7 | intechemistry
December 29, 2008 at 3:07 pm

@ twilightsaga December 29th, 2008 at 1:10 pm

Unit 1, topic 1.7, part c: “recall and explain that hydrogen halides are water soluble and acidic in solution”

The syllabus reveals what is important. I’ll explain it in terms of an answer that should be offered if asked to explain what is written in the syllabus.

“Polar molecules dissolve in polar solvents because of the similarity of their intermolecular forces. H2O is a polar molecule so dissolves polar H-X molecules. Going down the group of hydrogen halides, the H-X bond strength weakens allowing dissociation when added to water, producing acidic solutions of growing acidity”

That is the kind of answer expected of AS standard.

Sure there’s a few other things at play here, but anything else is too much really. Esp. for Unit 1.

11 | intechemistry
December 30, 2008 at 12:32 pm

Tomorrow I will try and address twilightsaga’s question, including the questions from the past year papers questions that prompted the inquiry (I don’t have them to hand right now)

17 | intechemistry
December 31, 2008 at 3:10 pm

@ comment 9, twilightsaga December 30th, 2008 at 2:24 am

Jan 02, Unit 1, Q2 a (iii):
Explain why all isotopes of magnesium have the same chemical properties.(2 marks)
MS says: Same number of electrons (in all magnesium isotopes) (1 mark)
MS says: outer electron structure determines chemical properties (1 mark)

contrast with.

June 06, Q1 e) iv):
Why do isotopes of the same element have the same chemical properties? (1 mark)
MS says: same number of electrons, OR, same electronic configuration/pattern/structure. (1 mark)
NOT same number in outer orbit.
IGNORE “same number of protons”

In both cases the same no. of e- is mentioned and is of course the fundamental starting point to answering the question, hence always got one mark. My guess is an examiner reviewed the Q later, and thought the accepted answer wa a bit sloppy, hence needed more specificity (gosh!). The second mark coming from giving a stronger, more satisfying answer.

It’s not just the number of e- but ALSO how they are arranged.!!

E.g. Mg = 3s2 would have very different chemical properties to a Mg 3s1 3p1 configuration despite them having the same number of valence e-.

I would answer “They have the same electron configuration in the valence shell” That answer automatically gives the same number of e- and say the arrangements are the same. I think my answer is more powerful. Also, given it’s simplicity, it may well appear as the expected wording in future mark schemes.

19 | intechemistry
January 2, 2009 at 6:33 am

1) Using starch iodine paper for the brown NO2 gas worries me, because if a reaction produced bromine under hot conditions, it might release a significant amount of Br2 vapour (would look a bit brown) and Br2 vapour would oxidise iodide ions in the starch iodine paper to blue-black also.

It’s very unlikely they will give you a reaction in which Br2 (or Cl2) is given off becasue of their toxicity, even if you are supposed to do it in a fume cupboard, but it is possible – esp if you are using c.H2SO4 and some unknown (which may be a metal halide).

So startch iodine paper isn’t so good.

And actually damp blue litmus paper can be used to distinguish Br2 from NO2 (if you had any confusion that is). The Br2 vapour reacting with the water in the damp paper would bleach the litmus paper whereas the NO2 reaction product with water (mixture of HNO3 and HNO2) would not bleach the paper as the oxidising acid: HNO3, isn’t a strong enough oxidising agent to cause the paper to bleach.

You ask “Is it acceptable?”….. Hummmm I can’t say for certain. Edexcel sometimes “vary” a bit on certain issues but this one does leave me feeling uncomfortable. My advice: stick with the damp blue litmus paper.

2) The Fe3+ violet colour is due to the organic compound forming a complex with the Fe3+. I think it’s violet colours are typical of Fe(III) complexes with phenol type organics. One thing is for sure, it cannot be from our standard hexa aqua Fe(II) or (III) complexes as we’ve seen, and know the colour of them.

Marks are only awarded for reporting the colour. No inference was required at that stage. So it looks like it’s meant to support the ‘phenol type’ evidence of the test given by the white ppte forming when aq. Br2 was added to a solution of unknown C. And most likely the mark is a ‘reward’ for successfully doing the oxidation of Fe2+ to F3+ using Pb(IV) oxide.

I Don’t feel so happy about this however, as students are not required to have seen/know Fe(III)/phenol complexes, so they may get the mark for the pure observation but they won’t be able to use it to reinforce the ‘must be a phenol’.

Oh well…

3) Nucleohholic substitution e.g. halogenoalkanes to nitriles. Potassium or sodium cyanide is usually used. It ionises in the ethanol solvent producing the nucleophile (:CN)- I don’t think there is a need to buffer it as there isn’t any source of acid(or base) present. I don’t know of any references to that. I meant to say… I don’t know of any references to buffering in these kinds of reactions.

I think the buffering is only used in the raction with carbonyls – when Hydrogen cyanide reactions involving carbonyls are used, I think you get the best conditions if slightly alkaline or slightly acidic conditions are used. Too acidic or too alkaline and you get hydrolysis of the nitrile to the carbox. acid or its carboxylate anion respectively. But it needs to be a bit acidic or a bit alkaline to get the some HCN(acidic conditions) and (:CN)- ions(alkaline condition). You get better conditions by doing this balancing act.

Hope that helps.

Good Q’s as usual.

23 | intechemistry
January 5, 2009 at 1:18 am

@ comment 20 neeesssaa January 3rd, 2009 at 2:44 pm

—[-CH2-CH(CH3)-]n— seems fine to me. The question is ‘disguising’ itself from the usual way it’s asked, which is: “Give the repeating unit of polypropene”.

I think they’ve “disguised” the question to see if you appreciate the ‘core structure’ of the polymer and that it repeats. Asking bluntly for ‘repeating unit’ is more of an exercise in how good you are at impersonating a parrot!

You did draw a structure rather than a formula right? I know the above is the sent you can offer here, but I’m just checking. The word ‘structure’ should not be ignored.

I may move your comment to the PAst Years Questions here please section.

25 | intechemistry
January 6, 2009 at 4:25 pm

That should do. There’s no indication that you should draw two (or more) units

29 | intechemistry
January 7, 2009 at 9:45 am

Don’t worry. It was me. Nobody could copy my poor spelling LOL

31 | intechemistry
January 7, 2009 at 5:37 pm

CO2 does dissolve in water. It forms carbonic acid

CO2(g) + H2O(l) CO2(aq) – dissolving.
CO2(aq) + H2O(l) H2CO3 {carbonic acid} – reacting
H2CO3 + H2O(l) HCO3- & H3O+

The blood makes exensive use of that last equilibrium and buffers the blood to pH 7.4

More CO2 will dissolve as the pressure increases. At 1 atm, the eqm of the first equation lies heavily to the left. You can see this in everyday situations with bottles of fizzy drinks (BM: air gas). It doesn’t dissolve much as the CO2 is non polar (despite having polar bonds!).

http://en.wikipedia.org/wiki/Carbonic_acid

H2 only has 2e- and is non-polar. Even the temporary dipoles ar incredibly weak, as shown by the fact it boils at about -250oC or thereabouts.

As for techniques… as it’s likely these questions are being asked in relation to trying to prepare for the planning exercise, it would not be appropriate for me to comment.

35 | intechemistry
January 12, 2009 at 4:56 pm

@ comment:33 Natasha January 12th, 2009 at 2:01 pm

Natasha asked: “Why is it just because HCl is covalent so it REACTS?”

Because in HCl, the e- in the covalent bond are shared. If when addded to water, HCl goes on to form H+ and Cl- (which it does!), then the H-Cl bond breaks and the electrons in that bond redistribute. Redistribution of e- is a the description of a chemical eaction.

In an ionic compound, the electrons have already separated when the ionic solid was formed. In NaCl, the Na give an electron to a Cl atom. So adding ionic NaCl to water doesn’t change the arrangement of e- in either the Na or the Cl, hence no reaction occurs when added to water.

Does that answer it?

39 | intechemistry
January 21, 2009 at 2:10 am

Akmar asked: in conjugate base n acid equ, if they giv equ n ask 2 identify acid n conjugate base, acid 1 react with base 1, or acid 1 wil produce conj base 1. Thx!

Answer will follow later… I have to go now.

45 | intechemistry
March 8, 2009 at 5:14 pm

Hi.

1) Experiemntal LE’s are more exothermic than theoretical LE’s, because the theoretical LE is based upon a purely ionic model.

In the real life lattice, you get a degree of covalency also, which acts as ‘additional’ bonding. To overcome this additional bonding, i.e. break the bond, more enthalpy must be put in. Meaning, the enthalpy released when the bond first formed (i.e. the LE) must be greater in magnitude, and bond making is exothermic, so the LE is more exothermic than for purely ionic systems (which don’t actually exist!)

——————————

2) The rate equation tells you how many reactant MOLECULES were involved up to and including the slow step.

if rate = k[A]^2 [B]^1 that means the slow step won’t ‘happen’ until 2 A (the two comes from the order of the specific reagent) and 1 B were used.

If the (overall) chemical equation was 3A + B -> 2D
then a proposed mechanism would be

A + B -> X + Y
X + A -> D (SLOW STEP – I’ve used 2A and 1B already)
Y + A -> D

You have to make the individual mechanism ‘fit’ the rate quation. As soon as it fits (like in the 2nd part) then you say that is the slow step.

In your example, as soon as I’ve used up one HA and one M that write ‘slow rate determinign step’ next to the equation where HA and M had been ‘consumed’. I can easily do this in a single step:

If (overall) chemical equation was:

HA + 2M -> P (note: I meant to write HA + 2M -> 2P)
then I’d write

HA + M -> T (slow)
T + M -> 2P

for my proposed mechamism.

——————————

3) It depends on the reagent you are using.

e.g. if you have C=C and COOH
If you add Br2(in organic solvent) the Br2 will
add across the double bond.
If you add NaOH(aq) then the COOH will react
to form COONa(aq)

So look at the reagent.

——————————

Does that help? this should also answer comment 44 by “intec studnt” March 8th, 2009 at 3:45 pm

P.S. You can sms me and tell me there is a Q waiting
to be answered if it’s urgent.

53 | intechemistry
March 15, 2009 at 5:30 pm

hi #52, intec studnet, March 15th, 2009

Good question and a lot of students get tripped up by this one. It kind of demonstrates how sometimes mark schemes can be a bit tricky for students to use as it’s designed for the smarty pants lecturers

Nitric acid is a strong acid. when first added to water it completely dissociates. HNO3(aq) –> H+(aq) + (NO3)-(aq)
so a solution of nitric acid already contains H+(aq)

we are told to consider reaction with Zn with H+ from nitric acid solution to produce H2. So, from H+(aq) to H2(g), what reaction is that??…. It’s the standard hydrogen electrode!

aaaaaaah!

and we know the standard hydrogen electrode has an E(std) of +0.00 V. for H+(aq) + e- –> 1/2 H2(g)

So, couple(i.e. join) this the Zn half-equation with the std. hydrogen equation to give Zn(s) + 2H+(aq) –> Zn(2+)aq + H2(g) and you will get +0.76V

As for “the NO3- is more positive E than H+” it means the reaction of NO3- with Zn compared to the reaction of H+ with Zn. You know E for reaction of NO3- with Zn becasue you already calculated it as +1.57V. I think because you were missing the stepping-stone involving the realisation of the std. hydrogen electrode as mentioned earlier, then this caused the problem here.

If a reaction is more +’ve then electron transfer between the reacting species, i.e. in the reaction overall will be more thermodynamically favourable. I think you are only thinking about an isolated half cell.

