GCE A Level H2 Chem Paper 1
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(drinking more chocoloate... wait... )
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lazy to type my answers, haha, theyre prolly wrong anyway..dont wanna think about it. :(
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Originally posted by ixamus:
oo Ultima Online posted first, saying 12 is D o.O since teacher i suppose im wrong then >_>
>>> oo Ultima Online posted first, saying 12 is D (o.O) since teacher i suppose im wrong then >_> <<<No, I didn't say it's D. What I (mistakenly) said, was that Chocoholicxx got 3 out of 3 correct. I correct myself : he/she got 3 out of 4 correct.
But this is where the real difficulty actually begins. There is actually a heck lot more to this question that almost all H2/H3 candidates (and probably many JC teachers as well, and possibly even some of the SEAB-Cambridge folks themselves!) realize.
Ok, here... we... GO!
BOTH Options A and C are BOTH correct and wrong, simultaneously!
(Disclaimer : Read everything below, all the way to the end. Do NOT expect this question to be nullified by SEAB-Cambridge and everyone to get 'free marks'.)
Why option A appears to be correct.
Equilibrium constant is, by it's mathematical definition, the ratio of the rate of forward reaction (compared) to the rate of the backward reaction. If both rates are independent of temperature, equilibrium constant must therefore be independent of temperature.
Why option C also appears to be correct.
If the forward reaction is endothermic, increasing temperature increases equilibrium constant (ie. higher yield at equilibrium).
If the forward reaction is exothermic, increasing temperature decreases equilibrium constant (ie. lower yield at equilibrium).
If the enthalpy change of the reaction is zero (ie. both forward and backward enthalpies are equivalent), increasing temperature has no effect on equilibrium constant. It only increases the rate of reaction (for both forward and backward reaction).
Why option A appears to be wrong :
Can such a reaction (in which the rate constants for either the forward or backward reaction, or both, are truly independent of temperature) actually exist in this Universe? When by definition, the rate of any reaction is dependent on how fast the particles move, which is in turn dependent on temperature!
Why option C also appears to be wrong :
What if we have an equilibrium reaction in which enthalpy change is totally zero, but let's say there is an increase in entropy for the forward reaction? Gibbs Free Energy predicts that increasing the temperature would shift the position of equilibrium to the right and hence increase the equilibrium constant!
So is this question unsolvable? Let's take a closer look.
From the above discussion ("why option C also wrong"), if we have a reaction in which enthalpy change is zero and entropy change is not zero, that means you can never reach equilibrium! But equilibrium certainly exists in real life.
So what's the deal? For all reactions that have achieved equilibrium, Gibbs Free Energy must be equal for both forward and backward reaction. Which means that enthalpy change vs entropy change must be zero. Which means that if your enthalpy change is zero, your entropy change must also be zero (for a non-zero equilibrium constant to exist).
Otherwise (if not both of them are zeros), equilibrium can only be reached if enthalpy change is non-zero, and entropy change is non-zero, and at that particular temperature, they cancel out exactly. But changing temperature would therefore change equilibrium constant!
Note that equilibrium does NOT mean we have equal moles of reactant vs product. What it does mean, is that the ratio of concentration of products to reactants, is a fixed value (but there will almost always be either more products for some reactions, or more reactants for other reactions).
A Gibbs Free Energy change which favours the forward reaction (under specified conditions including temperature), will of course result in greater concentration of products over reactants. But as long as the reaction is reversible, then there will come a point in time, where too much products will cause Gibbs Free Energy to favour the backward reaction, and there'll be a natural see-saw until equilibrium is reached.
Let's consider the following reversible reaction :
CaCO3(s) <---> CaO(s) + CO2(g)
Under 'standard conditions' and at 298K, enthalpy change = 177.8 kJ/mol, and entropy change = 160.5J/K.mol. Therefore (standard) Gibbs Free Energy at 298K for this reaction = 130kJ/mol.
In other words, equilibrium lies to the left under 'standard conditions' and at 298K.
At point of equilibrium, where the rate of formation of limestone = rate of decomposition of limestone, Gibbs Free Energy = zero = Enthalpy Change - Temperature (Kelvins) x Entropy Change
Such a temperature (Kelvins), for the limestone reaction to be at equilibrium, can then be calculated to be 1113K (or 840 deg C).
