Post by: Mao on November 13, 2008, 12:25:42 pm
ANYWAYS, here it is:
VCE Chemistry 2008 Unit 4 Exam
Suggested Solutions
Multiple Choice
For Q 1 to 3:
Question 1
A
To increase the proportion of reactants converted to products (yield), volume should be increased and temperature be increased also.
Question 2
B
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 3
A
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 4
D
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
Energy profile, with X being forward activation energy, Y being
Question 5
D
A reaction occurs when reactants collide with energy at or greater than activation energy.
Question 6
C
Adding a catalyst will affect activation energy only, i.e. X and Z
Question 7
A
The
Question 8
B...?
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]>[H2]>[I2], hence the rise in [HI] should be greater than the rise in [H2] which is greater than the rise in [I2]. 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
Question 9
B
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 10
B
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 PCl5
Question 11
C
pretty straight forward equilibrium constant question.
Question 12
A
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 13
A
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 14
C
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 15
A
numerical value of heat of combustion of 1-propanol in kJ/g, is
Question 16
D
Electrolysis of molten NaF would produce sodium metal and fluorine gas, whereas electrolysis of aqueous solution would just be the water reacting
Question 17
C
Cr2+ and Pb2+ is the only pair that will react spontaneously
Question 18
D
The order of increasing strength as reductants [i.e. top right to bottom right]
Code: [Select]
+ve electrode | -ve electrode | V
P | Cu | 0.46
Cu | Q | 0.57
Cu | R | 1.10
Q | R | 0.53
Question 19
D
ethane-1,2-diol (C2H4(OH)2) produce the greatest amounts of CO2 per electron.
the others are methanol, ethane, ethanol.
Question 20
C
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 question
Question 1
a) [2 marks]
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 2
a) [4 marks]
1.87 kJoC-1
marks 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 3
solution of 0.10 M acid of each
Code: [Select]
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
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
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]
or if you have memorised the superphosphate formula =\
Question 6
Electrolysis of CO2 to produce methanol
Cathode -
Anode -
a) i) [1 mark]
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 7
a)
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:
iii) [2 marks]
Two properties of salt bridge - readily soluble and does not react with reactants
b)
Code: [Select]
Cd(s)------------------------------[ - ]
Cd(OH)2(s) + KOH(aq)
Porous layer, KOH(aq)
Ni(OH)2(s) + KOH(aq)
NiO(OH)(s)
Steel------------------------------[ + ]
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:
iv) [1 mark]
Half equation connected to the negative terminal during recharge (at negative electrode, cathode):
Question 8
Fuel 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,
ii) [1 mark]
at the cathode,
b) [1 mark]
H2PO4- [anion] moves to the anode to balance the loss of negative charge. right to left.
c) i) [2 marks]
using
ii) [1 mark]
62% efficiency
d) [2 marks]
Advantage - higher efficiency than petrol motor
Disadvantage - more expensive to make (fuel more expensive, etc)
Question 9
a) [1 mark]
8.2 x 1015 kg
b) [2 marks]
3.8 x 1019 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
Post by: Synesthetic on November 13, 2008, 12:27:31 pm
Ignore red font, these answers are incorrect according to above.
Possibly correct...
EXTENDED RESPONSE
1)a) Mg + 2HCl -> MgCl2 + H2
b) Powdered => greater surface area => more frequent collisions => increased rate of reaction
Steeper graph attaining maximum V(H2) over less time
2)a) Calibration Factor = 1.87 kJ/°C
b) Final Temperature = 69.8 °C
3)a) Acid II, because Ka is proportional to 10^(-pH); substituting values shows that II has the lowest [H+]
b) Acid III, since pH < 1;
Quote from Pandemonium: [ http://vcenotes.com/forum/index.php/topic,7616.msg94599.html#msg94599 ]
We know that the [acid] was all 0.1M
therefore, assuming complete ionisation of the acid, we'd have [H+] = 0.1M
pH would therefore be equal to 1 at the minimum if it were monoprotic.
acid III i think had pH 0.7, therefore it had to be at least diprotic.