Fix me sir. LOL. I’ve heard that expression before. Are you fixed now?

55 | intechemistry
March 16, 2009 at 4:40 am

Yes

57 | intechemistry
March 16, 2009 at 3:49 pm

A multidentate (meaning ‘many toothed’) ligand, is a molecule or ion that contains more than one atom with a lone pair which can be used to form a coordinate (or dative) bond to a metal ion.

e.g. H2N-CH2-CH2-NH2 (abbreviated to ‘en’) is a multidentate ligand (specifically it’s bidentate). Each N has a lone pair allowing for a total of two coordinate bonds per ‘en’ molecule.

The EDTA complex you mentioned is more stable due to it displacing water molecules as the increase in ENTROPY is greater. EDTA + hexaaqua complex {2 species} –> [Metal(EDTA)] complex + 6 H2O molecules {now 7 species}

7 species can exist in more configurations, i.e. it can ‘mix more’ than 2 species can therefore 7 species mixture is more disordered o chaotic, so has greater entropy.

The syllabus is a bit vague:
U5, Topic 5.2 e) “understand simple ligand exchange processes” and
f) recall the formation of hydroxide precipitates on the addition of aqueous solutions of sodium hydroxide or ammonia, and that some hydroxide precipitates react with an excess of strong alkali, and some react with an excess of ammonia; limited to cations with formula
Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+, Cu2+, Zn2+
…• the concepts of deprotonation and ligand exchange should be applied to these reactions

As a conclusion, I very much doubt such specific stuff will appear.

If you know this:

If a reaction happens it is because the product is thermodynamically more favourable that the reactant. EDTA will displace ligands if it is thermodynamically favourable to do so. 1 EDTA can replace 6 monodentate ligands or 3 bidentate ligands or 2 tridentate ligands. The structure of EDTA is… then you should be fine.

59 | intechemistry
March 17, 2009 at 1:56 pm

Waalaikum assalam.

Hi Nurul. I’m not sure exactly that parts of optical activity is puzzling to you, so here is just a little bit:

Optical isomers are molecules (or molecular ions) which cannot be superimposed on it’s mirror image.

e.g. CH3-CH3-CH(CH3)-COOH

If you built 100 models of the molecule above, and then tried to put one model directly on top of the remaining 99 (i.e. try and superimpose the molecules), you may notice that approximately 50% of the time, no matter how much you rotate, invert or flip the model, you cannot superimpose them. It’s not that you made the models wrong (I hope!) but because for this particular molecule, The 3-dimensional arrangement of the same sequence of atoms is different

If you place two of these non-superimposable molecules side by side, you may notice each is the mirror image of the other.

Generally, they have the same chemical and physical properties, but show one big difference… the effect the have on “the plane of polarisation of monochromatic plane polarised light.”

A pure sample containing molecules of the same 3D arrangement, will rotate the plane of rotation of monochromatic plane polarised light in one direction, lets say 12o to the right (i.e. +12 degrees) . Whereas a pure sample of the other 3D arrangement will rotate the plane of rotation of monochromatic plane polarised light in the opposite direction by the same amount to the left, i.e. -12 degrees.

The molecule and it’s non-superimposable mirror image are called a pair of enantiomers. So an enantiomer is a single molecule in that pair. An enantiomer is not superimposable on it’s mirror image.

what more do you need to know?

Have a quick look here: http://jchemed.chem.wisc.edu/jcesoft/cca/CCA5/MAIN/1ORGANIC/ORG09/TRAM09/A/MENU.HTM

63 | intechemistry
March 18, 2009 at 12:16 pm

nuur

I didn’t know what exactly was causing us ‘stress’ about optical and geometric isomers, so I just covered what I thought might have been the problem. If I missed the thing that yor stuck on, please let me know.

Re: planning exercises. I don’t think there is a ‘technique’, that is why Edexcel love them. The best thing you can do is read as many planning exercises from past year papers as you can, and read the marks scheme. Then after reading the m.s. (not memorising it!!) then try and write out the answer yourself. If you have good composition skills, it will be easier for you to get your points across and also assist you to get through the logic in a smooth(er) fashion, both mentally and on writing it on the paper.

Go Here ( http://www.edexcel.com/migrationdocuments/GCE%20Curriculum%202000/331858_User_guide_2003_version.pdf ) go to page 14 /34 on the pdf (or document page 12) to get an idea of the scope of things they may ask you for planning exercises.

I advise you make a summary table like that pdf above, based on what you see in your past years papers.

—————————————-

Q1:

rxn 1 is an acid base rxn
rxn 2 is a redox reaction.

I don’t know the specific values, but an educated guess would lend me to say it means Sn4+ is thermodynamically and kinetically more stable compared to the Sn2+ state (visualised on an reaction profile diagram), meaning when going from Sn+4 to Sn+2 you need to put more energy in and the enthalpy of reaction would be less exo than the energy put in going from Sn2+ to Sn4+ which would be a more exo reaction.

It is energetically more likely that Sn will stay in it’s most stable +4 sate, a bit like CO2 is the more thermodynamically stable compound compared to CH4. To prove the Sn examples, you could physically do the reactions and saw which one happened, or you could look at the thermodynamic data, and do a rates experiment to calculate dH(rxn) and Ea.

Q2: explain the solubility of hydroxide n sulphates of group 2. Oh boy! That is perhaps the toughest thing on the A-lev syllabus, and I’m not too sure exactly what aspect of that is problematical to you. Let me recommend you read the Edexcel A2 student conference slides (go to slide 4/31) ( http://www.edexcel.com/migrationdocuments/GCE%20Curriculum%202000/264938_chemistry.pdf ) Then after reading that, please tell me what you need clarification of.

The AS part is here ( http://www.edexcel.com/migrationdocuments/GCE%20Curriculum%202000/264934_chemistry_session_1.pdf )

If you were more specific about Q2 I could give the answer more rapidly.

Please write in again if things still aren’t clear.

67 | intechemistry
March 18, 2009 at 3:12 pm

Please see the diagram at the top of this page.

could you re-type out the whole Q for part 1? You may get marks for fully labeling the diagram. I don’t know how ‘filled in’ it is.

In the cases you mentioned: 0.75 mole fraction and 50:50, these numbers tell you where to start drawing your lines from on te horizontal (or x-asis). I’ve given a diagram above using 50:50.

[ You probably know this, but a compound with 0.75 mole fraction means it is present in a 75% amount by moles,
Mole fraction = moles of a particular compound / total moles of all compounds present in a mixture. ]

A single ‘mole fraction value’ alone gives no info about which is the more volatile. But with additional info, it can reveal which is the most volatile. For example: we are told “after condensing, the resultant liquid has mole fraction of substance “P” in the mixture. This tells us “P” was the most volatile.

In the diagram above to get pure hexane, You would have to draw the red lines, the blue lines and according to the way I drew my diagram, there should be another set of lines that you would have to daraw. I didn’t show them as the diagram would be too crowded. Drawing the third set of lines means you should ‘hit’ the y axis and get 100 of the particular compound when the gas is condensed.

So do full diagram as shown above. Explain:

“the 50:50 mixture is bouled. The composition of the gas at that given temp {i.e. we ‘take a reading’ going horizontally across} is given by the upper curve. The composiiton of the gas is richer in the more volatile component. The gas cooled and condenses. This process is illustrated using by the red line. The condensate eventually boild again after absorbing more heat. The composition of this gas is again richer (compared to the liquid it came from) in the more volatile component, getting closer to becoming a pure substance. This is shown by the blue lines. The second condensate is boiled again and this time the composition of the gas consists purely of one substnace given once more by the upper curve. the gas is condensed into a pure liquid”

Something like that la. That should get you full marks. I think a fully labeled and clear diagram would probably be acceptable for most marks. If the Q said “explain” then you shoudl give some words to accompany the diagram.

Does that help? Basically – just describe what is happening with the arrows!

one and a new gas

71 | intechemistry
March 19, 2009 at 12:17 pm

68 | nurul, March 19th, 2009 at 8:28 am

Something that is NOT optically active [O.A.] isn’t necessarily planar. E.g. methane=tetrahedral & isn’t O.A also CO2=planar & isn’t O.A. either. What you can claim is: An optically active molecule cannot be planar.

It seems like you are thinking about planar molecules, or ‘reaction sites’ in molecules which are planar. These planar molecules are not themselves O.A., but react to produce O.A. products.chiral compounds e.g.:

ethanal + KCN/HCN –> CH3-C(OH)(CN)H

The carbonyl carbon, i.e. the C with the =O in ethanal is trigonal planar, but ethanal is not O.A. (or sometimes called chiral), but the product of the reaction… 1-hydroxyethanenitrile is chiral.

The planar C=O can be attacked from either side Attack from one side will give one enantiomer, attack from the other side will produce the other enantiomer. As the CN- has an equal chance of attacking C=O from either side, then a 50/50 mixture (called a racemic mixture) of both enantiomers.

ttp://www.btinternet.com/~chemistry.diagrams/nucleophilic_addition.swf The two products shown cannot be superimposed.

Try and make 3D models. This usually is great in helping people get to grips with it.

Let me recap: A racemic mixture can form from the reaction of some trigonal planar functional groups. The result will be a mixture of products called a racemic mixture.

75 | intechemistry
March 20, 2009 at 3:51 am

69 | nurul, March 19th, 2009 at 8:58 am
Sorry, but I didn’t have time to answer the other points you asked.

I corrected a v. minor error on the diagram (should have been 2 S’s)

Re: General formula. I can’t really say there is a fixed formula as it depends on the data given, and hence the cycle that gets constructed from that data. However, I will say you should always try thinking about summing (+) the enthalpy values for each individual arrow {once the cycle is worked out and you’ve identified the alternative route etc.}

Think about introducing negative’s ONLY as ways to swap the direction of an arrow. I.e. try not to get into the situation where you think things like “I should subtract that enthalpy value from this other enthalpy value…” and so on.

Stay in positive ‘addition method’ mode. If you do that, your mind should not get confused when -’ves occur (exothermic reactions and for ‘arrow direction changes’) It should lessen your mistakes and increase your ability to do these Q’s.

You are right. S(s) H2(g) and O2(g) are ELEMENTS in their standards states and are taken as having zero relative enthalpy ‘cos they are used as a standardised starting point by which other chemical deltaH values are measured. So only data from the elements to make H2O and H2SO4 were necessary in this case.

We can use the formula dH(rxn) = Sum of (dHf products) – SUM of (dHf reactants) here because we used formations in 2/3 parts of the cycle. The reactants were 2S(s)+ 2H2O + 3O2, the products were 2H2SO4. We must take into account the # of moles. Using the equation we get (2x-814)-(2x-286) giving us the -1056. Perhaps the slightly tricky bit is deciding which are the products and which is the reactants. Note this formula stated deltaH FORMATION. Students have used this formula in other cases where formation values are NOT given, and luckily they get the right answer, but it sometimes doesn’t work. Only use this formula for dH(f) values. Actually, doing a Hess cycle can usually cancel the need for confusing formula.

The Bonds broken – bonds formed formula is usually used when you have the bond enthalpy terms (or average bond enthalpy values). Those values are all positive ‘cos the ‘bond enthalpy term’ is the energy needed to break one mole of the bond mentioned. The formula is therefore constructed to give you the proper value at the end (+ or -). BEst not use this formula for anything other than when bond enthalpy terms are discussed.

If you want to see me personally to ask about these things feel free. Jims book is good for this. His website doesn’t really discuss it.

