Originally posted by MapPwner:Um u can ignore my query,as my school already uploaded the chem specimen p2 answers and after checking my answers my doubts are clarified already.
After checking I have lost approximately 12-15 marks total.
The difficulty of the paper is considered normal right?
A BedokFunland JC H2 Chemistry Challenge Question for you :
Amos Yee drinks Teh-O-Ice-Limau according to 1st order kinetics...
The amount of CN−(aq) ions in an analyte solution can be determined by titration against Ag+(aq), during which the soluble dicyanoargentate(I) coordination complex is first generated, until sufficient Ag+(aq) is added from the burette to generate the insoluble white precipitate of the silver(I) dicyanoargentate(I) coordination compound. If x mol of Ag+(aq) had been added at first sighting of the white precipitate, what was the amount of CN−(aq) ions present in the analyte solution?
A) 0.5 x
B) 1.0 x
C) 2.0 x
D) None of the above
(Every year, all the Singapore JCs are fiercely competing against each other to set the most impossibly sadistic Prelim papers, tis a traditional badge of pride & honor, you understand).
https://en.wikipedia.org/wiki/Histidine
Q1. Explain fully the relative basicities of the 3 N atoms in Histidine (ie. explain which N atom on the R group imidazole ring is more basic, and explain whether the alpha N atom or the R group imidazole ring is more basic).
Q2. Given pKa1 = 1.60, pKa2 = 5.97 and pKa3 = 9.28, calculate the isoelectric point of Histidine.
Q3. Calculate the pH at all significant points (ie. initial, 1st maximum buffer capacity, 1st equivalence point, 2nd maximum buffer capacity, 2nd equivalence point, 3rd maximum buffer capacity, 3rd equivalence point) in
(i) the titration of the fully deprotonated form of Histidine against a strong monoprotic acid added in excess, as well as in
(ii) the titration of the fully protonated form of Histidine against a strong monoprotic base added in excess.
Use 0.05M as the molarities for Histidine and both titrants, and 25cm3 as the analyte volume of the Histidine solution for both titrations. Include a sketch of both titration curves, with all significant pH values labeled.
Q4. By considering the enthalpy and entropy changes during decarboxylation, explain how temperature affects the thermodynamic feasibility and Kc value for the decarboxylation reaction.
Q5. Suggest and draw out the curved-arrow electron-flow reaction mechanism for the (non-enzyme catalyzed) decarboxylation of Histidine.
hello, with the h2 chem prac coming this thursday, i would like to ask how does one prepare for the inorganic planning section of the paper? All of the answers suggested have very very detailed amounts / conc of the reagents and steps required to be taken as well as the 'pre-calculation'. Considering that i would most likely be short of time in the actual exam, how am i suppose to reproduce it?
Originally posted by Zzzfrendzzz:hello, with the h2 chem prac coming this thursday, i would like to ask how does one prepare for the inorganic planning section of the paper? All of the answers suggested have very very detailed amounts / conc of the reagents and steps required to be taken as well as the 'pre-calculation'. Considering that i would most likely be short of time in the actual exam, how am i suppose to reproduce it?
Anyone want to recommend any particular JC's 2017 Prelim Paper? Eg. that you found particularly challenging or educational or unique? (in a good way hor, not uniquely atrocious).
On an unrelated note, after or alongside RJC, DHS also has one of the highest quality 2017 Prelim Papers amongst all the Singapore JCs' 2017 Prelim Papers. If I come across any more top quality 2017 Prelim Papers (which all Singapore JC students are recommended to learn from), I'll mention them (though I probably won't go through all of the Singapore JCs' papers, so if you've a particular JC to recommend, please make a post here to share your opinion with us).
Can i ask for the prac enthalpy change of decomposition,must we take into account stoichiometric coefficient,whereby -(heat change)/amount of h2o2 x2
Then fe(h2o)6 3+ catalyst role is intermediate(i forgot the compound) formed in step 1,it decomposes in step 2 and then fe(h2o)6 3+ regenerated in step 3.
Fe(h2o)6 3+ +H2o->fe(h2o)5(oh) +2 +h3o+
Fe3+ has high charge density than polarise.(I careless,said H2O act as base)
H2o2 is first order
Planning didnt bother doing
Then inorganic qa observation
Naoh soln turns purple to colourless aft addin kmno4,and turns blue aft some time.Deionised remains purple.