c) Percentage Ionisation = 7.9%
d) Ratio of [OH-] values = 100
e) Same change in pH, +1 (tick the third box)
Since dilution by a factor of 10 <=> divide [H+] by 10,
new pH = -log10{[H+]/10} = old pH - log10(.1) = old pH 1 + 1 => same change in pH for both acids (+1)
f)i) [HCOOH] = 7.9 x 10^-3 M
f)ii) Exothermic, heated => K decreases => [H+] decreases => pH increases
4)a) Kc = 0.48 M
We are told 0.45mol N2O4 is initially placed in the container, and 0.36mol NO2 forms;
Thus due to the 1:2 stoichiometry, 0.18mol N2O4 reacts to form 0.36mol NO2; and 0.27mol N2O4 remains.
Thus Kc = (0.36)^2 / (0.27) = 0.48 M
b) Endothermic reaction, since the lower temperature results in a lower Kc value.
5)a) Sulfuric Acid: circle O2 [I would like to make this blue...but I suspect VCAA do want more than one reactant circled...]
b) 2SO2 + O2 <-> 2SO3
c) In the converter, heated gas is recycled between runs through catalyst beds, hence reducing heating (energy) costs
d)i) Superphosphate [fertiliser]
d)ii) Did not expect this to be assumed knowledge...alternately you can write the equation for production of ammonium sulfate, also a fertiliser.
Superphosphate: Ca3(PO4)2 + 2H2SO4 + 4H2O -> Ca(H2PO4)2 + 2CaSO4.2H2O ... lol
( I was stymied and wrote the equation for dehydration of glucose ;D )
6)a)i)2CH3OH + 3O2 -> 2CO2 + 4H2O
a)ii) deltaH = -1450 kJ/mol
b)i) 2.20 x 10^6 C
b)ii) 122 g (3 s.f.)
b)iii) Gaseous CO2 escapes into the air thereby lowering expected yield of methanol
c) Overall neutrality - CO2 is produced by methanol combustion but consumed by electrolysis
7)a)i) electron flow direction: <----
a)ii) Cd -> Cd2+ + 2e-
a)iii) Inert / does not form precipitate with electrolytes
b)i) - (upper circle), + (lower circle)
b)ii) Products of discharge reaction remain in contact with electrodes in a convertible form.
b)iii) Anode equation [discharge]: Cd + 2OH- -> Cd(OH)2 + 2e-
b)iv) Cathode equation [recharge]: Cd(OH)2 + 2e- -> Cd + 2OH-
8)a)i) H2 + 2OH- -> 2H2O + 2e- I can't believe I did that...
a)ii) O2 + 4H+ + 4e- -> 2H2O
b) H2PO4- flow direction: <-------
c)i) 1.8 x 10^2 kJ
c)ii) Efficiency = 62%
d) Advantage: Fuel cells are more efficient due to their facility of directly converting chemical energy into electrical energy.
Disadvantage: Because fuel cell technology is still developing, they are more expensive than internal combustion engines.
9)a) m(CO2) = 8.24 x 10^15 kg
b) Energy produced = 3.69 x 10^19 kJ
^ [3 significant figures instead of what it should be (2, from 0.42% etc.), but VCAA allows one decimal place either way]
So after all...my paper is more like 71/79, which hopefully secures a low A+.
Post by: onlyfknhuman on November 13, 2008, 12:29:30 pm
Post by: vce01 on November 13, 2008, 12:32:09 pm
Post by: Synesthetic on November 13, 2008, 12:32:48 pm
Post by: Synesthetic on November 13, 2008, 12:33:41 pm
aww 16/20. oh well. are you sure the last two multi choices are D and C?
19 was D, divide n(CO2) by n(e-), in the ethane-1,2-diol equation this gives 1/5 > {1/6,1/7} in the other choices
20 was C, electrochemical series does not give information about reaction rates
Post by: L on November 13, 2008, 12:34:21 pm
Post by: onlyfknhuman on November 13, 2008, 12:35:54 pm
i expect a more detailed solution from mao for extended section rather than wat the other guy posted =X. 18/20 for multi-choice. as if i got the 1st question wrong T_T >.<
at least he was kind enough to actually put something up?