However, (as if everything wasn't complicated enough), as it turns out, the 'enthalpy change' and 'entropy change' values themselves (for all reactions), also change with temperature, pressure and molarity! Hence the above calculations are actually invalid and inaccurate (but to apply all of these considerations correctly in practice, would be beyond the scope of the H2/H3 syllabus).
But it is important to appreciate that enthalpy (change) and entropy (change) values, are themselves subject to change at different temperatures, pressures and molarities. This is at the crux of the reason why equilibriums exist. To go into a proper discussion of these, however, is beyond the H2/H3 syllabus.
For irreversible reactions, enthalpy certainly needn't be zero (and it is never actually zero), and entropy certainly needn't be zero (and it is never actually zero), yet equilibrium constant is unaffected by temperature! Because this is a non-reversible reaction, which can never reach a non-zero or non-infinity equilibrium constant.
Notice the question states that GIVEN that this equilibrium constant is independent of temperature. So there we have it - such a reaction MUST be an irreversible reaction.
But look at the options again :
A. The rate constants for the forward and reverse reactions do not vary with temperature.
C. The enthalpy change is zero.
For irreversible reactions, increasing the temperature would increase the rate constant for the forward reaction (since temperature increases the speed of which particles move, hence rate of forward reaction would increase). But the rate constant for the backward reaction remains zero (since it is irreversible). So option A is wrong.
For non-reversible reactions, the enthalpy change isn't necessarily zero either. (Just think of any irreversible reaction that has non-zero enthalpy change, and you've proven your case). So option C can't always be correct, so therefore it must be wrong too.
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Q12) Which statement is correct about a reaction for which equilibrium constant is independent of temperature?
A. The rate constants for the forward and reverse reactions do not vary with temperature.
B. the activation energies for both forward and reverse reactions are zero.
C. The enthalpy change is zero.
D. There are equal numbers of moles of reactants and products.
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Finally, back to my initial disclaimer :
The above 'lecture', 'discourse', 'treatise' or 'thesis', is really to encourage a deeper appreciation in the student for the wonderful complexity of the Universe. It discusses Q12 of the 2008 H2 Chem P2, from the author's (that's me) choice of interpretation of the question and the question's available 4 quirky options (or at least 2 of them are).
SEAB-Cambridge reserves the right to interpret their own question the way they see fit, and by their own reckoning, there can only be one 'right' answer for which they will give credit only to this 'right' option.
It's obviously either A or C. Which of these 2 options SEAB-Cambridge chooses to be the 'correct' answer (ie. depending on how they choose to interpret the question, from their own point-of-view), we'll know soon enough (or at least your JC teachers will, when sometime next year they receive an official document from SEAB-Cambridge examiners/markers, detailing the MCQ answers and comments on the common errors that students of this 2008 cohort made).
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Also, ixamus, just to say that I unfortunately can't comment on your list of answers (a lot of effort there, for you to type it all out), because I've not seen the actual paper myself. This was the last paper, so this was naturally also the last I've seen of my students.
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my answer for that is C actually.. but since u said something bout enthropy i tot i was wrong haha.. =x so put D wif a question mark :p
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I like the discussion on Q12 so far: Here's my input.
Q12) Which statement is correct about a reaction for which equilibrium constant is independent of temperature?
A. The rate constants for the forward and reverse reactions do not vary with temperature.
B. the activation energies for both forward and reverse reactions are zero.
C. The enthalpy change is zero.
D. There are equal numbers of moles of reactants and products.
UltimaOnline mentioned that the correct answer is EITHER A or C...
But if we assume the rate constants for the forward and backward reactions (kf and kb) respectively to follow the Arrhenius Law, where k = k0*exp(-Ea/RT), the only way that K (equilibrium constant) remains invariant with temperature is as mentioned in option A, that the rate constants themselves are invariant with temperature.
From the Arrhenius dependency rule, that can only happen if the activation energies of the reactions are 0. This causes the exp(-Ea/RT) term to be reduced to 1, and k = k0.
And this is applicable to reversible reactions, where K does not have to be assumed to be infinity but can be any finite number greater than 0.
