General questions
General chemistry questions can be posted here. This is to reduce ‘clutter’ of the main blog. I predict a day when putting all questions here may problematic, but for now I think it’s the right thing to do.
I would respectfully ask that you please include a brief summary of your understanding on the matter at hand – It is helpful for me to know your persepective on what you would like to discuss.
Note: The original diagram had only one S(s). It was an error but didn’t affect the calculation and final answer.
192 Responses to "General questions"
November 4, 2008 at 6:52 am
Hello Haziq and thanks for your question.
Discussions of NaCl and NaI boiling point is usually asked in a relative context i.e. comparing the bpt of one substance with a different substance, and is normally answered by identifying the differences between the substances, then describing how that difference is responsible for the change in bpt. But your question is an ‘absolute’ one, i.e. probing the fundamental reasons as to the magnitude of the boiling point, of which both NaCl and NaI are described as having a high bpt, and so requires a more detailed answer.
NaCl
Melting Point (°C)=808
Boiling Point (°C)=1465
data
NaI
Melting Point (°C)=661
Boiling Point (°C)=1304
data
All mpt’s and bpt’s are determined by forces of attraction.
If a substance has a high mpt or bpt it means the force of attraction in that substance must be strong.
When something melts, the forces of attraction are weakened (note: there is still some attraction) to the point where individual particles making up the substance, can move over each other – i.e. they can ‘flow’
When something boils, the forces of attraction between the individual particles in the substance are completely overcome and the particles move apart. How does this happen? The heat supplied causes the particles to vibrate. On boiling, the energy of vibration is greater than the energy of attraction so the particles are able to fully move apart.
In the case of NaCl, we have forces of ionic attraction between Na+ and Cl- ions. In ionic substances, the size and the charge of the ions determines the physical attraction.
The greater the charge the greater the attraction and the smaller the size, the greater the attraction.
There is an additional factor of how the ions are arranged in their solid form (i.e. the lattice or crystal – but both are Face Centred Cubics) but discussion of that is beyond the scope of this blog.
NaCl and NaI have the simplest of all charges (+1 and 1) but the size of Na+ is quite small. If a main reason was to be given for a high bpt, in this case it would be small size of the anions.
Here’s a simple illustration of ion sizes which you might like to see…
http://www.avon-chemistry.com/p_table_ion_rad.jpg
Does this help answer your enquiry?
December 29, 2008 at 7:25 am
for practical exam-identifying the unknown compound:
the appearance of the solid is blue.
through tests we identified that copper(II) ions and sulphate ions are present.
when we suggest the formula of the compound, is CuSO4 acceptable? or it must be written as CuSO4. 5H20?
December 29, 2008 at 7:47 am
I don’t think you are expected to know the degree of water of crystalisation in a hydrated salt, certainly not on the Edexcel syllabus.
CuSO4 is usually going to be sufficient.
However, if you were supposed to calculat the amount of water per mole of CuSO4, then in that case, you are very likely to be expected to add the .5H2O (or whatever number you calculated) in your answer.
December 29, 2008 at 1:10 pm
Salam~
sir , i hv one ques. why is the hydrogen halides are water soluble? can we say that it is because the hydrogen halides is polar?
Thanx!
December 29, 2008 at 1:43 pm
Sir,
Could we have the exact definition for Relative Isotopic Mass and Relative Molecular Mass as there are numerous different definitions for them and we would like to know the exact one.
Thanks..!
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.
December 29, 2008 at 3:27 pm
@ 8m2 December 29th, 2008 at 1:43 pm
Sure thing.
The definitions you find usually mean the same thing, just expressed slightly differently. Having said that the definitions in Edexcels ‘the moles’ book is rather basic and I think they would lose a mark in one of their own exams!!!
Try and stick with these:
The definition of relative atomic mass Ar is:
The weighted average mass of atoms of an element on a scale on which the mass of an atom of carbon12 has a mass of 12 atomic mass units.
The relative atomic mass does not have units.
The definition of Relative Molecular Mass Mr (also referred to as Molar Mass) is:
The average mass of a particular type of molecule on a scale on which the mass of an atom of carbon12 has a mass of 12 atomic mass units. The relative molecular mass does not have units.
The definition of Relative Isotopic Mass is:
The mass of an atom relative on a scale on which the mass of an atom of carbon12 has a mass of 12 units.
December 30, 2008 at 2:24 am
salam~
sir , if we want to explain about isotopes having the same chemical properties , can we say that it is the outermost electron which determine it? or is it enuf if we only say that is because they have same num of electrons? *confused*
thanx!
December 30, 2008 at 10:43 am
sir..for Qs like:
name the reagent and condition needed to convert X to Y.
if it involves a catalyst, is the catalyst in the category of “condition”?
if it involves solvent, for example conversion of halogenoalkane to Grignard reagent, the dry ether is supposed to be one of the reagents?
hmm…second thing..i’m quite confused with the definition of standard enthalpy of combustion, the part “…is burnt completely in excess oxygen under standard conditions”–when a substance is burning the temperature can’t be 298K..or..am i wrong?