Then inorganic qa,observation is redox,ligand exchage reaction,reducing agent for h2o2.
Compared my answers with many sources(classmates,friends from other schs like AJ/HCJC),excluding planning my score is about 30/41 for the 15% of practical,which is supposedly rather decent among those i asked so far(my friends from AJ/HCJC supposedly did worse than this score out of 41,and some skipped planning altogether).
I counted my score rather stringently so yeah,somewhat disappointed but oh well,at least i didn't do any worse as the practical really looked simple yet tricky,which was maybe why a lot of people lost a lot of marks due to careless mistakes,which I did too.
Lost approximately 4 marks in energetics component(-1 for failure to write down o2 evolved relights glowing splint,-1 for not taking into account x2 stoichiometric coefficient for enthalpy change of decomposition in both cases-assuming there is ecf,-1 for possibly inefficiently explaining how fe(h2o)6 3+ acts as catalyst and -1 for giving wrong reason for fe(h2o)6 3+ forming fe(h2o)5(oh) 2+ and h3o+ due to high charge density)
Kinetics lost 3 marks due to calculation misunderstanding,otherwise was relatively manageable,such as quoting of 2 consecutive t1/2 to prove whether is it first order or not(for me it is yeah).
Inorganic QA probably lost 3 marks,-1 due to failure of writing down blue ppt formed for test 2 but just wrote blue soln instead,-1 for not accurately explaining why the second test tube with naoh evolves o2,merely stated redox and mno4- reduced while OH- oxidised to O2 which im unsure whether thats correct so assumed wrong.Also -1 for incomplete observation of compound formed and reasoning for part ii,only said Co(NH3)6 2+ formed due to ligand exchange reaction,never accounted for Co(OH)2.
On top of those mistakes would also want to ask the following questions:
In part iii of role of H2O2,its a reducing agent if one deduces that from equation 2 of page 8 and quoting oxidation states of elements affected?Then explaining fe(h2o)6 3+ as a catalyst,is this the structure as follows:
In step 1,Fe(H2O)6 3+ reacts to form an intermediate Fe(H2O)5 something (1m)
The intermediate complex decomposes to form 2 products in step 2,which subsequently reacts with h2o2 to regenerate Fe(h2o)6 3+ in step 3(1m)
technically,such an explanation is 2/3m only right?I assume the last mark is with respect to colour change of solution during the steps.
Paper 4 practical was a disaster. I couldn't finish the entire planning question, only managed to write the first 5 steps, 20 marks gone.
First experiment was on catalysis of H2O2 decomp using Fe3+ as catalyst. As i was calculating the enthalpy change of decomposition: mcT/n, I wasn't sure whether to multiply by 2, because according to the equation, they wanted us to calculate the enthalpy change of this reaction 2H2O2 -> 2H2O + O2, where the stoic is 2. I decided not to multiply in the end, hence another 2 marks lost.
Next was a kinetics experiment for the decomp of H2O2, continuous method, extracting aliquots during timed intervals from the reaction. Drew a graph, deduced first order wrt H2O2.
Planning was a similar kinetics experiment but to investigate how temperature affects catalysed decomp of H2O2. (COULDN'T reach past 5 steps)
Next it was QA for redox of transition metals.
In the first test, there were 2 tubes.
1st tube: Deionised water + KMnO4
2nd tube: NaOH, heat, + KMnO4
The 2nd tube, small amts of O2 produced, purple turned blue then green. I wrote that the MnO4- was reduced to green MnO42-.
In the second test, Co2+ + NH3(aq) + H2O2
Effervescence of O2 gas, insoluble green/blue ppt, that turned to brown ppt on addition of H2O2. Co2+ in Co(OH)2 must have been oxidised to brown Co(OH)3. But i wrote that it was reduced to Co(OH) instead sobs.
Overall tons and tons of mistakes, i'm just a soppy sobbing mess. Am from HCJC and i must admit, our prelim Paper 4 was 10 times more manageable IMO than the A levels Paper 4. I just don't know if my Paper 1, 2, 3 will be able to make up for this.