79 | adilah
April 2, 2009 at 5:53 pm

{note: this comment was originally in Planning Exercises but was moved here for reasons of ‘housekeeping’}. Tq

Salam, Sir. i want to ask regarding the recent a-level trial exam. practical paper, 6A, Question 2b.
the question, to 2cm3 of solution H in test tube, add 6 drops of universal indicator solution.
identify cation in H.
(in the previous question, the cation detected is Zn2+, Pb2+ and Al3+.)

the observation, using universal indicator, red solution is formed.
so it is acid.
i got confused, why the answer is only Al3+,
but not Al3+ and Pb2+…
as both oxides are amphoteric.

thanx sir!

Reply:
Two ways to look at it…

Pb2+, is amphoteric yes, but isn’t a strong enough acid to react with water, so doesn’t consume OH ions causing an excess of H+ ions, so the solution isn’t acidic.

The other way is what you should be more familiar with:

Al3+ cation exists as the hexaaqua ligand in aq. soln. The small and highly charged Al3+ ion attracts and withdraws e- density from the OH bond in the attached H2O ligand. the OH bond is therefore weakend to the exent that water (a weak base) is then able to deprotinate the hexaaqua complex.

the same happens for [Cr(H2O)6]3+ and [Fe(H2O)]3+ the metal ions there are also small and highly charged. The Pb+II and Zn2+ don’t have sufficient charge/size ratio (i.e. charge density) to enable that deprotonation in water.

83 | intechemistry
April 11, 2009 at 12:16 pm

Salam siti. I’m confused too I’m afraid. I’ve tried assuming what the question was but I’ve not made any progress. I’m going to have to ask you to give me the full question.

The answer also looks a bit incomplete too. Is the beginning of the answer this: “repeat experiment, double the [ ] of one reactant, but keeping the other…”

But the main point of the answer is still strange to me ‘cos as all orders are two, then doubling conc should lead to a 4 times increase in rate.

I’ll wait for the Q. Could you please tell me where the Q came from also?

Thanks.

85 | intechemistry
April 11, 2009 at 3:36 pm

#81, nurul April 10, 2009 at 3:37 pm | edit

Use of the word “partially” is dominant. But the words are pretty much the same. I guess the word “slightly” is avoided as it may be mor elikely to make people think that the compound at the end is delta+ or delta-

In an exam however, I don’t think you would be penalised if you said slightly, but best avoid it I’d say.

93 | intechemistry
April 18, 2009 at 2:31 pm

fas. I corrected the error in my reply above.

97 | intechemistry
April 24, 2009 at 6:12 pm

#94 fas, April 24, 2009 at 7:51 am
(I got sleepy last night, so this whole comment is updated)

June 08 U4 Q3 Kp
Q3c) (c) (i) The effect on Kp by raising the temp (1 mark)
– ans: Kp inc.
(ii) use answer in (i) to explain T inc. on posn of eqm (2 marks)
– ans: quotient must increase as Kp goes up. Eqm shifts to RHS
Fraction/quotient MUST be mentioned

Jan 08 U4 Q3 (b) Kp simple Q

Jan 08 U6b Q4 (d) Section BTrivial Q

June 07 U4 Q4 (c) (iii) Kc
(d) (i) T is lowered. Explain effect on value of eqm const hence the yield. (4 marks)
– ans: Rxn is endo, Kc decreases, numerator in Kc must inc, yield inc.
Comment on either: numerator OR denominator OR fraction (overall) must be made to get the third mark

June 07 U6B Q1 Kc
Trivial Q

Jan 07 U4 Q2 Kp
(b) (i) State effect (if any) on Kp of increasing the Temp (Press = const) (1 mark)
– ans: Kp value decreases.
(ii) Use ans in (i) and the kp expresssion (as written in the paper) to explain the effect of the posn of eqm when inc T and keeping press const. (2 marks)
– ans: Quotient must decrease, eqm shifts to LHS
Quotient had to be mentioned to get a mark – (could have been more specific and focused on numerator or denominator also)

June 06 U4 Q6 Kc <<
(c) P in rxn chamber was Inc. With ref to changes in Kc, explain effect on the eqm posn (3 marks)
– ans: Kc unchanged. P^ increases the value of numerator more than denom(NOTE: more gaseous moles on ‘top’ in numerator, – pressure ‘squeezes’ the gases together whereupon prods react to go back to reactants again which occupy a lower volume hence causing the pressure to decrease somewhat), OR value of quotient has increased, eqm shifts to LHS.
Some discussion about the quotient must be referenced.

June 06 U6B Q4 section B, Kp
() (ii) State the effect (if any) on the value of Kp when T is increased.
Justify your answer. (2 marks) !!!!!!!!!!!!!!!!!!
– ans: Kp decreased reason given: rxn was exo.
Comment: Answer does not make reference to the quotient. This is because the ‘sequence’ of answering is as follows: 1) Kp (or Kc) up or down, 2) Reason why: rxn endo or exo respectivbely 3) The quotient (num & denom) adjust to get to the Kp value. In a 2 mark question there isn’t the ‘capacity’ to get to the third point. Beware: The effect on Kp may be a separate Q – the following reason may be worth two marks so then the endo/exo nature of the reaction and then the adjustment of the quotient would score the two marks.

Jan 06 U4 Q3 Kp
(c) Looking ar change in Kp (if any)happens to the equilibrium mixture as the T is increased (3 marks)
– ans: Rxn endo, Kp inc., eqm shif to rhs.
Examiners Report: In (c)(i), it was rare to see all three marks awarded as very few candidates firstly established the change in the value of Kp in order to then explain the shift in the position of equilibrium.
Comment: OK here no reference is made to the quotient, but that is because we are specifically told to address the eqm mixure. Actually, I don’t like this mark scheme as it doesn’t specifically state what happens to the mixture. For that point I’d have stated the eqm mixture becomes richer in products. I think examiners would have realised this (at least I hope so!) and marked accordingly.

June 05 U4 Q4 Kp
(b) (iii) Give the effect of an inc, in Temp on the value of Kp. Justify your answer. (2 marks)
– ans: rxn=exo, so Kp will inc. No mention of Quotient.
(iv) Hence suggest in which direction the position of equilibrium when temperature is increased. Justify your answer. (2 marks) << Aaaah! Quotient time
– ans: The quotient must increase and so eqm shifts to the right.

June 05 U6b Q3 simple Kc

Jan 05 U4 Q4 Kc
(a) (iii) simple Kc and T question (delta H was zero)

U6b Q 4 (a) (ii) simple Kp with pressure change only

Can you see better now? My advice if you have doubt as to when you are supposed to discuss the quotient, then always do so when you are asked about Kp or Kc when [ ], P or T changes. Not catalysits!

If at eqm and [ ] or P changes, the numerator and denominator values suddenly alter and the fraction(or quotient) no longer evaluates to the Kc or Kp present before the change was made. The eqm will shift and the numerator and denominator will change to once again give the same value.

If at eqm the T changes, A NEW value of the eqm constant is established. The quotient now not equal to the new Kp or Kc. The numerator and denominator shoft accordingly to get to the NEW value of Kp or Kc and the posn of the eqm shifts as a result. The way it shifts depends on whether it’s endo or exothermic.

Using quotient arguements shouldn’t be too hard. You know how to figure out which way the eqm shifts, so you look at the quotient: a/b if the overall value gets larger, the numerator (pressure or cond of products) must increase and the denom must decrease (pressure or conc of reactants)

Kp or Kc (Ka aswell actually but discussion of Ka with T is quite rare as Kp and Kc considerations are important industrial processes, in which ratr(T) and cost of high pressure eq(total P) is important)

Does that help?

99 | intechemistry
April 25, 2009 at 5:49 am

#95, N April 24, 2009 at 11:07 am

The A-level answer is:

An orbital is a region of space where there is a high probability of finding an electron. An orbital can hold only two electrons.
Plural of orbital is orbitals – but you won’t find the word in dictionaries. Also, don’t get confused with orbits – the circular path things!.

A subshell is an orbital of a certain principle energy value (n) of a certain orbital angular momentum value (l as in lemak). e.g. 2p(x) 2p(y) and 2p(z) They are all similar in that n=2 and l=1. I can’t imagine they will ask you to define subshell, although of course it is essential you understand what a subshell is. Plural of subshell is subshells.

Problem solved?

101 | intechemistry
April 25, 2009 at 9:18 am

“the ligands split the 3d orbital”, is wrong. There is no “3d orbital“. The thing that gets split is the subshell.

“the ligands split the 3d orbitals”, is not very good wording as it could be interpreted as meaning one or more of the orbitals themselves (i.e. the volumes of space containing up to two electrons) can/are being be split.

Best wording is ‘ligands split the 3d subshell into two sets of orbitals, each set having a slightly different energy.‘

If you want more info: In an octahedral complex the higher energy set contains two orbitals, the lower energy set contains three. note: in a tetrahedral comples that situation is reversed

“the orbital is the 1st, 2nd and 3rd…etc” – This is incorrect. This is the energy or SHELL.

“the subshells are within the orbital” – This is incorrect too. The obitals are inside the subshell.

Magnesium has 3 (occupied) SHELLS
for the second SHELL, there are two SUBSHELLS
the two subshells are 2s and 2p.
or if you want to put the e- in, you can say 2s2 and 2p6

Maybe this helps?
The world = the atom.
The countries making up the world are the energy SHELL.
Malaysia is just one energy shell.
In Malaysia we have Selangor, Kuantan etc. The states are the subshell.
In the states, e.g. Selangor, we have things like MBSA, MBPJ, MBSepang etc. These MB’s are the orbitals.
In the MB’s we have people. People are the electrons.

Getting back to the actual atom:

1s is the first SHELL
2s and 2p subshells make up the second shell
3s 3p and 3d subshells make up the third shell
4s 4p 4d 4f subshells make up the fourth shell

each s subshell contains only one orbital
each p subshell contains three orbitals 2p(x) 2p(y) and 2p(z)
each d subshell contains five orbitals 3d(x2-y2) 3d(z2) 3d(xy) 3d(xz) 3d(yz)

Quantum numbers:

n=1,2,3,4,5,etc
l=0 up to and including (n-1)
ml= -l up o and including to +l
ms= each ml +/- 1/2

e.g. n=3
all values of n are 1,2 and 3, i.e. three values. This tells us there are 3 energy shells in this atom.
When n=1, l=0 the first energy shell isn’t split into any subshell.
When n=2, l can be 0 or 1; two values. This second shell is therefore split into 2 subshells.
When n=3, l can be 0.1 or 2; three values. The third energy shell therefore split into 3 subshells.

Any valid set of n and l values can have their corresponding ml value calcualted which reveals the number of orbitals in that subshell.

e.g. when n=3 and when l=2 (i.e. we are looking at the 3d subshell). ml for that l value can be -2, -1, 0, 1 and 2. 5 values. Therefore the 3d subshell is split made of 5 orbitals.

You might be wondering why I’m saying all this stuff. It’s to give you more exposure to the terms shell subshell and orbital so that your mind gets the feeling of how to use the terms propely.

If you now need a panadol, then you are free to ignore it.

Remember Venn diagrams? A subset is drawn as a small circle inside a bigger circle. The subset is a constituent part of the bigger set.