“fix” me sir.
thanks.
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)
December 30, 2008 at 1:45 pm
@ 10 | wen December 30th, 2008 at 10:43 am
“fix me sir” – LOL!
A reagent is a rather loosely used term.
The word reactant almost always used with regards to a specific compound.
It seems to me that the word reagent came about to describe a mixture which contains at least one reactant, which were used to test for some compound.
E.g. Tollen’s reagent (ammoniacal silver nitrate solution}, Brady’s reagent {a mixture of methanol, sulphuric acid, and 2,4-dnp}. Having said that, “Grignard reagent” – probbly the most frequently encountered use of the word reagent at A-level, isn’t really used as a chemical test. If anyone knows better, please let me know!
Edexcel don’t seem to make any distinction, using them interchangeably, not many people I know, actually bother to make the distinction either.
However, a Malaysian Lecturer who did her studies in the US, seemed surprised when I dismissed any difference between reactant and reagent, when I said they were “essentially the same thing”. Maybe they are more a bit more precise in the US.
Catalysts I think come under the category of conditions (as they are not consumed), although often I think Edexcel have asked about catalysts outside the scope of conditions (but I have a horrible feeling they have asked about a catalyst under the heading of conditions also). It’s something I have to check out if I have time.
Solvents are a condition, not reagents, as they are not consumed.
If there isn’t a part of the question specifically talking about catalysts yet they can be discussed somewhere in the question (e.g. reaction conditions are mentioned only when talking about the Haber process) then obviously the thing to do is mention the iron catalyst in the conditions part.
Re: combustion and 298K. Good question. I wondered about this too when I was a youngster. When you burn methane the temperature quite obviously isn’t 298K. The equipment that measures the heat change in combustions, like a bomb calorimeter, measures ALL the heat given off from the start (298K) to the point where 298K is attained again by the finished reaction, so you start and 298, and finish recording heat output when you get back to 298K. That’s how they measure combustions under standard conditions. Neat, huh?
Are you fixed now?
December 31, 2008 at 12:09 pm
Hi Mr Allan, [name corrected by Admin]:
i got another question for you. =)
The question is on explaining why HF acts as a weak acid in dilute solution compared to other hydrogen halides. Should we explain in terms of covalent bonding in HF or hydrogen bond between HF molecules – that will cause less dissociation.
December 31, 2008 at 12:10 pm
Hi Mr Allan,
i got another question for you. =)
The question is on explaining why HF acts as a weak acid in dilute solution compared to other hydrogen halides. Should we explain in terms of covalent bonding in HF or hydrogen bond between HF molecules – that will cause less dissociation.
December 31, 2008 at 2:46 pm
It’s the high strength of the H-F bond inside the HF molecule itself that primarily determines the acid strength. There may be some secondary arguements that may modify the compounds acidity (such as identity of the ‘delta+ H’ attracting species etc.), but to treat this question properly, one should really examine the thermodynamics of the process and I don’t have time for that (plus my thermodynamics is in need of some revision), so please forgive me if I say bonding other than the INTERmolecular H-F covanent bond are not relevant.
{your mention of dilute solutions makes me raise an eyebrow!}.
It’s ok about the name. The ‘n’ and ‘h’ are close on the keyboard. But I’m sure you can understand why I had to change it.
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.
January 2, 2009 at 2:56 am
Hi Mr Allan,
Happy new year!!!
For question of identifying NO2 gas, the identification should be ” brown gas which turns damp blue litmus paper red” right? But i had came across ” turns starch- iodine paper blue-black”. Is that an acceptable answer?
And another thing, i came across two colours of Fe3+ in a copy of analytical investigation. It says brown or violet. The MS for U6A 21Jan 2004 Q3a)iii) also said the solution formed is violet. I am confused… Is it something to do with ligands?
And the last question is…
what is the condition for nucleophillic substitution of HCN on carbonyls? Is it acidic or alkaline? I came across both pH5 and pH8.
thanks a lot!!=)
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.
January 3, 2009 at 2:44 pm
salam..
i’ve a question..its frm jan 06,Q4 (b) i unit 2..” draw enough of the chain of poly(propene) to make its structure clear”
should we drawn as one repeating unit or a continuous chain?
ex : —[-CH2-CH(CH3)-]n— or
—CH2-CH(CH3)–CH2-CH(CH3)–CH2-CH(CH3)–
which one should i choose? still confuse til now..
January 4, 2009 at 8:31 am
hi mr allan,
im not sure whether this question falls under this category. anyway, i have one simple question regarding the flame test. Usually in da exams or practicals, we luckily only got lilac flame for almost all experiments. But, what if we got red coloured flame? I know there are 3 metal ions whuch have 3 red colours, which are calcium(brick-red), strontium(crimson) n lithium (scarlet).. However, it is definitely not easy to defferentiate the 3 colours. So, what is the most safest answer to write if we got a red flame?
btw, thanx for correcting my mistake about the lab practical we did. I think the mark scheme I have is for the other gp.
