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Chemistry A LEVEL > Common Qs > Flashcards

Common Qs Flashcards

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1
Q

Why the compounds formed in the following equations can be described as salts

A

The hydrogen ions in the (_____) acid have been replaced by ____ ions

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2
Q

Explain how the ‘Electron Pair Repulsion Theory’ can be used to predict the shape of a molecule

A

Electron pairs repel each other to get as far apart as possible
• Lone pairs of electrons repel more strongly than bonding pairs
• Shape is determined by the number and type of electron pairs (bond pairs / lone pairs) around the
central atom
• YOU MUST STATE THE NUMBER OF BOND PAIRS AND LONE PAIRS OF ELECTRONS

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3
Q

State and explain the shape and bond angle in CH4

A

Shape is tetrahedral and bond angle 109.5o
• There are 4 bonding pairs of electrons around the C atom
• Bonding pairs repel equally

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4
Q

State and explain the shape and bond angle in NH3

A
  • Shape is pyramidal and bond angle 107o
  • There are 3 bonding pairs of electrons and 1 lone pair around the N atom
  • Lone pair of electrons repel more strongly than bonding pairs
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5
Q

Explain why water, H2O is a polar molecule

A
  • O is more electronegative than H
  • Makes a permanent dipole across the O-H bonds
  • (water is not symmetrical so dipoles don’t cancel)
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6
Q

Explain why carbon dioxide, CO2, is non-polar although the C-O bond is polar

A
  • CO2 is symmetrical (linear)

* Dipoles cancel

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7
Q

Hydrogen bonding drawing

A

Draw

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8
Q

Describe and explain two anomalous properties of water, resulting from hydrogen bonding

A

• The boiling point of water is higher than expected (or the melting point of ice), as hydrogen bonds
between molecules are the strongest intermolecular force and more heat energy is needed to break
them
Liquid
• Ice is less dense than water, as hydrogen bonds are longer than covalent bonds, so they hold the
molecules apart in an open lattice

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9
Q

Diamond

•

A

Giant covalent lattice
• Very high melting/boiling point; all strong covalent bonds between atoms so lots of heat energy needed
to break them
• Does not conduct electricity as there are no ions or electrons; all electrons are fixed in covalent bonds

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10
Q

Graphite

A

• Giant covalent lattice
• Weak induced dipole-dipole interactions between layers of atoms
• High melting/boiling point as there are many strong covalent bonds between atoms so lots of heat
energy needed to break them
• Conducts electricity as there are delocalised electrons which are free to move between the layers and
carry charge

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11
Q

Graphene

•

A

Giant covalent lattice
• High melting/boiling point as there are many strong covalent bonds between atoms so lots of heat
energy needed to break them
• Conducts electricity as there are delocalised electrons which are free to move and carry charge

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12
Q

Sodium chloride (true for any compound of metal and non-metal)

A

• Giant ionic
• High melting/boiling point as there are strong electrostatic attraction between oppositely charged ions
(ionic bonds) so lots of heat energy needed to break them
• Does not conduct electricity when solid as the ions are fixed in the lattice; does conduct electricity
when molten or in solution as the ions are free to move and carry charge

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13
Q

Iodine (true for any element or compound of non-metals)

A

• Simple covalent / molecular
• They have induced dipole-dipole interactions between molecules
• Low melting/boiling points are little heat energy is needed to overcome intermolecular forces
• Does not conduct electricity as there are no mobile ions or electrons (charge carriers); all electrons are
fixed in covalent bonds

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14
Q

Sodium (true for any metal)

A

• Giant metallic lattice
• High melting/boiling point as there are strong electrostatic attraction between the lattice of positive
metal ions and delocalised electrons (metallic bonds) so lots of heat energy needed to break them
• Conducts electricity as the delocalised electrons can move and carry charge

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15
Q

Explain, in terms of bond breaking and forming, why a particular reaction is exothermic.

A
  • Breaking bonds requires/absorbs energy
  • Forming bonds releases energy
  • (In an exothermic reaction) more energy is released than required
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16
Q

Explain, in terms of bond breaking and forming, why a particular reaction is endothermic.

A
  • Breaking bonds requires/absorbs energy
  • Forming bonds releases energy
  • (In an endothermic reaction) more energy is required than released
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17
Q

Why is the value of an enthalpy change of combustion, calculated from a calorimetry experiment,
different to its data book value?

A
  • Incomplete combustion
  • Heat loss
  • Non-standard conditions
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18
Q

Why is the value of an enthalpy change, calculated using mean bond enthalpies, different to its data
book value?