105 | intechemistry
April 26, 2009 at 7:39 am

#103, intec studnt April 26, 2009 at 5:39 am

I have the feeling the Q is likely to have said the water was pure, and it was about 50oC (greater than 25oC)

A solution is classified as:

acidic if [H+] > [OH-]
neutral if [H+] = [OH-]
and basic if [H+] < [OH-]

You will notice there is no mention the the -lg[H+] must be 7 for a neutral solution

Theoreticaly, in a particular solution:
The conc of H+ could be 0.1M therefore pH=1
and conc of OH- could also be 0.1
but it would be neutral because [H+] = [OH-] even though the pH was 1.

When you heat pure water a greater proportion of the H-OH bonds break, releasing equal quantities of H+ and OH- into the solution. so no matter what the [H+] is, there are the same number of OH-’s present so it’s neutral.

Our minds have become familar with talking about water and it’s autoionisation at 25oC where the Kw for water is approximated to 1×10^-14 and so neutral pH will be 7 at that point.

That’s what our minds reset themselves to when we start learning about pH.

OK with this now?

113 | intechemistry
April 29, 2009 at 3:28 pm

#110, lucky guy April 29, 2009 at 2:59 pm

1) What’s in the conical flask? Acid or base? This will tell you what region (to start your curve in)

2) What type of acids or bases are being used – strong or weak?
The answer to that gives you what general shape to draw in the relevant region.

- strong components (HCl, H2SO4, H3PO4, HNO3) have a vertical section that should be drawn at least 4 pH units tall.
- weak components have NO vertical in their relevant ‘acidic’ or ‘basic’ regions.

3) Calculate and plot the exact initial pH (if data is available)

4) Calculate and plot what volume of reactant in the burette is needed to get to the end point and use that volume (x-axis) as the place where the vertical section occurs. You may need data to calculate this.

5) Calculate and plot the end pH if possible.

Label axes if necessary.

The start pH, vertical section(likely) and end pH will give enough info for you to join everything up. LOOK at a number of titration curves to see their shape.

Go here: http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html

It’s unlikely they will give you the weak-weak (acid-base) version as usually they want you to use the graph to select a suitable indicator.

Be prepard for if they give you a dibasic acid. e.g H2SO4, H2CO3 or worse case = tribasic acid like phosphoric acid. It’s unlikely you will have to accurately plot a dibasic and tribasic acid however.

145 | intechemistry
May 7, 2009 at 8:56 am

#139-141, lucky guy, May 7, 2009 at 5:36 am

Your Q was vague, so I thought I should cover more than one possibility in my answer. But no worries.

You asked: “can we leave earlier if we finish earlier? or what we should do if in the half way of titration we just found out the burrette leaks?”

Reply: Good, some specifics. If you are in the early session, you cannot leave early as there will be no lecturer/invigilator to take you to quarantine. If you’re a ‘last session’ student then I think you can leave early but I’ll have to check it out.

If titration is 50% done and you realise there is a problem with the burette, you will have to decide yourself whether it is worthwhile continuing (is the leak so small? is it the first approx titration? etc), or perhaps you should start again with a better burette – I’d recommend you start again using the spare burette from the extra apparatus. The burettes are new as far as I know show expect it to be unlikely that there’s a leak problem.

I’m trying to get both papers for Jan09 and mk.sch. Can’t promise anything however. Would have been better to ask earlier.

As for sig. fig. can you be more specific? I don’t have enough time to write so much on the subject. You’re lecturer is almost certain to have addressed this issue in class. What do you need to know about s.f.?

147 | intechemistry
May 7, 2009 at 10:37 am

If no data is given, then usually 3 s.f. is standard.
e.g. 0.100 M, or 5.43 M or 0.000312 M (usually written as 3.12×10-4 M.

If data is given, e.g. you are told that a 25.5 cm3 of a 2.0 M solution present…, then your answer should have the same number of s.f. as the data that was used in your question. If you used the vol and the molarity (or just the molartiry by itself), then the answer should be to two s.f.
If you only used the volume data, the answer should be to three s.f.

Page 51 of the full syllabus says:
Significant figures: Students should always quote their final answer to an accuracy consistent with the accuracy of the figures given in the question.

Sometimes Q will tell you how many s.f. to write. In those cases you MUST follow those instructions.

155 | intechemistry
May 8, 2009 at 4:37 pm

Halogenoalkanes are poorly soluble in water, but they are soluble in ethanol. NaOH(aq) is also soluble in ethanol. Using aqueous NaOH in ethanol therefore allows for a larger degree of OH-(aq) ions to interact with halogenoalkane molecules.

If NaOH(aq) an aqueous solution of NaOH alone was added to the halogenoalkane, then they could only interact at the interface (where the two immiscible layers meet) and so reaction rate would be slow. It would still happen, but adding ethanol is better.

Why does the m.s. disallow non-ethanol solutions?? I’m not sure. Do you have any reference that I can look up?

Usually 1) Add small amount of NaOH(aq) 2) Acidify with HNO3(aq) 3) Add AgNO3(aq) is what I tell students. Doing as you say: a) halogenoalkane then ethanol then AgNO3 seems unusual.

Please can you give a reference.

When only ethanol is used, this suggests/points towards an elimination reaction (producing HX and an alkene) but it still requires a conc. base, but you’re pure ethanol reaction doesn’t involve hydroxide.

Humm… Reference here is very much needed. Tq.

161 | intechemistry
May 11, 2009 at 12:33 pm

The upper element (e.g. Cl2) will displace the lower halogen anion e.g. Br-.
F2 will displace Cl-

Br2 will displace I-

etc.

I guess you’re asking in U3A in mind. I don’t know if this answer is too late.
respect

171 | intechemistry
May 23, 2009 at 1:58 pm

Sorry, rushing right now. will try & reply in a couple of hours.

179 | intechemistry
June 4, 2009 at 12:32 pm

Damian Riddle{Damian is representing Edexcel}
20/04/2009 01.44 PM
Dear Sir,

Damian Riddle

199 | intechemistry
April 24, 2010 at 1:20 am

Hello “The smiley little girl”
Actually I’m not too sure what Nuffield really is. It’s ‘done’ by edexcel and is kind of A-level standard but like a bit on extra hard chemistry. I think they scrapped it now.

I wonder what the paper looked like! If you are around campus, sms me and give me a look. In return I’ll buy you some food and a drink.

201 | intechemistry
May 2, 2010 at 1:53 am

# 200 Superman;-D(who has been bothering you in sms)
May 2, 2010 at 12:33 am

Dear Superman hehehe…

1. The A2 George Facer book I have on p 163 talks about Emission spectra under “Chapter 9 Spectroscopy and chromotography”, and I can’t see any mention of such a combination if ions on pages 208-233 which covers redoxII. Perhaps you have a different George Facer book? Let me know the ISBN.

The I-/(1/2)I2 half cell is in the data book p15, halfcell #50 Its E(std) = +0.54V.
for MnO4- to Mn2+ is E(std) = +1.51V

so The rxns are MnO4–> Mn2+ and I—> (1/2)I2 Ecell=+0.97 V
Notation puts the most -’ve species on LHS {were not using the SHE, which if present ALWAYS goes on the LSH} so we get the I-/(1/2)I2 half cell and this is where oxidation occurs i.e. the anode and should therefore as you rightly say be on the LHS.

I would (quickly)write the cell as Pt| {5I-(aq) | 2(1/2)I2(s)} || {MnO4-(aq), Mn2+(aq)} |Pt E(std)cell = +0.97V. If the your Facer book has it the other way around, is his Ecell=-0.97V ? If it is, then his ‘swapped around’ cell is OK I guess, but it doesn’t obey the conventions, which, although not strictly necessary, make life easier.

As for which Mn species goes where, the most important thing is next to the “||” you put the most oxidised species in the halfcell. In the Mn 1/2 cell MnO4- is +VII whereas Mn2+ is only +II so MnO4- wins and is put . In the RHS elecectrode, put the Pt last, so it will indeed be next to Mn2+(aq). It’s not that important where the Pt goes in this halfcel as both ions are aqueous.

2. “does it mean that we do have to know and memorise the exact procedures of the process that include salt extraction, salting out..etc?” YIP, it sure does. This is most likely to be asked in Unit 6B, the written paper of A2 labwork under systhesis/preparation

3. YIP (as far as I can tell) I think edexcel will like fuel cells because fuel cells are portrayed (laughably) as green.

I’ll try and answer your other Q’s after some sleep.

205 | intechemistry
May 2, 2010 at 5:08 pm

# 200 Superman;-D(who has been bothering you in sms)
May 2, 2010 at 12:33 am //PART 2\\

4. Again my Facer A2 book p259 seems quite different from yours, and I’m surprised this would appear in an A2 book. My A2 G.F. book talks about organic nitrogen on that page, so I’m not too sure what your point of reference is. I think you meant AS but I don’t have that book at home so I can’t check it out until monday. However in AS Pearson and AS Edexcel revision guides, the following step is shown:
clO(dot) + O3 –> Cl(dot) + 2O2

It is of course possible that ClO(dot) can react with O(dot) but the conc of radicals is quite low. The chance of these two radicals to come together is less likely than a radical to react with a more abundant non-radical such as O2 or O3. Maybe ClO(dot) + O(dot) is describing a step in a different process.

5. Enthalpy of neutralisation involved the formation of one mole of water when an acid is neutralised by a base. That has to be the definition as other definitions are likely to be ambiguous. I really don’t like the AS Pearson definition. there are a few things about Perason that I’m not happy with. {doubtless I would say some things that ‘Pearson’ would not be happy with! LOL}. In the Edecxel revision guide it uses the one mole of water answer. So stick with that.

6. In Pearson AS p 201, The horizontal line between the two humps in the profile (or in SN1 profiles) is just a ‘style’. It has no meaning (you will notice there is no X axis in that case). If in fact if “reaction coordinate” is on the X axis then the horizontal line would actually be incorrect as the increasing ordinate cannot be a straight line! But that’s a side issue. It’s common to draw the intermediate as higher than the reeactants initial energy level. But the important point is to the left or the right of that line (or point) you have to go UP in energy again. Drawing the intermediate at the higher level allows one to draw endo and exo reactions quite easily. to have the line or point below the energy of the reactants makes it ‘weird’ to draw the endo reaction profile. I have no doubr that some REAL intermediats have a higher energy and others a lower energy. As you are sketching a graph with no actual data as to the energy of the intermediate, then you would be forgiven for drawing either illustration. Perhaps Pearson p201 is just trying to show you both can be used.

Once again, (point 6) my Facer A2 book on p271 talks about polymers. I have a strong feeling you are meaning to say G.F. AS but are accidentally saying A2, althoug the ISBN you gave is exactly the same as my A2 book. Hummm…

207 | intechemistry
May 2, 2010 at 5:38 pm

#203, zereda May 2, 2010 at 3:35 pm

Unfortunately I don’t have the answers. Please take a look at the 2nd paragraph on point 7, comment #206. We CAN get answers (I’d like them too! and hence I said I would contribute towards the students quest to purcahse those answers) but this MUST be done VERY QUICKLY. Perhaps you can act with “Superman” to get the collection needed. The more students that contribute the cheaper these useful answers can be obtained. I am left with a very bad feeling at the blatent commercialisation of student learning. It must be a very difficult struggle for those students who are financially struggling. It’s terrible.

You are welcome to make an appointment with me to look over the answers. It might take me a while to get going, not being psyched up for exams, I find these days I’m a bit slow to get into the ‘mood’ of the questions. Perhaps you could leave them with me for a day before meeting to discuss them? What do you think?

213 | intechemistry
May 4, 2010 at 12:25 am

#211, yeah it’s Superman’s time!, May 3, 2010 at 10:12 pm

1. In general, the usual # of SigFigs is 3. The reason why at A-level 3 SigFigs is generally used, is that it means the greatest error will be 1 out of 100 (100 is the first 3 SigFig number) which would give an error of 1%. The 1% error margin is regarded as tolerable or ‘OK’ for A-level.