January 5, 2009 at 12:01 am
If it’s not a past years Q, then don’t worry, this is as good a place as any to put it.
If you got red, there would either be information in the question to help you decide which one it was, e.g. Unknown X is the salt of a group 1 metal., or the thermal stability of this is the lowest for the s-block… …Something like that
OR
At some stage, you may have done chemical tests. E.g. add hydroxide ions or carbonate ions to the solution of the unknown, try and thermally decompose it… etc.
My advice: Try and make a plan how to identify whether an unknown solid contains LiCl, CaCl2 or SrCl2
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.
January 6, 2009 at 4:01 pm
ok thanks.. yup i draw a structure.. so if this type of q’s appear i should draw only ONE repeating unit with the n at the end of the structure eh..
January 6, 2009 at 4:25 pm
That should do. There’s no indication that you should draw two (or more) units
January 7, 2009 at 1:26 am
Salam sir..
1) What is the difference btw “initial rate of reaction” and
“rate of reaction”? Is there any difference? or merely to make the former sounds nicer?
2) I came across a Q (actually it’s in one of our topic test, dated 15Sept08) with the equation
CH4(g) + 6H20(s) –> [CH4(H20)6] (s)
and we hv to write Kp expression.
Kp involves gases but in the above equation, water, and methane hydrate exist as solid
Shd we include them in Kp?
Or shd the Kp be
Kp = 1 / p(CH4) ?
January 7, 2009 at 4:26 am
Salam 2u2.
Initial rate is a special ‘once only’ value. ‘Rate’ without being proceeded by the word ‘initial’ is a gereral term of any rate at any particular moment in time.
As a reaction proceeds, the rate changes – becoming smaller. The initial rate is when the rate of the reaction is at its fastest, and from this data, we can work out order of reaction. See initial rates mothod.
Do NOT put solids in Kp expressions. Kp talks about gas pressure, so only terms relating to gases will be in a Kp expression.
Kp = 1 / p(CH4) is correct.
January 7, 2009 at 8:10 am
May i know, is the answer above by the real Mr Allan? Or an impostor? Juz wondering, as the name used is not same as previous names..
And there had been impostor cases b4, as faced by LC.
Bid forgiveness if it’s the real Mr Allan.
January 7, 2009 at 9:45 am
Don’t worry. It was me. Nobody could copy my poor spelling 
LOL
January 7, 2009 at 5:18 pm
Hi sir, will CO2 dissolve in water assuming we put CO2 and water in say, an inverted burette held up in a beaker of water? I mean will it dissolve significantly enough that that technique will not be a good one to see how much co2 is produced? what about H2?
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.
January 11, 2009 at 9:20 am
Sir,
For chemistry equilibrium, why is it that the equal forward and backward rates causes the equilibrium concentration to remain constant?
Rate is defined as change in concentration over change in time so…
if we were to set it over a specified amount of time eq 3 secs:
Thus is the amount of reactants used for the forward reaction and the amount of products formed (which are the reactants for the reverse reaction) equal?
Thus why are the amounts (thus concentration) equal at equilibrium as rate is defined as change in reactants ( for both forward reactions and reverse reactions respectively)… which would not mean that the products formed are equal to the amount of the corresponding reactants which formed it….
Which would lead the amounts of both reactants and products not being constant thus not at equilibrium!
Your help sir would be TRULY appreciated and i hope this could be replied ASAP as it is always at the back of my mind.
Thanks sir!!!
January 12, 2009 at 2:01 pm
hello sir,
i’ve come across a number of questions which are very similar to each other..and after a while they confuse me.
basically, the question asks us to give equations for the reaction of NaCl and HCl with water. So in these equations, obviously the NaCl dissociates in water to give it’s Na+ and Cl- ions right.. and, HCl to give Cl- and H3O+ ions.
what confuses me is that in several mark schemes that i checked, for these kinda questions that ask us to EXPLAIN what happens, they say the NaCl is ionic so it dissociates in water, but HCl is covalent so it REACTS with water to form hydrated ions. why is it just because HCl is covalent so it REACTS?
is it because the energy liberated during covalent bond breaking of HCl in the water need to be compensated for by forming new bond with water molecules?
tq sir~
January 12, 2009 at 4:37 pm
@ comment:32 alm January 11th, 2009 at 9:20 am
Hi ‘alm’
We’ve already discussed this face to face, but this reply is given for the benefit (?) of all.
For any given reaction, the equilibrium equation is indeed a ratio of the rate for the forwards reaction and the reverse reaction. i.e. the rate equations divided by eachother according to a convention.
At eqm, the concentrations of reactant and prod are constant but don’t have to be equal! Most equilibria under standard conditions point to the left(reactants side) or to the right) products side)
I tried today to see if I could try and show this mathematically in a page or two. I couldn’t. The mathematics is quite detailed, and it’s been quite some time since I’ve played around with it, that I’m quite rusty.
