okay, because of the excitement that it's all over [see
http://vcenotes.com/forum/index.php/topic,7613.0.html#new ], I couldn't sleep [even though I am sleep deprived
]
ANYWAYS, here it is:
VCE Chemistry 2008 Unit 4 ExamSuggested Solutions
Multiple ChoiceFor Q 1 to 3:
+H_2)(g) \leftrightharpoons CO(g) + 3H_2 (g),\; \Delta H = +206\; kJ\; mol^{-1})
Question 1A
To increase the proportion of reactants converted to products (yield), volume should be increased and temperature be increased also.
Question 2B
Unrelated to the stimulus, but it tells of a solid (powder) catalyst
Catalyst increase the reaction rate because reactants can become attached to its surface where they can meet and undergo reaction [this surface attraction somehow manage to lower the activation energy,
don't ask me why =S]
Question 3A
Equal amounts of reactants are added initially, when equilibrium is established, hydrogen is added.
A and B are both "correct" in showing the general trends, but B violates the mole ratio concept completely [its rise/fall should be thrice that of methane]
Note the way in A, the two species actually crossed, VCAA have taken the left axis seriously, which will be a consideration in question 8.
Question 4D
The biggest effect on the increase in reaction rate due to temperature rise is because the proportion of particles with high Ke increases, hence more fruitful collisions.
For Q 5 and 6
+OH^-(aq) \to CH_3OH (aq) + Cl^- (aq))
Energy profile, with X being forward activation energy, Y being
and Z being the activation energy of reverse reaction.
Question 5D
A reaction occurs when reactants collide with energy at or greater than activation energy.
Question 6C
Adding a catalyst will affect activation energy only, i.e. X and Z
Question 7A
+O_2(g) \to 2CuO(s)\; \Delta H = -312 \; kJ\; mol^{-1})
[1]
+\frac{1}{2}O_2(g) \to Cu_2O(s)\; \Delta H = -170 \; kJ\; mol^{-1})
[2]
The
for
 \to 2Cu_2O (s) + O_2 (g))
would be
)
(use energy profiles as supplied. +284
Question 8B...?
 \leftrightharpoons H_2(g) + I_2(g))
, initially in 2.0L vessel at equilibrium, then suddenly decreased volume to 1.3L.
Seeing the mole ratio of reactants to products is 1:1, no subsequent change occurs. In B, the concentration of the three species changed differently, whereas in C, they changed by the same amount.
Remembering back to Q3, that VCAA actually used the y axis as a measurement rather than an "indicator of change", the rise in concentration should be
proportional to the concentration of each species. In both graphs, the order was [HI]>[H
2]>[I
2], hence the rise in [HI] should be greater than the rise in [H
2] which is greater than the rise in [I
2]. Hence, B. [note that this has got NOTHING to do with mole ratios in the equation]
Though I am unsure if VCAA will actually test students with this kind of tricky stuff... maybe C will be accepted due to conventions [of ignoring the actual values, and just going by the change in amounts relative to each species].
For Q 9 and 10
 \leftrightharpoons PCl_3 (G) + Cl_2 (g))
Question 9B
working out the concentration quotient for all four options, B is the only one that is different. We also learn that K is 0.4
Question 10B
Unchanged, conservation of mass... This question was trickily placed [got me
], especially just after Q9 which indirectly gives the equilibrium constant. the key word was "the mass of the
gas mixture", which is the entire vessel, rather than PCl
5Question 11C
pretty straight forward equilibrium constant question.
Question 12A
Production of sodium propanoate from propanoic acid (100mL, 0.16M) and sodium hydroxide (100mL, 0.08M)
I - the pH of resulting solution will be less, which is false: dilution AND reaction with base will increase the pH, not decrease
II - the resulting solution contains equal amounts of propanoic acid and its conjugate base, technically not entirely correct, but it is close enough [there will be slightly less propanoic acid than conjugate base]
III - Before NaOH was added there were no propanoate ions present, wrong, some ionisation would occur, though to a very small extent, and there would be a tiny amount of propanoate ions.
hence, only II is correct.