Post by: monicak on November 13, 2008, 12:36:45 pm
Post by: Synesthetic on November 13, 2008, 12:47:14 pm
hmm hopefully they accept 8C and B, if it could be either based on VCE knowledge.. Could you put up Q17 and explain it please?
17)
From the given equations you can determine that the Cr3+ reduction equation lies below Co2+ and above Fe2+, because Cr2+ reduces Co2+, and Cr3+ oxidises Fe
Pb2+ is above Co2+ hence Pb2+ reacts spontaneously with Cr2+
None of the other choices result in a spontaneous reaction
Post by: nrisme on November 13, 2008, 12:48:27 pm
Post by: cara.mel on November 13, 2008, 12:48:42 pm
i expect a more detailed solution from mao for extended section rather than wat the other guy posted =X. 18/20 for multi-choice. as if i got the 1st question wrong T_T >.<
People are not your solution writing slaves...
Post by: Synesthetic on November 13, 2008, 12:50:43 pm
i expect a more detailed solution from mao for extended section rather than wat the other guy posted =X. 18/20 for multi-choice. as if i got the 1st question wrong T_T >.<
at least he was kind enough to actually put something up?
i expect a more detailed solution from mao for extended section rather than wat the other guy posted =X. 18/20 for multi-choice. as if i got the 1st question wrong T_T >.<
People are not your solution writing slaves...
;D Thanks...I'll try to retain my altruism!
Post by: L on November 13, 2008, 12:51:57 pm
GO MAO!
but yeh ty to the other guy 4 putting up the answers
Post by: Pandemonium on November 13, 2008, 12:53:57 pm
the step from reactant to products actually consumes slightly more energy than is released from products to reactants.
so you'd have a net gain of CO2 actually.
Post by: Synesthetic on November 13, 2008, 12:59:49 pm
6c...My understanding was, the question involves the production of methanol by electrolysis (thus consuming CO2), balanced against its subsequent use in combustion engines (thus producing CO2) => overall carbon neutrality.
the step from reactant to products actually consumes slightly more energy than is released from products to reactants.
so you'd have a net gain of CO2 actually.
Post by: monicak on November 13, 2008, 01:01:17 pm
17)Thank you :)
From the given equations you can determine that the Cr3+ reduction equation lies below Co2+ and above Fe2+, because Cr2+ reduces Co2+, and Cr3+ oxidises Fe
Pb2+ is above Co2+ hence Pb2+ reacts spontaneously with Cr2+
None of the other choices result in a spontaneous reaction
Post by: bucket on November 13, 2008, 01:03:55 pm
6c...My understanding was, the question involves the production of methanol by electrolysis (thus consuming CO2), balanced against its subsequent use in combustion engines (thus producing CO2) => overall carbon neutrality.
the step from reactant to products actually consumes slightly more energy than is released from products to reactants.
so you'd have a net gain of CO2 actually.
I wrote the same thing =\
Post by: Pandemonium on November 13, 2008, 01:05:08 pm
Post by: riadnicolas on November 13, 2008, 01:05:23 pm
6c...
the step from reactant to products actually consumes slightly more energy than is released from products to reactants.
so you'd have a net gain of CO2 actually.
My understanding was, the question involves the production of methanol by electrolysis (thus consuming CO2), balanced against its subsequent use in combustion engines (thus producing CO2) => overall carbon neutrality.
wasnt it just about methanol production in fuel cell
Post by: doboman on November 13, 2008, 01:06:27 pm
Post by: Synesthetic on November 13, 2008, 01:07:59 pm
3b was which one had more than one proton didn't it?
well we know that the [acid] was all 0.1M
therefore, assuming complete ionisation of the acid, we'd have [H+] = 0.1M
pH would therefore be equal to 1 at the minimum if it were monoprotic.
acid III i think had pH 0.7, therefore it had to be at least diprotic.
(from the other thread)
Thanks, that sounds like the required explanation!