Originally posted by Ellipsisss:Paper 4 practical was a disaster. I couldn't finish the entire planning question, only managed to write the first 5 steps, 20 marks gone.
First experiment was on catalysis of H2O2 decomp using Fe3+ as catalyst. As i was calculating the enthalpy change of decomposition: mcT/n, I wasn't sure whether to multiply by 2, because according to the equation, they wanted us to calculate the enthalpy change of this reaction 2H2O2 -> 2H2O + O2, where the stoic is 2. I decided not to multiply in the end, hence another 2 marks lost.
Next was a kinetics experiment for the decomp of H2O2, continuous method, extracting aliquots during timed intervals from the reaction. Drew a graph, deduced first order wrt H2O2.
Planning was a similar kinetics experiment but to investigate how temperature affects catalysed decomp of H2O2. (COULDN'T reach past 5 steps)
Next it was QA for redox of transition metals.
In the first test, there were 2 tubes.
1st tube: Deionised water + KMnO4
2nd tube: NaOH, heat, + KMnO4
The 2nd tube, small amts of O2 produced, purple turned blue then green. I wrote that the MnO4- was reduced to green MnO42-.
In the second test, Co2+ + NH3(aq) + H2O2
Effervescence of O2 gas, insoluble green/blue ppt, that turned to brown ppt on addition of H2O2. Co2+ in Co(OH)2 must have been oxidised to brown Co(OH)3. But i wrote that it was reduced to Co(OH) instead sobs.
Overall tons and tons of mistakes, i'm just a soppy sobbing mess. Am from HCJC and i must admit, our prelim Paper 4 was 10 times more manageable IMO than the A levels Paper 4, and I only attained 35/55. I just don't know if my Paper 1, 2, 3 will be able to make up for this.
If i'm not wrong,the brown one is Co(NH3)6 3+,using data booklet equations,and 35/55 is already very good,excluding planning i only got 30/41.
But i think u are probably correct.
Answers i compiled:
Energetics: For enthalpy of decomposition: -(heat change)/amount of h2o2 x2 Rest such as graph and table is fine
Student plotted graph,range of temperature change:10.5-11.5 degrees(most people i asked got 11-11.3 range).
T2 max is more accurate as use of a graphical method accounts for the compensation of heat loss to the surroundings as the temperature of the mixture becomes greater than the environment(1m)
Then fe(h2o)6 3+ catalyst qn(3m) It forms an intermediate complex(name the compound too also which I forgot)in Step 1(1m),it decomposes in step 2 and then one of the products reacts with h2o2 to regenerate fe(h2o)6 3+ in step 3.(1m) Idk how the other 1m is allocated,maybe stating it is homogenous catalyst etc.
Fe(h2o)6 3+ +H2o->fe(h2o)5(oh) 2+ +h3o+ Fe3+ has high charge density then polarises,weakens OH bond in H2O of the complex to release H+ ions,which is accepted by H2O to form H3O+.(1m)
Why is there a change in colour?as there is a difference in energy gaps,thus different wavelength of light absorbed,resulting in different colours observed(1m)
Kinetics:
Gradient: negative 3.3-4.2 something cm3min-1(well I mod my gradient,so it became positive,idk whether that's correct)
Rate of change of KMno4- = concentration of KMnO4 solution given x gradient of curve/1000cm3=___molmin-1
Last 2 marks idk
H2o2 is first order(agree with student)(1m).As seen from graph as t 1/2=7.2min and 2nd t 1/2=6.7min,which is approximately constant(accept within 20% difference of each other).It is thus first order.(1m)
Planning I not sure.
Inorganic QA: Observations: Naoh soln turns purple to colourless aft addin kmno4,and turns blue aft some time.(1m)Deionised water remains purple/pale pink after addition of KMnO4.(1m)
Addition of NH3 dropwise produces a blue precipitate in solution which produces gas with turns moist red litmus blue.Gas evolved is NH3(1m)
Upon addition of H2O2,solution turns dark brown with a dark brown precipitate formed.Effervescence is also observed and gas evolved which relights a glowing splint.Gas evolved is O2(2m).
Part (I):The second test tube containing boiling NaOH.(1m)Redox reaction occurs which reduces MnO4- and oxidises OH- to O2(1m)[Or better phrasing].
