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chemical synthesis Flashcards

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1
Q

What is green chemistry?

A

Defined as the principles developed by chemists and the chemical industry to enact a more sustainable industry

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2
Q

What are biofuels

A

renewable energy sources produced from biomass which have limited environmental and health impacts as well as producing less emissions

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3
Q

What conditions are required for fermentation

A

As the second step is an exothermic process, cooling is required to prevent a denaturing of enzymes

Slighly acidic pH (3-5) is also used

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4
Q

In what circumstance can glucose be converted into ethanol

A
  • During the initial phase of fermentation, yeast grows and reproduces rapidly, respiring aerobically and quickly using up all available oxygen
    • If reaction vessel is sealed, this then causes anaerobic respiration to take place where glucose is then converted to ethanol
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5
Q

What nutrients can be added to promote growth + reproduction of yeast enzymes

A

diamond hydrogenphosphate

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6
Q

Equation for synthesis from ethene

A

CH2=CH2(g) + H2O(g) ⇌ CH3CH2OH(g) ΔH = ‒45 kJ (phosphoric acid catalyst)

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7
Q

What molar ratio is required for synthesis from ethene and why

A

Molar ratio of ethene to water is 1:0.6 as 1:1 ratio will cause steam to dilute and washes away phosphoric acid

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8
Q

1st step of Fermentation name and equation

A

Hydrolysis of Sucrose

C12H22O11(g) + H2O(l) —Invertase» 2C6H12O6(aq)

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9
Q

Conditions, Pressure and % Yield for Synthesis via Ethene

A
  • Temperature of around 300C and pressure of 6-7 MPa
  • Yield of 95%
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10
Q

2nd step of Fermentation name and equation include states

A

Fermentation

C6H12O6(aq) —Zymase» 2C2H5OH(aq) + CO2 (G)

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11
Q

Overall Fermentation Equation + States

A

C12H22O11(aq) + H2O(l) —Yeast Enzymes» 4C2H5OH(aq) + 4CO2 (g)

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12
Q

Haber Process Equation

A

N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH = ‒92 kJ

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13
Q

Haber Process Purpose, Temperature, Pressure and Catalysts

A
  • Purpose is to produce ammonia on an industrial scale
  • Temperature conditions of 350-550C
  • Pressure conditions 15-35MPa
  • Catalyst of MgO and SiO2
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14
Q

The ammonia produced (with a yield of 15-30%) is condensed so that the ammonia is liquefied and can be drained off whilst the remaining N2 and H2 can be recycled. How can the NH3(g) be liquefied and the N2 and H2 remain gaseous?

A
  • NH3 possesses stronger intermolecular forces of hydrogen bonding, dipole-dipole and dispersion forces
  • N2 and H2 possess only dispersion forces
  • Thus, NH3 has a higher boiling point as its bonds require more energy to overcome and will liquefy first
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15
Q

How are the reactants for Haber Process sourced

A
  • N2 is obtained from the atmosphere
  • H2 is produced through 2 reactionsCH4(g) + H2O(g) ⇌ CO(g) + 3H2(g) ΔH = +206 kJ - Steam ReformingCO(g) + H2O(g) ⇌ H2(g) + CO2(g) ΔH = -41 kJ - Shift Reaction
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16
Q

Contact Process Purpose and examples of produce

A
  • Produces sulfuric acid on an industrial scale
    • Can be used to produce fertilisers, detergents and other acids
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17
Q

Stage 1 of Contact Process name and equation include states

A

Stage 1: Production of Sulfur Dioxide

S(l) + O2(g) → SO2(g) ΔH = -297 kJ

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18
Q

Stage 2 of Contact Process name and equation

A

Conversion of Sulfur Dioxide to Sulfur Trioxide

2SO2(l) + O2(g) ⇌ 2SO3(g) ΔH = -198 kJ

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19
Q

Stage 2 of Contact Process Temperature, Pressure and Catalysts

A
  • Uses a temperature of 450C
  • Uses a pressure of 100-200kPa
  • Uses a V2O5 catalyst to increase the reaction rate
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20
Q

Explain using collision the effect of removing SO3 product on reaction rate and yield

A
  • Does not affect reaction rate
  • However, increases yield as partial pressure of SO3(g) is reduced, this increases the distance between SO3(g) particles which decreases the frequency of successful collisions between them
    • As a result this decreases the rate of the reverse reaction relative to the forward reaction
    • As the forward reaction rate is now greater relative to the reverse reaction rate, the yield of SO3(g) is increased
21
Q

Give Stage 3 of contact process name of process and equations

A

H2SO4(l) + SO3(g) → H2S2O7(l) - Oleum Production

H2S2O7(l) + H2O(l) → 2H2SO4(l) - Water Added

H2O(l) + SO3(g) → H2SO4(l) - Overall

22
Q

Explain how soap can clean greasy dishes

A
  • The strength of the dispersion forces between thenon-polartail of the soap molecule (or surfactant ion) and thenon-polargrease are strong enough to overcome the soap-soap dispersion forces and grease-grease dispersion forces
  • The strength of theion-dipole forcesbetween the anionic head and water are strong enough to overcome the soap-soap dispersion forces and water-water hydrogen bonding, dipole-dipole bonding and dispersion forces.
  • The soap molecule forms amicellewhich lifts the grease from the plate and allows it to forman emulsionin the water
    • Hot water and agitation further improve the cleaning action of soap
23
Q