Post by: nathankb on November 13, 2008, 01:08:47 pm
Post by: bucket on November 13, 2008, 01:09:20 pm
i thought it was getting rid of CO2
Yeah but then the methanol was used in a combustion engine, hence producing more CO2.
Post by: Synesthetic on November 13, 2008, 01:09:34 pm
6c...
the step from reactant to products actually consumes slightly more energy than is released from products to reactants.
so you'd have a net gain of CO2 actually.
My understanding was, the question involves the production of methanol by electrolysis (thus consuming CO2), balanced against its subsequent use in combustion engines (thus producing CO2) => overall carbon neutrality.
wasnt it just about methanol production in fuel cell
Oh - yeah, it WAS a fuel cell wasn't it...
i thought it was getting rid of CO2CO2 was on the reactants side of the equation producing methanol
Post by: monicak on November 13, 2008, 01:10:17 pm
Post by: charm on November 13, 2008, 01:11:36 pm
I said it was neutral, since 1 mol of methanol produced 1 mol of CO2 in the combustion and 1 mol of CO2 was used to produce 1 mol of methanol...
yes me too...
Post by: riadnicolas on November 13, 2008, 01:14:19 pm
6c...
the step from reactant to products actually consumes slightly more energy than is released from products to reactants.
so you'd have a net gain of CO2 actually.
My understanding was, the question involves the production of methanol by electrolysis (thus consuming CO2), balanced against its subsequent use in combustion engines (thus producing CO2) => overall carbon neutrality.
wasnt it just about methanol production in fuel cell
Oh - yeah, it WAS a fuel cell wasn't it...i thought it was getting rid of CO2CO2 was on the reactants side of the equation producing methanol
wat did the question say exactly?
Post by: bucket on November 13, 2008, 01:15:41 pm
Post by: Mao on November 13, 2008, 01:23:25 pm
what's this about fuel cell...? i don't see it on that page at all6c...
the step from reactant to products actually consumes slightly more energy than is released from products to reactants.
so you'd have a net gain of CO2 actually.
My understanding was, the question involves the production of methanol by electrolysis (thus consuming CO2), balanced against its subsequent use in combustion engines (thus producing CO2) => overall carbon neutrality.
wasnt it just about methanol production in fuel cell
Oh - yeah, it WAS a fuel cell wasn't it...i thought it was getting rid of CO2CO2 was on the reactants side of the equation producing methanol
wat did the question say exactly?
just "electrolytic cell"
Post by: Synesthetic on November 13, 2008, 01:31:46 pm
Quote
b) cell operates at 24.0 hours at 25.5 A
i) [1 mark]
Q = 2.20 x 10^6 A
ii) [3 marks]
12.2 g of methanol
2.20 x 10^6 C :P
Hmm... I got 122 g for m(CH3OH)
Post by: lwine on November 13, 2008, 01:33:56 pm
Question 6 b) ii.
I got 122g, and so did Synesthetic.
Operating a fuel cell for 24 hours at 35.5A with 12.2 grams of methanol feels just a little unrealistic, but I can't remember the specifics of the question.
I was hoping you could check that.
Thanks.
Post by: We Want Peace. on November 13, 2008, 01:37:40 pm
some MC .
Post by: Mao on November 13, 2008, 01:38:57 pm
To Mao:
Question 6 b) ii.
I got 122g, and so did Synesthetic.
Operating a fuel cell for 24 hours at 35.5A with 12.2 grams of methanol feels just a little unrealistic, but I can't remember the specifics of the question.
I was hoping you could check that.
Thanks.
arghhhh i hate not being able to read my own scribbles. [very messy indeed]
fix'd
Post by: We Want Peace. on November 13, 2008, 01:39:27 pm
To Mao:
Question 6 b) ii.
I got 122g, and so did Synesthetic.
Operating a fuel cell for 24 hours at 35.5A with 12.2 grams of methanol feels just a little unrealistic, but I can't remember the specifics of the question.
I was hoping you could check that.
Thanks.
me too.