A
  • Mean bond enthalpies are average values; they do not measure the exact value of the bond
  • Bonds have different strengths in different environments
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19
Q

Why is it not possible to measure the enthalpy change of formation directly, for the following reaction:
N2(g) + 1⁄2O2(g)→N2O(g)

A
  • Activation energy is too high

* Other products may be formed, e.g. NO, NO2

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20
Q

Explain the effect of increasing concentration (of solution) on the rate of reaction

A
  • More particles / molecules per unit volume
  • More frequent successful collisions
  • Rate increases
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21
Q

Explain the effect of increasing pressure (of gas) on the rate of reaction

A
  • More particles / molecules per unit volume
  • More frequent successful collisions
  • Rate increases
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22
Q

Using Boltzmann distribution diagrams, explain the effect of an increase in temperature on the rate of
reaction

A

• (As temperature increases) there is a greater proportion of molecules with energy greater than or equal
to the activation energy (as shown by the increased shaded area on the graph)
• More frequent successful collisions
• Rate of reaction increases

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23
Q

Using Boltzmann distribution diagrams, explain the effect of a decrease in temperature on the rate of
reaction

A

• (As temperature decreases) there is a smaller proportion of molecules with energy greater than or
equal to the activation energy (as shown by the decreased shaded area on the graph)
• Less frequent successful collisions
• Rate of reaction decreases

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24
Q

Using a Boltzmann distribution diagram, explain the effect of adding a catalyst on the rate of reaction

A

• (With a catalyst) the activation energy is lower
• There is a greater proportion of molecules with energy greater than or equal to the activation energy
(as shown by the increased shaded area on the graph)
• More frequent successful collisions
• Rate of reaction increases

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25
Q

Using an enthalpy profile diagram, explain the effect of adding a catalyst to the rate of reaction

A
  • (With a catalyst) the activation energy is lower
  • There is a greater proportion of molecules with energy greater than or equal to the activation energy
  • More frequent successful collisions
  • Rate of reaction increases
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26
Q

State the benefits of using catalysts in industry

A

• Lower energy demand: lower temperature can be used, which reduces CO2 emissions / burning fossil
fuels
• (Different reactions could be used with) higher atom economy

27
Q

State the benefits of using enzymes to catalyse reactions

A
  • Lower energy demand: lower temperature can be used, which reduces CO2 emissions
  • Enzymes are very specific
  • Higher atom economy
  • Less use of toxic reactants / solvents
28
Q

Describe and explain, using le Chatelier’s principle, the effect of an increase in temperature on the
position of equilibrium on the following reaction: (exothermic)

A

Forward reaction is exothermic
• (When temperature is increased) the position of equilibrium shifts to the left
• To minimise the effects of the increase in temperature by absorbing energy

29
Q

Describe and explain, using le Chatelier’s principle, the effect of an increase in pressure on the position
of equilibrium on the following reaction:
e.g. 2A(g) + B(g) C(g) + D(g)

A
  • There are more gaseous moles of reactants than products (more gaseous moles on the left hand side)
  • (When pressure is increased) the position of equilibrium shifts to the right
  • To minimise the effects of the increase in pressure by reducing the pressure
30
Q

Describe and explain, using le Chatelier’s principle, the effect of the addition of a catalyst on the position
of equilibrium:

A
  • Catalyst increases the rate of the forward and reverse reactions equally
  • No effect on the position of equilibrium
31
Q

What is the advantage of a high percentage yield?

A
  • Shows that there is little waste in the starting materials

* Efficient conversion of reactants to products

32
Q

What is the advantage of a high atom economy?

A
  • Shows that there is little waste products
  • More of the desired product
  • 100% atom economy when there is only one product (addition reaction)
  • Lower atom economy when there is a by-product (substitution reaction)
  • Atom economy can be increased by finding a use for the by-product
33
Q

Describe, using a labelled diagram, the bonding in a C=C double bond.

A

Draw

34
Q

Give the mechanism for the reaction between, e.g. 2-methylbut-1-ene with hydrogen bromide,
explaining which of the two possible organic products is more likely to be formed.

A
  • Major product (2-bromobutane) is formed via the tertiary (3o) carbocation
  • Minor product (1-bromobutane) is formed via the primary (1o) carbocation
    • The tertiary carbocation is more stable than the primary carbocation
35
Q

Explain how E/Z (or cis/trans) isomers occur.

A

• Carbon-carbon double bond
• Each carbon atom of the double bond is attached to two different groups
Restricted rotation about the double bond

36
Q

Describe the problems with disposing of polymers, e.g. PVC.

