic tails will form dispersion forces with the non-polar grease and will embed themselves in the grease particle. The hydrophilic heads will form ion-dipole interactions with the water. With agitation, a micelle is formed around the grease which is lifted off a surface and becomes suspended in the water. When the water is removed the cleaning is complete.

Question 20

How soap cleans clothes

Answer 20

Soap molecules consist of two parts: a hydrophobic ‘tail’ consisting of fatty acids; and a hydrophilic, charged ‘head’. The grease on dirty clothes will attract the hydrophobic tails which will embed themselves in the grease particle. The hydrophilic heads will be attracted to the water and will lift the grease off the clothes and suspend it in the water as shown in the diagram.

Question 21

how Spectrometer works

Answer 21

The positively charged ions are accelerated through an electric field before passing into a magnetic field. This causes the ions to travel in a curved path where the curvature is dependent upon the mass to charge ratio and the speed at which the ions enter the magnetic field. Lighter ions have less momentum and are deflected more strongly than heavier ions.

Question 22

Mass spectrometer analysis

Answer 22

The mass spectrometer can quickly identify components of a mixture or components
in a compound by their mass/charge ratio. Mass spectrometers are able to identify different isotopes of elements in a compound which allows the compound to be matched to a sample. Mass spectrometry is both qualitative and quantitative, requires only small quantities and is accurate, fast and sensitive.

Question 23

Ethanol as a fuel

Answer 23

Ethanol is a biofuel grown from crops such as sugar cane and corn. It is blended with petrol and used to run cars such as E10 which is made of 90% petrol and 10% ethanol. An advantage of blending fuels is that our non-renewable oil reserves will last longer, and we will be less dependent on foreign countries for our energy. Another advantage of using ethanol is that combustion releases less CO2 emissions than petrol and growing the crop absorbs CO2, hence ethanol is claimed to be carbon neutral. CO2 emissions are blamed for causing anthropogenic climate change.
The crop grows by photosynthesis which absorbs CO2 from the atmosphere: 6CO2 +6H20 + light  C6H12O6 +6O2
The sugar then needs to be fermented to obtain ethanol: C6H12O6  2C2H5OH +2CO2 + heat
Combustion of ethanol releases energy to power the vehicle: 2C2H5OH +6O2  4CO2 +6H2O + heat
Therefore, the amount of CO2 extracted by the crop from the atmosphere is equal to what is emitted. However, the claim of being carbon neutral does not consider harvesting, processing and transport requirements. Also, huge amounts of land that once produced food for human consumption are now growing crops for biofuels. This has led to a sharp increase in world food prices, particularly ion the developing world. This is leading to malnutrition and starvation. Combustion of ethanol also releases less energy than a similar mass of petrol resulting in less distance that can be travelled between refills. Ethanol blended fuels can only be used in cars designed for them as they can corrode and put the fuel systems in non-compatible cars. Overall, the use of ethanol as a fuel is problematic with the ever-increasing world population there is need for increased food production. This requires agricultural land producing food instead as biofuels as these already have a cheap and reliable source.

Question 24

Hydrogenation

Answer 24

Hydrogenation is the process of adding hydrogen atoms across a double bond in an alkene (or across a triple bond in an alkyne).

Ethene can be hydrogenated to produce ethane. A metal catalyst, eg Ni or Pt, is often used.

Question 25

Halogenation

Answer 25

Halogenation is the process of adding halogen atoms (F, Cl, Br, I) across a double bond in an alkene (or across a triple bond in an alkyne).

Ethene can be halogenated with chlorine to produce 1,2-dichloroethane.

Question 26

Hydrohalogenation

Answer 26

Hydrohalogenation is the process of adding one hydrogen and one halogen atom (F, Cl, Br, I) across a double bond in an alkene (or across a triple bond in an alkyne).

Question 27

Hydration

Answer 27

Hydration is the process of adding water molecules, or the equivalent of water molecules, to a substance

Ethylene can be hydrated to produce ethanol when heated with a dilute sulfuric acid (H2SO4) catalyst.

Question 28

Dehydration

Answer 28

Dehydration is the process of removing water molecules, or the equivalent of water molecules, from a substance.

