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AQA A-Level Chemistry Year 2Student Book Answers
CHAPTER 1
Assignment 1A1. a. ΔfHƟ[MgCl(s)] = 148 + 738 + 121 − 349 − 815 = −157 kJ mol−1
b. ΔfHƟ[MgCl3(s)] = 148 + 738 + 1451 + 7733 + 121 − 349 − 5540 = +4302 kJ mol−1
A2. Its enthalpy of formation is highly endothermic.
A3.
Assignment 2A1. 1000 g
A2. 98 g
A3. 98/18 moles
A4. 2.4 × 100 kJ
A5. (2.4 × 100)/(98/18) = 44.1 kJ mol−1
A6. 373 K
A7. (44.1 × 1000)/373 = 118 J mol−1 K−1
A8. For example, reduce energy loss by improving the kettle’s insulation; make sure that no water vapour condenses and returns to the kettle; try experiments with different quantities of water and compare results.
Assignment 3A1. a.
ΔfHƟ [KCl(s)] = (+89) + (+112) + (+419) + (−348) + (−711)
= −439 kJ mol−1
b.
ΔfHƟ[KCl2(s)] = +(+89) + (2 × +122) + (+419) + (+3051) + (2 × −348) + (−2350)
= +757 kJ mol−1
Generally, a transfer of energy from the surroundings is needed to overcome attractive forces, and energy is transferred to the surroundings (released) when particles are attracted to each other and move towards each other.
Atomisation is an endothermic change. A transfer of energy from the surroundings is needed.
In metals, a transfer of energy from the surroundings is needed for atoms to break free from the lattice, where positive ions are held together by a sea of mobile, negative electrons.
In covalent molecules, a transfer of energy from the surroundings is needed to form separate atoms by breaking covalent bonds.
The ionisation enthalpy is endothermic. The negative electrons are attracted by the positive nucleus. A transfer of energy from the surroundings is needed to remove them.
The electron affinity is exothermic for the first electron and endothermic for any further electrons. Electrons are attracted towards the positive nucleus, so energy is released. If more electrons are introduced then the electron–electron repulsion means that a transfer of energy from the surroundings is needed.
The enthalpy of lattice formation is exothermic. There are attractive forces between the positive and negative ions, so energy is released. ΔfH for KCl is −429 kJ mol−1 (exothermic) and ΔfH for KCl2 is +757 kJ mol−1 (endothermic); therefore, KCl is the more favoured in terms of enthalpy changes.
A2.
ΔlattHƟ (NH4Cl) = 705 kJ mol−1
Therefore ΔsolHƟ = (+705) + (−307)+(−381)
= 17 kJ mol−1
A3. For a reaction to occur spontaneously, the Gibbs free energy must be zero or negative, and ΔG = ΔH − T ΔS. The process proceeds spontaneously because there is an increase in entropy as the ordered crystal arrangement changes to a disordered system in solution. Even though ΔH is positive, the increase in entropy (and, combined with the temperature, −T ΔS) is sufficiently negative to make ΔG negative.
A4. ΔH = mc ΔT = 200 g × 4.2 J g−1 K−1 × 16.5 K = 13 860 J = 13.86 kJ
A5. The molar enthalpy change for ammonium nitrate is +26.5 kJ mol−1, so the number of moles needed is
5 = 0.52 mol
Practice questions1a
Mg2+(g) + 2e− + 2Cl(g) (Note: this is the only answer for the top line)
Mg2+(g) + 2e− + Cl2(g)
Mg+(g) + e− + Cl2(g)
Mg(g) + Cl2(g)
1b
642 + 150 + 736 + 2 × 121 = 2 × 364 + 2493 = = +1451 kJ mol−1
1c
ΔH = −ΔH(lattice formation) + ΣΔH(hydration)
= 2493 – 1920 – 2 × 364 = −155 kJ mol−1
1d(i)
Increase in disorder on dissolving or ΔS positive
ΔG is negative or TΔS > ΔH
1d(ii)
Moles of NH4Cl = 2/53.5 = 0.0374
Heat absorbed = 15 × 0.0374 = 0.561
Q = m c ΔT
ΔT = Q/mc = (0.561×1000)/(50×4.2) = 2.6°C
Final temperature = 20 – 2.6 = 17.4°C
2a
ΔHϴ = ΣΔH(formation products) - ΣΔH(formation reactants)
= 3 × −111 – (−1669) = +1336
ΔSϴ = ΣS(products) - ΣS(reactants)
= 2×28 + 3×198 – (51 + 3×6) = +581
ΔGϴ = ΔH – TΔS
= 1336 – (298×581)/1000 = 1163
ΔGϴ is positive (or free energy (G) increases)
2b
When ΔGϴ = 0 or T = ΔHϴ/ΔSϴ
= (1336×1000)/581 = 2299 K
2c
Enthalpy too high or reaction too slow
2d
Method: electrolysis
Conditions: molten or high T or 500–1500°C or dissolved Cryolite
3a
Particles are in the maximum state of order (entropy is zero at 0 K by definition)
1
3b
Ice melts (or water freezes)
3c
The increase in disorder is bigger (at T2)
3d(i)
Moles of water = 1.53/18 = 0.085
Heat change per mole = 3.49/0.085 = 41.1 kJ mol−1
3d(ii)
ΔG = ΔH – TΔS
3d(iii)
ΔH = TΔS or ΔS = ΔH/T
ΔS = 41.1/373 = 0.110 kJ K−1 (mol−1 (or 110 (J K−1 (mol−1))
4a
1s2 2s2 2p6 3s2 3p6
4b
S–(g)
4c
The negative S– ion repels the electron being added
4d(i)
Enthalpy of atomisation of sulfur
4d(ii)
Second ionisation enthalpy of calcium
4d(iii)
Second electron affinity of sulfur
4e
Electron more strongly attracted nearer to the nucleus or attracted by Ca+ ion
4f
+178 + 279 + 590 +1145 – 200 + ΔH –3013 + 482 = 0
ΔH = 539
5a
ΔS = 238 + 189 – 214 – 3 × 131 = –180 J K–1 mol–1
ΔG = ΔH – TΔS
= –49 – 523 × (–180)
1000
= +45.1 kJ mol–1
5b
When ΔG = 0, ΔH = TΔS
Therefore, T = ΔH/ΔS = –49 × 1000/–180 = 272 K
5c
Diagram of a molecule showing O–H bond and two lone pairs on each oxygen
Hydrogen bonding is a strong enough force to hold methanol molecules together in a liquid.
CHAPTER 2Assignment 1A1. The temperature of the reaction mixture.A2. A pipette and safety filler. The size of the pipette depends on the volume required, but 10.0 cm3 is a reasonable choice. Smaller pipettes will be slightly less accurate; larger pipettes require a larger volume of reaction mixture to allow a sufficient sample to be taken.A3. Sodium hydroxide solution (concentration depends on initial concentration of ester).A4. Alkalis are corrosive. It is best to work with concentrations as low as the experiment allows. Eye and hand protection should be used and a safety filler used to fill the pipette. Should any alkali get onto a person’s skin, it should be washed off immediately with cold running water.
A5.
A6. a. At 0.200 mol dm−3, rate = 2.3 mol dm−3 s−1 (note: to calculate rate, time must be converted from minutes to seconds) At 0.150 mol dm−3, rate = 1.9 mol dm−3 s−1
At 0.100 mol dm−3, rate = 1.2 mol dm−3 s−1
At 0.050 mol dm−3, rate = 0.7 mol dm−3 s−1
Because tangents are difficult to draw with accuracy, the rates are given to just 2 significant figures)b.
Since the graph is a straight line, the order of the reaction with respect to methyl ethanoate is 1.
c. The gradient of the rate against the ester concentration gives the rate constant k for the hydrolysis of methyl ethanoate as 4 s−1.
Assignment 2A1. A light probe in, for example, a colorimeter can accurately record the moment at which the dark blue colour appears. Linking the probe to an electronic timer allows the time for the colour change to be measured accurately, avoiding human error when trying the judge the endpoint.
A2. TemperatureA3. The final volume of reaction mixture in all experiments is 100 cm3, so the concentration of sodium thiosulfate is 0.005 mol dm−3.
Reaction 1: 2.00 × 10−4 mol dm−3 s−1. Reaction 2: 1.61 × 10−4 mol dm−3 s−1. Reaction 3: 1.22 × 10−4 mol dm−3 s−1. Reaction 4: 0.81 × 10−4 mol dm−3 s−1. Reaction 5: 0.48 × 10−4 mol dm−3 s−1.A4. Reaction 1: 0.05 mol dm−3. Reaction 2: 0.04 mol dm−3. Reaction 3: 0.03 mol dm−3. Reaction 4: 0.02 mol dm−3. Reaction 5: 0.01 mol dm−3.A5.
A6. Straight-line graph, so first order with respect to hydrogen peroxide.
Assignment 3A1. a. Changing the concentration of OH− has no effect on rate, so zero order with respect to OH−. Halving the concentration of C4H9Br halves the rate, so first order with respect to C4H9Br.
b. Rate = k[C4H9Br]
c. From experiment 1, k = 0.001/2.5 = 0.0004 (or 4 × 10−4) s−1.
d. SN1 since the rate-determining step involves just one species (C4H9Br)
A2. a.
b.
[OH−] (mol dm−3) Rate (mol dm−3 s−1)0.241 3.10.195 2.40.155 1.50.118 10.099 0.9Note: Values for rate are not accurate because of the difficulty in drawing tangents.
c.
d. Since the graph in c is not a straight line, it appears that both 1-bromobutane and sodium hydroxide affect the rate of reaction and so an SN2 mechanism is likely.
Required practicalP1. a. Initial rate method
b. i. 10 cm3 pipette, ii. 25 cm3 pipette, iii. dropping pipette, iv. 15 cm3 pipette or, since these are rare, a 50 cm3 burette.
c. It is simply an indicator to show the presence of iodine.
d. Instead of making a judgement by eye, a light meter can measure the precise moment when a certain depth of darkness appears.
P2. a. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
b. Hydrogen – measure the volume produced.
c. Temperature.
d. Measure the volume of hydrogen produced, either using a gas syringe or by the downward displacement of water.
e. Plot a graph of the volume of hydrogen produced against time.
If the graph is a straight line, the reaction is zero order and the rate constant is given by its gradient.
If the graph is a curve, measure gradients at various points to calculate the rate of reaction at different stages of the reaction. Plot rate against concentration, and if a straight-line graph is produced the reaction is first order.
Practice questions1a Initial rate of reaction is proportional to [(CH3)3Br]
1b Initial rate of reaction is not affected by changes in [OH–].