But at equilibrium the two rates are equal, which is why neither concn changed any more.
The constant value the rate ratio’s evaluate to, does depend somewhat on the stoichiometries of the reactants in the equation, but that is simply because the parameters which make up the order of reaction simplify at equilibrium, and effective only leave the parts that are linked to the number of species in the coefficinets in the balanced chemical equation, which then go on to appear in the equilibrium equation.
for A-level equilibrium is almost completely discussed without reference to rate, other than appearing in a ’super duper’ version of the definition of dynamic equilibrium.
When discussion equilibrium, its v. rare to have to treat eqm in a mathematical sence that involves the rate equations.
If noone has a clue what I’m talking about, it doesn’t matter one bit. Treat kinetics and eqm exactly as you did before and there won’t be any conceptual problems.
I don’t like saying things like that but as being of limited knowledge in regards to every single thing that ever is, it is often (but not always) better for us to take the simplest way around a problem.
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? 
January 19, 2009 at 7:36 am
Can we, or can we not, use pencil for drawing structural formulae, drawing mechanisms, and labelling stuff (e.g. Ea) on graphs?
The front page says DO NOT USE PENCIL
But will they actually penalise if not allowed for drawings, and the pencil is really black, after all they are scanning it in and it looks just like pen?
January 20, 2009 at 6:13 am
Once more apologies for the late reply. Got loads of things to do.
Pencils are allowed for sketches and graphs (joining data points). The instruction DO NOT USE PENCIL refers to written (as opposed to sketched) answers.
I’ve never seen an Edexcel directive saying ‘penalise if they use a pencil; so it shouldn’t be a problem.
They don’t like pencil as it may reflect/scatter the light of the scanner and give poor sanned images.
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.
January 21, 2009 at 8:45 am
Acid1 and base1 appear on the LHS, acid2 and base2 on the RHS, even though there is nothing ’sacred’ about this terminology.
e.g.
CH3COOH(aq) + NH3(aq) = CH3COO-(aq) & NH4+(aq)
CH3COOH(aq) is acid1
NH3(aq) is base1
CH3COO- is base2
NH4+ is acid2
These days, Edexcel have moved away from this rather silly/clumbsy way of asking the question and seem to have replaced it with “name the acid/base conjugate pairs
Using the example above CH3COOH(aq) is the acid CH3COO- is its conjugate base
NH3(aq) is the base and NH4+ is its conjugate acid
January 22, 2009 at 2:29 pm
“Acid1 and base2 appear on the LHS, acid2 and base2 on the LHS, even though there is nothing ’sacred’ about this terminilogy.”
Sir. i guess u were in a rush.
Anyway, the “easy approach” to it is..
base 1 react with acid 1
base 2 react with acid 2
that means acid 1 produce conjugate base 2
acid 2 produce conjugate base 1.
Is it correct?
January 22, 2009 at 2:52 pm
UMPH!!! Yes… *aaagh*! I was in a rush… I typed the wrong number. When the heck are we going to get around to telaphy? I think light years quicker than I can type, or even worse.. write!
Acid 1 and Base 1 on LHS produce Base2 and Acid2 on RHS
conjugates on RHS, ‘cos the equation is written as them appearing as products.
acid 1 produce (conjugate) base 2
base 1 produce (conjugate) acid 2.
(I’m going to edit the error after posting this)
March 8, 2009 at 3:30 pm
sir,
i was wondering
1) lattice enthalpy values in practical is less exothrmic compared to the theoretical one due to the covalent charactristic…why does this happened? why covalent chracter makes energy less being released?
2) how can we determine the slow step(rate determining step) from the rate equation….is it from the order of the reaction? for example :
rate = [HA]”[M]‘….in rate determining step, do they both will involve or there’s other factors to determine it? i’m very weak in this topic…
3) in organic chem, sometimes there are more than 1 functional group…what is the easiest way of determining which funtional group will be invovled?
help me….
thank you…
March 8, 2009 at 3:45 pm
just now i asked u about covalent character affects the lattice entalphy value….in my understanding, less energy release is due to less strength of the bond….is the covalent character makes the ionic bond weaker? so less compact to form solid….therefore less energy release…is it?
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.
March 9, 2009 at 12:40 am
sir,
according to your answer for rate determining step, the last example…
the overall equation is HA + 2M -> P
then the steps suppose to be
HA + M -> T (slow)
T + M -> P (its suppose to be single P, right?)
March 9, 2009 at 4:34 am
Aaagg! I left off the 2.!
In my mind the chemical eqn was: HA + 2M -> 2P
in which case my mechanism was correct.
But if, as I actually wrote, the chem eqn was HA + 2M -> P, then your correction for the last part of the mechanism is correct i.e.
T + M -> P
Sorry for that.error. Good that you spotted it. It shows you have good understanding.
March 14, 2009 at 11:02 am
hi….
can i ask u some Qs abt aluminium compound?