Ethanol can be dehydrated to produce ethylene and water when heated with a concentrated sulfuric acid (H2SO4) or phosphoric acid (H3PO4) catalyst

Question 29

How to make a buffer solution

Answer 29

1. Chop a small red cabbage leaf into pieces
2. Grind up the piece in a mortar and pestle
3. Boil for a few minutes in 100ml distilled water
4. Decant the cabbage extract into a clean 100ml beaker

Question 30

How could a natural indicator be tested?

Answer 30

Pipet 10 mL of the buffer prepared in the first part of the experiment into a large test tube. Measure the pH. Add 0.10 mL of 0.10 M HCl and measure the pH again. Pipet 10 mL of the buffer prepared in the previous step into a large test tube. Measure the pH. Add 0.10 mL of 0.10 M NaOH and measure the pH again.

Question 31

Explain why the salt, sodium acetate forms a basic solution when dissolved in water

Answer 31

Sodium is a basic salt. Sodium is a neutral conjugate of a strong base. Acetate is the strong base of a weak acid, acetic acid. A strong base will react with water, accepting a proton and producing hydroxide.
CH3COO-(aq) + H20(1)   CH3COO-(aq) + OH- (aq)

Question 32

A solution is prepared by using equal volumes and concentrations of acetic acid and sodium acetate. Explain how the pH of the solution would be affected by the addition of a small amount of sodium of sodium hydroxide solution.

Answer 32

The pH of the solution will not be affected by adding sodium hydroxide. The acetic acid/ sodium acetate solution will act as a pH buffer.
CH3COOH (aq) + H20(l)   CH3OO- +H30+ (aq)
When hydroxide is added to the buffer, it will neutralise the hydronium ion, recording its concentration. This equilibrium will shift to the right, according to Le Chateliers principle, to practically counteract this change and release the pH to what it was originally.

Question 33

Primary alcohol chemical test

Answer 33

butanol can be oxidised by heating excess ethanol in acidified permanganate or dichromate. This forms an aldehyde (butanal). The aldehyde can be further oxidised to produce butanoic acid by heating under reflux. (can be separated from reaction mixture. Test with pH probe or litmus. A pH <7 would be expected.

Question 34

Secondary alcohol chemical test

Answer 34

A secondary alcohol can be oxidised by heating excess ethanol in acidified permanganate or dichromate, this forms a ketone (butanone) with the c=o bond on the second C. The reaction cannot proceed further.

Question 35

Tertiary alcohol chemical test

Answer 35

A tertiary alcohol will have the following structure and will not readily undergo oxidation reactions.

Question 36

Crystalline regions

Answer 36

In crystalline regions the chains are closer together, so the intermolecular forces are stronger. This leads to greater rigidity, higher softening and melting points, greater density, opaque appearance and resistance to air, moisture and chemicals.

Question 37

Amorphous regions

Answer 37

In amorphous regions, there are gaps between chains and weaker intermolecular forces. This leads to greater flexibility, lower softening and melting points, lower density, transparency and permeability to air, water and chemicals.

Question 38

Stability and biodegradability

Answer 38

Most of the bonds in polymers are strong covalent C-C and C-H bonds, so polymers are fairly stable and not biodegradable.
However, PVC has weaker C-Cl bonds which can be broken by ultraviolet light if it is left out in the Sun.
Special additives, called stabilisers are added to prevent it from becoming cracked and brittle in sunlight.
To increase the biodegradability of synthetic polymers, chemists have copolymerised them with natural polymer segments, such as those in starch. This causes them to break down into many smaller segments.

Question 39

Condensation polymers

Answer 39

If organic molecules have a functional group at each end, a functional group such as -COOH, can undergo condensation reaction with a functional group e.g. -NH, on another molecule to form a condensation polymer
Condensation polymers can be:
- Synthetic, e.g: polymers and polyamides (nylons)
- Natural e.g: proteins

Question 40

Natural polymers

Answer 40

hormones (insulin) and enzymes +antibodies are proteins
Proteins are large polymers (polyamides or polypeptides) formed when many smaller monomer molecules (amino acids) link together in a condensation polymerisation reaction.