Post by: riadnicolas on November 13, 2008, 01:48:13 pm
what's this about fuel cell...? i don't see it on that page at all6c...
the step from reactant to products actually consumes slightly more energy than is released from products to reactants.
so you'd have a net gain of CO2 actually.
My understanding was, the question involves the production of methanol by electrolysis (thus consuming CO2), balanced against its subsequent use in combustion engines (thus producing CO2) => overall carbon neutrality.
wasnt it just about methanol production in fuel cell
Oh - yeah, it WAS a fuel cell wasn't it...i thought it was getting rid of CO2CO2 was on the reactants side of the equation producing methanol
wat did the question say exactly?
just "electrolytic cell"
my bad, and also for the methanol question i f we said it reacted with oxygen and therefore lost some mass is that correct
Post by: We Want Peace. on November 13, 2008, 01:53:13 pm
why increase the pressure??. ( 2 particles on left. 4 particles on right?)
Post by: Synesthetic on November 13, 2008, 01:55:28 pm
Post by: We Want Peace. on November 13, 2008, 01:56:21 pm
Post by: Synesthetic on November 13, 2008, 02:29:16 pm
Quote
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.
Right...so they wanted a value from the data booklet for 2 marks??
I calculated [H+] from pH of Acid IV, and then let that equal [HCOOH] = 7.9x10^-3 M ...since it is monoprotic
In supposition this is completely off, but in the exam I could not imagine the question to be asking us to look up that figure in the data booklet!
*edit* Well, now I see that you added [H+] to that value, took me a while :P
---
While I'm at it I'd appreciate elucidation on this:
Quote
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.
So, we needed to observe Ka [equilibrium] principles? I thought dilution would affect each acid in the same way, adding 1 to the pH value of both.
*edit* Is it incorrect to assume pH=2.1 => strong acid? Although, I am aware that acidity strength relates to Ka
It almost seems...unreasonable to me that they should expect us to take that into consideration. I thought they were testing us on the formula for interchanging pH / concentration, hence in my solution:
e) Same change in pH, +1 (tick the third box)
Since dilution by a factor of 10 <=> divide [H+] by 10,
new pH = -log10{[H+]/10} = old pH - log10(.1) = old pH 1 + 1 => same change in pH for both acids (+1)
Post by: 205ism on November 13, 2008, 02:31:48 pm
There goes 40, damn
Post by: username on November 13, 2008, 02:33:53 pm
Post by: cosmicgate on November 13, 2008, 02:38:32 pm
I'm such a spastic, 67 or 68 /79wat u mean? last year u could drop 11 marks and get an A+
There goes 40, damn
u should have a good chance at 40+ if u had A+ mid year, A+ sacs and HIGH A or A+ end of year... otherwise no idea.
Post by: Synesthetic on November 13, 2008, 02:43:34 pm
It might just be me but the ambiguity of these acid equilibria questions is terrible...I'd be surprised if the majority of students were able to get most of the marks available.
It's these types of questions that make me really frustrated with chem...
Post by: Mao on November 13, 2008, 02:44:55 pm
for acid IV,
hence, when it is diluted to 0.01M, assuming the H+ from self ionisation from water is insignificant
solving this quadratic,
so change in pH for IV is about 0.5, whereas it is 1.0 for I.
Post by: Mao on November 13, 2008, 02:46:53 pm
*referring to my post above*
It might just be me but the ambiguity of these acid equilibria questions is terrible...I'd be surprised if the majority of students were able to get most of the marks.
completely agree
i would not have known this if i hadn't ask coblin about this earlier this year.
Post by: Synesthetic on November 13, 2008, 02:59:31 pm
This is not in the course, but fuck, the course is over, so I'll just do it anyways:
for acid IV,
hence, when it is diluted to 0.01M, assuming the H+ from self ionisation from water is insignificant
solving this quadratic,, pH=2.64
so change in pH for IV is about 0.5, whereas it is 1.0 for I.
LOL a quadratic
Thanks a lot for that explanation; now I can see your logic...and wishing I had known of it beforehand too :)
Nothing prepared me for this type of question, and I do a lot of work...