1. why AlCl3 is covalent and Al2O3 is ionic?Al is a metal and chlorine is non-metal..how can al-cl form covalent bond?
2. why anhydrous AlCl3 is ionic and hydrated AlCl3 is covalent?
3. how AlCl3 can form Al2Cl6 when it is heated?
March 14, 2009 at 1:36 pm
mia. Hello.
1) a) data:
dH(f for Al2O3) = -1675.7
dH(atom for Al) = +326
sum of 1st IE, 2nd IE and 3rd IE for Al = +5138.9
dH(atom for O) = +249
sum of 1st IE and 2nd IE for O = +657
LE for Al2O3 = -15,263.5 kJ mol-1 !!!! This number is hugely exothermic!!! indicating it’s v.favourable to be ionic. A covalent structure isn’t likely to be anywhere near this. Now we need to think about polarisability.
O2- is about 0.16 nm in ionic radius
Al(3+) is about 0.05 nm in ionic radius – which is v. v. small!
Despite Al(3+) being incredibly polarising, O2- is very small and is very poorly polarisable, O2- is arguably the 2nd most small common ion (F- is smaller), so Al(3+) can hardly polarise O(2-). Some polarisation occurs, but not isn’t dominant, so a high degree of charge separation still remains. Al2O3 is ionic.
For AlCl3
dH(atom Cl2) = +121 kJ mol-1
dH(e.a. for Cl) = -351 kJ mol-1
dH(f AlCl3) = -352.1 kJ mol-1 -703.4
A quick play with these numbers gives the LE of AlCl3 to be about -5235.4 kJ mol-1. A lot less reason to be ionic than Al2O3.
Now, considering AlCl3 ionic or AlCl3 covalent, we have just calculated LE for AlCl3 (an ionic model) and we must now consider polarisation: The Cl- ion, because of its size, is polarisable, and Al3+ is highly polarising. It results in a large amount of e- density between both the Al and Cl giving it dominat covalent character.
2) anhydrous AlCl3 (i.e. covalent AlCl3) isn’t as thermodynamically as stable as hydrated AlCl3 (i.e. ionic AlCl3), which is why, over time, anhydrous AlCl3 reacts with trace amounts of water and turns into hydrated AlCl3.
When water is present, the lp on O in a water molecule reacts with the Al (Al-Cl bond breaks and e- go onto Cl – you should know the mechanism for this reaction) and with just the right amount of water will give you [Al(H2O)6]3+(Cl-)3 i.e. a hexaaqua Al cationic complex surrounded by anionic Cl- ions. The Cl- anions aren’t polarised by the Al3+ anymore as the hexaaqua complex has the 3+ charge delocalised over the whole complex [sometimes called the 'ion sphere' as the complex is kind of 'ball shaped'] dramatically reducing its polarising power towards anions. So overall it has turned into ionic hydrated AlCl3
*phew*
3) The dimerisation of AlCl3 into Al2Cl6 seems from my experiences to vary according to which book you read. I think the thermal energy causes a greater degree of vibrational freedom and so the shape from trigonal planar (120o Cl-Al-Cl bond angles) can change. This exposes the e- deficient Al to Cl atoms from a different AlCl3 molecule allowing the lone pair on Cl to attack the e- deficient Al (dative bond forms) making a 4 bonded Al. The complementary attack by Cl on the first molecule to the Al in the second molecule happens (as there all in close proximity) forming your dimer.
That is just a guess from my knowledge of chem. It may be completely wrong. But heheheh, hurray – it’s not on the syllabus!!
Does that help?
March 15, 2009 at 1:37 pm
sir,
this is Qs jun 07 unit 5 no.3…
i still dont get it….
E ○/V
Zn2+ + 2e– —–> Zn – 0.76
Cu2+ + 2e– —–> Cu + 0.34
NO3– + 2H+ + e– ——> NO2 + H2O + 0.81
(i) Use the half equations given above and the values of E ○ to calculate the standard electrode potential for the reaction between zinc and nitric acid and derive the equation.
i got d answer which is +1.57V…..
(ii) Suggest why zinc does not produce hydrogen with nitric acid
according to the marking scheme :
Ereaction for the production of hydrogen is (+) 0.76 (V) (1)
smaller than reaction in (i) so is less likely (1)
OR
NO3- being the oxidised form of a redox couple with a
more positive E o than E o H+/½ H2 (1)
is a stronger oxidising agent than H+ (1)
i dont understand the answer….why it said the production of hydrogen is +0.76? and how cud i know that the NO3- is more positive E than H+? if NO3- is more positive, then its hard for it to release e, isnt it? then it will accept e and itself being reduced….so is that why it’s a stronger oxidising agent? fix me, sir…..
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?
March 16, 2009 at 4:30 am
if it’s more positive, more thermodynamically favourable, then the reaction is more likely to happen. which means Zn and NO3- more likely to react to form H2O rather than with H+ to form H2….am I correct, sir?