Which poses an anomaly: This means that B leads to the same conclusion as A. And usually in these scenarios, if 2 options lead to the same conclusion, both are wrong.
On the other hand, if we consider chemical thermodynamics:
d(ln K)/dT = (-delta_H)/RT^2
If ln K were to remain invariant with temperature it would mean that delta_H = 0. Hence C looks like a more probable choice.
JY
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Originally posted by ixamus:
based on discussions with friends, IN NO WAY CONFIRMED.
1. B (no. of mol = 1 / 24000, then multiply by Avogadro's constant)
2. A (as said above, only option more molecularly massive than N2)
3. C (data booklet)
4. D (proton number increases, while nucleon number remains constant)
5. C (N-: 1s2 2s2 2p4, remove one to get 2p3, i.e. half filled)
6. C (trivalent N (2 and 4) have a lone pair for dative; the other two have -ve charge, ionic)
7. A (discussed above)
8. D (I2 exist as diatomic molecules, only possibility)
9. A (combustion is an exothermic process)
10. C (heat change must be +ve to absorb heat (cold pack); spontaneity => delG -ve)
11. D (Ka is solely temperature dependent)
12. C (not sure. the fact that LCP doesnt come into play means neither forward nor backward reactions are exo / endothermic. heat change should thus be zero)
13. D (10 halflives later, should halve 10 times, i.e. (1/2)^20)
14. D (1mol of cpd -> 3 mols of Cl-; either Al2Cl6 or PCl3; but liquid, fumes => PCl3)
15. A (discussed above; anyway, B results in no change in titre value; both C and D increase it (if any of them were to happen))
16. C (dispersion forces increases => bp increases; shielding increases=> E.A. decr)
17. B (Fe: [Ar] 3d6 4s2; Fe (II) remove the valence 2)
18. C (as discussed above; highest E0 anyway)
19. A (both CH are sp2 hybridised because 3 sigma bonds, only A possible)
20. B (B results in random subs at any position, i.e. low yield)
21. B (NOT SURE calc using bond energies; B results in most exothermic rxn, i.e. most stable products with respect to reactants)
22. A (duh)
23. A (Cl attached to aryl ring is 'stuck' due to strong bond with partial double bond char)
24. C (NOT SURE from the pattern you're supposed to infer from the given reaction, the O radical part attacks the C atom of epoxyethane, i.e. no O-O bonds so not B or D. A is quite clearly wrong due to the (CH3(CH2)10O) recurring units (10 of them, too O_O ))
25. C (conc H2SO4 for dehyd; NaOH for removing any HCl carry-over)
26. B (homogeneous catalysis with small amt of base or aq KCN)
27. C (phenolic group the only nucleophile to attack electron deficient acyl chloride C)
28. C (+ve iodoform => alcohol part is ethanol. only possibility is C)
29. C (low protonating pH of 2 => both amine groups get protonated to NH3+ groups)
30. D (only non-betaaminocarboxylic acid)
(i hate section Bs for chem mcq)
31. B (C=O bond is v polar. COS overall less polar means C=S bond counters the C=O polarity to a greater extent => C=S more polar than C=Se)
32. D (i put C >_< ) (2 is not even balanced; 3 is nonstandard due to gaseous H2O)
33. B (i put D >_> must have been cock-eyed read left as right) (left-hand electrode is standard hydrogen electrode, so E0cell is E0 of Fe(II)/Fe(III) i.e. +0.77V; right hand electrode higher E0 => reduction occurs => electrons flow to it => it is the positive electrode)
34. B (should be obvious. 3 is wrong because rate constant is 2 x 10^-2. usually cambridge papers dont have this type of cunning thing >_> )
35. D (NOT SURE 1 is definitely correct. (the two are Br- and Sr2+) Sr has more protons => greater nuclear charge => pulls electron cloud tighter; 2 SHOULD be wrong because same electronic config => same shielding effect; 3 looks correct but i think its wrong cos the two have same electronic config, its correct but doesnt explain the difference in size)
36. C (X2 accepts electrons most readily, so it is the strongest oxidising agent => 1 is wrong)
37. A (SN2 mech as inferred from a single activation energy. 1 is correct. 2 is correct cuz bond forming and bond breaking (of C-Hal) occur simultaneously, i.e. bond lengthening; 3 is correct also)
38. B (7 chiral centres is correct; hot acidified KMnO4 cleaves lower left double bond to give ketone on right and carb acid on left, upperright primary OH becomes the other carb acid group, so 2; hot acidified K2Cr2O7 does not oxidise double bond, upperright OH gets oxidised to carb acid, tertiary OH on right side is immune to oxidation; that leaves 3 carbonyl groups)
39. C (aldehyde group, 1 only possible with ketone group; 2 is possible with HOOCCH=CHCO(OCH3)H as the aldehyde, add H2; 3 possible with CH3CHO)
40. A (VERY NOT SURE i saw cross-chain link = lolwut ok 2 and 3 should be correct, as the respective double bonds can break their pi bond to form single bonds with the doubly bonded in the adjacent chains, i.e. C in one with N in another for 2; C in one and C in another for 3; 1 can only form hydrogen bonds, wiki says cross linkages are IONIC OR COVALENT but my friends and I think it refers to any discrete or distinct bond stabilising / bonding side chains in any manner, that would mean H bonds too right? i don't know for sure.....)