1c Rate=k ×[ (CH3 )3Br] so either by using a graph plot or by using values directly in equation:
k= rate[ (CH3 )3Br ]
=6.0×10−3
2.0×10−3 =3.0 s– 1
1d Rate=k × [ (CH3 )3 Br ]=3.0×4.0×10−3=12mol dm−3 s−1
2a Table of data
Experiment Initial [A] /mol dm–3
Initial [B] /mol dm–3
Initial rate of reaction /mol dm–
3 s–1
1 0.020 0.020 1.2×10−4
2 0.040 0.040 9.6×10−4
3 0.010 0.040 2.4×10−4
4 0.060 0.030 8.1×10−4
5 0.040 0.035 7.2×10−4
2b k= rate[A ][B ]2
= 1.2×10−4
0.020×(0.020)2 =15.0mol−2dm6 s–1
2c
Order with respect to A = second, order with respect to B = first
2d
Order with respect to A = zero, order with respect to B = zero
3a Table of data
Experiment Initial [X] /mol dm–3
Initial [Y] /mol dm–3
Initial rate of reaction /mol dm–
3 s–1
1 3.0×10−2 4.0×10−2 1.6×10−5
2 6.0×10−2 4.0×10−2 6.4×10−5
3 3.0×10−2 16.0×10−2 6.4×10−5
4 1.5×10−2 16.0×10−2 1.6×10−5
3b k= rate[Y ][X ]2
= 1.6×10−5
4.0×10−2×(3.0×10−2)2 =0.444mol−2dm6 s–1
3c Rate constant increased as temperature is increased.
3d Rate constant unchanged as concentration of Y is increased at a fixed temperature.
4a(i)
Expt 2 2.68 ×10−4
Expt 3 10.7(2) ×10−4
Expt 4 2.08 ×10−3
4a(ii)
= 186 mol−1dm3s−1
4bIncreases (exponentially): straight line, not a curve
5a(i)k = 0 65/(0 15)(0 24)2 = 75.23 to 74.7 mol−2dm6s−1
5a(ii)0.081 (min sig. figs required)
5b(i)2
5b(ii)0
6a(i)26a(ii)1
6a(iii)0
6bk = rate/[D]2[E] or
8 36×10−4/(0.84)2(1.16)= 1.0(2) × 10−3 to 1.05 × 10−3
mol−2dm6s−1
7ak = rate/[CH3CH2COOCH3][H+]or= 1⋅15×10−4/(0.150)(0.555)= 1.38×10−3 to 1.4×10−3
mol−1dm3s−1
7bans = rate constant × ( ½ × 0.150) × (½ × 0.555)= rate constant × 0.02082.88 × 10−5
8aConsider experiments 1 and 2:[B constant], [A] increases × 3: rate increases by 32
Therefore, 2nd order with respect to A
Consider experiments 2 and 3:[A] increases × 2: rate should increase × 22 but only increases × 2Therefore, halving [B] halves rate and so 1st order with respect to B Rate equation: rate = k[A]2[B]
8brate = k [C]2[D]; therefore, k = rate / [C]2[D]
k = 7.2 × 10−4/ ((1.9 × 10−2)2(3.5 × 10−2)) = 57.0
mol–2 dm6 s–1
8crate = 57.0 × (3.6 × 10–2)2 × 5.4 × 10–2 = 3.99 × 10–3 (mol dm–3 s–1)
8dReaction occurs when molecules have E≥Ea
Doubling T causes many more molecules to have this E Whereas doubling [E] only doubles the number with this E
8eEa = RT(lnA – lnk)/1000 Ea = 8.31 × 300 (23.97 – (–5.03))/1000 = 72.3 (kJ mol–1)
CHAPTER 3Assignment 1A1. Only a small amount of product is expected, so the reaction can be considered as unlikely to occur.
A2. At the higher temperature, the value of Kp will be higher and the yield of NO(g) will increase.
A3. Endothermic because Kp increases with an increase in temperature, where
so the yield of NO (right-hand side of equation) increases. Le Chatelier’s principle predicts that if temperature is increased, a reaction will shift in the endothermic direction.
A4. At midnight (the start of the graph), the concentration of NO is quite low. As traffic starts to build up, the concentration of NO increases. Following the production of NO, it is
then converted to NO2 by oxidation by oxygen in the atmosphere. The NO is being removed as it is formed, so its concentration reaches a maximum value. Also, as the light intensity increases, the concentration of NO2 decreases as it undergoes a series of photochemical reactions (one of the products is ‘ground-level ozone’). As the level of traffic decreases later in the morning, the concentration of NO2 decreases. There is then a slight increase in the level of NO from increased traffic in the afternoon (but this is removed as it is formed, by the ground-level ozone present in the atmosphere from the previous reactions). From then the levels of NO and NO2 decrease.
A5. Arguments supporting opinion in favour of LEZs: Fossil-fuel-powered vehicles produce pollution. The photograph shows a brown haze of nitrogen dioxide (which in turn produces low-level ozone). This can cause breathing problems. Arguments against LEZs: People may have to drive around cities to carry out their legitimate work, for example taxi drivers and delivery drivers.
Practice questions1a
1b
K p=(C2H 2)(H2)
3
(CH ¿¿4 )2 ¿
K punits=kPa×kPa×kPa×kPa
kPa×kPa=kPa2
Total number of moles = 0.44 + 0.28 + 0.12 = 0.84 moles
P(C H 4)=200× 0.44
0.84=104.76 kPa
P(C2H 2)=200× 0.12
0.84=28.57kPa
P(H 2)=200× 0.28
0.84=66.67 kPa
K p=(28.57)(66.67)3
(104.76)2 =771.45 kPa2
1c
To right or to product(s) or forwards; Increase
1d
To left or to reagent or backwards; no effect
2a(i)
Increase because higher P gives lower yield or moves to left, equilibrium shifts to reduce P or equilibrium favours side with fewer moles
2a(ii)
Endothermic because increase T increases yield or moves to right, equilibrium shifts to reduce T or equilibrium favours endothermic direction
2b(i)
Moles of iodine = 0.023
Moles of HI = 0.172
2b(ii)
2b(iii)
= 0.0179 or 1.79 × 10−2
3a(i)
Moles of C2F2 = 0.40
Moles of HCl = 0.80
3a(ii)
3a(iii)
mol dm−3
3b(i)
Increase
3b(ii)
Decrease
3c
Addition or radical
4a
mol Cl2 = 1.2
total mol = 3.8
4b
mol fraction PCl5 = 1.4/3 8 = 0.368
mol fraction Cl2 = 1.2/3 8 = 0.316
4c(i)
pp = mol fraction × total P
4c(ii)
min: pp PCl5 = 0.368 × 125 = 46(.0) 0.37 × 125 = 46.3
max: pp Cl2 = 0.316 × 125 = 39.47 0.32 × 125 = 40.0
4d
4e(i)
No effect
4e(ii)
Increase
4f
42 62 / 36.9 = 49.2
kPa
CHAPTER 4Assignment 1A1. CH3CH(OH)COOH(aq) + H2O(l) ⇌ CH3CH(OH)COO−(aq) + H3O+(aq)
A2. CH3CH(OH)COOH(aq) + OH−(aq) ⇌ CH3CH(OH)COO−(aq) + H2O(l)
A3. Ka = 1.38 × 10−4 mol dm−3
A4. Lactic acid in venous blood = 0.5 × 10−3 mol dm−3 to 2.2 × 10−3 mol dm−3
[H+] range = to
2.63 × 10−4 mol dm−3 to 5.51 × 10−4 mol dm−3
The pH range would be from 3.58 to 3.26.
Lactic acid in arterial blood = 0.5 × 10−3 mol dm−3 to 1.6 × 10−3 mol dm−3
[H+] range = to
2.63 × 10−4 mol dm−3 to 4.70 × 10−4 mol dm−3
The pH range would be from 3.58 to 3.23.
A5. a. The pH values are much lower than blood pH.
b. Blood is buffered to resist changes in pH when acid is produced in blood.
Assignment 2A1. Ideas that may be included: CO2 causes warming by absorbing infrared. CO2 is released into the atmosphere by burning, respiration and decomposition. The amount of CO2
released has increased rapidly in the last 200 years. CO2 dissolves in the oceans, sets up an equilibrium and lowers the pH. CO2 reacts and forms HCO3
− and CO32−, which react to form
shells of sea creatures and eventually rocks. Carbonates act as a buffer. The resulting change in pH and the ability of oceans to buffer this must be fully understood. The impact of climate change on weather and marine life is not fully understood.
Required practicalP1. Apparatus and instruments for measurement have either a graduation mark or a scale. Regular checks are needed and, if necessary, adjustments are made using standard materials (ones whose properties are known to an appropriate level of accuracy). For a pH meter, the standards used are buffer solutions of known pH. The checks and adjustments are called calibration.
P2. a. A pipette has a single graduation mark (though some graduated pipettes have more than one mark). It is used to measure the required volume of the weak acid into a beaker because this is a fixed volume that needs to be measured out accurately.
b. A burette is used to measure quantities of the strong base because the volume needed to react completely with the acid is being measured. Since this is not known, a burette has many graduation marks and, therefore, allows the volume of solution dispensed from it to be measured accurately and to the appropriate precision.
P3. A measured volume of the weak base is placed in a conical flask. A strong acid of known concentration is added from a burette, with the pH being recorded. The endpoint is identified from the change in the shape of the pH curve.
We need the equation for the reaction between the weak base and the strong acid, as this gives the stoichiometry of the reaction.
The number of moles of acid added to reach the endpoint is calculated from the volume of the acid and its concentration. The equation for the acid–base reaction enables the number of moles of base to be calculated. Now, both the volume and the number of moles are known and the concentration of the base can be calculated.