March 16, 2009 at 1:33 pm
what is multidentate ligand and unidentate ligand, sir?
and it is stated that multidentate showing by EDTA(4-) is more stable than unidentate due to it’s ability to displaces more H2O than unidentate….why it’s more stable when more water being displaced? and can this topic b asked as detail as this in our exam, sir?
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.
March 17, 2009 at 11:25 am
assalamualaikum sir (^_^)
i have a Q which might be simple 4 u,but not 4 me. could u explain whats is optical isomers is all about and also the geometric isomers..this topic is very famous in exam,but i don’t really understand bout it.thanks sir..
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
March 17, 2009 at 2:09 pm
Geometric isomers are molecules with the same sequence of atoms but have a different 3D arrangement due to restricted bond rotation and each atom in the restricted bond has two different groups attached to it An isomer can be superimposed in its mirror image (distinguishing it from optical isomers).
For edexcel, the restricted bond is almost always C=C carbon based molecules, but they could also use a ring structure, e.g. 1,2-dichlorocyclohexane (you can get cis and trans for these molecules also despite them not having a C=C!)
March 18, 2009 at 4:49 am
salam sir.
i’m working out with my planning..
is there any technique on answering these type of questions.?
does edexcel have any specific type or pattern of the question..?
exp : if i see any gas, i shud relate it with gas sringe etc.
somehow, i am not good on planning an experiment.
p/s: its so hard to study at home. too much distraction. >_<
tq.
March 18, 2009 at 5:40 am
sir, i always come acroos these questions until i can remember the answer, but, the thing is, i dunt really understand why it happens.
Q1:
rxn 1 – SnO2 + 4HCl –> SnCl4 + 2H20
rxn 2 – SnO2 + 4HCl –> SnCl2 + Cl2 + 2H20
Which rxn is more likely to happen.? explain. (2mrks)
Answr : rxn 1. bcause for Sn, +4 oxdtion state is more stable than +2 oxdtion state. Sn(IV) will not oxidise Cl- to Cl2
what does it mean by “more stable”? n y is it more likely to happen?
Q2:
explain the solubility of hydroxide n sulphates of group 2.
thanx..
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.
March 18, 2009 at 12:32 pm
“exp : if i see any gas, i shud relate it with gas sringe etc.”
collecting a gas in a frictionless syringe in a horizontal orientation is often used in quantitative chem (moles/conc/mass determination etc) as well as kinetics and also for equilibrium experiments.
But because of the variability of planning exercises, it’s hard to make the absolute statement: “when you see gas think syringe, syringe, syringe” but thinking of using those techniques for planning exercises seems reasonable. to me.
I would advise you think it through at least twice BEFORE you start writing. As your thinking, try and imagine if there is any flaws in your plan. Also write in bullet points rather than ‘Shakespeare’ sentences and paragraphs – it may be easier.
March 18, 2009 at 2:11 pm
sir,
there are some questions about the boiling point diagram for particular organic compounds… (which often related to Raoult’s law)
i am not sure how to answer Q when i’ve been asked to
1) explain how fractional distillation separates the mixture
containing 0.75 mole fraction of 2-methylpropan-1-ol.
2) explain how fractional distillation of a mixture of 50% octane and
50% hexane can be used to produce pure hexane (lower boiling point)
both Qs asked us to refer to the diagrams given, so i need to draw at least two lines in the diagram…do the values (0.75 mole fraction and 50:50) give any clue of answering these Qs? or I can just draw the line on the diagram wherever i like?
0.75 mole fraction is the vapour pressure contribute by 2-methylpropan-1-ol, am I right sir? and does this give us clue that it is more volatile than the other one? is that why it is more richer?
50:50 represents the composition of the mixture, do I have to use it to answer the Q or just focus on which is more volatile and more richer in vapour?
let say if it is 3 marks,
then, the final mark is of course the condense, reboil steps until i get the pure product…..
so, please help me how to get the first and the second marks…
which are the important points to be included?
March 18, 2009 at 2:22 pm
do you have a second compound for part 1? 2-methylpropan-1-ol and ____?
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
March 19, 2009 at 8:28 am
thanks sir 4 the optical n geometrical isomers explanation sir. so,we can say that if a molecule is not optically active, means that it have planar structure,isn’t it?
thus the molecule can be equally attacked at both side and produced a racemic mixture….
(im not very clear with what the racemic mixture is all about,but its just another name for mixture of molecule that can be attacked at both side,am i right sir??)
March 19, 2009 at 8:58 am
hmm,sorry sir,but there is another Q that keep on hunting me since i was a baby..hehehe
what is the specific formula to count the enthalpy change when the Q give the data for each enthalpy change of the reactants n product?coz i always got the wrong answer for this type of Q.
for ex: Qjune08 2C.