A
  • Non-biodegradable

* (With PVC) toxic fumes of hydrogen chloride, HCl, are released when burned

37
Q

How are scientists working to minimise the problems associated with disposal of polymers

A
  • HCl can be removed by dissolving in water then reacting with a base, e.g. NaOH
  • (Scientists are developing) biodegradable or compostable polymers
38
Q

State ways, other than landfill and combustion, of processing waste polymers

A
  • Sorting and (mechanical) recycling

* Feedstock recycling (or cracking)

39
Q

Compare the reactivity of e.g. chloroalkane to bromoalkane in nucleophilic substitution

A
  • Bromoalkane is more reactive than chloroalkane
  • C-Br bond is weaker than C-Cl bond (C-X bond gets weaker down the group)
  • Less energy is needed to break the C-Br bond
40
Q

Describe and explain the difference in boiling point between an alcohol and a hydrocarbon of similar
molecular mass

A

• Alcohol has a higher boiling point
• There is hydrogen bonding between molecules
• Hydrogen bonding is a stronger than induced dipole-dipole interactions between hydrocarbon
molecules
• More heat energy is needed to break the intermolecular forces

41
Q

Give a use for infrared spectroscopy

A

• Breathalysers or to monitor air pollution

42
Q

Outline the use of infrared spectroscopy in monitoring air pollution

A
  • Identifies functional groups

* Match spectra to spectra of known pollutants

43
Q

Describe and explain the effect of an increase in temperature on the composition of the following
equilibrium mixture and on the value of Kc
• The forward reaction is exothermic

Kc = [NH3]2 [N2] [H2]3

A
  • If temperature increases the position of equilibrium shifts to the left
  • The reverse (endothermic) reaction takes in the heat
  • The composition of the reaction mixture: concentration of reactants increases (bottom of Kc)

concentration of products decreases (top of Kc)

• The value of Kc decreases

44
Q

Describe and explain the effect of an increase in concentration of N2 on the composition of the following
equilibrium mixture and on the value of Kc

N2(g) + 3H2(g) —> 2NH3(g)

A

• If concentration of N2 increases the position of equilibrium shifts to the right
• To minimise the effects of the increase in concentration
There is no effect on Kc
• When the concentration of N2 increases, the bottom/denominator of Kc increases (or the ratio decreases)
• So the concentration of products increase to make the top/numerator of Kc increase (or to increase the ratio)
Value of Kc is restored

45
Q

Describe and explain the effect of an increase in pressure on the composition of the following
equilibrium mixture and on the value of Kp

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

A

• The reaction has 4 moles gas→2 moles gas
Kp = p(NH3)2 p(N2) (H2)3

• If pressure increases the position of equilibrium shifts to the right
• There are fewer gaseous moles on the right
• The pressure decreases
There is no effect on Kp
• When the total pressure increases, all the partial pressures increase
• There are more partial pressures of reactants so the bottom/denominator of Kp increases (or the ratio
decreases) more than the top/numerator of Kp
• So the partial pressures of products increase to make the top/numerator of Kp increase (or to increase the
ratio)
• The value of Kp is restored

46
Q

Why is it difficult to predict the effect of an increase in temperature and an increase in pressure on the
composition of the following equilibrium mixture:

N2(g) + 3H2(g) —> 2NH3(g)
Exo

A

If temperature increases the position of equilibrium shifts to the left (forward reaction is exothermic)
• If pressure increases the position of equilibrium shifts to the right (fewer gaseous moles on the right hand side

The relative effects of temperature and pressure are not known

47
Q

Explain what is meant by a buffer, giving an example, and describe how it is prepared.

A

• A buffer is a mixture which minimises pH changes when small amounts of acid or alkali are added
• It is a mixture of a weak acid (e.g. propanoic acid, CH3CH2COOH) and its salt (e.g. sodium propanoate,
CH3CH2COONa)
• It is prepared by mixing a small amount of a strong base (e.g. NaOH) with an excess of the weak acid; this
produces the salt but also leaves some acid remaining

48
Q

Explain how a buffer works.

A
  • Weak acid: CH3CH2COOH CH3CH2COO- + H+
  • Salt: CH3CH2COONa → CH3CH2COO- + Na+
  • If acid (H+) is added: CH3CH2COO- + H+ → CH3CH2COOH position of equilibrium shifts to the left
  • If alkali (OH-) is added: H+ + OH- → H2O position of equilibrium shifts to the right
49
Q

How is a suitable indicator chosen for a particular acid-base titration?