Practice questions 1a(i)
Kw = [H+][OH–]
1a(ii)
2.34 ×10−7
1a(iii)
2.34 ×10−7
1a(iv)
5.48 to 5.50 ×10−14
1b
[H+] = 10−14 / 0.136
= 7.35 ×10−14 OR pOH = 0.87 1
pH = 13.13
2a
Ka =
[H+] = √(1.35 ×10−5 × 0.169)
pH = 2.82
2b(i)
CH3CH2COOH + NaOH → CH3CH2COONa + H2O
or CH3CH2COOH + OH– → CH3CH2COO– + H2O
2b(ii)
mol propanoic acid = 0.250 – 0.015 = 0.235
mol propanoate ions = 0.190 + 0.015 = 0.205
2b(iii)
(= 1.548 ×10–5)
4.81
3a Rearranging the dissociation constant equation gives
¿¿
So ¿ and pH=– log10¿¿
3b(i) Amount of propanoic acid = 0.0165 mol
3b(ii) Amount of sodium hydroxide = 0.002 30 mol
3b(iii) Amount of propanoic acid in buffer solution = 0.0165 – 0.00230 = 0.0142 mol
3b(iv) pH of buffer − log10 ¿¿
4a(i)
Proton donor - alone
4a(ii)
Completely dissociated
4b(i)
7.05 ×10−3 × 103/50 = 0.141
4b(ii)
−log [H+] or log 1/[H+]
4b(iii)
0.85
4b(iv)
pH = 1, [H+] = 0.10 (mol dm−3 )
(7.05 ×10−3)/0.10
vol = 7.05 ×10−2 dm3 or 70.5 cm3
4c(i)
4c(ii)
4d(i)
[H+] = 1.66 ×10−4
= 7.22 × 10−5
pKa = 4.14
4d(ii)
Effect = none/ negligible/v small decrease/v small change
Salt or Y− reacts with extra H+ or
equilibrium shifts to LHS or
H+ is removed as equilibrium shifts to LHS
Therefore, [H+] or ratio [HY]/[Y-] or ratio [Y-]/[HY] remains almost constant
5a
−log [H+]
4.57 × 10−3
5b(i)
[
5b(ii)
= 1.39 × 10−4
mol dm−3
5b(iii)
pKa = 3.86
5c(i)
1000
30/1000 × 0.480 = 0.0144 or 1.44 ×10−2
5c(ii)
18/1000 × 0.350 = 0.0063 or 6.3 × 10−3
5c(iii)
0.0144 × 2(0.0063) = 1.80 × 10−3
5c(iv)
1.80 × 10-3 × 1000/48 = 0.0375
5c(v)
10−14 / 0.0375 = 2.66 × 10−13
or pOH = 1.426
pH = 12.57
6a(i)
[H+][OH−]1
–log [H+]
6a(ii)
[H+] = [OH−]
6a(iii)
(2.0 × 10−3) × 0.5 = 1.0 × 10−3
6a(iv)
pH = 10.40
6b(i)
[H+] = √(1.35 × 10−5) × 0.125 ( = 1.30 × 10−3)
pH = 2.89
6c(i)
(50.0 × 10−3) × 0.125 = 6.25 × 10−3
6c(ii)
(6.25 × 10−3) – (1.0 × 10−3) = 5.25 × 10−3
6c(iii)
Mol salt formed = 1.0 × 10−3
pH = 4.15
7a(i)
B
C
A
7a(ii)
Cresolphthalein or thymolphthalein
7b(i)
-log[H+]; 1
7b(ii)
[H+] = 1.259 × 10−12 or pOH = 14 – pH
or = 2.10
= 7.94 × 10−3
7c(i)
Ka = [H+]2/[CH3CH2COOH] or [H+]2/[HA] or [H+] = [A−]
[H+] = √1.35 × 10−5 × 0.117 = 1.257 × 10−3
pH = 2.90
7c(ii)
Ka = [H+] or pKa = pH
pH = 4.87
8a
pH = −log[H+]
[H+] = [A−]
[H+] = √ 1.74 × 10−5 × 0.15
pH = 2.79
8b(i)
Solution which resists change in pH /maintains pH despite the addition of (small amounts of) acid/base (or dilution)
8b(ii)
CH3COO− + H+ → CH3COOH
8c(i)
= 2.61 × 10−5
pH = 4.58
8c(ii)Moles H+ added = 10 × 10−3 × 1.0 = 0.01 1
Moles ethanoic acid after addition = 0.15 + 0.01 = 0.16 1
Moles ethanoate ions after addition = 0.10 – 0.01 = 0.09
pH = 4.51
9a
Concentration of acid: m1v1 = m2v2
Hence 25 × m1 = 18.2 × 0.150
or
moles NaOH = 2.73 ×10−3
m1 = 18.2 × 0.150 / 25 = 0.109
9b(i)
Ka = [H+][A−] / [HA]
9b(ii)
pKa = –log Ka
9b(iii)
[A−] = [HA]
Hence Ka = [H+][A−] / [HA] = [H+]
and –log Ka = –log [H+]
9c
Ratio [A−] : [HA] remains constant
Hence as [H+] = Ka [HA] / [A−], [H+] remains constant
10
[H+] = √Ka[HA] = √(1.15×10−4 × 0.5)
=7.58×10−3 mol dm−3
pH = -log10[H+] (or log or lg)
pH = 2.12
11aBurette, because it can deliver variable volumes11b
The change in pH is gradual / not rapid at the end pointAn indicator would change colour over a range of volumes of sodium hydroxide
11c
[H+] = 10–pH = 1.58 × 10–12
Kw = [H+] [OH–]; therefore [OH–] = Kw
/ [H+]Therefore, [OH–] = 1 × 10–14/1.58 × 10–12 = 6.33 × 10–3 (mol dm–3)
11d
At this point, [NH3] = [H+]
Therefore,
[H+] = 10–4.6 = 2.51 × 10–5
Ka = (2.51 × 10–5)2/2 = 3.15 × 10–10 (mol dm–3)
11e
When [NH3] = [NH4+], Ka = [H]Therefore – log Ka = –log [H+]Therefore pH = – log10(3.15 × 10–10) = 9.50
12a Ethanedioic acid dissociates more readily than ethanoic acid, partly because there are two –COOH groups per molecule. Electronegativity measures the tendency of an atom to attract a bonding pair of electrons. Oxygen atoms are more electronegative than carbon atoms. –COOH groups tend to attract electrons onto the oxygen atoms, whereas –CH3 groups tend to push electrons away. Therefore ethanedioic acid has a greater tendency to pull the electrons towards the oxygen atoms within the molecule, meaning that the first proton more readily dissociates than with ethanoic acid, so ethanedioic acid is a stronger acid than ethanoic acid.
12bMoles of NaOH = Moles of HOOCCOO– formed = 6.00 × 10–2
Moles of HOOCCOOH remaining = 1.00 × 10–1 – 6.00 × 10–2
= 4.00 × 10–2
Ka = [H+][A–]/[HA]
[H+] = Ka × [HA]/[A–]
[H+] = 5.89 × 10–2 × (4.00 × 10–2/V)/(6.00 × 10–2/V) = 3.927 × 10–2
pH = –log10(3.927 ×10–2) = 1.406 = 1.41
CHAPTER 5Assignment 1A1. A volumetric flask can measure a specified volume precisely.
A2. The measured rotation of sample 4 is anomalous. The mass of sugar was not weighed accurately and was too high, the volume of water was not measured accurately and was too small, or the polarimeter was not calibrated properly.
A3. 0.5 g dm−3
A4. Average measured rotation = 3.33°
Specific rotation =
A4. D-(+)-sucrose
A5. The measured rotations are small and close to the precision of the instrument, which gives high percentage errors from the equipment. Larger values of measured rotations would give lower percentage errors. Increasing the concentration of the sugar solution by increasing the mass of sucrose or decreasing the volume of water would increase the measured rotation.
A6. Nine.
A7. Specific rotation = 66.6° =
Concentration c = 2.25 g dm−3
Assignment 2A1. −102°
A2. It contains a 50 : 50 ratio of D-(+)-limonene and L-(-)-limonene
A3. Polarimeter
A4. IR spectroscopy, 1H NMR, 13C NMR, mass spectroscopy, HPLC
A5. The enantiomers have identical chemical and physical properties, including spectra.
Assignment 3 A1.
A2.
The type of isomerism is optical isomerism.
A3. It is a 50 : 50 mixture of D- and L-enantiomers.
A4. Both contain a chiral centre. For both, the D-isomer is biologically active. They both have a benzene ring and a carboxylic acid functional group. Parts of the structures are identical, as shown in the blue boxes.
A5. In both drugs, the D-isomer is the active compound. In ibuprofen, the L-isomer is not biologically active but it is not toxic, and so it can be sold as a racemic mixture of isomers. For naproxen, because the L-isomer is toxic, it is important to only sell the D-isomer. Naproxen will have to be prepared as a single enantiomer and so it will be more expensive to prepare than the racemate of ibuprofen.
Practice questions1a4 (monochloro isomers)
1b
2
3The major product exists as a pair of enantiomersThe third isomer is 1-bromobutane (minor product)Because it is obtained via primary carbocation
4Compounds/molecules with same structural formula but with bonds/atoms/groups arranged differently in space or in 3D. Use plane-polarised light, which will be rotated by samples in opposite directions.
5. B. Four different atoms or groups attached to a carbon atom
6. A. CH3CH2CH(OH)CH3
7. C. Three
8. B. Rotation of plane-polarised light
9. B. One compound is optically inactive and the enantiomers of the other compound are present in a 50 : 50 ratio.
CHAPTER 6Assignment 1A1. Nucleophilic addition
A2.
A3.
A4. No, it is a racemate because the addition of the cyanide anion can occur equally from either side of the planar aldehyde to give a 50 : 50 mixture of enantiomers.
A5. It will be faster if ionic KCN is used (a source of cyanide ion), as hydrogen cyanide is a very weak acid. The degree of ionisation in solution, producing hydrogen ions (H+) and the reactive nucleophilic cyanide ions (:CN−), is very small (Ka for HCN = 4.0 × 10−10 mol dm−3).
A5. Potassium cyanide is a highly toxic material that causes respiratory failure, convulsions and death if inhaled or absorbed through the skin. Disposable gloves should be used and all reactions undertaken in a fume cupboard, where vapours can be removed. As the reaction involves the addition of acid in a second step, then any unreacted potassium cyanide will react to give hydrogen cyanide gas, which is also very toxic and might be accidently inhaled.
Assignment 2A1.
A2.
This is a reduction or a nucleophilic addition.
A3.
A4.
A5.
Required PracticalP1.
Sample preparation
For each of the solids, add about 0.5 g to a clean test tube and dissolve it in approximately 5 cm3 of deionised water. If the sample does not fully dissolve, warm in a water bath until dissolved. Label each test tube with a permanent marker in order not to confuse the samples.
Fehling’s solution
For each sample to be tested, make up a fresh sample of Fehling’s solution by mixing about 1 cm3 of Fehling’s A and Fehling’s B solutions in a clean test tube with gentle swirling. When transferring the solutions, use a clean pipette for each solution. Check the Fehling’s test solution is blue in colour. Next add a few drops of the solution of one of the unknown samples to be tested, swirling the tube gently. Warm the reaction mixture in a water bath (at approximately 60 °C) for a few minutes. Note any colour change in your lab book. Repeat the whole process (making fresh Fehling’s test solutions each time) for the other four samples.
P2.
Compound tested Outcome of test Oxidation state of copper
No reactionSolution remains blue
Cu(II)
ReactionBrick-red precipitate
formed
Cu(I)
No reactionSolution remains blue
Cu(II)
ReactionBrick-red precipitate
formed
Cu(I)
No reactionSolution remains blue
Cu(II)
P3. You can heat the sample by placing the test tube in a bath of hot water.
P4. a. You could use either Tollens’ reagent or Fehling’s solution. These distinguish between aldehydes and ketones.
b. You would require two clean test tubes. For Fehling’s solution, you would require Fehling’s A and B solutions, three clean pipettes and a hot water bath at 60 °C. For Tollens’ reagent, you would require Tollens’ reagent, two pipettes and possibly a water bath at 60 °C (the reaction may go at room temperature).
c. You would add a few drops of each material to the test solution (Fehling’s or Tollens) in a clean test tube. Then warm the tube and observe if any colour change occurs.
d. In the Tollens test, the aldehyde would be oxidised to a carboxylic acid and the Ag(I) reduced to an Ag(0) mirror. This would be observed coating the test tube. The ketone would not undergo any reaction and the colourless test solution would remain. In the Fehling’s test, the aldehyde would be oxidised to the carboxylic acid and the blue Cu(II) complex reduced to an insoluble Cu(I) brick-red precipitate. The ketone would not undergo any reaction and the blue test solution would remain.
P5. a. For every 1 mole of AgNO3 we require 2 moles of NH3.