I suppose 70/79 isn't a disaster, probably my worst paper of the exam period though.
Post by: Mikey123 on November 13, 2008, 02:59:46 pm
Post by: sb3700 on November 13, 2008, 03:10:03 pm
This is not in the course, but fuck, the course is over, so I'll just do it anyways:
for acid IV,
hence, when it is diluted to 0.01M, assuming the H+ from self ionisation from water is insignificant
solving this quadratic,
so change in pH for IV is about 0.5, whereas it is 1.0 for I.
That is of course assuming that acid IV is monoprotic, which is not told either way in the question. Otherwise, denominator changes as appropriate.
Post by: nathankb on November 13, 2008, 03:23:11 pm
Post by: Mao on November 13, 2008, 03:31:09 pm
Post by: cosmicgate on November 13, 2008, 03:34:40 pm
dont have scanner =S sorrymassive ask, but maybe take photos of it???
or too much :P?
u'd make thousands of people happy... literally.
Post by: csq7 on November 13, 2008, 03:36:06 pm
And, for qn 9b, i got 3.7x10^ 16.. you sure your answer is in Kj??
Thanks for the solutions :)
Post by: Synesthetic on November 13, 2008, 03:38:09 pm
Hey mao, for qn 2b and the 69.8 degrees, shouldnt it be to 2 sig figs, cos it said 60% of the energy.. and thats to 2 sig figs..
And, for qn 9b, i got 3.7x10^ 16.. you sure your answer is in Kj??
Thanks for the solutions :)
I got 3.69 x 10^19 for 9b, I initially had 10^16 then realised it wasn't the correct units;
Don't worry too much about sig. fig.'s, VCAA allows one decimal place either way (as far as I know, and I believe that's how they marked my mid-year paper too), so whether it's 2 or 3 shouldn't make a difference.
Post by: csq7 on November 13, 2008, 03:39:46 pm
Post by: nathankb on November 13, 2008, 03:40:36 pm
Post by: csq7 on November 13, 2008, 03:44:36 pm
Post by: Mao on November 13, 2008, 03:49:36 pm
dont have scanner =S sorrymassive ask, but maybe take photos of it???
or too much :P?
u'd make thousands of people happy... literally.
my phone camera is crap :P
Post by: username on November 13, 2008, 03:51:26 pm
Post by: Synesthetic on November 13, 2008, 04:03:46 pm
Could you have written lack of standard conditions or something really vague like that for the methanol question, 6bii?
Actually that's probably the safest answer...instead of what I wrote (gaseous CO2 escapes => less methanol than expected produced)
Post by: username on November 13, 2008, 04:20:13 pm
Could you have written lack of standard conditions or something really vague like that for the methanol question, 6bii?
Actually that's probably the safest answer...instead of what I wrote (gaseous CO2 escapes => less methanol than expected produced)
Oh that's a relief. But its okay, you aced it anyway, unlike I did.
Post by: Synesthetic on November 13, 2008, 04:38:09 pm
I lost 9 marks, I hardly aced it :PCould you have written lack of standard conditions or something really vague like that for the methanol question, 6bii?
Actually that's probably the safest answer...instead of what I wrote (gaseous CO2 escapes => less methanol than expected produced)
Oh that's a relief. But its okay, you aced it anyway, unlike I did.
If 87% is the cut-off for A+ again I might get a score in the low 40's, though.
Post by: ilovevce on November 13, 2008, 04:43:53 pm
Also, for the disposal question in MC, did it say NaCl? I thought it said unused NaOH.
Post by: charm on November 13, 2008, 04:46:49 pm
Also, for the disposal question in MC, did it say NaCl? I thought it said unused NaOH.
it said NaCl im pretty sure...
Post by: username on November 13, 2008, 05:02:20 pm
I lost 9 marks, I hardly aced it :PCould you have written lack of standard conditions or something really vague like that for the methanol question, 6bii?
Actually that's probably the safest answer...instead of what I wrote (gaseous CO2 escapes => less methanol than expected produced)
Oh that's a relief. But its okay, you aced it anyway, unlike I did.