A

• The colour change (end-point) of the indicator must lie within the pH range of equivalence (equivalence
point) (quote this range from the pH curve – the vertical section)
• The colour change must also be sharp and distinct

50
Q

compare the lattice enthalpies of NaCl and MgCl2

A
  • Mg2+ ion has a higher charge than the Na+ ion
  • Mg2+ ion is also smaller than the Na+ ion
  • Greater attraction between Mg2+ and Cl- ions
  • MgCl2 has a more exothermic lattice enthalpy
51
Q

compare the lattice enthalpies of KCl and CaO

A
  • Ca2+ ion has a higher charge than the K+ ion
  • Ca2+ ion is smaller than the K+ ion
  • O2- ion has a higher charge than Cl- ion
  • O2- ion is smaller than Cl- ion
  • Much greater attraction between Ca2+ and O2- ions!
  • CaO has a much more exothermic lattice enthalpy
52
Q

compare the lattice enthalpies of MgS and Na2O

A
  • Mg2+ ion has a higher charge than the Na+ ion
  • Mg2+ ion is smaller than the Na+ ion
  • O2- ion is smaller than S2- ion
  • Mg2+ attracts negative ions more strongly than Na+
  • But O2- attracts positive ions more strongly than S2-
  • Cannot predict which lattice enthalpy is the most exothermic!
53
Q

compare the hydration enthalpies of Na+ or K+

A
  • Na+ ion is smaller than K+ ion
  • Greater attraction between Na+ and H2O molecules
  • Na+ has a more exothermic hydration enthalpy
54
Q

compare the hydration enthalpies of Cl- or O2-

A
  • O2- ion has a higher charge than the Cl- ion
  • O2- ion is also smaller than Cl- ion
  • Greater attraction between O2- and H2O molecules
  • O2- has a more exothermic hydration enthalpy
55
Q

Explaining feasibility

A

Look at sheet

56
Q

Describe, using diagrams, the bonding and structure of benzene

A

Draw

57
Q

What is the evidence for the delocalised model of benzene structure?

A

• The delocalised electrons make benzene resistant to reaction; when it does react it is by substitution rather than
addition
• It only reacts with bromine (or chlorine) in the presence of a halogen carrier
• in benzene all the carbon-carbon bond lengths are equal and intermediate between single and double bonds
• the enthalpy of hydrogenation is less exothermic than expected

58
Q

Why is benzene less reactive than cyclohexene?

A
  • Benzene has (6) delocalised pi electrons spread over the six carbon atoms
  • This has lower electron density
  • Cyclohexene has (2) localised pi electrons spread over the two carbon atoms
  • This has higher electron density
  • Electrophiles are less readily polarised and less strongly attracted to benzene
59
Q

Why is phenol more reactive than benzene?

A

In phenol the lone pair of electrons on the oxygen atom (p orbital) is donated
• Becomes (partially) delocalised into the -system (8 electrons)
• This has higher electron density (than benzene)
• Electrophiles are more readily polarised and more strongly attracted to phenol

60
Q

How could you identify an unknown carbonyl compound (aldehyde / ketone)?

A
  • Add 2,4-DNPH (acid solution in methanol)
  • Orange crystalline precipitate
  • Purify the precipitate by recrystallization
  • Measure its melting point
  • Compare to known melting points
61
Q

Why do amines behave as bases?

A
  • Lone pair of electrons on nitrogen atom is donated
  • Accepts proton
  • By forming dative covalent (coordinate) bond
62
Q

What information about a molecule can be deduced from a 1H NMR spectrum?

A

• The number of peaks gives the number of proton environments in the molecule
• The relative areas (intensities) of the peaks gives the number of protons in each environment
• The chemical shift value gives the type of proton environment, e.g. d 1-2 ppm R-CH
• The splitting pattern gives the number of protons on the adjacent carbon; the ‘n + 1’ rule: if there are n protons
on the adjacent carbon, the peak is split into n+1 peaks
• e.g. if there are 2 protons on the adjacent carbon, the peak is split into a triplet

63
Q

Describe the use of TMS in NMR spectroscopy

A
  • It is a standard – chemical shift delta set as zero
  • Give large singlet peak at higher frequency than other organic compounds
  • Also chemically inert (does not react with sample)and volatile (can be removed from sample)
64
Q

Describe the use of deuterated solvents, e.g D2O in 1H NMR spectroscopy

A

• Solvents dissolve the sample
• In 1H NMR hydrogen atoms present in a solvent would give peaks on spectrum
• All hydrogen atoms are replaced by 2H, deuterium, which does not give peaks
• If a sample is mixed with the deuterated solvent D2O (only), labile protons on O-H and N-H groups are replaced by
deuterium and O-H and N-H peaks disappear from the sample
• This is useful to confirm the presence of O-H and N-H groups

Chemistry A LEVEL (10 decks)

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