What is cross linking and why does it occur

A
  • Effectively makes many polymers into one giant polymer through formation of covalent bonds between different polymer chains
  • Improves strength, chemical resistance and raises melting point
24
Q

N2 + 3H2(g) –> 2NH3(g) ^H = -92kJ
Endothermic or Exothermic

A

Exothermic

25
Q

If heat is a product, is it exothermic or endothermic

A

Exothermic

26
Q

Effect of cataylst on yield

A

NO IMPACT
Rates of forward and reverse reaction are increased equally

27
Q

Effect of a catalyst on reaction rate

A

Increases the rate of reaction because - Adding a catalyst provides an alternate reaction pathway with lower activation energy, so a larger percentage of collisions have enough energy to meet activation energy
- This increases the rate of successful collisions thus increasing reaction rate

28
Q

Explain an increase in pressure having an increase/decrease in yield

A

An increase in pressure will decrease the distance bteween the particles, increasing the frequency of successful collisions which increases the rate of both the forward and reverse reaction
As the ratio of reactants to products is greater/less (give specific example), the forward reaction will be greater relative to the revrse reaction
Results in an increase/decrease in yield

29
Q

Explain an increase in pressure having an increase in reaction rate

A

An increase in pressure, decreases the distance between particles, increasing the frequency of successful collisions which increases the reaction rate both of the forward and reverse reaction

30
Q

Compromise statement for temperature

A

Moderate temperature ensures acceptable yield at a reasonable rate

31
Q

Compromise statement for pressure

A

The pressure is limited by dangers associated with operating at high pressures and cost of specialised machinery and generating high pressure

32
Q

Explain an increase in temperature having an increase in reaction rate

A

Increasing temperatures increases the average kinetic energy of the particles, so a larger percentage have enough energy to meet the activation energy, it also increases their velocity, so the rate of successful collisions increases, increasing the rate of both the reverse and forward reactions

33
Q

Explain how an increase in temperature can increase/decrease yield

A

Increased temp increases the rate of both the forward and reverse reactions
The endothermic reation however, is favoured and increased more
Thus, the endothermic reaction rate will be greater relative to the exothermic reaction rate thus yield can be increased or decreased

34
Q

Explain how side chains would affect a polymer’s melting point

A

The presence of side chains would reduce how close together the molecules can pack, reducing their total intermolecular forces, reducing the energy needed to overcome them, reducing the melting point

35
Q

Are ethenes saturated or unsaturated

A

Unsaturated

36
Q

Lipase Enzyme Catalyst Advantages

A
  • Mild temperature (less energy required)
  • Can tolerate feedstocks high in free fatty acids
  • Can convert both triglycerides and fatty acids to biodiesel
  • No side reactions
37
Q

Lipase Enzyme Catalyst Disadvantages

A
  • Slower reaction rate - higher concentration of enzymes needed
  • Expensive
  • Recovery from final mixture is difficult
  • High concentration of alcohol inhibits its use
38
Q

Two methods in which biodiesel can be produced

A
  • Pretreatment with methanol and sulfuric acid
  • Lipase enzymes for whole process
39
Q

Pre-treatment with methanol and sulfuric acid Advantages

A
  • Faster
  • Cheaper
40
Q

Pre-treatment with methanol and sulfuric acid Disadvantages

A
  • 2 Steps
  • High temperature - more energy
41
Q

Benefits of using a biological catalyst

A
  • Lower Temperature
  • Lower pressure
  • Saves money
  • More environmentally friendly
42
Q

Explain Properties and Uses of Polyethene

A

Low Density Polyethene
- Relatively low melting point, soft, flexible and insulating
- Used for packing film, tubing, bottles, toys

High Density Polyethene
- Relatively higher melting point, hard, rigid
- Used for chopping boards, pipes, toys and crates

43
Q

Explain Forces, Properties and Uses of Polytetrafluorothylene (PTFE)

A
  • High strength C-F bponds give it high chemical resistance, high electrical resistance, high melting point and is non-stick
  • Used for cooking pans, semi-permeable membranes in fuel cells and coatings on bearings
44
Q

Explain Forces, Properties and Uses of Polythylene Terephthalate (PET)

A
  • Dipole-Dipole forces give it a Strong, rigid and transperent properties
  • Thermal blankets, plastic bottles, fibres, solar cells
45
Q

Explain Forces, Properties and Uses of Nylon

A
  • Ability for hydrogen bonding between lone pair of electrons of oxygen in carbonyl group and hydrogen atom from amine group give it strong tensile strength, high melting point, high chemical resistance
  • Used in fishing lines, rope, carpet, parachutes, tents
46
Q

2 Examples of condensation polymers

A

Nylon
Polyethylene terephthalate (PET)

47
Q

2 Examples of addition polymers

A

Polyethene
Polytetrafluorothylene (PTFE)

48
Q

What happens as branching of polymers increase

A

This decreases intermolecular forces as molecules are unable to pack as closely together compared to their unbranched couterparts
Thus, with this decreased intermolecular forces, the bonds require less energy to overcome decreasing their strength, rigidity and melting/boiling points

chem again (7 decks)

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