If 87% is the cut-off for A+ again I might get a score in the low 40's, though.
Don't worry, I know you'll get >40s ^_^. Goodluck.
Post by: julezian on November 13, 2008, 05:06:55 pm
there goes my 40
Post by: hamtarofreak on November 13, 2008, 06:02:08 pm
I did pretty good in the end it seems. See, it does pay to neglect Chemistry. =P
Post by: lanvins on November 13, 2008, 06:28:55 pm
Post by: ed_saifa on November 13, 2008, 06:30:38 pm
See, it does pay to neglect Chemistry. =PMy Chem teacher would strike you down for saying that :P
Post by: hamtarofreak on November 13, 2008, 06:33:53 pm
See, it does pay to neglect Chemistry. =PMy Chem teacher would strike you down for saying that :P
Yeah lol, my Chem teacher gave me the coldest look when I said I'd studied for every exam but my 2 today =P
Post by: Synesthetic on November 13, 2008, 07:00:30 pm
For an advantage of a fuel cell would it be acceptable to say becuase it supplies a continuous supply of electricty as long as a continous supply of reactants is provided or was it necessary to make a comparison?
The comparison is somewhat implied here. I would accept that...
Post by: pfftlah on November 14, 2008, 12:16:10 am
im quite happy, i think i can get an A+ [hopefully!] does anyone remember for MC question 8 which graph had the curved lines rather than the straight ones? i think it was D..damn i think i got that wrong.
is incomplete reaction a suitable answer to the lower methanol output?
thanks mao and synesthetic for putting up the answers.
yaaaaaaaaaaaaaaaay, FREEDOM =D
we can enjoy ourselves without feeling guilty now [well for those of us who've finished exams that is]
Post by: Synesthetic on November 14, 2008, 09:14:17 am
Synesthetic: i got the same answers as you regarding the change in pH and oxygen being the only reactant [the answers you had different to mao's]. i doubt that we were required to think that in depth for the pH question - if they had wanted us to consider the relative weakness and strength of the acids, they would have guided us through it and it would have been worth more than 2 marks.
im quite happy, i think i can get an A+ [hopefully!] does anyone remember for MC question 8 which graph had the curved lines rather than the straight ones? i think it was D..damn i think i got that wrong.
is incomplete reaction a suitable answer to the lower methanol output?
thanks mao and synesthetic for putting up the answers.
yaaaaaaaaaaaaaaaay, FREEDOM =D
we can enjoy ourselves without feeling guilty now [well for those of us who've finished exams that is]
Well it's somewhat a relief to know that I was not the only one :)
I'm hoping you are right about the change in pH, as I said before it seems unreasonable for - as you mentioned - a 2 mark question. Still would rather have known about Mao's answer, that way you cannot be penalized.
Quite a few people I know also only chose O2, as we did - I think my problem was that I was too fixated on the materials for the Contact Process, rather than industrial production on a broader scale. But that question was farcical, we spent the entire year learning about the equilibrium considerations of the converter reaction and all we needed for the exam was a reaction above room temperature!
Yeah incomplete reaction might be acceptable, especially if there were stoichiometric considerations in the question parts beforehand (can't recall...) I really can't give a more definite answer! But I will maintain that username's answer - absence of standard conditions - is probably the safest answer, given that it is an electrolytic reaction. Although the only problem with that is the methanol synthesis reaction is not in the assigned electrochemical series....but it shouldn't really matter.
And yes MC #8 the graph with curved lines was D, I chose it too and felt like an idiot when I read his solutions ;D
Post by: Synesthetic on November 14, 2008, 09:20:28 am
c) In the converter, heated gas is recycled between runs through catalyst beds, hence reducing heating (energy) costs
Hmm, perhaps this was not as far off the mark as I thought...
"To conserve energy, the mixture is heated by exhaust gases from the catalytic converter by heat exchangers."