Part (II) Co(OH)2(blue ppt) and Co(NH3)6 2+(dark brown ppt) is formed(1m).CO2+ reacts with OH- produced by dropwise addition of aqueous ammonia to form Co(OH)2 precipitate.Addition of H2O2 result in the formation of Co(NH3)6 2+ through a ligand exchange reaction as NH3 molecules displaces h2o ligands in Co(H2O)6 2+(1m).
50% observation+50% explanation correct for the 2 mark explanation question(part ii)=1m For observation questions(2m in energetics and 4m in inorganic QA,a deduction of 1 mark per missing/incomplete information will probably be meted out).
Part(iii) Role of H2O2 is a reducing agent then quote O.s states using eqn 2 on page 8.
Those are supposedly the answers to most of the questions.But i suspect the part(ii) explanation of inorganic QA is wrong as chemguide uk said is Co(OH)3.
UltimaOnline,do you find the practical today easy?I felt it was doable but probably due to complacency and negligence i lost quite some marks.
On a side note, here is the link to my answers to the questions above. Can i trouble you to point out the mistakes? Thank you!!
https://imgur.com/a/qS4uU
So I'll leave it to your good hands, MapPwner, to lead the forum discussions on P2, P3 and P1. And since I won't have seen any of the Papers, I can't comment on the difficulty of the questions. If you'll like me to address any particular question, you'll have to type it out from memory (for P2) or upload an image of the question (for P3 and P1).
For inorganic QA,we were tasked to add 1cm depth of CoCl2,followed by addition of NH3(aq) dropwise,forming a blue ppt and blue solution.Then thereafter,we are supposed to add H2O2(aq),resulting in formation of dark brown solution and dark brown precipitate.
I identified dark brown solution as co(nh3)6 2+ but is probably wrong as it is co(oh)3 i think.Failed to identify co(oh)2 as i only saw a blue solution in my test tube.
We were also tasked to compare role of h2o2 between the reaction mentioned above and role of h2o2 in reaction between MnO4- and H2O2.
For energetics the practical was on determining enthalpy change of decomposition of H2O2,given by the equation 2 H2O2->2H2O+O2 Hdecomp=to be determined
Many people could calc heat change and amt of h2o2 used,but failed to take into account the stoichiometric coefficient in energetics in calculating Hdecomp of H2O2,which is 2 moles,so -1/2 marks at least.
Originally posted by Ellipsisss:Paper 4 practical was a disaster. Am from HCJC and i must admit, our prelim Paper 4 was 10 times more manageable IMO than the A levels Paper 4. I just don't know if my Paper 1, 2, 3 will be able to make up for this.
Originally posted by UltimaOnline:
Damn, this makes me all the more regretful that I'll have to be skipping the A levels this year. If this year's A levels P2, P3 and P1 are also concordantly so much tougher than the Singapore Prelim Papers (as appears to be the case for P4), I would really have liked to take them. If this really turns out to be the case (which would be rather shocking to Singapore JC school teachers), don't feel too badly or worry too much : after Cambridge finishes marking and sends the raw scores to SEAB, SEAB will apply the bell-curve to assign grades after moderation.
I also felt that the chem prac was tough,but oh well i will probably end up with 31/55 at best,including planning,like my prelim p4 score of 31/55.
From my discussion with my AJ/HC friends,some got the same score as me or way worse.
Just quoting a Whatsapp message from my teacher in my chemistry class group:
''Hi all
Mr Lee who did the practical today says it is "creative"... and creative for us in another words means the difficulty is high for students.
No fear. Others will also find it the same
Cheers.''
Originally posted by Ellipsisss:On a side note, here is the link to my answers to the questions above. Can i trouble you to point out the mistakes? Thank you!!
https://imgur.com/a/qS4uU
Originally posted by Ellipsisss:On a side note, here is the link to my answers to the questions above. Can i trouble you to point out the mistakes? Thank you!!
https://imgur.com/a/qS4uU
Here are my comments on your attempted answers.