Question 13A
Disposal of ethyl ethanoate should be in "ORGANIC LIQUIDS ONLY", NaCl (salt) can be tipped straight down the drain [or all kitchen should purchase "AQUEOUS WASTE ONLY" disposal units], and Solid lead compounds should be put in "DRY SOLIDS ONLY"
Question 14C
foam cup calorimeter containig 100mL of water would have slightly greater than 418 as the calibration factor [heat up water as well as container]
Question 15A
numerical value of heat of combustion of 1-propanol in kJ/g, is
Question 16D
Electrolysis of molten NaF would produce sodium metal and fluorine gas, whereas electrolysis of aqueous solution would just be the water reacting
Question 17C
Cr2+ and Pb2+ is the only pair that will react spontaneously
 + Co^{2+}(aq) \to 2Cr^{3+}(aq) + Co(s))
 + Pb^{2+}(aq) \to Co^{2+}(aq) + Pb(s))
 + 2Cr^{3+}(aq) \to Fe^{2+}(aq) + 2Cr^{2+}(aq))
Question 18D
The order of increasing strength as reductants [i.e. top right to bottom right]
+ve electrode | -ve electrode | V
P | Cu | 0.46
Cu | Q | 0.57
Cu | R | 1.10
Q | R | 0.53
Question 19D
ethane-1,2-diol (C2H4(OH)2) produce the greatest amounts of CO2 per electron.
the others are methanol, ethane, ethanol.
Question 20C
The electrochemical series cannot predict anything about reaction rates, however it can predict equilibrium constants to some extent, i.e. spontaneous redox pair has a reasonable K, non-spontaneous redox pair has an extremely low K [D is true, reaction between Sn2+ and Cu2+ has a much greater K than between Sn2+ and Zn2+, try it on the electrochemical series].
Short answer questionQuestion 1a) [2 marks]
 + 2HCl(aq) \to MgCl_2(aq) + H_2(g))
b) [2 marks]
higher rate of reaction, same yield. i.e. it flattens out to the same value, but just get there faster.
This is because having powdered form as opposed to ribbon increases the surface area, hence more collisions, greater rate of reaction. but same amount --> same yield
Question 2a) [4 marks]
1.87 kJ
oC
-1marks allocated to
- number of moles
- energy (using the data booklet for
)
- calibration factor
- correct significant figures
[Data given: 2.09g of ethanol, increase of 33.2 degrees of bomb calorimeter]
b) [3 marks]
69.8
oC
allocated to
- energy (60% of the second mark in previous question)
- change in temperature (SHC of water)
- final temperature
[Data given: 2.09g of ethanol, 60% efficiency of spirit burner, 200g of water, initial temp 25.3 degrees]
Question 3solution of 0.10 M acid of each
Acid | pH
I | 1.0
II | 3.0
III | 0.7
IV | 2.1
a) [1 mark]
Smallest Ka = II
b) [2 marks]
III has more than 1 acidic proton per molecule ([H+]=0.2 .... is it phailed sulfuric acid or super-spastic acid III?)
c) [1 mark]
% ionisation of IV = 7.9%, assuming it is monoprotic
d) [1 mark]
e) [2 marks]
diluting I and IV by a factor of 10. We note that IV is a weak acid (incomplete ionisation), hence dilution will increase the percentage ionisation. Also note that I is a strong acid, hence dilution will not increase the number of moles of H+. Hence the change in pH will be greater for I than IV.
f)
i) [2 marks]
Concentration of methanoic acid solution that has a pH the same as IV (2.1)
using Ka found from data booklet, we find the
equilibrium concentration of methanoic acid to be 0.350M
however, if we add the concentration of H+ at equilibrium, we will have 0.358M for initial concentration
so it's either
0.35M or 0.36M.
Which one VCAA accept/want will be a completely different matter. It's only 2 marks.
ii) [2 marks]
dissociation of methanoic acid is exothermic, hence heating it up will
increase the pH (less H+)
Question 4 \leftrightharoons 2NO_2(g))
at 100 degrees, 0.45 mol of N2O4 is placed in an empty vessel of 1.0L. When equilibrium is established there is 0.36 mol of NO2
a) [3 marks]
K at 100 degrees is 0.48M
b) [2 marks]
at 25 degrees K is 0.144
hence it is endothermic, as T increases K increases.