if my friends and i are correct, then i get 38! yay haha. hope ya'll did okay too! all the best for remaining papers :)
rgding qsn 30, why isn't the answer A? A contains a benzene ring, and i don't think that is manufactured in the body.
q35, yes, although both anion and cation are isoelectronic, it should not be forget that the anion has less protons compared to the cation, thus, there's more electron shielding comparatively to the anion. if option 2 make sense, by right option 3 shd be right as it too make sense.
q36. if X2 is the most reducing, isn't it the least oxidising?
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Originally posted by JY:
I like the discussion on Q12 so far: Here's my input.
Q12) Which statement is correct about a reaction for which equilibrium constant is independent of temperature?
A. The rate constants for the forward and reverse reactions do not vary with temperature.
B. the activation energies for both forward and reverse reactions are zero.
C. The enthalpy change is zero.
D. There are equal numbers of moles of reactants and products.
UltimaOnline mentioned that the correct answer is EITHER A or C...
But if we assume the rate constants for the forward and backward reactions (kf and kb) respectively to follow the Arrhenius Law, where k = k0*exp(-Ea/RT), the only way that K (equilibrium constant) remains invariant with temperature is as mentioned in option A, that the rate constants themselves are invariant with temperature.
From the Arrhenius dependency rule, that can only happen if the activation energies of the reactions are 0. This causes the exp(-Ea/RT) term to be reduced to 1, and k = k0.
And this is applicable to reversible reactions, where K does not have to be assumed to be infinity but can be any finite number greater than 0.
Which poses an anomaly: This means that B leads to the same conclusion as A. And usually in these scenarios, if 2 options lead to the same conclusion, both are wrong.
On the other hand, if we consider chemical thermodynamics:
d(ln K)/dT = (-delta_H)/RT^2
If ln K were to remain invariant with temperature it would mean that delta_H = 0. Hence C looks like a more probable choice.
JY
Good input, JY. Yes, having a closer re-look at option B, it is indeed equivalent to option A. Which means that A & B, as well as C, are all arguably correct explanations for the question.
And, as you pointed out, since A = B, and futhermore, based on the formula quoted (though the use of mathematical formulae in and of itself without elaboration on the concepts involved, does little to help JC students understand which is my concern here, but it cannot be helped, and it's certainly the most efficicient!), yes I agree that option C stands out as the most probable. -
And, as you pointed out, since A = B, and futhermore, based on the formula quoted (though the use of mathematical formulae in and of itself without elaboration on the concepts involved, does little to help JC students understand which is my concern here, but it cannot be helped, and it's certainly the most efficicient!), yes I agree that option C stands out as the most probable.
This is what happens when you study chemical engineering... After a while everything boils down to mathematical equations :P
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Originally posted by deathmaster:
rgding qsn 30, why isn't the answer A? A contains a benzene ring, and i don't think that is manufactured in the body.
q35, yes, although both anion and cation are isoelectronic, it should not be forget that the anion has less protons compared to the cation, thus, there's more electron shielding comparatively to the anion. if option 2 make sense, by right option 3 shd be right as it too make sense.
q36. if X2 is the most reducing, isn't it the least oxidising?