For Q1, your order of basicity is wrong, and your explanation has multiple errors (including an internal inconsistency, prolly just ur careless typo), and a word of advice : for A levels, never use "it", but specify exactly which, otherwise Cambridge will think you're trying to smoke your way through and you'll kena penalized (even if your answer is otherwise correct). Hint : your answer would be somewhat less wrong (but there are still errors) for a typical alpha-amino acid, but Histidine happens to be anomalous amongst alpha-amino acids (this is afterall a BedokFunland JC Challenge, ie. difficult even for A grade students, and for some private tutors and school teachers as well), but still totally explainable, if you understand your chemistry well enough. You can 'cheat' a little and google out where the 3 pKa values belong, and thus 'reverse engineer' the basicity of the 3 N atoms (the least basic N atom doesn't even have a Kb value, if Cambridge asks why, do you know how to explain?).
Q2. Correct, well done. More than half of all JC H2 students wouldn't know how to solve this.
Q3. Ok, I was admittedly being unfairly sadistic in setting this question, as the pH values for the more extreme ends of the titration cannot be correctly obtained by using the A level acid-base equilibria within-syllabus methods, which are only approximations that fail when dealing with fairly strong acids and bases. In addition, do note that if the initial molarity of the acid or base is not sufficiently larger than the change in molarity (ie. as indicated by the magnitude of the relevant Ka or Kb value), it becomes inaccurate and indeed invalid to neglect the change in molarity, and to simply approximate equilibrium molarity back to initial (as taught by Singapore JCs, and as you had done in some of your calculations). But no worries, Cambridge probably wouldn't be as sadistic, and would also usually set initial molarities to be much larger than the change, so you usually wouldn't need to solve quadratic or cubic equations for H2 Chemistry (though all students should still bring 2 calculators as a backup, a non-graphing and a graphing). So as far as A levels are involved, very well done, your titration pH values are mostly correct.
Q4. Correct, well done. Most Singapore JC students are confused by the effect of temperature change on the Gibbs free energy change value VERSUS the equilibrium constant Kc value, because they simplistically assume that the larger the Kc value, the more negative the Gibbs free energy change. This is only true for constant temperature. When temperature changes, you must consider both the enthalpy and entropy changes. The effect of temperature on Kc is controlled by the enthalpy change, while the effect of temperature on Gibbs free energy change is controlled by the entropy change. Interested A grade H2 students, H3 students and Olympiad students, can explore this aspect of thermodynamics deeper for themselves, including the Van 't Hoff equation.
Q5. Almost correct, but not quite. Did it not occur to you that the decarboxylation (both with and without enzyme catalyst) is kinetically feasible only due to the zwitterionic nature of amino acids? An intermediate with a uninegative formal charge on a C atom that is not delocalized by resonance is prohibitively destabilizing making your mechanism kinetically unfeasible. On a related note : writing C- or H- for Grignard reagents and LiAlH4, is an oversimplification, and does not actually exist as an intermediate, but only as a transition state (so to speak). On another related note : investigate the mechanism for decarboxylation of beta-keto-carboxylic acids (which Cambridge used in a Singapore TYS Qn) followed by prototropic tautomerism, notice there is no C- intermediate there either.
Overall, a very good attempt by Ellipsisss of HCJC, you should be able to get your A grade if you don't screw up your remaining P2, P3 and P1.
Based on yesterday's P4, you can expect that this year's 2017 A level's P2, P3 and P1 to be comparable with the RJC, DHS and VJC papers. So be sure you (ie. all JC students intent of scoring a distinction for your H2 Chem) go through these 3 JCs' 2017 prelim papers very thoroughly.
Disclaimer : I haven't, and probably won't, go through all of the other remaining Singapore JCs' 2017 prelim papers (I've gone through about 2/3 of all JCs, and find the quality of the RJC, DHS and VJC 2017 prelim papers in particular to be worthy of honorable mention). So it's entirely possible that some other JCs' 2017 prelim papers are every bit as good as RJC / DHS / VJC. And everyone is entitled to his/her own opinion. If you've any specific JC's 2017 prelim papers that you particularly enjoyed doing (helps if you're masochistic), do share your vote here.
Overall, I'm honestly glad to see a substantial increase in both the quality and difficulty of all JCs prelim papers this year. Very good work done here setting the 2017 prelim papers, all Singapore JC teachers. And based on yesterday's P4, Cambridge certainly seems to have the same idea (ie. raising the quality, creativity and difficulty of the A level papers for the new syllabus). So be mentally (and emotionally!) well-prepared.