5 cm3 of 1 M AgNO3 solution = 0.005 moles. We therefore require 0.01 moles of NH3. This is 5 cm3 of a 2 M aqueous NH3 solution.
b. None. The Tollens reaction occurs only with aldehydes. Propanone is a ketone.
c. For every mole of aldehyde oxidised, 2 moles of Ag(0) are formed. Therefore we would obtain 0.1 mol of Ag(0). The relative atomic mass of silver is 107.9 g per mole, so we would expect 10.79 g of silver.
Practice questions 1a
CH3CH2COCH3 + 2[H] → CH3CH2CH(OH)CH3
1b
The carbonyl bond is planar in shape, so the [H] has an equal likelihood of attacking from either side of the molecule. This means that when the product is chiral a racemic mixture will be formed and so will not rotate the plane of polarised light.
2
b. 2-methylbutan-2-ol
3a
Nucleophilic addition
2-hydroxy-2-methylpentan(e)nitrile
3b
Product from Q is a racemic mixture/ equal amounts of enantiomers
Racemic mixture is inactive or inactive
Product from R is inactive (molecule) or has no chiral centre
3a
CH3CH2CHO + HCN → CH3CH2CH(OH)CN or C2H5CH(OH)CN2-Hydroxybutanenitrile OR 2-hydroxybutanonitrile
3b
Nucleophilic addition
3c
(Propanone) slower or propanal faster
Inductive effects of alkyl groups or C of C=O less δ+ in propanoneor alkyl groups in ketone hinder attack or easier to attack at end of chain
4a
Nucleophilic addition
4b
Planar C=O (bond/group)
Attack (equally likely) from either side
Product is a racemic mixture formed, or a 50:50 mixture of each enantiomer equally likely
5
Nucleophilic addition
6a
Nucleophilic addition
6b(i)
6b(ii)
(Plane) polarized light Rotated in opposite directions (equally) (only allow if M1 correct or close)
6c
2-Hydroxybutane(-1-)nitrile
6d
Weak acid / (acid) only slightly / partially dissociated/ionised [CN–] very low
CHAPTER 7Assignment A1A1.
A2. Olive oil contains two esters of oleic acid and one of palmitic acid, so we would need
a ratio of glycerol : oleic acid : palmitic acid of 1 : 2 : 1.
A3. If we mix the two carboxylic acids oleic acid and palmitic acid with glycerol, statistically it is possible to make a range of triglycerides:
A4. Animal fat contains the saturated fatty acid stearic acid, while olive oil contains the unsaturated fatty acid oleic acid. Saturated fatty acids have higher melting points than unsaturated fatty acids because they have better packing in the solid state and greater van der Waals interactions. Consequently, animal fat has a higher melting point and is a solid at room temperature.
A5. Trimyristin is made up of saturated fatty acids so will be a solid at room temperature.
Assignment A2A1.
A2. Moles of methyl benzoate = 4/136 = 0.03 moles.
Moles of NaOH (20 cm3 of 2 mol dm−3) = 0.04 moles.
A3. The condenser was added so that the reaction mixture could boil without the loss of any of the reagents. Antibumping granules were added to ensure smooth boiling.
A4. An excess of NaOH was used, so initially the reaction mixture will be basic and red litmus would turn blue.
A5. In order to turn the carboxylate salt into the carboxylic acid. Blue litmus should turn red:
A6. For efficient recrystallisation, a saturated solution of the compound in the hot solvent is required. When cooled, the compound will start to crystallise from the saturated solution as the solubility of the compound in the solvent decreases. If too much hot solvent is used initially, the compound might not crystallise out of solution, even at room temperature.
A7. Impure compounds have wide melting point ranges which are lower than that of the pure compound. We can conclude that although the reaction worked, the recrystallisation did not provide us with a very pure product.
Assignment A3A1.
A2. Second generation
A3. The amount of carbon (usually carbon dioxide) released is equal to the amount of carbon dioxide captured in its formation.
A4. Particulates can cause asthma. Carbon monoxide is a poisonous gas which interferes with oxygen transport in the blood. Nitrogen oxides can convert oxygen to ground-level zone, which can affect breathing and the growth of plants. Greenhouse gases result in warming of the atmosphere and cause climate change. Sulfur oxides are acidic and can damage buildings and living organisms.
A5. Land that could be used for growing food could be taken for biodiesel production, thus increasing world hunger.
A6. Saturated fatty acid esters will have a high melting point because of the van der Waals interactions between the chains. Therefore the fuel could freeze in winter.
A7.
Assignment A4A1. Moles of salicylic acid used = 6.0/138 = 0.043 moles
Moles of aspirin produced = 6.1/164 = 0.037 moles
% yield = 0.037/0.043 × 100% = 86.0%
A2. It is cheaper, reacts more slowly and does not evolve HCl.
A3. % Atom economy = molecular mass of desired product/sum of molecular masses for all reactants × 100%.
a. Molecule mass: ethanoyl chloride = 78.5, salicyclic acid = 138, aspirin = 164
Atom economy = 164/216.5 = 75.7%
b. Molecule mass: ethanoic anhydride = 102, salicyclic acid = 138, aspirin = 164
Atom economy = 164/240 = 68.3%
The use of ethanoyl chloride has greater atom economy.
A4. Crush the sample into a powder using a spatula on a filter paper. Add a small amount of the sample to a melting point capillary by pushing the open end of the capillary into the powder. Tap the tube so that the powder settles to the bottom of the tube. Make sure that about 3–4 mm of the tube is filled with the sample.
A5. The ethanol solution contained both the aspirin and any unreacted impurities, such as salicylic acid. Aspirin is not very soluble in water at room temperature, whereas salicylic acid (being a carboxylic acid) is more soluble. Upon cooling in water, the aspirin recrystallises and the salicylic acid remains in the water.
Required Practical 10bP1. So that any pressure build-up can escape to the atmosphere.
P2. If the apparatus was disassembled or the water supply to the condenser was shut off then the hot vapours in the flask would escape, which would be a safety hazard.
P3. The reagents would evaporate and the reaction would fail.
P4. Pour all the liquid back into the separating funnel, wait for the two layers to separate and start again.
P5. a. The liberation of carbon dioxide, as well as the exothermic nature of the acid–base reaction causing some of the solvent to vaporise.
b. To remove the acidic ethanoic acid from the organic layer by neutralising it with a base to provide the water-soluble sodium ethanoate salt.
c. The ethyl ethanoate will be in the top (organic) layer.
d. The anhydrous calcium chloride removes any residual water carried over from the separation using the separating funnel.
P6. Filtration
P7. If the stopper is left in the top of the separating funnel and the tap opened, the liquid does not drain out easily. This is because as the liquid leaves the funnel, a partial vacuum is formed above the liquid, which stops the liquid draining.
Required Practical 10aP1. If the impurities are soluble, the solid compound should be highly soluble in the solvent at high temperature but relatively insoluble at room temperature. The impurity should remain soluble at room temperature.
If the impurities are insoluble, the solid compound should be highly soluble in the solvent at high temperature but relatively insoluble at room temperature. The impurity should remain insoluble at high temperature.
P2. For efficient recrystallisation, a saturated solution of the compound in the hot solvent is required. When cooled, the compound will start to crystallise from the saturated solution as the solubility of the compound decreases. If too much hot solvent was used initially, the compound might not crystallise out of solution, even at room temperature.
P3. a. 10 g of compound A in mixture, solubility at 90 °C = 20 g in 100 cm3 in water. So 10 g will be soluble in 50 cm3.
b. At 20 °C the solubility of A is 0.2 g in 100 cm3, so in 50 cm3 only 0.1 g would remain in solution.
c. The maximum recoverable amount is 9.9 g (or 99%). There is 1 g of B in the mixture. The solubility of B is 2.5 g per 100 cm3 of water, so in 50 cm3 a maximum of 1.25 g would remain in solution. As only 1 g of B is present, it will all remain in solution. Therefore recrystallised A will be pure.
P4. To allow sufficient time for the heat to transfer from the heating solution through the capillary to the sample. If the sample is heated too quickly, the heating block will be at a higher temperature than the sample. In addition, slower heating allows a more accurate melting range to be reported.
Practice questions 7
1(a)
1(b)
2a
(Nucleophilic) addition-elimination
2b(i)
To ensure the hot solution would be saturated / crystals would form on cooling
2b(ii)
Yield lower if warm / solubility higher if warm
2b(iii)
Air passes through the sample not just round it
2b(iv)
To wash away soluble impurities
2c
WaterPress the sample of crystals between filter papers
2d
Mr product = 135.0
3a
Propan(e)-1,2,3-triol or 1,2,3- propan(e)triol
3b
Soaps
3c(i)
Biodiesel
3c(ii)
3c(iii)
CH3(CH2)12COOCH3 + 21½ O2 → 15CO2 + 15 H2O
4
(Nucleophilic) addition-elimination
N-ethylpropanamide
5a
CH3CH2CH2COOH + CH3CH2OH (or C2H5OH) → CH3CH2CH2COOCH2CH3 + H2OH2SO4 or HCl or H3PO4
5b
H3CH2CH2CH2OH + (CH3CO)2O → CH3COOCH2CH2CH2CH3 + CH3COOH
5c
(Nucleophilic) addition-elimination
5d
6
Nucleophilic) addition-elimination
N-propylethanamide
7a(i)
3CH3OH
HOCH2CH(OH)CH2OH
7a(ii)
C17H35COOCH3 + 27½ or 55/2 O2 → 19CO2 + 19H2O
8
Electron pair donor or lone pair donor
(Acid) anhydride
9a(i)
9a(ii)
(Nucleophilic) addition-elimination or (nucleophilic) addition followed by elimination
9a(iii)
Any two from: ethanoic anhydride is • less corrosive • less vulnerable to hydrolysis
• less dangerous to use, • less violent/exothermic/vigorous reaction OR more controllable rxn • does not produce toxic/corrosive/harmful fumes (of HCl) OR does not produce HCl • less volatile
9b
CHAPTER 8Assignment 1A1. Accept 4-nitromethylbenzene, 1-methyl-4-nitrobenzene or 4-nitrotoluene.
A2. Electrophilic substitution by nitronium ion (NO2+).
A3.
A4. Increases it.
A5. Decreases it.
Assignment 2A1.
A2. The electrophile is NO2+:
A3. As the acids are corrosive, rubber gloves must be worn, as well as a lab coat and safety spectacles. Because concentrated acids fume in air, the reaction should be carried out in a fume cupboard.
A4. 14 cm3 of liquids and 4.00 g of solid are reacted in the conical flask, which must be stirred. As a conical flask should only be filled between a third and half full, either a 50 cm3 or 100 cm3 flask would be chosen.
A5. Both the formation of the nitronium ion and the nitration reaction are exothermic. Cooling the reagents ensures that the reaction mixture does not boil and the reaction does not become too violent.
A6. A Pasteur pipette.
A7. a. It is important to prepare a hot saturated solution of the product to be recrystallised. This ensures the maximum yield when the solution cools to room temperature and the product crystallises out.
b. You can heat the ethanol using a water bath.
A8. To assess the purity of the recrystallised product.
Assignment 3A1. A lone pair acceptor
A2.
A3.
A4.
Reduction reaction
A5. Another reducing agent that can be used is nickel and hydrogen.
A6. Number of moles of B = 1.34/134 = 0.1 mol. A 75% yield would give 0.075 mol of product C.