Question 5 - only got answers for Sulfuric Acid
a) [1 mark]
Sulfuric acid, O2 and FeS2
It is also possible to use H2 and O2 to produce water for the final step... but that's just unrealistic
b) [1 mark]
Reaction above room temperature
 + O_2 (g) \leftrightharpoons 2SO_3(g))
or any other ones [I think they are all above room temperature?]
c) [1 mark]
Waste heat from production can be used to heat water to drive a steam turbine to produce electricity, which provides energy to the plant and reduce energy costs.
d)
i) [1 mark]
Hydronium ion (H3O+) is one of the useful products... You could have said superphosphate or anything else
ii) [1 mark]
 + H_2O(l)\to H_3O^+ (aq) + HSO_4^-(aq))
or if you have memorised the superphosphate formula =\
Question 6Electrolysis of CO2 to produce methanol
Cathode -
 + 6H+(aq) + 6e^- \to CH_3OH(aq) + H_2O(l))
Anode -
\to O_2(g) + 4H^+ (aq) + 4e^-)
a) i) [1 mark]
 + 3O_2(g) \to 2CO_2(g) + 4H_2O(g))
ii) [1 mark]
b) cell operates at 24.0 hours at 25.5 A
i) [1 mark]
Q = 2.20 x 10^6 A
ii) [3 marks]
122 g of methanol
allocated to
- number of moles of electrons (Faraday's constant)
- number of moles of methanol (mole ratio)
- mass of methanol
iii) [1 mark]
Lower output of methanol may be due to CO2/H+ reacting to form other compounds
c) [1 mark]
Overall effect of using methanol generated from this method would be nil. CO2 released is balanced by the CO2 originally extracted. Also note that the electric power used to power the electrolytic cell is harvested from solar cells.
Question 7a)
Ni/Ni2+ and Cd/Cd2+ (-0.40V potential) is connected as galvanic cells
i) [1 mark]
direction of flow would be from Cd to Ni (right to left)
ii) [1 mark]
Half equation at anode:
 \to Cd^{2+}(aq) + 2e^-)
iii) [2 marks]
Two properties of salt bridge - readily soluble and does not react with reactants
b)
Cd(s)------------------------------[ - ]
Cd(OH)2(s) + KOH(aq)
Porous layer, KOH(aq)
Ni(OH)2(s) + KOH(aq)
NiO(OH)(s)
Steel------------------------------[ + ]
+2NiO(OH)(s) + 2H_2O(l) \to Cd(OH)_2(s) + 2Ni(OH)_2(s))
i) [1 mark]
oxidation numbers, Cd (0) --> Cd2+ (2+), is the anode.
ii) [1 mark]
The feature that enable secondary cells to be recharged is that the products of discharge remain in contact with the electrode(s).
iii) [1 mark]
Half equation at negative electrode during discharge:
 +2OH^-(aq) \to Cd(OH)_2(S)+2e^-)
iv) [1 mark]
Half equation connected to the negative terminal during recharge (at negative electrode, cathode):
_2(s) + 2e^- \to Cd(s) + 2OH^-(aq))
Question 8Fuel cell, H2(g) coming from the left, at the anode. O2(g) coming in from the right (cathode), H2O(g) going out. H+ and H2PO4- electrolyte
a) i) [1 mark]
at the anode,
\to 2H^+ (aq) + 2e^-)
ii) [1 mark]
at the cathode,
+4H^+ (aq) + 4e^- \to 2H_2O(g))
b) [1 mark]
H2PO4- [anion] moves to the anode to balance the loss of negative charge. right to left.
c) i) [2 marks]
using
, 180 kJ (2 sig figs)
ii) [1 mark]
62% efficiency
d) [2 marks]
Advantage - higher efficiency than petrol motor
Disadvantage - more expensive to make (fuel more expensive, etc)
Question 9a) [1 mark]
8.2 x 10
15 kg
b) [2 marks]
3.8 x 10
19 kJ
DONE
[in total, I'd have lost 3 or more marks I think.... but that's nothing to worry about....! I AM FREE!!!!!]
Congrats to everyone else, hope you've done well
GG
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