Amino acids contain the amine (NH2) functional group and the carboxyl (-COOH) functional group, giving them both basic and acidic properties.

Question 41

Addition vs condensation polymers

Answer 41

Addition polymers
- monomers must be unsaturated, containing a double or triple carbon-carbon bond
- no by-producers are produced during the reaction
- the polymer back bone is a long C-C chain
condensation polymers
- monomers must contain two functional groups that can react with those on neighbouring molecules
- small molecules, often water are produced
- polymer backbone contains functional groups, e.g: amides or esters

Question 42

need for monitoring the environment

Answer 42

Environmental monitoring measures risk factors that result from by-products created as industrial and agricultural wastes. Such by chemical by-products may contaminate the environment and cause adverse effects on the organisms within the environment

Agricultural fertilisers containing phosphates and nitrates may run off into rivers and lakes. This leads to toxic algal blooms where algae grows to such an extent that aquatic life is deprived of oxygen and die

Heavy metals such as lead and mercury are environmental. They can enter soil and waterways through mining and various industrial and agricultural activities. Contamination of aquatic and terrestrial ecosystems with toxic heavy metals is an environmental problem of public health concern. Being persistent pollutants, heavy metals accumulate in the environment and consequently contaminate the food chains.

Question 43

cooling condenser

Answer 43

The apparatus is a cooling condenser. It facilitates the cooling of the reactant gases condensing them to liquids so that continued heating can take place. The reactants In this reaction are volatile so as they are heated they readily turn into gas and will escape from the reaction mixture if they are not condensed back to liquids.

Question 44

hazard producing an ester

Answer 44

Producing an ester requires concentrated sulfuric acid as a catalyst which is corrosive to eyes and skin

Question 45

What should the conical flask in a titration be rinsed with

Answer 45

the conical flask would have been rinsed with distilled water because the 25ml aliquot of sodium carbonate pipetted into the flask was the standard used in the titration and as such had a known concentration and number of moles. Rinsing with distilled water cleans the flask without influencing the number of moles of the standard.

Question 46

polyethylene

Answer 46

Polyethylene can be produced in two forms, LDPE and HDPE.

LDPE is an amorphous polymer with a high degree of chain branching that prevents the polymer chains from packing closely together, so the dispersion forces between the polymer chains will be quite weak. This makes LDPE a soft and flexible polymer which can be used as glad wrap.

HDPE is a crystalline polymer with very minimal chain branching within its structure, so the polymer chains can pack closely together, resulting in stronger dispersion forces. This makes HDPE a very hard and rigid polymer which can be used as buckets.

Question 47

Atomic absorption (AAS)
Light wavelengths

Answer 47

It is used to measure the concentration of specific metal ions in solution but does not identify what ion it is.

• A hollow cathode lamp, with the cathode made of the metal to be tested for, emits lights through the vaporised sample.
• The degree of light absorption = the concentration of the metal
• The intensity of the light is measured by a photomultiplier tube.
• The absorbance is then compared to a series of diluted standard solutions to determine concentration.

matched with a concentration using the
Calibration graph.

Question 48

UV Vis

Uv and visible light.

Answer 48

• Give a light source: entire visible+ UV spectrum (200-800nm)
• Sample unit: contains the dilute solution of the substance being test and a reference containing a blank.
• Use A=log10I0/I to determine the absorbance of sample. Then uses A=ecl to determine the concentration of the sample

matched with a concentration using the Calibration graph.

Question 49

Mass Spec

*mass spec does not use EMR as ions are separated in a magnetic field

Answer 49

Every compound as a unique mass spectrum and so mass spec can identify compounds and their structure

• Ions are formed in an ionisation chamber from vapourised compound (exposed to high voltages)
• Ions are separated in a magnetic field (based on mass to charge m/z ration)
• The number of ions with different m/z values are measured by a detector and display data as a mass spectrum
• Look at the last molecular ion peak as it gives the compounds molar mass
• Individual peaks can be used to identify fragments

Question 50

Infra-Red (IR)

Infrared light - 780 nm to 1mm

Answer 50

To determine the types of bonds present in a molecule and therefore, what molecule is present in the sample.