0.075 mol C = 0.075 × 176 = 1.32 g
A7. Ethanoyl chloride will react vigorously with water, producing ethanoic acid and hydrochloric acid gas. Precautions must be taken so that moisture is not allowed into the reaction flask. Clean, dry apparatus must be used.
A8. It could be purified by distillation
Practice questions 1
2a
or C6H5NHCOCH3 + NO2+ → C6H4(NHCOCH3)NO2 + H+
2b
Electrophilic substitution
2c
Hydrolysis
2d
Sn/HCl
3a(i)
NO2+
HNO3 + 2H2SO4 →NO2+ + 2HSO4– + H3O+
or HNO3 + H2SO4 →NO2+ + HSO4– + H2O
3a(ii)
Electrophilic substitution
3b
H2/Ni or H2/Pt or Sn/HCl or Fe/HCl, dilute H2SO4
4a
CH3CH2COCl or CH3CH2CClO or propanoyl chloride or (CH3CH2CO)2O or propanoic anhydrideAlCl3 or FeCl3CH3CH2COCl + AlCl3 → CH3CH2CO+ + AlCl4–
4b
4c
Tollens or ammoniacal silver nitrate
5a
Benzene is more stable than cyclohexatriene
Expected ΔHϴ hydrogenation of C6H6 is 3(–120) = –360 kJ mol−1
Actual ΔHϴ hydrogenation of benzene is 152 kJ mol−1 (less exothermic) or 152 kJ mol−1
different from expected because of delocalisation or electrons spread out or resonance
5b
Conc HNO3, Conc H2SO4
2 H2SO4 + HNO3 → 2 HSO4– + NO2+ + H3O+ or H2SO4 + HNO3 → HSO4– + NO2+ + H2Oor via two equations:H2SO4 + HNO3 → HSO4– + H2NO3+
H2NO3+ → NO2+ + H2O
6a(i)
Conc HNO3
Conc H2SO4
6a(ii)
2 H2SO4 + HNO3 → 2 HSO4– + NO2
+ + H3O+
or H2SO4 + HNO3 → HSO4– + NO2
+ + H2O
Via two equations:H2SO4 + HNO3 → HSO4– + H2NO3+
H2NO3+ → NO2+ + H2O
6a(iii)
7a
Sn / HCl or Fe / HClEquation must use molecular formulae
C6H4N2O4 + 12 [H] → C6H8N2 + 4H2O
7b
H2 (Ni / Pt)
CH2
In benzene 120°
In cyclohexane 109° 28′ or 109½°
7c(i)
Nucleophilic addition
7c(ii)
Planar C=O (bond/group)
Attack (equally likely) from either side
About product: racemic mixture formed or 50:50 mixture or each enantiomer equally likely
CHAPTER 9Assignment 1A1.
Nucleophilic substitution
A2.
Reduction reaction
A3. Ethanoic anhydride and a base, or ethanoyl chloride and a base.
A4.
A5. The primary amine product is still nucleophilic and can react further, and mixtures of secondary and tertiary amines and quaternary ammonium salts are inevitably produced.
A6. The overall yield of A → B → C → D is [0.91 × 0.87 × 0.96] × 100% = 76%.
The overall yield of E → C → D is [0.65 × 0.96] × 100% = 62.4%.
Despite the second route being shorter, the overall yield is lower so A → B → C → D would be favoured.
This does not take into account the cost of the reagents and any cost of purification, and the safety considerations of using toxic KCN. When this is taken into account, the lower-yielding route might become more attractive.
A7. Drug molecules must interact with their biological receptors. The binding sites have a defined three-dimensional shape. H-bonding, ionic bonding or van der Waals interactions between drugs and their receptors in the binding site are important for binding. The drug molecule must fit into the binding site and its functional groups must be displayed so that binding interactions are maximised. Molecule D is similar in size and shape to benzedrex and has a group capable of H-bonding in a similar position.
Assignment 2A1.
0–5 °C
A2 a. NaNO2 + HCl → HNO2 + NaCl
b. Nitrous acid and the diazonium salt are unstable and decompose above 5 °C.
c. 4-Methylazobenzene
A3. The N=N double bond in the chromophore allows the delocalised electrons of one benzene ring to be extended into the second ring. The delocalised electrons in the chromophore can absorb certain wavelengths of visible light, and the light not absorbed gives rise to the observed colour.
A4. The NH2 group can hydrogen-bond with the surface molecules of the fabric.
A5. The CO2− group can make an ionic bond with the surface molecules of the fabric.
Ionic bonds are normally much stronger than H-bonds.
A6.
Practice questions 1
3
2a
Dimethylamine
2b
Nucleophilic substitution
2c
Quaternary ammonium salt
(Cationic) surfactant / bactericide / detergent / fabric softener or conditioner / hair conditioner
2d
3a(i)
Concentrated HNO3, concentrated H2SO4
HNO3 + 2 H2SO4 → NO2+ + H3O+ + 2HSO4
−
or HNO3 + H2SO4 → NO2+ + H2O + HSO4
−
3a(ii)
Electrophilic substitution
3b
Sn or Fe / HCl or Ni / H2
3c(i)
NH3
Use an excess of ammonia
3c(ii)
Nucleophilic substitution
3d
Lone pair on N less available, delocalised into the ring (Q of L)
3e
3f
4a
(Nucleophilic) addition-elimination
N-ethylpropanamide
4b
CH3CN or ethan(e)nitrile or ethanonitrileStep 1 Cl2UV or above 300 °CStep 2 KCNaqueous and alcoholic (both needed)Step 3 H2/Ni or LiAlH4 or Na/C2H5OH
5a
Proton acceptor
5b(i)
CH3CH2NH2 + H2O → CH3CH2NH3+ + OH–
5b(ii)
Reaction/equilibrium lies to left or low [OH–] or little OH– formed or little ethylamine has reacted
5c
Ethylamine Alkyl group is electron releasing/donatingor alkyl group has (positive) inductive effect that increases electron density on N(H2)or increased availability of lpor increases ability of lp (to accept H(+))
5d
CH3CH2NH3Cl
5e
Extra H+ reacts with ethylamine or OH–
or CH3CH2NH2 + H+ → CH3CH2NH3+
or H+ + OH– → H2O
Equilibrium shifts to RHS or ratio [CH3CH2NH3+]/[ CH3CH2NH2] remains almost constant
6a
(Nucleophilic) addition-elimination
Propanamide
6b
Nucleophilic substitution
Propylamine or propan-1-amine or 1-aminopropane
6c
Electron-rich ring or benzene or pi cloud repels nucleophile/ammonia
7a
Nucleophilic) addition-elimination
N-propylethanamide
7b(i)
Primary
Secondary
Tertiary
7b(ii)
Absorption at 3300–3500 (cm−1) in spectrum
N―H (bond) (only) present in secondary amine or not present in tertiary amine orThis peak or N―H absorption (only) present in spectrum of secondary amine or not present in spectrum of tertiary amine
7c(i)
Route A: stage 1 KCN
Aqueous or ethanolic
Route A Intermediate
CH3CH2CN or propanenitrileName alone must be exactly correct to gain M1 but mark on if name close
correct formula gains M1 (ignore name if close)
contradiction of name and formula loses markRoute A: stage 2 H2
H loses M4 but mark on
LiAlH4
Ni or Pt or Pd ether
Route B NH3
Excess NH3
7c(ii)
Route A disadv Toxic /poisonous KCN or cyanide or CN− or HCN
Expensive LiAlH4
Ignore acidified
OR lower yield because 2 steps
Route B disadv Further reaction/substitution likely
8
Lone pair on N labelled b more available / more able to be donated than lone pair on N labelled alp or electrons or electron density on N labelled a: delocalized into (benzene) ringlp or electrons or electron density on N labelled b: methyl/alkyl groups electron releasing or donating or (positive) inductive effect or push electrons or electron density
9a
Quaternary (alkyl) ammonium salt / bromide CH3Br or bromomethane Excess (CH3Br or bromomethane)
9b
Nucleophilic substitution
10a(i)
Ammonia is a nucleophile Benzene repels nucleophiles
10a(ii)
H2/Ni or H2/Pt or Sn/HCl or Fe/HCl
10a(iii)
Concentrated HNO3
Concentrated H2SO4
HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4–
or using two equations:HNO3 + H2SO4 → H2NO3+ + HSO4–
H2NO3+ → H2O + NO2+
10a(iv)
Electrophilic substitution
CHAPTER 10Assignment 1A1. a. Corrosive
b. Corrosive
c. Flammable, toxic
A2. The reaction should be undertaken in a fume cupboard; lab coat, safety spectacles and disposable gloves should be worn.
A3. a. 2.2 g of 1,6-diaminohexane (molecular weight = 116) = 2.2/116 = 0.02 moles, in 50 cm3 = 0.02/50 × 1000 = 0.4 M.
b. 1.5 g of decandioyl dichloride (molecular weight = 239) = 1.5/239 = 0.006 moles in 50 cm3 = 0.006/50 × 1000 = 0.12 M.
A4. Nylon 6,10
A5. HCl
A6. 1:1 ratio
A7. During the condensation polymerisation, HCl is produced. The excess 1,6-diaminohexane reacts with the HCl, making an ammonium salt.
A8. Unreacted monomers might be adhering to the fibre. As these are corrosive, they must be washed off the nylon before it is used or disposed of.
Assignment 2A1.
A2.
A3. Hydrogen bonds
A4.
A5. Both types of polyamides are capable of H-bonding between chains. But the more rigid nature of the linking aromatic rings in Kevlar® and other aramids compared with the flexible alkyl groups in nylons makes aramids stronger and tougher.
A6. Kevlar® because the 1,4 monomers cause the polymer chain to be less zigzag shaped, so the molecules lie closer together and the intermolecular forces can be more effective.
Practice questions1
Can be hydrolysed or can be reacted with/attacked by acid/base/nucleophiles/H2O/OH−
2a
Addition
2b
2c
Q is biodegradablePolar C=O group or δ+ C in Q (but not in P) Therefore, can be attacked by nucleophiles (leading to breakdown)
3
Kevlar is biodegradeable but polymerised alkenes are not
Kevlar has polar bonds / is a (poly) amide / has peptide link
Can be hydrolysed/attacked by nucleophiles/acids/bases/enzymes
Polyalkenes are non polar /have non-polar bonds
4
Advantages: reduces landfill saves raw materials lower cost for recycling than making from scratch reduces CO2 emissions by not being incinerated
Disadvantages: difficulty/cost of collecting/sorting/processing product not suitable for original purpose, easily contaminated
5a(i)
5a(ii)
In polyamides – H bondingIn polyalkenes – van der Waals forcesStronger forces (of attraction) in polyamides or H bonding is stronger
6a(i)
As a soap
6a(ii)
Biodiesel or biofuel or fuel for cars/lorries
6a(iii)
Cationic surfactant /detergent /fabric softener /germicide / shampoos /(hair) conditioners /spermicidal jelly
6b(i)
(Poly)ester Terylene or PET
6b(ii)
(Poly)amide Kevlar or nylons
6b(iii)
Hydrogen bonding in b(ii)Imfs in (b)(ii) are stronger or H bonding stronger than dipole-dipole/van der Waals/ dispersion/London forces in b(i)
7a
Methyl propanoate
7b(i)
Pentane-1,5-diol
7b(ii)
7b(iii)
(Base or alkaline) hydrolysisδ+ C in polyester reacts with OH– or hydroxide ion
8a(i)
Condensation
8a(ii)
Propane-1,3-diol
8b(i)
Addition
8b(ii)
9
Hydrogen bonding
10
d. NH2CH2COCl
11
b. Amide
12
c. There are no gas emissions from a landfill site.