You don’t state the data used in coming up with your dH value, so I can’t say for sure if your 18831 value is OK or not, however, in an actual lab setting, it’s probably the case that the temperature will have been be read to 1 dp (you’d probably have a 0.5oC or a 0.1 oC graduated thermometer) so the thermometer is probably being read to 3 SigFigs because the temp is probably going to be between 10oC and 99oC that’s 2 SigFigs and then add on the 0.5 or 0.1’s, e.g. temp was 25.6oC. That my reasoned guess anway.
All other data is probably read to 3 or more SigFigs. So, IF THE DATA SUPPORTS IT, reporting the dH value as 18800J mol-1 i.e. 18.8 kJ mol-1 would be OK.

The problem is that zero’s on the RHS of the decimal point are sometimes significant which can confuse things. In the event of confusion, just look back at the data. What’s the lowest number of SigFigs in the data you used to come up with your calculated value? Use that level of accuracy for your FINAL answer.

Intermediate answers or steps in a long calculation can be written as 3 SigFigs as an indication your value is ‘on the right path’ but keep the full value in your calculator and ONLY round up this possibly large number at the very end.

the original number 18831J mol-1 only has 5 SigFigs, no decimal places and is a J mol-1 value, so it might be tolerated if you left the number like 18831 J, but remember, it is the accuracy of the data which determines the accuracy of your final answer.

2. Use the original number of you have to take the answer from (a) and use it in (c).

3. Look at the SigFigs of the data. If various bits of data have different SigFigs then it’s always the data with the LOWEST number of SigFigs that your answer should be reported to.

e.g. (and I’m just making numbers up here, so don’t expect an actual calculation to work out) 1.5g of Ba(OH)2 was dissolved in 1000cm3. 25.0 cm3 of which was sampled and titrated against 0.3 M HCl solution. The Burette recorded a volume of 25.65cm3 HCl.
Here we have 4,3,2 and 1 SigFig! If the calculation used the concentration, i.e. 0.3M then your final answers should only be 1 SigFig too!. You would be unlikely to get this terrible standard of accuracy in an exam. The conc would probably say 0.300 which is 3 SigFigs.

Hope that helps.

217 | intechemistry
May 4, 2010 at 7:17 pm

#215, Superman, May 4, 2010 at 9:48 am

Sorry for overlooking the 1st part of your Q.

In the past, I penalised students for NOT writing a sign at that stage in the calculation. One ALG student who I penalised wasn’t very happy because in the mark scheme it wasn’t necessary to have a sign at that stage.

E.g. Jan 2006, Unit 3bB, Q 4 d i)
The Students had to do a dH = m c dT calculation. The reaction was exothermic. The Q said: “calc the heat change for this reaction.” and it gave the dH = m c dT type info. The answer on the mark scheme had no -’ve sign for the reaction. (and it didn’t have a ‘+’ sign either)
The next part of the question, part ii), said “Use your answer from (d)(i)… to calculate the enthalpy of neutralisation
per mole of nitric acid, HNO3. Include a sign and units with your answer.”. So I could understand why he was not so happy,. He did have a very valid point, and was probably correct, BUT I maintained the penalty because if that ‘remember to include a sign” statement was missing, which is I guess possible, then the student might have forgotten to write it. My logic was by doing it at the first opportunity, the chances the student would forget it would be less., if in some other paper there was no reminder about the sign. One could argue, as I did, that it was correct, irrespective of the mark scheme, to put the sign in at the stage. The reaction did liberate heat. The enthalpy change was exo.

It is true, the mark scheme for part i) only had a signless number in it {in the working out and in the answer}, so there is a chance the student might have LOST :s a mark if they did put a -’ve sign in at that stage.

All things considered then, it might be best to write something in brackets like: “this value calculated is the magnitude for the exothermic reaction” so you are still in agreement with the answer but you are addressing the sign issue early on, and letting the examiner know that you understand the sign stuff.

This is one of those slightly unclear areas.

Hope that helps.

219 | intechemistry
May 4, 2010 at 7:33 pm

p.s. I didn’t realise there were so many ‘“super heros” in INTEC. Wow! Must be something in the Bloc V canteen Nasi Lemak

223 | intechemistry
May 5, 2010 at 11:00 am

#221, zereda, May 4, 2010 at 10:57 pm

Pity they don’t want to get the ‘guide’, and I’m not spending that amount of money all by myself.

As for the Q’s, OK, leave it with me. I’ll try and get those Q’s done. Perhaps see me tomorrow (thursday) in the afternoppon. 3:00pm?

229 | intechemistry
May 9, 2010 at 1:09 am

#226, MY, May 8, 2010 at 11:30 am

1) Here’s the reasoning currently stuck in my mind…

Starch is a polysaccharide, and forms molecules with a distribution of lengths (depending on how many starch units happened to link up) Like a protein or enzyme, these long starch molecules tend to fold in on itself (intRAmolecular attractions).

Starch looks ‘cloudy’ as the large folded polymeric units it forms can get to a size around about the same as the wavelength of light, so the big starch particles interfere with the passage of light through the solution.

If the particles are very small much less than (10)^-6m as in the case of hydrated Na+ and Cl- ions, then light passes through the solution and it looks transparent.

But for starch , because of its size similarity to light, visible light interacts with the starch structures, which results in the light getting scattered – the light ‘bounces’ off and is deflected by the starch. This scattered light can reach the eye. All frequencies of the white incident (‘initial’) light are scattered which is why it looks a bit cloudy white.

The point here is that starch in solution form a BIG and folded/twisted structure. If you add iodine molecules to starch, the I2 binds to it forming a starch-iodine complex which the eye can easily see/detect. The more I2 you have, the more it can get into pores etc in the structures starch forms. These ‘pores’ and other ‘cave like’ areas on the molecule can ‘trap’ or ‘seal away’ I2 making it difficult for the added (S2O3)2- difficult to reduce the I2 molecules.

Therefore you end up adding more thiosulphate that you would otherwise need. You can think of it as more (S2O3)2- is needed to force the I2 out { but please don’t that that visualisation aid too literally}

By adding starch at the end, i.e. when there is v. little I2 present. the vast majority of the I2 will be bonded to the starch on the outside of the starch structure. making it easy and quick to react with the thiosulphate, so the moles of thiosulphate is closer to the appropriate number of I2.

The titration will be more accurate.

But remember starch is added to make the colour change easier to see. No starch means the colour would be brown and gradually gets paler and paler yellow. One or two drops before the end point, the paleness makes it harder for the human eye to see the disappearance of the colour. Whereas the eye can see the blue/black starch-iodine complex to colourless much more easily.

2) Step 1 and 2 make the crude solid. That’s why they have filtration in the second diagram. Quite often recrystalization is discussed from the point where you already have your crude solid already formed, but I agree. The rest after that is pretty poor recrystalization technique. What can I say? – “Typical Pearson” perhaps?? I know it ‘s not easy to write a book, but really for a high cost commercial product like this book, Pearson is a let down and seems to have skipped Quality control.

PROPER METHOD: Once the solid is made, filter it to get the crude solid or crystals Using a hotplate. dissolve the solid in the minimum amount of hot solvent. Filter the solution WHILE HOT in a pre-heated ordinary funnel with fluted filter paper. Allow the filtrate (the liquid left after filtration) to cool SLOWLY. Use a Buchner funnel to filter under vacuum collecting the solid . Wash solid with a small amount of ICE COLD solvent. Allow crystals to dry.

I don’t like Pearsons method. To me it seems to encourage small crystal (usually a bad idea as they can sometimes be so small that they pass through the filter paper! – so you lose them!) formation and the also encourage occlusion (holes), which can trap impurities and solvent!

It is possible that Edexcel have ‘relaxed’ the learning requirements in describing the recrystalization process. But the old system is FAR, FAR better.

231 | intechemistry
May 9, 2010 at 1:35 am

#228, Catwoman’s housemate, May 8, 2010 at 10:04 pm

There should be no “different version of facts on the colour of phenolphthalein at the end point”. If you spotted something, please let me know.

In acid solution it’s colourless, in basic solution it’s pink. The indicator is normally initially added to the ‘conical flask reactant’ giving a colourless solution. When the titration begins but acid and base, e.g. NaOH, the solution gets more and more basic. Finally a pink colour suddenly appears. Note: you MUST use a strong base for phenolphthalein. It won’t work is you use a weak base, as the colour will gradually appear, not appear suddenly. The eye would find it very difficult to spot colourless to “almost colourless, but the tiniest hint of pink”,

The base is sometimes (accidentally) put in the conical flask and phenolphthalein added. In this case the colour would be pink. Titrating with acid (in the burette) means the pink colour would slowly fade. The eye would find it hard to see the total disappearance of the very pale pink (just before the end point) to colourless AT the end point.

Titrations are sometimes done to give a colour rather at the end point, rather than just have a colour fade to colourless.

241 | intechemistry
May 10, 2010 at 10:06 pm

#238, Superman May 10, 2010 at 8:38 pm

1) Yes. The order of addition is important.

a) If you add HCl(aq) first, [Assuming no Pb4+, Pb2+ or Ag+] then the only time a ppte might occur is afte you add BaCl2(aq). If a white ppte then definately BaSO4 formed, hence (SO4)2- was present.

b) If BaCl2(aq) is added first [again assuming no Pb4+, Pb2+ or Ag+] then a white ppte may form if (CO3)2-, (SO3)2- or (SO4)2- was present. Add HCl(aq) now and if a white ppte remains then it was sulphate.

You should be able to see the difference here and understand what’s happening.

2) The spelling doesn’t matter, however it seems the powers that be decided to use the American spelling sulfate. The latter is more in line with malaysian phonetic spelling so should be easier for you guys, so use it.

249 | intechemistry
May 11, 2010 at 11:42 am

#247, The smiley little girl May 11, 2010 at 9:12 am

Well “reduce[ing] the solubility of phenylamine in aqueous solution” is a major part of it, but your wording doesn’t indicate how the solubility is decreased. Was the temperature changed? – one wouldn’t know. “Salting out” indicates the a salt is used to achieve the decrease in solubility.

I don’t have any advice as to whether the term should be used or not, so know the term and what it means.

259 | intechemistry
May 11, 2010 at 11:34 pm

SUGGESTED (!) ANSWERS to PEARSON A2 Exam Zone Questions available here:
http://intechemistry.wordpress.com/2010/01/01/file-depository/

Spread the word, for the word is good!

263 | intechemistry
May 12, 2010 at 12:11 am

#256, MY May 11, 2010 at 10:36 pm
Pearson (A2) p185 has a diagram of the setup needed and a description of what happens. Do you have that book? The book says the tap of the funnel is closed to get the pressure to force Cr2+ up the ‘delivery’ tube. After it’s added to sodium ethanoate solution. The bit about the “the cap of the seal is screwed shut ” does seem a bit weird, as though it should be talking about the funnel tap closing instead. If this is not shit, then there would not be any pressure build up. I can think of why there would a second ‘open’ place that needs a seal closed. Hummm. Sorry, but I am not familiar with that practical.

269 | intechemistry
May 12, 2010 at 10:45 am

#265, anon May 12, 2010 at 9:42 am
Salam. I don’t have any practical stuff with me right now on the main campus, but I’d say:

1- The main source of error is most likely to be the temperature readings.