[ http://en.wikipedia.org/wiki/Contact_process ]
And well, in this following image these heat exchangers seem to facilitate my 'runs through catalyst beds' :P
[ http://media-2.web.britannica.com/eb-media/59/7359-004-224E8A41.gif ]
Not an identical response but it makes me feel a whole lot better, I thought I had totally stuffed that one by conflating considerations for the exothermic reaction with the double absorption process :)
*edit* regarding, 5)a) circling O2 and (?) FeS2
"During WWII...The sulfur [from iron sulfide] was used in the production of sulfuric acid, an important chemical for industrial purposes. Now most sulfur production comes from H2S gas recovered from natural gas wells."
[ http://mineral.galleries.com/minerals/sulfides/pyrite/pyrite.htm ]
A bit out of date, VCAA; I thought the whole coursework was engineered towards rendering students as MODERN chemists ;D Maybe there were testing us on our knowledge of the chronology of the industrial chemical age...
Post by: jeremykleeman on November 14, 2008, 12:06:05 pm
I did the same pH calculation as you guys (same change), but Mao raised doubts in me. Ignoring the fact it is a weak acid and just given the pH of solution I think we would be right, but now I am not sure.
Good luck to everyone with your study scores!
Post by: Pandemonium on November 14, 2008, 03:34:59 pm
c) In the converter, heated gas is recycled between runs through catalyst beds, hence reducing heating (energy) costs
Hmm, perhaps this was not as far off the mark as I thought...
"To conserve energy, the mixture is heated by exhaust gases from the catalytic converter by heat exchangers."
[ http://en.wikipedia.org/wiki/Contact_process ]
And well, in this following image these heat exchangers seem to facilitate my 'runs through catalyst beds' :P
[ http://media-2.web.britannica.com/eb-media/59/7359-004-224E8A41.gif ]
Not an identical response but it makes me feel a whole lot better, I thought I had totally stuffed that one by conflating considerations for the exothermic reaction with the double absorption process :)
*edit* regarding, 5)a) circling O2 and (?) FeS2
"During WWII...The sulfur [from iron sulfide] was used in the production of sulfuric acid, an important chemical for industrial purposes. Now most sulfur production comes from H2S gas recovered from natural gas wells."
[ http://mineral.galleries.com/minerals/sulfides/pyrite/pyrite.htm ]
A bit out of date, VCAA; I thought the whole coursework was engineered towards rendering students as MODERN chemists ;D Maybe there were testing us on our knowledge of the chronology of the industrial chemical age...
Did you circle those two hydrocarbons as well?
Cause they produce water and heat. LAwlLALWLWlaLASWlalWLAwll you could circle anything as long as you made a justification.
Post by: Synesthetic on November 14, 2008, 04:29:34 pm
c) In the converter, heated gas is recycled between runs through catalyst beds, hence reducing heating (energy) costs
Hmm, perhaps this was not as far off the mark as I thought...
"To conserve energy, the mixture is heated by exhaust gases from the catalytic converter by heat exchangers."
[ http://en.wikipedia.org/wiki/Contact_process ]
And well, in this following image these heat exchangers seem to facilitate my 'runs through catalyst beds' :P
[ http://media-2.web.britannica.com/eb-media/59/7359-004-224E8A41.gif ]
Not an identical response but it makes me feel a whole lot better, I thought I had totally stuffed that one by conflating considerations for the exothermic reaction with the double absorption process :)
*edit* regarding, 5)a) circling O2 and (?) FeS2
"During WWII...The sulfur [from iron sulfide] was used in the production of sulfuric acid, an important chemical for industrial purposes. Now most sulfur production comes from H2S gas recovered from natural gas wells."
[ http://mineral.galleries.com/minerals/sulfides/pyrite/pyrite.htm ]
A bit out of date, VCAA; I thought the whole coursework was engineered towards rendering students as MODERN chemists ;D Maybe there were testing us on our knowledge of the chronology of the industrial chemical age...
Did you circle those two hydrocarbons as well?
Cause they produce water and heat. LAwlLALWLWlaLASWlalWLAwll you could circle anything as long as you made a justification.
Question 5)a) required you to circle the 'raw materials' without any mode of reasoning
And in any case my justification is that deriving S from FeS2 is a half-century old technique as according to that extract!