2S(l)+2H20(l)+3O2(g) ->2H2SO4(l)
the data given only :
H2O= -286
H2SO4= -814
what is the general formula to calculate this type of Q?
n yet,another confirmation is needed here,am i right that the data given is just for H2O n H2SO4 since the other compounds are in their standard state.so we dont need to include them. isn’t it?
im confused coz there is multiple formula being used by my friends such as
enthalpy change= bonds breaking – bond forming
or another one is
enthalpy change = products – reactants
(these formula make me confused to used which one n at last end up getting wrong answer!)
which one is correct? why they have different calculation but eventually same answer??? how they DERIVE THE FORMULA?
coz sometime in certain cases which i am consider to be lucky,my answer is correct,but mostly i got it wrong!!shame on me…i know this type of Q shouldn’t be a problem to most of the students,but sadly not for me..i do have problem with this type of Q..help me sir!
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.
March 19, 2009 at 1:47 pm
69 | nurul, March 19th, 2009 at 8:58 am
Please see the enthalpy cycle above, near the top of this post, for the reaction you mentioned. (sorry but the quality of the pic degraded when I saved it as .jpg I think it’s generally ok to get the points across. Some of the ’stems’ of the letters e.g. “H” in H2SO4(l) got ‘cut out’. Is it OK?)
1) Learn your definitions. This is VITALLY important. These definitions usually talk about one mole being formed or one mole being burnt. And the data given refers to those definitions.
But the questions often involve more than one mole, so you usually have to multiply up the values given.
Also in these types of Q’s there is usually a point in the cycle where all elements in their standard states at 298K are collected. The products given in the question form another point in the cycle as does some other point with an ‘intermediate’ molecule (don’t get that phrase mixed up with intermediate used in discussions of some organic reactions!) on the way to forming the product. In the case you gave, the intermediate is H2O.
The answer -1056 kJ mol-1 refers to the equation:
2S(s) + 2 H2O(l) + 3O2(g) –> 2H2SO4(l). The question may have various ‘twists’ in it for example by asking you how much energy would be liberated when one mole of the H2SO4 is formed using the equation given.
2 moles = -1056 kJ mol-1, 1 mole = -526 kJ mol-1
but the enthalpy LIBERATED (already assumes it’s negative) is the magnitude of that i.e. 526 kJ mol-1
They may also ask the Q in terms of grams instead of the ‘half mole’ bit I used.
Practice these cycles. With practice you WILL get really good. Sometimes you just need a ‘breakthrough’ moment to ‘crack’ any mental block you may have.
It’s almost 100% guaranteed that one of these WILL come up in the U2 exam. In U4 you will get a Q based on Born-Haber cycles and v.v. likely in addition to that a LE, HE and dH(solution) question.
March 19, 2009 at 3:42 pm
sorry sir, this is the full Q….
Propan-1-ol boils at 82°C and 2-methylpropan-1-ol at 109°C.
(a) Draw a labelled boiling point/composition diagram for the mixture of propan-1-ol and 2-methylpropan-1-ol.
(b) Use your diagram to explain how fractional distillation separates the mixture containing 0.75 mole fraction of 2-methylpropan-1-ol.
this gives us clue that propan-1-ol is more volatile as it has lower bpt….if 0.75 mole fraction of 2-methypropan-1-ol, then we hav to draw the first vertical line near pure 2-methylpropan-1-ol….and the horizontal line move towards the pure propan-1-ol….and so do the next vertical line….the only way to separate 2-methylpropan-1-ol is by getting the pure propan-1-ol as it is more volatile….is it true?
so what i understand here, the easier way to know which 1 is more volatile is by looking at the boiling point….the lower 1 is more volatile….
but sir, mole fraction is the mole of substance in the mixture…..the mixture u mention is the liquid, right? based on the diagram u’ve drawn, let say if we change the Q to ——-> the mixture containing 0.75 hexane….. so i’ve to draw the first vertical line near the pure octane and the horizontal 1 move towards the pure hexane?
March 19, 2009 at 3:57 pm
alhamdulillah. thanks a lot sir. ur diagram really help me to have better understanding bout the enthalpy change (^_^)
so,i just need to do more exercise on this type of Q ,stick to the definition n also,draw the hess cycle,rite?
fuh, no need any formula like b4 which i don’t really understand.
thanks again sir
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.
March 20, 2009 at 4:00 am
73 | intec studnt, March 19th, 2009 at 3:42 pm
“if 0.75 mole fraction of 2-methypropan-1-ol, then we have to draw the first vertical line near pure 2-methylpropan-1-ol {YIP, that’s right}….and the horizontal line move towards the pure propan-1-ol {YIP, right again}….and so do the next vertical line….the only way to separate 2-methylpropan-1-ol is by getting the pure propan-1-ol as it is more volatile….is it true?{kind of, you are allowed to say pure 2-methylpropan-1-ol is left behind!}
so what i understand here, the easier way to know which 1 is more volatile is by looking at the boiling point….the lower 1 is more volatile….{Yes, use te bpt given – or you may have deduce that info from the graph}
IF Q said “mixture containing 0.75 hexane” Then you would draw the first vertical line closer towards the hexane. The mixture is 75% hexane and so closer to being pure hexane!