2- The change in entropy

271 | intechemistry
May 12, 2010 at 11:02 am

#267, anonymous May 12, 2010 at 10:15 am

If it’s an observation then yes, say it. The solid does melt and the Na(l) eventually disappears too on completion of its reaction with water. Say all observations that are valid and relevant.

273 | intechemistry
May 12, 2010 at 12:30 pm

#272, girLnexTDooR May 12, 2010 at 11:59 am

a) C6H5-NH2 + NaNO2 + 2 HCl –> C6H5-N(+)NCl- + NaCl + 2H2O
I put in the + of C6H5-N(+)NCl- to tell you the positive charge is on the N bonded to the ring, and NOT the end N.

b) The HCl is aqueous. I dont think using concs gives any advantage advantageous so the aqueous (i.e. dilute) HCl is used instead.

c) Asprin [acetylsalicylic acid] is probably a white ppte. I know it’s poorly soluble which is why some versions of it are sold as the sodium salt to make it soluble (COOH part is turned into COONa).

d) so the temperature of the heater and the solid are the same. A high heating rate will take some some time to heat to enter the solid and warm it. By thich time the heater will be much hotter as it’s still heating up!. Slow heating allows equilibration of the heat.

p/s Your welcome.

275 | intechemistry
May 12, 2010 at 1:43 pm

Nitrous acid needs to be made fresh from NaNO2 and HCl. It decomposes. If Nitrous acid only is asked for then OK, just write HNO2, but that strikes me as being unusual! Best remember how to make HNO2 too.
HCl(aq) + NaNO2(s) –> HNO2(aq) + NaCl(aq)

So the equation would be:
C6H5-NH2 + HNO2 + HCl –> C6H5-N(+)NCl- + 2H2O

281 | intechemistry
May 12, 2010 at 5:17 pm

#276, Superman May 12, 2010 at 3:39 pm

1. I’ve not seen the need for adding ‘+’ or ‘-’ to the error. I was always under the impression absolute error as asked for, hence no need for sign. I’ve seen no guideline requiring a sign. My advice is to see the lecturer who didn’t give you the mark. There may be another reason why the mark wasn’t awarded e.g. did you clerly show your working?

2. You don’t need to shade the layer as long as you clearly show there are layers. In exams NEVER shade using ‘solid colour’. Instead use diagonal lines instead to illustrate/draw attention to some area.

3. A water condenser is IMHO, always better. My advice esp. A-level, is always use a water condenser. I’ve never seen a NEED for an air condenser at A-level. If the heating is mild then perhaps an aircondenser could be used, but I don’t like it. You are vertainly not expected to suggest an air condenser.

283 | intechemistry
May 12, 2010 at 5:39 pm

# 278, Superman May 12, 2010 at 4:05 pm

You are right. A lot of heat is generated, so adding too much nitrating mixture would therefore make the temperature go >55oC and wen that happens, two nitro groups may add into the ring!! By doing it slowly the heat has a chance to dissapate (spread) throughout the mixture so the temperature doesn’t get too high. You are told to swirl or mix it as well, yes?

After a while the whole mixture doesn’t generate much heat and so needs heating to keep it close to 50oC to improve the rate. The Ea may well be fairly high as the aromatic stability is (temporarily) lost.

Overall, I think you’ve got a good understanding of whats happening here.

Re: Q’s given out and no answers. Sorry about that. We are human, but you (Superman) already knew that right – which is why you look after us

There are however a few reasons why we may not give out answers (so casucally at least!)

285 | intechemistry
May 12, 2010 at 6:08 pm

#280, Superman May 12, 2010 at 4:35 pm

You could well be asked that question. The answer would be to remove any (OH)-, (HCO3)- or (CO3)2- impurities which may ppte with the Ag+(aq) and hence make it difficult/impossible to deduce whether halide anions were present.

293 | intechemistry
May 13, 2010 at 7:49 am

#286 girLnexTDooR May 12, 2010 at 8:27 pm

Ok then, just use Br2(aq) is orange coloured.

Other Question:
Re: June 04 U3B Q5, we were testing for the realtive rate of hydrolysis (related to the relative C-X bond strengths) and NOT the identity of X.

In that question, the iodo ws faster than the bromo which was faster than the chloro – but how long did it take the chloro to form??? You might have done this in the lab and might remember that it probably didn’t form! That it didn’t form was good enough because we were measuring time taken for the ppte to appear relative to the iodo and bromo. The lack of ppte for chloro was distinguishable from the others.

So imagine in a new experiment, you are trying to find the identity of some compound. You decide to test for a haloalkane (but it could be anything) If you decided to test for an R-X by hydrolyzing it and you only used water, you might think that ‘oh no ppte’ so it’s not R-X , but it could have been the chloro compound, so when testing for R-X, (OH)- should be added as it will ensure all C-X bonds hydrolyze and it will do so quickly.

Does that answer your question?

295 | intechemistry
May 13, 2010 at 8:21 am

#287, Superman May 12, 2010 at 8:39 pm

There are a number of ways you can discuss it.

Fix the mechanism type. say Sn1.
In SN1, intermediate carbocations form. The ease at which the form is:
(most difficult to form) 1o < 2o < 3o (forms most easily)

For SN2, the steric hinderance factor is important. In SN2, the Nu attacks from behind the C-X bond, so it is very difficult for 3o R-X's to react because of the large/bulky side groups causing steric hinderance. The rate of reactivity is
(most difficult to react) 3o < 2o < 1o (reacts most easily)

Actually for ANY R-x, Sn1 and Sn2 are possibilities and they are in ‘competition’ with one another, so you could ‘weigh’ up the significane/importance/dominance of each type. Doing this, according to George Facer AS book page 271 (the version I have!), OVERALL 3o is the quickest followed by 2o followed by 1o (slowest) at rate of reaction. So the CARBOCATION factor, Sn1, will give the general overall trend.

But I must say, I think such ‘overall’ considerations is very much a simplification!

As for explanations of the ‘greatest charge density’ – ummm, it gives me an uncomfortable feeling to explain it in such a way. It’s certainly ‘unconventional’ to do so – not that conformity is always so great, but the conventional reasons/explanations are pretty much accepted and established already.

297 | intechemistry
May 13, 2010 at 8:57 am

#289, Superman May 13, 2010 at 12:14 am

A 25ml (i.e. 25 cm3) pipette typically has an accuracy of +/- 0.03ml. they are classified as ‘accurate’ pieces of equipment. You can use them for volumetric studies.

Measuting cylinders accuracy depends but they are NOT classified as being ‘accurate’ I think they have a typical accuracy of about 1%. They are fine for adding solutions that are in excess or whose conc doesn’t matter.

I’m struggling to understand what that guide means as well LOL. Don’t worry Q’s re accuracy will give you the necessary numbers for you to calculate their accuracy.

299 | intechemistry
May 13, 2010 at 9:22 am

# 294 girLnexTDooR May 13, 2010 at 7:53 am

Personally for the determination of rate of R-X according to their degree of substitution (3o, 2o, 1o), then for

R-I, I’d use acidified(HNO3) water, aq. ethanol and AgNO3(aq)

R-Br I’d use the same as R-I above

R-Cl I’d be very tempted to use the (OH)- instead of water. as I’d expect 1o and 2o R-Cl’s to hydrolyze slowly.

303 | intechemistry
May 13, 2010 at 10:17 pm

#301 anonymous May 13, 2010 at 3:24 pm

Exexcel made a very vague ‘document’ supposedly about U3 and U6 written alternatives. I must say Edexcel disappoint me with the ‘document’ but here it is…
http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/International-spec-Chemistry.pdf

After, get a better flavour by analyzing the specimen and past years ‘new specification’ papers
http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/Chemistry-Practical-Alternative-SAMs.pdf

Please understand my reservation about ‘predicting’ what will come up on this paper. I’ve pushed it to the ‘back burner’ and haven’t really analysed it myself, but of course it is designed to test practical elements of A2 chem. The labs you did last year are likely to be vey relevant.

The edexcel page for Chem is at:
http://www.edexcel.com/quals/gce/gce08/chemistry/Pages/default.aspx

307 | intechemistry
May 14, 2010 at 7:29 pm

#304 Superman May 13, 2010 at 11:41 pm

Prac 9:
Q4: You remove impurities (soluble and insoluble) by doing the recrystalization. The step at when you filter off the hot solution (see comment 229) using fluted filter paper is the step where the insoluble impurities are removed.

Q5: My guess would be that the bpt of aldehydes and ketones cover a smaller range than the mpts of their 2,4-dnp derivitives. An an example (I think) of this the data given shows the two isomers of pentanone with the same bpt: 102oC. Bpt determination at A-level (the inverted capilliary tube method) may also not give such a narrow range for bpt, unlike mpt for the 2,4-dnp derivative.

Prac 10:
Qi: I don’t have a copy of that prac. The esterification step is likely to exhibit a signigicant equilibrium and hence not all product will be converted to the methyl ester. The recrystalisation process will also cause loss of material as not all recrystalises out of the solution and there is some mechanical loss (sticking to filter paper etc)

Qb. You should apply the answer in reply #283. The ‘cool’ temps are different but seeing as a different compound is to be nitrated, it’s not surprizing the ‘cool’ temp is also different. The Ea’s will be different.

309 | intechemistry
May 14, 2010 at 8:47 pm

#306 Superman May 14, 2010 at 1:07 am

H2O2 as a reducing agent? It has a Std Redn Potential of +1.77 V in acidic soln, so is quite a strong O.A. It’s not usually employed as a R.A.

6a(2) 1(a) Here you make the sandy/beige ppte. Indicating Mn(OH)2(s) formed from [Mn(H2O)6]2+. The interface between the solution and indeed the ppte itself gets darker over time. The H2O2, an O.A., accelerated the process of oxidation of the Mn(OH)2 ppte to dark brown/black solids i.e. MnO2(s). I think the gas is O2 as I believe some disproportionaltion of H2O2 was also occuring.

1(d) Pb and Sn are not on the syllabus, therefore are VERY unlikely to appear in tests and exams. (Some student may never have come across the properties of Pb and Sn before!).

For 1(d), you were supposed to add NH3 until excess – which always means after the hydroxide ppte has been made,: ‘try and make a soluble amine complex’

Aluminium can therefore be dosciunted as it won’t form a soluble amine complex but Zn will.

313 | intechemistry
May 16, 2010 at 8:32 pm

#310, #311 & #312, Superman May 16, 2010 at 4:18 pm

Hi Superman

I don’t think you have to erase all underlined stuff in the question. Certainly I’ve never heard of such a thing before!

You are not supposed to highlight it or use gel pens or fountain(ink)pens as the highlighter (etc) can get through to the other page which MIGHT cause problems when the sheet is scanned.

Underlining reasonably lightly in pencil or biro {i.e. ball point pen} should be OK.

I can’t promise anything but I’m pretty sure you will NOT get a separate sheet for the mcqs. Instead you will probably have to use the ‘cross boxes’ given at the end of the question in the exam script itself.

INTEC give the detachable sheet to make marking easier for us.

As for rough work do it anywhere that isn’t going to ‘overlap’ with where your answers will go. If you need extra space for your answers then make it clear on the script where the other part of your answer can be found.

They usually hold a briefing for students in which all of this is mentioned and you can ask Q’s. Were you not briefed about all this?

As for crossing any wrong(etc) answers out a line or double line is ok. Even free hand. You are not going to lose marks because your freehand line isn’t straight.