• Infrared light is shone through an organic sample
• Organic compounds absorb specific frequencies of IR due to chemical bonds present
• Bonds vibrate (Bend or stretch) and absorb energy resulting in an ‘excited state’
• Absorbed frequencies (wavelengths) determined by detector and used to identify specific atoms within molecules

looking up these wavenumbers on the data table.
Distinguish between narrow and broad peaks to differentiate between bonds

Question 51

Nuclear magnetic resonance (NMR)

Radio waves

Answer 51

To determine the structure or a molecule and the number of chemical environments.
Also, the number of hydrogen neighbours.

Apply external magnetic field, spin of nuclei causes alignment with field

• Application of radio waves excite particles to un-align with field
• nuclei return to original state, emitting measurable EMR. Wavelength of radio waves emitted are unique for each chemical environment.
• Few electrons surrounding atom ➡ Deshielded environments ➡ greater chemical shift
  Many electrons surrounding atom ➡ Shielded ➡ lesser chemical shift

1. Number of peaks: number of different chemical environments
2. Location of peaks: shows how shield or de-shielded the hydrogen/carbon nucleus is
3. Intensity of peaks (ONLY FOR H-NMR): height of peaks can be used as a ratio to determine the number of hydrogens in each chemical environment
4. Splitting of the peak (ONLY FOR H-NMR): ‘n + 1’ neighbour rule

Question 52

Colourimetry

Visible light

Answer 52

Colourimetry is used to test the concentrations of unknown COLOURED solutions

• A light source is shone through a filter to produce a certain wavelength
• The filtered light passes through the solution and the amount absorbed is measured
• The coloured filter is used to select colour of light that will be strongly absorbed by sample which is complementary (e.g. blue solution use orange filter)
• The machine is calibrated by measuring the light absorbed by distilled water, which should be zero

calibration curve can be created from the data, and a line of best fit can be used to calculate the concentration of an unknown substance.

Question 53

Polyvinyl chloride

Answer 53

the large chlorine atoms along the carbon chain cause the chain to be quite stiff and rigid. the polarity of the carbon-chlorine bonds also enables the formation of dipole-dipole interactions between the polymer chains which also impact the strength and rigidity of the polymer. this means that drain pipes and gutters made of PVC will maintain their shape in outdoor environments.

Question 54

Polystyrene

Answer 54

The polymer is mainly amorphous because of the way it sticks out from the carbon chain and their size. there are no polar bonds only London dispersion forces but the large ring groups restrict movement of the chain so it tends to be clear, hard and brittle. Ps makes dispossible kitchenware, home insulation and DVDS.

Question 55

Account for the differences in pH between an acetic acid solution and a hydrochloric acid 1
solution of the same concentration.

Answer 55

HCl is a strong acid that completely ionises in water to produce H+ ions, whereas acetic acid is a weak acid that only partially ionises in water to produce H+ ions. Therefore, HCl will have a higher [H+] and thus a lower pH compared to acetic acid.

Question 56

The equilibrium constant for this reaction decreases when the system is heated. Predict whether the reaction is endothermic or exothermic and justify your answer.

Answer 56

At higher temperatures, Le Chatelier’s principle predicts that the equilibrium will favour the heat-absorbing endothermic reaction in an attempt to decrease the temperature. Since K decreases when the system is heated, this indicates the equilibrium shifts left (more reactants and fewer products at equilibrium) at higher temperatures. Therefore, the reverse reaction is endothermic and so the forward reaction is exothermic.

Question 57

BP of alkanes, alcohols and carboxylic acids

Answer 57

The alkanes have low BPs because they are non-polar molecules that are held together by weak dispersion forces which require little energy to overcome. Alcohols, however, have higher BPs since they are polar molecules due to their polar hydroxyl (OH) group, and are capable of forming strong hydrogen bonds which require significant energy to overcome.

It can also be seen that when molecular mass increases, the BP of substances in each homologous series increases. This occurs because larger molecules have stronger dispersion forces, since they contain more electrons and there is a higher probability of temporary dipoles forming.