13
a. Poly(ethene)
CHAPTER 11Assignment 1A1. a. Negative electrode
b. Positive electrode
A2. No movement.
A3. The structures are:
Alanine: HOOCCH(CH3)NH2
Aspartic acid: HOOCCH(CH2COOH)NH2
Lysine: HOOCCH(CH2CH2CH2CH2 NH2)NH2
A4.
They are chemically and physically the same except that they cause plane polarised light to rotate in opposite directions.
A5. Lysine has an extra basic NH2 group, causing it to be the most alkaline. Aspartic acid has an extra acid COOH group, causing it to be the most acidic.
A6.
a
b
c
A7. Perform an electrophoresis experiment at pH 6.1. Use a buffer to fix the pH.
At pH 6.1, glycine will be in its zwitterionic form and will not move in an electric field.
At pH 6.1 (above the isoelectric point), cysteine will be protonated on the nitrogen and move to the negative terminal.
At pH 6.1 (below the isoelectric point), aspartic acid will be deprotonated at the carboxylic acid group and move to the positive terminal.
Assignment 2A1. 5 mol dm−3 hydrochloric acid for 24 hours.
A2. Glycine – no carbon atom is bonded to four different groups.
A3. Dipeptides = A and C. Tripeptide = B.
A4.
A5.
Assignment 3A1. The sequence of α-amino acids linked together in the protein chain
A2. Yes. The α-helix is elastic and can be stretched but will hold its shape owing to the H-bonds. If it is stretched too far then the H-bonds will break.
A3. α-Helix and β-sheet
A4. The cysteine residues could covalently link adjacent α-helices of α-keratin together.
A5. When hair is wetted the H-bonding of the α-helices is disrupted, meaning that the hair can be stretched.
Assignment 4A1.
A2. a.
b.
A3. H-bond
A4. The active sites of enzymes are stereospecific and the two enantiomers interact with different enzymes, thereby producing different biological effects.
A5. The structures of thalidomide and pomalidomide are very similar (differing only in one amine group). They both might be able fit into the active site of the target enzyme and so both inhibit it.
A6. By forming a salt with a carboxylic acid active group in an enzyme
A7. Owing to the similarity to thalidomide, the patient must not be pregnant and must agree to use contraception during the treatment.
Assignment 5A1. AZT has an N3 group at the 3′-position on the sugar, while 2′-deoxythymidine-5′-monophosphate has a hydroxyl group.
A2.
It gets incorporated into the growing DNA strand as it binds to the complementary nucleotide during DNA replication.
A3. The N3 group at the end of the growing chain cannot bond with another nucleotide, so the chain stops growing and the DNA strand cannot finish replicating.
A4. In high doses AZT can stop human DNA replication as it can also be incorporated into human DNA, so it will be toxic. However, it is 100 times more toxic to the HIV virus than to human cells.
Practice questions1a
Secondary
1b
Nitrogen and oxygen are very electronegativeTherefore, C=O and N–H are polar, which results in the formation of a hydrogen bond between O and H in which a lone pair of electrons on an oxygen atom is strongly attracted to the δ+H
2a
A nucleotide is the monomer that makes up DNA. Each nucleotide consists of three components linked together, a phosphate, a sugar and a base.
2b
2-deoxyribose does not have an alcohol group at the 2-position whereas ribose does.
2c
H-bonds between bases are being broken. A high temperature is required as there are a large number of H-bonds to break in the DNA strand.
2d
TCTCATGCAA
3a
2-Deoxyribose
3b
Base A:Top N–H forms hydrogen bonds to lone pair on O of guanine The lone pair of electrons on N bonds to H–N of guanine A lone pair of electrons on O bonds to lower H–N of guanine
3c
Either of the nitrogen atoms with a lone pair NOT involved in bonding to cytosine
3d
Use in very small amounts / target the application to the tumour
4a(i)
4a(ii)
4b
4c
5a
5b
Ionic bonding in aminoethanoic acid
Stronger attractions than hydrogen bonding in hydroxyethanoic acid
6
7
(a) Glycine does not contain a chiral C atom i.e 1 with 4 different groups bonded to it and alanine and phenylalanine do.
(b)(i)
(ii)
(c)(i)
(ii)
(d)
8
Facts (b) and (d) are correct
9
(a) A cation
10
Facts (b) and (d) are correct
CHAPTER 12Assignment 1A1.
A2. Concentrated nitric and concentrated sulfuric acid. Electrophilic substitution (nitration).
A3. Phenylamine
A4. N-methylphenylamine
A5. CH3Cl in ethanol solvent; heat
A6. Ethanoic anhydride. It is cheaper, reacts slowly and does not evolve HCl.
A7. Only ibuprofen contains a carbon atom attached to four different groups and so exhibits optical isomerism.
A8. Aluminium trichloride and 2-methylpropanoyl chloride.
A9.
A10.
The electrophile is NO2+.
Practice questions1a
Diethylamine or ethyl ethanamine or ethyl aminoethane
1b
Three valid routes: A, B, C
Route A Route B Route CF CH3CH2Br or CH3CH2Cl C2H6 CH3CH2OH
G CH3CH2NH2 ethylamine OR ethanamine OR aminoethane
CH3CH2Br OR CH3CH2Cl
CH3CH2Br OR CH3CH2Cl
1c
Route A Route B Route C
Step 1 Reagent(s) HBr OR HCl H2/ Ni (Not NaBH4) H2O & H3PO4 OR H2O & H2SO4
Mechanism Electrophilic addition
addition (allow electrophilic OR catalytic but not nucleophilic) ignore hydrogenation
Electrophilic addition
Step 2 Reagent(s) NH3 Cl2 OR Br2 HBr OR KBr & H2SO4 OR Pcl3 OR Pcl5 OR SOcl2
Mechanism Nucleophilic substitution
(free) radical substitution Nucleophilic substitution
Step 3 Reagent(s) CH3CH2Br OR CH3CH2Cl
CH3CH2NH2 OR NH3 but penalise excess ammonia here
CH3CH2NH2 OR NH3 but penalise excess ammonia here
Mechanism Nucleophilic substitution
Nucleophilic substitution Nucleophilic substitution
1d
Tertiary amine or triethylamine or (CH3CH2)3NQuaternary ammonium salt or tetraethylammonium bromide or chloride or ion or (CH3CH2)4N+ (Br– or Cl–)Further substitution will take place or diethylamine is a better nucleophile than ethylamine
2a [First part of question 2]
Nucleophilic addition
2b
2-bromobutanenitrile
2c
Reagent: ammonia, NH3
Conditions: excess ammonia, heat
Mechanism: nucleophilic substitution
2d(i)
2d(ii)
The electrostatic attraction between the oppositely charged parts of the ion accounts for the relatively high melting-point values of amino acids.
2 [Second part of question 2]
Compound L
Compound M
Step 1 NaBH4 or LiAlH4, (nucleophilic) addition Step 2 conc H2SO4 or conc H3PO4 or Al2O3, eliminationStep 3 HBr, electrophilic addition
3a(i)
3a(ii)
3a(iii)
3b
CHAPTER 13Assignment 1A1. The sample contains paracetamol but it is impure; the impurity is a minor component of the mixture.
A2. Silica
A3. a. Compound B (there are four peaks between 100 and 160 ppm)
b. C=O
c. Compound B because it has an aromatic group, whereas A does not.
A4. 4
A5. Z. This is the only structure that has two CH3 groups, which will give singlets. The triplet is from the CH2 next to the carbonyl group; it is coupled to the neighbouring CH2 group, which is at higher ppm (being next to the electronegative O atom).
Required practicalP1. 0.25
P2. Ibuprofen
P3. D contains both paracetamol and ibuprofen. E contains a third painkiller that is not paracetamol or ibuprofen.
P4. Paracetamol
P5. I hour: mainly alcohol, a little aldehyde formed. 2 hours: mainly product aldehyde, a little alcohol remains.
P6. After three hours, as pure aldehyde is produced.
P7. Another product is forming, probably the carboxylic acid.
Practice questions1a
5
1b
a, singlet QWC
b, triplet QWC
2a
or Si(CH3)4
Inert/non -toxic/volatile or low bp
2b
2
2c(i)
a = quartet or 4
b = triplet or 3
2c(ii)
3230 – 3550 (cm−1)
3a(i)
3a(ii)
3a(iii)
CH2CH2 or two adjacent methylene groups
3a(iv)
or CH3COCH2CH2OCH3
3b(i)
OH in acids or (carboxylic) acid present
3b(ii)
4
IR:Absorption at 3360 cm–1 shows OH alcohol presentNMR:There are 4 peaks which indicates 4 different environments of hydrogen The integration ratio = 1.6 : 0.4 : 1.2 : 2.4 The simplest whole number ratio is 4 : 1 : 3 : 6 The singlet (integ 1) must be caused by H in OH alcohol The singlet (integ 3) must be due to a CH3 group with no adjacent H Quartet + triplet suggest CH2CH3 group Integration 4 and integration 6 indicates two equivalent CH2CH3 groups
5a
Chromatography: GLC, TLC, GC, HPLC
5b
5320.0 or 322.0
5c
Use of excess air/oxygen or high temperature (over 800 °C) or remove chlorine-containing compounds before incineration
5d(i)
Si(CH3)4
5d(ii)
3
6a
6b
6c
E CH3CH2COOCH3F CH3COOCH2CH3
6d
6e
7a
J acid amide K secondary amine or amino
7b
δ = 3.1–3.9Doublet or duplet
7c(i)
Solvent must be proton-free or CHCl3 has protons or has H or gives a peak
7c(ii)
CDCl3 is polar or CCl4 is non-polar
7d
11
7e(i)
Si(CH3)4 or SiC4H12
7e(ii)
A single number or range within 21-25
7e(iii)
CHAPTER 14Assignment 1A1. The first fuel cells powered a vehicle for only 30 seconds, but now with technological advances more than 300 miles is possible. The voltage of the fuel cell and the chemical reactions – which depend upon the scientific principles – are just the same.
A2. No pollutants of carbon dioxide and nitrogen compounds are produced while the bus is using its fuel cell. Fuel cells are quiet.
A3. Air is a mixture of nitrogen and oxygen; when the fuel burns at high temperatures, these two combine to form nitrogen oxides.
A4. Fuel cells are nonpolluting when they are being used, and they have no moving parts, so they are reliable. But there needs to be a store of hydrogen, which is flammable and has to be pressurised.
Assignment 2A1. a. Zn2+(aq) + 2e− ⇌ Zn(s)
b. Mn4+(aq) + 2e− ⇌ Mn2+(aq)
c. ½O2(g) + H+ + 2e− ⇌ OH−(aq)
Reduction is more likely because the equilibrium goes from left to right. Therefore, the cell potential will be less negative.