Electronegativity is the attraction for a pair of electrons in a covalent bond. Electropositivity is a measure at which a pair of electrons is ‘pushed away’ in a covalent bond.. Note: the electrons are still there as the covalent bond is still there! Electronegativity for isolated atoms is meaningless… where is the covalent bond in an isolated atom?

Electronegativity and it’s definition is v. important and it makes discussion of it’s ‘kind of inverse’ i.e. ‘electropositivity’, rather unnecessary.

Things that affect it are nuclear charge and inner electron shells which shield out the nuclear charge. Similar considerations as to effective nuclear charge and things that effect ionization energy and to a degree electron affinity etc. The books should have a good discussion on electronegativity.

Chemguide is good for this kind of thing. I advise you give it a read http://www.chemguide.co.uk/atoms/bonding/electroneg.html

321 | intechemistry
May 18, 2010 at 7:13 pm

Hummmm
My Virus scanner is blocking me from getting onto http://www.freeexampapers.com/

323 | intechemistry
May 19, 2010 at 7:53 am

#322 Superman May 19, 2010 at 12:53 am

Dear Superman.

1. At the top of this post I’ve put up a ‘snapshot’ of a typical draw a dot-cross diagram…” type question. This question has the word outer in bold. As the old shell, n=3, has gone, you should therefore show what is now the new outer shell, i.e. n=2. Thats the usual form of the Q as far as I remember. You can see from the mark scheme that it also accepts Mg2+ with no electrons, which is what you seem to say you did (yet got no marks for it -boo hoo-). Maybe your Lecturer disagreed with the M.S. on that point. I have some disagreement with it, so I could understand why it was rejected. Maybe your lecturer was in a bad mood? Hopefully you will realise that if you did show the n=2 shell you will not be at risk of losing the mark (according to the wording in this Q, so I’d advise you to stick to doing that, but remember, follow the wording of the particular Q given in front of you, it may result in you having to do somthing different to what we’ve discussed here!)
Actually I advise you to discuss the Q and answer with your lecturer, as here, I’m having to try and read their mind somewhat. Please go see them.

2. If it says ALL the electrons then do exactly that – show (where necessary) the n=1, n=2 etc… up to the point there you’ve reached the new outer shell.
They will probably give you small atoms or one small atom and one medium atom as they don’t want you to waste time drawing loads of crosses and dots.

3. Standard conditions and thermodynamically stable state are not the same. Say you were at 500K and 1atm. The thermodynamically stable state for water would be H2O(g). At 20K and 1atm, the thermodynamically stable state for water would be H2O(s). But at 298K and 1atm, i.e. standard conditions, then the state under standard conditions the thermodynamically stable state for water would be H2O(l).

327 | intechemistry
May 20, 2010 at 8:47 pm

#325 | The smiley little girl May 20, 2010 at 11:31 am

pKa of methanoc acid (formic acid) = 3.75 [1]
pKa of ethanoic acid = 4.74 [2]
pKa of butanoic acid = 4.82 [2]
pKa of propanoic acid = 4.87 [2]
The smaller the pKa the stronger the acid.

Data sources:
[1]. Source:http://www.cem.msu.edu/~reusch/VirtualText/acidity2.htm
[2]. L.D.Wade Jr, Organic Chemistry Fourth Edition ISBN: 013010339-x

First off, methanoic acid, like all first members of holologous series has somewhat unusual properties (due to its v. small size).
In general, the more e- withdrawing substituents (e.g. Cl) on the hydrocarbon chain after the COOH group, the more acidic the acid the carbox acid will be. This is because it can help delocalize the charge on the resulting COO- by spreading some of its -’ve charge down the chain of the molecule onto the e- withdrawing substituent (e.g. as before, Cl). Remember alkyl groups display some e- releasing properties? e.g. 3o carbocations more stable than 1o carbocations?, so the longer the alkyl chain, the harder it is for the COO- to push e- onto the (weaklY) electron releasing chain. Methanoic acid only has a H coming off the COO- and so it is the least resistant against the COO- pushing its electons away.

Relative acidity amongst carboxylic acids isn’t on the specification.

331 | intechemistry
May 31, 2010 at 12:04 am

A good question, one that students often ask.

If you draw one H2O molecule, you can draw 4 H-bonds to other water molecules nearby. Now, look at just one of those H-bonds. It’s shared between two molecules yes? on average then, for that partucular bond, that’s half a H-bond PER MOLECULE. Each of the 4 drawn bonds are like that. So 4 x 1/2 = 2. Water can make 2 hydrogen bonds per molecule.

When I tell students this they often don’t believe me! I’m not sure why. I tell them to draw 10 H2O molecules make the max number of H-bonds. divide the number of H-bonds by the number of molecues and you should get two. They rarely do seemingly prefering not to try it themselves but simply disbelieve me. Strangely, it only seems to happen on this particular issue.

NH3 and HF can do one H-bond per molecule.

335 | intechemistry
June 2, 2010 at 1:09 pm

#332 Sharon June 1, 2010 at 5:51 pm

Try and draw some HF molecules and correctly draw in all the H-bonds you can. You will see on average, there is one H-bond per molecule. NH3 turns out to be the same.

One way to think about it (Note this is a memory aid, NOT a correct explanation) is to use the conditons necessaty for H-bonding to occur, i.e. you need a lone pair AND a H bonded to N,O or F.

lp= lone pair

HF: 3 lp, 1 H
take a lp and the H away, leaves you with
2 lp. Therefore you can’t make any more H-bond with just 2 lp only.

NH3: 1 lp, 3 H
take the lp and the H away, leaves you with
2 H. Therefore you can’t make any more H-bond with just 2 H only.

H2O: 2 lp, 2 H
Twice, you can take a lp and a H away, therefore 2 H-bonds.

337 | intechemistry
June 2, 2010 at 1:21 pm

#334 Superman
June 1, 2010 at 9:24 pm

Is CaCl2 (aq) basic? Would it react with any acid present?

339 | intechemistry
June 3, 2010 at 5:56 pm

#338 Superman 3 June 2010

1. To be honest I’m not sure about the thermal props of diamond and graphite. I have no (trusted) data.

Metals are said to be good conductors of heat due to their their delocalized e- being able to transmit kinetic energy (i.e. heat) throughout their structure.. In graphite those e- are pretty much stuck within the planes so perhaps like the electrical conduction, graphite may direct heat in one particular direction. OR the hear may dissipate by ‘waving’ of the planes. I simply don’t know and I can’t find it specifically mentioned in the AS syllabus, although Topic 1.6 (Bonding) 2.a.i might allow them to ask about it, but it would be extra mean and strange of them to ask. Topic 2.3 (Shapes of molecules and ions) part e makes me think it is unlikely that thermal properties will be asked for esp (as you can see below) as many books don’t mention it. Just in case however, read the Hodder entry (again below).

I had heard that diamond was a good conductor of heat which in the case of diamond HAS to be because of it’s lattice (that’s all there is in diamond!)

Pearson AS P150
“It [diamond crystal] is a poor conductor of heat”

Hodder AS P123
“Diamond conducts thermal energy very well – 5 times better than copper”

Richard Parsons AS Revision Guide (ISBN: 978-1-84762-124-5) P66
“…it’s a good thermal conductor”

AS Revision guide – no mention.

George Facer AS (2nd Ed) No mention

George Facer AS & A2 Revision Guide: No mention.

I think Pearson is wrong (another error in the book)

As for graphite, there’s even less mention of its thermal properties.

Generally thermal properties of the elements is not important, or should I say not as important as the electrical properties {of graphite}. I guess there are more important things that Edexcel would like to test you on with regards to bonding and structure properties.

2. As for the profile of a catalysed rxn, you have to show the energy level of the intermediate (catalysed or not) as it is central to the discussion and whole point of drawing the profile. Profile diagrams DO have an x-axis. As you correctly identified, it’s called “reaction coordinate” (or ‘reaction progress’ in baby talk). As for which hump of two (or more!) is the highest, it is usual to make the first hump the highest although really, the point is that any one of the humps in the catalysed reaction, which ever you draw as the highest, will be lower than any other of the highest humps in the uncatalysed reaction. That may sound quite horrible so to make life easy, draw the first hump in a cat. rxn. as the largest – unless the Q suggests otherwise {— which is not very likely

341 | intechemistry
June 3, 2010 at 6:33 pm

#340 Superman June 3, 2010 at 6:21 pm

Anything with uneven e- distribution will have polairy. The db in the alkene is e- rich acts as a delta(-) centre compared to the rest of the molecule. You wouldn’t just look at the C=C but the whole molecule. The regions of different e- density take place over relatively small molecule so even if just weakly polar does lead to an observable effect. Remember studying oils and fats? The significance of a db dimished when the molecules gets bigger so the oil/fat is said to be unpolar.

You called cyclohexene polar but it is probably very poorly polar relative to other polar solvents.

347 | intechemistry
June 15, 2010 at 12:06 pm

Hi Sharon.

When you refer to the blue solution, I think you are refering to Fehlings solution, a [Cu]2+ complex. It is different from the Tollens silver mirror test which uses a silver diammine complex, so don’t mention anything to do with blue. As for Tollens silver mirror test in the presence of a ketone “No siilver mirror” produced will be ok.

If you were using Fehlings solution and you had a ketone (while say, testing an unknown carbonyl) then “No red precipitate formed” would probably be fine. For edexcel U6B you would probably be expected to add “carbonyl cannot be oxidised” as well.

349 | intechemistry
June 15, 2010 at 1:22 pm

#345 Superman June 13, 2010 at 6:10 pm

From what you say, you might be right. I don’t have that Pearson book right now, but I’ll try and check it out ASAP. Perhaps it was because if the order w.r.t. the other 2 reactants was 2 each (i.e. overall order was 4) then the least number would have been three. The number of steps up to and inc. the RDS isn’t necessarily the least number of steps in the mechanism (as a whole)

353 | intechemistry
August 4, 2010 at 9:11 pm

66.66 cm^3 ???

361 | intechemistry
August 7, 2010 at 11:33 am

Aaaah, sleep.

OK. Divide the % by mass by the atomic number as one would do for empirical formula calculations.

Ca 25.64%, C 15.29%, N 17.93% and for S 41.14%

divide by: 40.1, 12.0, 14.0, 32.1 respectively gives…
0.639 1.27, 1.28, 1.28
divide by the smallest number (0.639) gives:
1, 1.99, 2.00, 2.00
i.e.
Ca 1, C is 2, N 2, S 2

Thats empirical formula. The rest of the steps to molecular formula are trivial.
The molecular formula is Ca(SCN)2 Calcium dithiocyanate.

The key step is getting the % by mass for the relevent atom. You do this for each of the samples for the particular atom.

363 | intechemistry
August 11, 2010 at 10:38 pm

I forgot to offer you the spreadsheet that I used to calculate the initial % by mass numbers. Did you manage to do that or do you think you would want the spreadsheet?
p.s. It would porbably take some time to get to understand the spreadsheet I made.

365 | intechemistry
August 13, 2010 at 1:20 am

Here ya go. You’ll have to un-embed it. WordPress doesn’t like stand-alone spreadsheets!

http://intechemistry.files.wordpress.com/2010/08/pandoras-questions-1-spreadsheet-embedded.doc

367 | intechemistry
August 15, 2010 at 3:15 pm

1) The physical driving force for the way the universe works involves all physical entities (or objects) ‘giving off’ a special energy called entropy, symbol S. In the process, the ‘noticeable’ energy’ i.e. the enthalpy, symbol H, of these objects usually decreases*. There is a relationship between these two energy types.