Originally posted by UltimaOnline:
Ellipsisss of HCJC,Here are my comments on your attempted answers.
For Q1, your order of basicity is wrong, and your explanation has multiple errors (including an internal inconsistency, prolly just ur careless typo), and a word of advice : for A levels, never use "it", but specify exactly which, otherwise Cambridge will think you're trying to smoke your way through and you'll kena penalized (even if your answer is otherwise correct). Hint : your answer would be somewhat less wrong (but there are still errors) for a typical alpha-amino acid, but Histidine happens to be anomalous amongst alpha-amino acids (this is afterall a BedokFunland JC Challenge, ie. difficult even for A grade students, and for some private tutors and school teachers as well), but still totally explainable, if you understand your chemistry well enough. You can 'cheat' a little and google out where the 3 pKa values belong, and thus 'reverse engineer' the basicity of the 3 N atoms (the least basic N atom doesn't even have a Kb value, if Cambridge asks why, do you know how to explain?).
Q2. Correct, well done. More than half of all JC H2 students wouldn't know how to solve this.
Q3. Ok, I was admittedly being unfairly sadistic in setting this question, as the pH values for the more extreme ends of the titration cannot be correctly obtained by using the A level acid-base equilibria within-syllabus methods, which are only approximations that fail when dealing with fairly strong acids and bases. In addition, do note that if the initial molarity of the acid or base is not sufficiently larger than the change in molarity (ie. as indicated by the magnitude of the relevant Ka or Kb value), it becomes inaccurate and indeed invalid to neglect the change in molarity, and to simply approximate equilibrium molarity back to initial (as taught by Singapore JCs, and as you had done in some of your calculations). But no worries, Cambridge probably wouldn't be as sadistic, and would also usually set initial molarities to be much larger than the change, so you usually wouldn't need to solve quadratic or cubic equations for H2 Chemistry (though all students should still bring 2 calculators as a backup, a non-graphing and a graphing). So as far as A levels are involved, very well done, your titration pH values are mostly correct.
Q4. Correct, well done. Most Singapore JC students are confused by the effect of temperature change on the Gibbs free energy change value VERSUS the equilibrium constant Kc value, because they simplistically assume that the larger the Kc value, the more negative the Gibbs free energy change. This is only true for constant temperature. When temperature changes, you must consider both the enthalpy and entropy changes. The effect of temperature on Kc is controlled by the enthalpy change, while the effect of temperature on Gibbs free energy change is controlled by the entropy change. Interested A grade H2 students, H3 students and Olympiad students, can explore this aspect of thermodynamics deeper for themselves, including the Van 't Hoff equation.
Q5. Almost correct, but not quite. Did it not occur to you that the decarboxylation (both with and without enzyme catalyst) is kinetically feasible only due to the zwitterionic nature of amino acids? An intermediate with a uninegative formal charge on a C atom that is not delocalized by resonance is prohibitively destabilizing making your mechanism kinetically unfeasible. On a related note : writing C- or H- for Grignard reagents and LiAlH4, is an oversimplification, and does not actually exist as an intermediate, but only as a transition state (so to speak). On another related note : investigate the mechanism for decarboxylation of beta-keto-carboxylic acids (which Cambridge used in a Singapore TYS Qn) followed by prototropic tautomerism, notice there is no C- intermediate there either.
Overall, a very good attempt by Ellipsisss of HCJC, you should be able to get your A grade if you don't screw up your remaining P2, P3 and P1.
Hi UltimaOnline thank you for taking the time to look over my answers!
Q1. Do excuse my disaster of an answer to this question. This is why I am so worried, there are so many gaps in my concepts thanks to all the sleeps I had during lessons, I dont even know if doing papers will be able to fill in these gaps. With regards to the question, would it be correct to deduce that the ability of the lone pair of electrons on the N to delocalize into the double bond would depend on the hybridization of the N atom and the orientation of its orbitals? I would guess the least basic N would be the sp3 hybridized one in the ring, as its lone pair of electrons is delocalized into the ring, so not available for protonation, hence no Kb value?
Q3. I was very unfamiliar with drawing the graph the other way around (protonation of histidine). I am so sorry I was just rather confused in general!
Q5. Apologies for my horrendous mechanism. Wow I will definitely go look it up! Thank you for the tips!