A2. a. i. Zn(s) | Zn2+(aq) ||
ii. Li+(aq) | Li(s) ||
iii. Pt(s) | H2(g) | H+(aq) ||
b. Under standard conditions this is the standard hydrogen electrode. Its potential is assigned a value of zero and all other standard electrode potentials are measured relative to this.
c. Hydrogen pressure = 1 bar, temperature = 298 K, [H+] = 1 mol dm−3.
A3. a. Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
b. Pt(s) |H2(g) | H+(aq) || Zn2+(aq) | Zn(s)
A4. a. The zinc/silver cell gives a larger e.m.f. (1.56 V) than the zinc/copper cell (1.10 V).
b.
c. −0.76 + (+0.80) V = 1.56 V
d. It has a higher cell potential. The cell reactants have a lower density.
A5. a. Silver will be displaced. For zinc to displace silver from aqueous silver solution, the following two half-reactions must occur:
Zn(s) ⇌ Zn2+(aq) + 2e−, EƟ = −(−0.76) V
Ag+(aq) + e− ⇌ Ag(s), EƟ = +0.80 V
The sum of these two half-reactions is
Zn(s) + Ag+(aq) ⇌ Zn2+(aq) + Ag(s), EƟ = +1.56 V
For this reaction to be spontaneous, EƟ must be positive. Therefore zinc does displace silver from solution.
b. No reaction. This is the reverse of the reaction in part a. EƟ is negative. For this reaction to be spontaneous EƟ must be positive, so no reaction occurs.
c. No reaction. For iodine to displace bromine from bromide solution, the following two half-reactions must occur:
I2(s) + 2e− ⇌ 2l−(aq), EƟ = +0.54 V
2Br−(aq) ⇌ Br2(l) + 2e−, EƟ = −1.07 V
The sum of these two half-reactions is I2(s) + 2Br−(aq) ⇌ 2l−(aq) + Br2(l).
For this reaction to be spontaneous, EƟ must be positive. But (−1.07 V) + (+0.54 V) = −0.53 V. Therefore, iodine does not displace bromine from solution.
d. This is the reverse of the reaction in part c. EƟ is positive. +0.54 V − (−1.07 V) = +0.53 V. Therefore, bromine does displace iodine from iodide solution.
Required practical 8P1. The metals are cleaned with abrasive paper to remove the layer of metal oxide and then with propanone to remove any grease coming from handing them
P2. Stay away from flames and sources of heat, use protective glasses and gloves, perform the operation in a fume hood
P3. The salt bridge maintains the charge neutrality in the solutions
P4. Temperature needs to be 25 degrees Celsius, pressure 1 Bar (100 KPa)
P5. Take a number of readings at different temperatures e.g. every 10 degrees from 10 to 90 and plot them against the e.m.f.
P6. Assemble an electrochemical cell consisting of two different metals, A and B, as described in Required Practical 8. Test each metal of the four in turn against the others in order to determine the cell with the highest e.m.f. Alternatively, test each metal against a standard electrochemical cell, in order to determine their standard electrochemical potential. The two metals with the highest electrochemical potentials will form the cell with the highest e.m.f.
Practice questions1a(i)
None or no reaction
E(Zn2+/Zn) more negative than E(Fe2+/Fe)
1a(ii)
Fe2+
Cr3+
E(Fe3+/Fe2+) more positive than E(Cr3+/Cr2+)
1b
E.m.f. = -0.41 – (-0.76) = 0.35
Zn + 2Cr3+ → Zn2+ + 2Cr2+
2a(i) Iron(II), Fe2+
2a(ii) Oxygen difluoride, F2O
2a(iii) Fe2+, Cl–
2b(i) E.m.f. = E (right) – E (left) = 1.52 – 0.77 = 0.75 V
2b(ii) Fe2+ Fe3+ + e–
2b(iii) Decrease Fe3+ concentration to increase e.m.f. of cell. Equilibrium shifts in favour of more Fe3+, so electrode potential for Fe3+/Fe2+ becomes less positive.
3a
Most powerful reducing agent: Zn
3b(i)
Reducing species: Fe2+
3b(ii)
Oxidising species: Cl2
3c(i)
Standard electrode potential 1.25 V
3c(ii)
Equation: Tl3+ + 2 Fe2+ → 2Fe3+ + Tl+
4a(i)
0.60 V
4a(ii)
H2O + H2SO3 → SO42− + 4H+ + 2e−
4b(i)
2IO3− + 2H+ + 5H2O2 → 5O2 + I2 + 6H2O
4b(ii)
The concentration of the ions change or are no longer standard or the e.m.f. is determined when no current flows
4b(iii)
Unchanged
4b(iv)
Increased
Equilibrium IO3− /I2 displaced to the right
Electrons more readily accepted or more reduction occurs or electrode becomes more positive
CHAPTER 15Assignment 1A1.
[Cu(H2O)6]2+ ⇌ [Cu(H2O)5(OH)]+ + H+;
[Fe(H2O)6]3+ + H2O ⇌ [Fe(H2O)5(OH)]2+ + H3O+
A2. These are hydrolysis reactions because the ion reacts with water and releases hydrogen ions, giving an acidic solution.
A3. The Fe3+ ion has a higher ionic charge than Cu2+, so the attraction for the lone pairs on the water molecule is greater and this weakens the O–H bond more.
A4. The equilibrium for Fe3+ lies further to the right, the solution is more acidic and therefore the pH is lower for Fe3+ than for Cu2+.
A5. Any test from the following.
Reaction with aqueous sodium hydroxide gives a blue precipitate of copper hydroxide:
[Cu(H2O)6]2+ + 2OH− → [Cu(H2O)4(OH)2] + 2H2O
Reaction with aqueous ammonia gives a blue precipitate of copper hydroxide:
[Cu(H2O)6]2+ + 2NH3 → [Cu(H2O)4(OH)2] + 2NH4+
A6. Fe2+ ions. The green precipitate is caused by the reaction
[Fe(H2O)6]2+ + 2OH− → [Fe(H2O)4(OH)2] + 2H2O
The iron(II) slowly oxidises in air to give a brown precipitate of iron(III) hydroxide.
Assignment 2A1. The water is acidic and contains copper ions. In acid conditions the equilibrium
favours [Cu(H2O)6]2+, which is the blue ion, and does not favour the insoluble hydroxo complex, which would make the water cloudy.
A2. a. A green–blue precipitate of CuCO3
b. [Cu(H2O)4(OH)2]
c. Experiment 1:
[Cu(H2O)6]2+ + CO32− → CuCO3 + 6H2O
Experiment 2:
[Cu(H2O)6]2+ + 2NH3 → [Cu(H2O)4(OH)2] + 2NH4+
which gives a blue precipitate, and then reaction with more aqueous ammonia to give [Cu(NH3)4(H2O)2]
d. In low concentrations it is an irritant, and at higher concentrations it is corrosive and toxic. Use with gloves in a fume cupboard.
A3. There may be hidden pollution problems, and companies may be liable for compensation claims in the future.
A4. Plants growing in different areas will be exposed to different levels of copper ions in the soil.
A5. Plant A, because the average amount of copper extracted from the soil is greater. With Plant A, the average amount of copper is 498 mg per kg of leaves, while the value for plant B is 67.4 mg per kg of leaves.
A6. The hydrolysis reactions of transition metal aqua complexes make the soil acidic. Different levels of metals will give rise to different levels of pH.
A7. If the post was made from iron, then the hexaaquairon(III) ions in solution in the soil could undergo hydrolysis,
[Fe(H2O)6]3+ + H2O → [Fe(H2O)5(OH)]2+ + H3O+
so as to acidify the soil. This would produce blue flowers.
A8. Hydrolysis of the aluminium aqua ion in soft water lowers the pH of the water. Copper is more soluble at low pH values: the equilibrium [Cu(H2O)6]2+ ⇌ [Cu(H2O)5(OH)]+ + H+ lies further to the left. In tap water, the copper comes from the copper pipes.
Required practicalP1. At low concentration it is an irritant; at a higher concentration (>2 M) it is corrosive.
P2. Tap water will contain dissolved metal ions, which can affect the analysis.
P3. A = FeSO4
[Fe(H2O)6]2+ + 2OH− → [Fe(H2O)4(OH)2] + 2H2O
B = CuCl2
[Cu(H2O)6]2+ + 2OH− → [Cu(H2O)4(OH)2] + 2H2O
C = Fe(NO3)3
[Fe(H2O)6]3+ + 3OH− → [Fe(H2O)3(OH)3] + 3H2O
P4. Experiment a suggests the salt contains sulfate ions:
Ba2+(aq) + SO42−(aq) → BaSO4(s)
Experiment b suggests the salt contains Cr(III) ions as CO2 is produced. Metal(III) complexes are more acidic than metal(II) complexes and enough acid is liberated to turn carbonate into CO2 and water:
2H3O+ + CO32− → CO2 + 3H2O
The precipitate is [Cr(H2O)3(OH)3]:
2[Cr(H2O)6]3+ + 3CO32− → 2[Cr(H2O)3(OH)3] + 3CO2 + 3H2O
Therefore the compound is Cr2(SO4)3.
Practice questions1a(i)
MgO: ionic
P4O10: covalent
1a(ii)
Electronegativity difference small or electronegativities similar or big difference in electronegativity leads to ionic bonding
1b
Na2O + H2O → 2Na+ + 2OH− (or 2NaOH)
SO2 + H2O → H2SO3
1c
MgO + 2HCl → MgCl2 + H2O (or MgO + 2H+ → Mg2+ + H2O)
1d
P4O10 + 12NaOH → 4Na3PO4 +6H2O (or P4O10 + 12OH− → 4PO43− +6H2O)
2
Na2O: vigorous or violent or exothermic reaction; or forms a colourless solution
pH of solution formed = 13 or 14
Na2O + H2O → 2NaOH
P4O10 or P2O5: vigorous or violent or exothermic reaction; or forms a
Colourless solution; 1
pH of solution formed = 0 or 1
P4O10 + 6H2O → 4H3PO4
3 From sodium on the left-hand side of the Periodic Table, to sulfur on the right, the trend of the pH in solutions of the oxides is to decrease. Oxides on the left of the Period are alkaline, oxides on the right are acidic.
Sodium oxide is ionic and contains the oxide ion, O2–, that readily combines with hydrogen ions.
Na2O + H2O 2NaOH
Magnesium oxide is ionic but reacts less strongly with water than sodium oxide.
Aluminium oxide is amphoteric, so can react as either an acid or a base. It is ionic but the oxide ions are too strongly held in the lattice to react with water.
Silicon dioxide has a giant covalent structure which is difficult to break apart, and so does not react with water.
Oxides of phosphorus are covalent and exist in different forms, they form weak acids in solution. For example
P4O10 + 6H2O 4H3PO4
Oxides of sulfur are covalent and exist in different forms: sulfur dioxide and sulfur trioxide. Sulfur dioxide forms a weak acid but sulfur trioxide forms a strong acid
SO3 + H2O H2SO4
4a(i) Na2O + H2O 2NaOH
pH 14
4a(ii) SO2 + H2O H2SO3
But the main species present in solution is hydrated sulfur dioxide SO2.xH2O
pH approximately 1
4b Generally, ionic bonded oxides are basic, so pH is high. Covalent bonded oxides are acidic, so pH is low.