As a very simple and general statement, we can say ‘things like to be low energy’. Let go of an object – see it fall and lower it’s energy. Objects already on the floor don’t suggenly start floating upwards an increase their energy. Hence when a chemical rection takes place, the enthalpy (noticeable energy) usually falls with the effect that more entropy is given off.

2) The answer is actually above. The products of Endothermic recactions have absorbed heat energy (i.e. enthalpy). They have the potential to release that energy again somehow; Often by coming into contact with a different molecule inducing the reaction, or by the supply of heat – inducing a decomposition reaction. Supplying heat overcomes a compounds ‘resistance’ called the ‘Activation Energy, symbol Ea) to undergo decomposition or/and reaction.

3) Well, yes, if it’s talking about the energy change, i.e. the difference in enthalpy, Delta(H). Delta meaning “a (large)change in”. The Delta H value is responsible for the therodynamic stability (I don’t think I’ve heard thermodynamic stability called energetic stability before which is very misleading). Thermo=heat, dynamic=in motion i.e. change.

4) If you supplied the Delta(H) to a compound it will not revery back its elements. This is because of the activation energy for the reverse reaction. It is bigger than the actication energy for the forwards reaction. Forwards or backwards, this energy must be present (sometimes you have to supply it) for a reaction to work. For an exothermic (forwards)reaction, the enthalpy needed to be put in to allow for the possibility for the compound to revert to its elements [actually a new compound is often formed instead!] is greater than the enthalpy given off in the first place. this is called kinetic stability. It can be discussed for forwards or reverse reactions. You should look at AN ENERGY PROFILE DIAGRAM to ‘see’ all this.

5) A bond is an attraction between the nuclei and electrons. IF the bond is longer, the e- are distributed over a longer distance. The attraction between the +’ve nucleus and the -’ve electrons is therefore decreased.

any errors/typos in this will have to be rectified later. I’ve not no free time left now.

* An abject can move to a higher increased ‘noticeable energy’, i.e. to a higher (internal) enthalpy, but the result will be that somewhere, this will be paid for by the release of a greater amount of entropy.
Note: You will study entropy in year 2.

369 | intechemistry
September 1, 2010 at 11:09 pm

It’s not a ridiculous question so dun wori
The x,y & z are axes simply act as a fixed reference point so that the shapes of the orbitals relative to each other can be discussed with ease.

If you wanted to, you could indeed orientate the atom so that the usual dz2 orbital could lie along the x or the y axis, therefore calling it the dx2 or dy2 orbital respectively. If you did this, all the remaining orbitals would therefore have to be relabelled as their electron density would now be orientated along different axes.

So we stick with one way of naming them for the sake of clarity.

371 | intechemistry
December 11, 2010 at 12:58 pm

Salam Salha. I’m fine tq as I hope you are.

I’m guessing the author of that sentence is thinking something but didn’t really say enough about it to enable the reader to really grasp the thought.

The previous bit was explaining that reactions took place and ions were the product, so I think the opening sentence in para2 means ions don’t always have to be the product of a chemical reaction, that you can get covalently bonded products as well.

Glad to see you here. Please take maximum opportunity to use this blog over the next month to get all your queries resolved

373 | intechemistry
December 11, 2010 at 4:14 pm

If you live locally, drop in and we’ll go through the test.

379 | intechemistry
May 9, 2011 at 1:38 pm

Salam.
anhydrous Na2SO4 is usually used for drying liquids (like what we did (or some of ud did) after liquid-liquid separations) and is good for that purpose.

Anhydrous CaCl2 is better used for drying gases. Does that case match with when it’s accepted/rejected?

If, perhaps, CaCl2 is rejected when drying a liquid, it might be because the Ca ion or Cl ions that would be produced when absorbing the water, might react with something and form a ppte, e.g. AgCl or Ca(OH)2) or some other case like that. If you give me the Past years papers this comes from then I’ll check it out.

387 | intechemistry
May 20, 2011 at 8:11 pm

Hi Xin Ling.

1) Yip. Those tests look good to me.

2) You can do mass or moles (being a ‘moley’ kinda person, I much prefer moles), sometimes one way is more efficient than the other. Possibly, the Q might steer you to answer by a particular way, but it would be rather trivial to do so.

3) Mark schemes have been in the past ‘flexible’ on this in the past, but please do what is correct, i.e., use Roman numerals for oxidation numbers and ‘arab/english’ numbers for ionic charges. My guess is that Edexcel are not too worried about as it may be confusing (?) for some Iodine and Vanadium oxidation states.

4) Use the one on the left, it’s better (no rubber or cork to get eaten away by solvent or degrade with heat) but it’s not really that important an feature. Why do you say you are NOT required to draw one-piece apparatus? What do you mean by one-piece apparatus and where did you come across that? Do you mean apparatus that ‘looks’ like it’s one piece? I’m pretty sure Margaret Cross (ex-chief examiner for chemistry) said drawing like that – the left hand side diagram – is OK even though “it looks a bit funny”.

389 | intechemistry
May 21, 2011 at 12:58 pm

#3
I mean like the Oxidation state of iodide is minus one, i.e. O.S. of I- is -I and oxidation state of vanadium +5 oxoion is plus five, i.e. O.S. of VO2+ is +V, here the symbols have similar appearance to the chemical species.

#4 The best way I’ve seen to try it make like different pieces of equipment are shown HERE or at the top of this page. I’ve used colour coding to help you see what’s going on. Notice the female necks widen out and the male parts narrow in.
I think the “examiner’s tips” is there just to tell students at the already/potentially ‘funny’ looking diagram in the book that the diagram doesn’t show one piece but it may look like it. I don’t think it’s a kind of “if you draw it looking like one piece you won’t get any marks” so no need to worry. showing it as one is fine as its a quite well established way of doing it; The examiners will accept it.. Also when everything is fitted together with no air gaps, in effect, it does become ‘one’ piece of glassware.

As for the drying agent, it kind of depends on what you are trying to do and how much water is present. For example, if you had a non-polar organic compound that may have some small quantities of water in it, then you could use anhydrous MgSO4 as that substance (ionic) will not dissolve into the organic phase. If you were trying to dry a gas, you might want to bubble it through c.H2SO4 as it’s a liquid. You wouldn’t pass the gas over a solid, but you could use a solid drying agent in a desiccator, e.g. silica gel. If you are doing elimination of water (i.e. removing the elements of water from a compound and not removing ‘free’ independent H2O molecules) you could pass it overheated Al2O3 in a sealed tube. etc.,,

It might benefit you if you look look for cases where H2O is to be removed (chemically (e.g. elimination) or physically (e.g. drying)) and note the drying agent responsible.

391 | intechemistry
May 22, 2011 at 3:35 pm

Extract from Advanced Practical organic chemistry, 2E, By J. Leonard, B.Lygo and G. Procter Pub: Nelson thornes . ISBN: 0-7487-4071-6

5.3 Drying Agents, p55-59

Drying agents fall into two categories, those used for preliminary drying and the drying of extracts [At A-level, drying agents are almost 100% used for drying extracts. Other techniques to remove (large)quantities of water will be physical techniques, i.e. distillation, decantation, or liquid liquid extraction], and [secondly] those used for rigorous drying [< not covered at a-level]. The pre-drying agents are largely interchangeable with each other and the choice is usually limited only by the chemical reactivity of some of the reagents.

…

CaCl2
Both the powder and pellet forms are effective for pre-drying hydrocarbons and ethers. It reacts with [and hence isn't suitable for drying:] acids, alcohols, amines and some carbonyl compounds

CaSO4
It is only suitable for drying organic extracts.

Mg
Suitable for methanol and ethanol

MgSO4
The monohydrate [MgSO4.H2O] is fast acting and has a high capacity (forms a heptahydrate), making it the dessicant of choice for organic extracts. It's slightly acidic so care is required with very acid sensitive compounds.

Molecular sieves
These are sodium and calcium aluminosilicates which have cage-like lattice structures containing pores of various sizes, depeding on their constitution… after activation… they are probably the most powerful dessicants available.

P2O5 (or P4O10)
A rapid and efficient dessicant but limited by it's reactivity. It reacts with alcohols, amines, acids and carbonyl compounds. useful for drying acetonitrile and may be used for drying hydrocarbons.

KOH
A good drying agent for amines and pyridines but inferior to calcium hydroxide. Should not be used with base sensitive solvents.

Na2SO4
a weak drying agent, suitable only for drying of extracts. It is preferable to magnesium sulfate for drying very acid sensitive compounds

393 | intechemistry
May 22, 2011 at 11:39 pm

399 | intechemistry
November 7, 2011 at 11:35 pm

Salam.
It can act as a buffer but a very poor one.
For a weak acid, you know this: HA (H)+ & (A)-
the very small amount of A- that’s produced when the weak acid ionizes to a low degree, can actually react with any added H+.

On addition of H+ ions, the eqm will shift slightly to the LHS so we have less ‘free H+ ions’ than would be the case if there was no A- to absorb (some of) these protons, Therefore the pH will not fall as much as might be expected,.

But because the amount of A- is tiny, the small amount of added H+ is easily going to be overwhelmed/consume the A-. Once the A- is consumed, then there is no more buffering effect. The buffering ability here so small that has no real significance.

Hence buffers require a LARGE RESERVOIR of HA and the salt (which produced A- on dissolution i.e. on dissolving)

p.s. remember a good buffer has equal large amounts of salt and it’s weak acid (or base).

Hope that helps,

407 | intechemistry
April 28, 2012 at 3:09 pm

Salam erly.

I prefer “chemistry” questions. That one they hit you with isn’t really a “chemistry question”. As I’m a chemistry lecturer I can only guess at this, but don’t’ forget ordinary healthy food also contains iron! So the (single)tablet is acting as a booster to get to the minimum amount. A similar reason exists for why they don’t make a pill that gives the maximum dose as intake of ordinary feed after that will push you beyond the maximum level.

Getting back to the minimum level again. one can increase the dose by simply taking another table if one desires. to go above the min recommended. So in general single tablets contain a dose lower than the minimum recommended dose. Having a lower amount per tablet allows greater flexibility in prescriptions.

When you take tablets, you often take two or more, yes?

Hope that helps.

Note: This question is FAR better than the “Is testing pharmaceuticals on animals?” question, which if you said “No, because it is cruel to animals” you would NOT get all the marks – penalizing you if you sense if morality is greater than your sense of making a fast buck.

409 | intechemistry
April 29, 2012 at 1:10 am

I totally understand what you say.

“to answer this kind of question seriously I think we need general knowledge,” – Very much agreed. It’s not a question suitable for A-level chem IMO.

The specimen papers are a bit weird. In some ways I think it’s a bit harder than the actual papers. There is a LOT of ‘explain’ type questions on the specimen – which actually isn’t such a bad thing for practice and the (kind of)model answers made available for those kinda Q’s is something to welcome.
So, sweet and sour I guess.

411 | intechemistry
April 30, 2012 at 8:54 am

They may change style true, but we can be sure that the stuff they will be asking about will not change – it’s still going to the content of the specification.

A different emphasis of the same material shouldn’t cause much problem if understanding of the knowledge is fairly sound. It’s why students should be able to explain and understand things from different perspectives, which doing millions of Q’s will have given you exposure to.

Best blessings to you for the forthcoming exams.

413 | intechemistry
May 4, 2012 at 12:32 pm

Essentially it’s how the theoretical knowledge you learn gets applied. It’s best to expose yourself to as many of these applications as possible so you bridge the gap between “theory and application” more easily. E.g. redox fuel cell and I.R. and reduction of acidified dichromate all used in breathalyzers.