>>> q35, yes, although both anion and cation are isoelectronic, it should not be forget that the anion has less protons compared to the cation, thus, there's more electron shielding comparatively to the anion. if option 2 make sense, by right option 3 shd be right as it too make sense. <<<If the two species are isoelectronic, it means that they have exactly the same shielding effect (mostly from inner electrons shielding (shielding = inter-electron repulsion reducing the attractive effect of nuclear charge) the outer or valence shell electrons, and also a little shielding shielding effect from inter-electron repulsion between electrons in the same or valence shell). Therefore, any difference in ionic radius, must only be due to the only difference between the two species - the no. of positively charged protons in the nucleus.
>>> q36. if X2 is the most reducing, isn't it the least oxidising? <<<
Distinguish between halogen and halide ion.
F2 is the strongest oxidizing agent, and F- is the weakest reducing agent.
I2 is the weakest oxidizing agent, and I- is the strongest reducing agent.
The reason is due to the number of electron shells, hence the distance between the +vely charged nucleus and the valence shell, and therefore the ease of gain of electrons (for halogens) or the ease of loss of electrons (for halide ions).
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Originally posted by JY:
This is what happens when you study chemical engineering... After a while everything boils down to mathematical equations :P
I suspected you were a chemical engineer - you guys have an unavoidable intimate love relationship with thermodynamics.
Is your work also equally involved with organic chemistry? Do you lean more towards thermodynamics or organic chemistry (academically, professionally as well as by personal preference)? -
I'm not even done with it yet! 3 more semesters to go, haha! But what concerns us mainly would be the entire process to mass produce something useful. Can be organic or inorganic. Chemists decide the route of synthesis in test tubes. Chemical engineers upsize it to mega huge reactors.
In the final analysis of the process we have to take into consideration many factors. Thermodynamics decides process feasibility at optimal temperatures and pressures and is only 1 part of it. Kinetics determines reactor size and type. And a lot more... It's much more physics than chemistry.
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Originally posted by JY:
I'm not even done with it yet! 3 more semesters to go, haha! But what concerns us mainly would be the entire process to mass produce something useful. Can be organic or inorganic. Chemists decide the route of synthesis in test tubes. Chemical engineers upsize it to mega huge reactors.
In the final analysis of the process we have to take into consideration many factors. Thermodynamics decides process feasibility at optimal temperatures and pressures and is only 1 part of it. Kinetics determines reactor size and type. And a lot more... It's much more physics than chemistry.
I've always felt that researchers-&-engineers (physical, chemical, biological, medical, all disciplines) contributed much more to human technology and progress, than those involved in the finance/banking/investment/economics/management industry (no offence to anyone from those industries, this is just a personal opinion).The unfortunate trend of developing societies, in Asia, in Singapore, is that more youths are attracted to the 'glamourous image' or 'big quick bucks' of finance/banking/investment/economics/management, and we've seen a dramatic decline in number of applications to the engineering sciences (in NUS, NTU, and universities across the world), and a corresponding dramatic increase in applications for the finance/banking/investment/economics/management courses and professions (SMU, SIM, etc).
In part, this misguided, delusionary mindset ('glamourous image' or 'big quick bucks'), contributed (both directy and indirectly) to the massive world economic & financial collapse of 2008. But my concern is that it does little to change the unfortunate mindset of young students, who see the finance/banking/investment/economics as a get-rich-quick-big-bucks route, while the research-&-engineering sciences lose much needed talent as a result, and consequently the technological progress of humanity is compromised and tragically slowed down.
Again, no offence meant to anyone. I just hope to encourage passion (for the sciences) in students on these forums, that they may consider, at least as equivalent to, if not moreso, in terms of satisfaction and meaningfulness, the professions of medicine, teaching, research, engineering, etc, which contribute more directly to humanity's progress, as compared to simply dedicating your attention, time, life to the science of making money (essentially what the finance/banking/investment/economics/management industries are about), and usually the (real) motivation in these industries is for making money for oneself (one's clients, one's company, but ultimately for oneself), as opposed to contributing technological progress for humanity.
Merry Christmas!