5a
MgO (is a white solid that) forms a suspension (or slightly soluble)
MgO + H2O → Mg(OH)2 (or → Mg2+ + 2OH−)
pH is 8 to 10
SO2 dissolves (or forms (colourless) solution)
SO2 + H2O → H2SO3 (or → H+ + HSO3−) (or → 2H+ + SO3
2−)
pH is 1 to 4
5b
Al(OH)3 + OH− → Al(OH)4 (or forms Al(OH)63−)
Al(OH)3 + 3H+ + 3H2O → Al(H2O)63+ (or forms [Al(H2O)5(OH)]2+, Al3+, AlCl3
6(i)
Green solution (not blue-green or grey-green)
[Cr(OH)6]3− (or Cr(H2O)(OH)5]2− or Cr(H2O)2(OH)4]−)
6(ii)
Green precipitate
Bubbles (or gas or fizzing or effervescence)
Cr(H2O)3(OH)3 (or Cr(OH)3)
7a
The number of protons increases (across the period) / nuclear charge increases Therefore, the attraction between the nucleus and electrons increases
7b
S8 molecules are bigger than P4
molecules
Therefore, van der Waals / dispersion / London forces between molecules are stronger in sulfur
7c
Sodium oxide contains O2– ions
These O2– ions react with water forming OH– ions
7d
P4O10 + 12OH– → 4PO43– + 6H2O
8a
An electron pair on the ligand Is donated from the ligand to the central metal ion
8b
Blue precipitate
Dissolves to give a dark blue solution
[Cu(H2O)6]2+ + 2NH3 → Cu(H2O)4(OH)2 + 2NH4+
Cu(H2O)4(OH)2 + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 2OH– + 2H2O
9a
MgO is ionicMelt itMolten oxide conducts electricity
9b
Macromolecular Covalent bonding Water cannot (supply enough energy to) break the covalent bonds / lattice
9c(Phosphorus pentoxide’s melting point is) lowerMolecular with covalent bondingWeak / easily broken / not much energy to break intermolecular forces or weak vdW / dipole-dipole forces of attraction between molecules
9dReagent (water or acid) Equation e.g. MgO + 2HCl → MgCl2 + H2O9e12NaOH + P4O10 4Na3PO4 + 6H2O
10aYellow / purple (solution) Brown precipitate / solid [Fe(H2O)6]3+ + 3OH– → Fe(H2O)3(OH)3 + 3H2O
10bBlue (solution) Dark / deep blue solution [Cu(H2O)6]2+ + 4NH3 → [Cu(H2O)2(NH3)4]2+ + 4H2O
10cColourless (solution) White precipitate / solid Bubbles / effervescence / gas evolved / given off 2[Al(H2O)6]3+ + 3CO3
2– → 2Al(H2O)3(OH)3 + 3CO2 + 3H2O
11Moles NaOH = 0.0212 × 0.5 = 0.0106Moles of H3PO4 = 1/3 moles of NaOH (= 0.00353)Moles of P in 25000 l = 0.00353 × 106 = 3.53× 103
Moles of P4O10 = 3.53 × 103/4Mass of P4O10 = 3.53 × 103/4 × 284 = 0.251 × 106 g = 251 kg
CHAPTER 16Assignment 1A1. Since the concentration is only given to one significant figure, precise measurements are not needed. Calculate (a) the mass of 1 mol of CuCl2.2H2O(s) and (b) the number of moles to be dissolved in 10 cm3 to make 0.1 mol dm−3 of solution. Weigh out the calculated quantity and dissolve it in 10 cm3 of deionised water. Note: this is the preparation of a reagent solution, not a standard solution for volumetric analysis.
A2. Both are corrosive and can cause burns. In both case the bottles should be stoppered when not in use. They should be stored in separate places.
A3. Octahedral aqua complex
A4. a. Water ligands are displaced by chloride ion ligands. The green colour is caused by a mixture of blue [Cu(H2O)6]2+ and yellow CuCl42− complex ions, and when there is little or no [Cu(H2O)6]2+ remaining the solution is yellow.
b. Similar to a. except that [Co(H2O)6]2+ is pink and CoCl42− is blue.
c. Successive replacement of water ligands in [Cu(H2O)6]2+ by ammonia to give a mixture of ammine complex ions; the most abundant one in concentrated ammonia solution is [Cu(NH3)5H2O]2+.
d. Successive replacement of water ligands in [Co(H2O)6]2+ by ammonia to give [Co(NH3)6]2+.
Assignment 2A1. 60 s, c = 0.049 mol dm−3; 120 s, c = 0.101 mol dm−3; 180 s, c = 0.169 mol dm−3; 240 s, c = 0.223 mol dm−3; 300 s, c = 0.261 mol dm−3
A2.
A3. This is given by the slope of the graph, 9.1 × 10−4 mol dm−3 s−1.
A4. 0.22 = 9.1 × 10−4 t; t = 242 s
A5. kr = 0.44/9.1 × 10−4 = 483 s−1
Assignment 3A1. MnO4
−(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
A2. a. Tablet 1b titre 17.1 cm3, therefore number of moles of MnO4− = (0.02 × 17.1)/1000
= 3.42 × 10−4
b. 0.00171 mol of Fe2+
c. Mass = 56 × 0.00171 = 95.76 mg
d. 95.76/845 = 0.113, or 11.3% w/w
A3. The expression is percentage of iron = titre (cm3) × 0.56/mass of the tablet (g).
Tablet 1a: 17.4 × 0.56/0.850 = 11.5%
Tablet 2a: 16.5 × 0.56/0.845 = 10.9%
Tablet 2b: 16.9 × 0.56/0.855 = 11.1%
A4. Only batch 1
A5. a. ±0.135%
b. The same apparatus, equipment and method are used for all tablets.
A6. By taking more sample tablets from a batch and calculating the mean.
Assignment 4A1. The unused feedstocks can be recycled to save costs and are cracked because the amounts in the feedstocks will not necessarily match the demand for the products.
A2. A build-up of carbon will reduce the surface area of the catalyst available for reaction. Burning off the carbon will regenerate the catalyst.
A3. To expose the maximum amount of catalyst to the reactants.
A4. Catalytic converters remove hydrocarbons, carbon monoxide, sulfur oxides and nitrogen oxides. They do not remove carbon dioxide, a greenhouse gas.
A5. The use of nonrenewable resources, dust hazards, health risks to miners and damage to the landscape due to mining.
A6. A catalyst in a different phase from that of the reactants is considered heterogeneous.
A7. To provide a greater surface area for the reactants.
A8. The presence of double bonds increases the distance between adjacent molecules and this reduces the van der Waals forces of attraction between the molecules, reducing the melting point.
A9. Animal fats contain a greater proportion of saturated fats, which, it is claimed, contribute to heart disease.
Practice questions1a(i)
SO2 + V2O5 → SO3 + V2O4
V2O4 + 1/2O2 → V2O5
V(IV) or 4 and V(V) or 5
1a(ii)
MnO4– + 8H+ + 4Mn2+ → 5Mn3+ + 4H2O
2Mn3+ + C2O42– → 2Mn2+ + 2CO2
Mn(III) or 3 and Mn(II) or 2
1b
[Co(NH3)6]2+ formed
Complex easy to oxidise
H2O2 (or air or oxygen)
1c
Moles of dichromate = (29.2/1000)×0.04 = 0.001168
Moles of Q2+ = (25/1000)×0.140 = 0.00350
Each mole of dichromate needs 6 electrons or half equation with 6 e−
Moles of electrons = 6×0.001168 = 0.007008 or moles Q2+:moles dichromate = 3:1
Moles of electrons per mole of Q = 0.007008/0.0035 = 2.002 = 2
Q(IV) or Q4+
2a(i)
Fe + 2HCl → FeCl2 + H2
or Fe + 2H+ → Fe2+ + H2
2a(ii)
PV = nRT, n = PV/RT
= 4.53 × 10−3 mol
2a(iii)
Moles of iron = 4.53 × 10−3 mol (or = 4.25 × 10−3)
Mass of iron = 4.53 × 10−3 × 55.8 = 0.253 g
2a(iv)
0.253 × 100/0.263 = 96.1%
2b(i)
Fe2+ → Fe3+ + e−
Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O
Cr2O72− + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+
2b(ii)
Moles of dichromate = moles Fe2+/6
= 4.53 × 10−3/6 = 7.55 × 10−4
Volume of dichromate = moles/concentration
= (7.55 × 10–4 × 1000)/0.0200
V = 37.75 cm3
2b(iii)
KMnO4 will also oxidise (or react with) Cl– (or chloride or HCl)
3a
Initially slow because reaction is between two negative ions (or between two negative reactants or two negative species) which repel each other.
Then Mn2+ (or Mn3+) ions are formed acting as an autocatalyst (or Mn2+ ions formed in the reaction act as a catalyst)
2MnO4− + 16H+ + 5C2O4
2− → 2Mn2+ + 8H2O + 4CO2
MnO4− + 4Mn2+ + 8H+ → 5Mn3+ + 4H2O
C2O42− + 2Mn3+ → 2Mn2+ + 2CO2
3b
Active sites are where reactants are adsorbed onto a catalyst surface (or bind or react on a catalyst surface).
Number of active sites increases if the surface area is increased (or catalyst is spread thinly or on honeycomb or powdered or decreased particle size.
Active sites blocked by another species or poison (or species adsorbed more strongly or species adsorbed irreversibly or species not desorbed).
4a
Effect on reaction rate: catalyst provides an alternative reaction route with a lower Ea, more molecules able to react or rate increased
Equilibrium: forward and backward rates changes by the same amount ; hence concentration of reactants and products constant or yield unchanged
4b
Heterogeneous: catalyst in a different phase or state to that of the reactants
Active site: place where reactants adsorbed or attached or bond; where reaction occurs
Reasons: large surface area; reduce cost or amount of catalyst
Catalyst poison: lead adsorbed; lead not desorbed or site blocked
4c
Reaction slow as both ions negatively charged or ions repel
2Fe2+ + S2O82− → 2Fe3+ + 2SO4
2–
2Fe3+ + 2I− → 2Fe2+ + I2
5a
3d7
5b
[Co(H2O)6]2+
Pink
5c(i)
[Co(NH3)6]2+
Pale brown or straw
5c(ii)
[Co(H2O)6]2+ + 6NH3 → [Co(NH3)6]2+ + 6H2O
5d
[Co(NH3)6]3+
An oxidising agent
6a
Iron
Heterogeneous; catalyst in a different phase from that of the reactants
Poison; a sulphur compound
Poison strongly adsorbed onto active sites / blocked
Poison not desorbed or reactants not adsorbed or catalyst surface area reduced
6b
Pale green solution
Green precipitate formed
Insoluble in excess ammonia
Equation:
e.g. [Fe(H2O)6]2+ + 2NH3 → [Fe(H2O)4(OH)2] + 2NH4+