March 20, 2009 at 8:13 am
thnk u sir, im clear about the solubility of g2 now. and, about the stable state thingy.
the simple xplanation had done the understanding job much easier.
*releived*
perhaps, i need to enhance my english. ?
March 21, 2009 at 6:30 am
Salam sir.
Yes, here i am, asking a simple chem question.
we can say a species is oxidised when it acquires O.
then it is reduced if it acquires H+
i hope i dont get that wrong..
the q here is,
can we say the spesies is oxidised when it loses H+ or reduced when it loses O?
means, i’m taking the opposite of oxidation as reduction and vice versa.
thx!
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.
April 10, 2009 at 11:04 am
salam sir, i have questions to ask u. this is Q from Unit 4 june 2005 q no. 4 b (iii, iv)
for q iii, it wants us to state whether the Kp increase or decrease. the ans is the Kp increase coz forward reaction is endothermic
n for q iv, it want us to state the position of equilibrium. the answers is the equilibrium shift to right coz the value of Kp goes up, the value of quotient must also goes up.
im fully understand bout how they get the answers, but im confius with the reasons for the answers.
i thougt it to be the other way round.
coz should we say that for q (iii),the Kp increase bcoz the quotients increase.
n for q (iv) the equilibrium shift to rigjht bcoz the forward reaction is endothermic.
but why is my answers is wrong sir? did i miss understanding any fact about this topic? really hope that u can help me. thanks sir
April 10, 2009 at 3:37 pm
salam sir, whats is the differences between partially n slightly? is it true that we cant used partially ionised for the definition of weak acid?bcoz it will have different meaning,isnt it?
April 11, 2009 at 11:56 am
This question was moved from “Resources – Chem Videos from the web.” to here for househkeepig purposes.
From siti (April 11, 2009 at 4:48 am)
salam.sir.i’m a bit confused about the answer given.
Q: I : rate=k[S2O8]2
II : rate=k[H3AsO3]2
how could the experiment be adapted to distinguish between these 2 rate equations?
A: repeat experiment double,
but keeping the other concentration unchanged
if inital rate double,rate equation I is correct
hurm..
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.
April 11, 2009 at 3:21 pm
#80, nurul April 10, 2009 at 11:04 am
Sorry for the delay in replying.
This is a Q and mark scheme answer that students have difficult to get to grips with. So I’m gonna explain both answers and explain what’s happening.
(iii) State the effect of an increase in temperature on the value of the equilibrium constant, Kp. Justify your answer. (2 mks)
Ans: As T increases, the equilibrium constant Kp will increase because the reaction is endothermic.
Comment: Simply comment on whether the rxs in endo or exo. If endo then Kc (or Kp) will increase, if exo then Kc (or Kp) will decrease.
If you can’t remember this, then you can use Le Chatelier MENTALLY only – to get the answer!
Lets rewind a bit and talk about conc changes. When conc changes, we still have the old value for the eqm constant, but we are no longer at eqm so we say the “position of the eqm shifts”. It does so to get back to the old value of Kc or Kp.
example: Imagine a + b c
K= c / (a x b) Say the system has got to eqm. [ ] of a, b and c is constant at this point. Now if we increase [c], then at that very instant, c / (a x b) is greater than the old eqm value. So the position of the eqm shifts to the left to lower the numerical value of [c] and to increase the numerical concs of a and b. the value of K never changed. A similar thing happens for pressures for gaseous species. The partial pressure changes this time, not(!) the conc! i.e. don’t use [ ] when talking about partial pressure. Use round brackets instead: ( )
We have the same reaction again and it gets to eqm. This time the T changes. Lets say T it goes up. Now would now have a NEW equilibrium constant (the reasons why are beyond the scope of this blog. Just remember TEMP is the ONLY thing that changes the value of K). and the c / (a x b) expression does not have the same value as the new K. So once again the concs of reactants and products change when the position of tto get to the new equilibrium value. How they change is revealed by Le Chatelier thinking.
(iv) Hence suggest in which direction the position of equilibrium moves when the temperature is increased. Justify your answer. (2 mks)
The position of the equilibrium will shift to the right. The value of (p NO)2 and (p Cl2) will increase and p( NOCl)2 will decrease to attain the new higher value of K.
NOT THIS: The eqm shifts to the right so the value of Kc increases!, not least due to the fact we saw before that the position of eqm can shift left {in the example} while the value of K remained.
Summary:
category a) The eqm responds to new values in K
and NOT:
category b) K responds to eqm shifts.
When you asked
“should we say that for q (iii),the Kp increase bcoz the quotients increase.”
that dosn’t address the exo/endo info necessary to answer the question here and it mixes info from question (iv) which incidentally would come under category b) above and be wrong.
“n for q (iv) the equilibrium shift to rigjht bcoz the forward reaction is endothermic.” – the eno/exo bit is for answering q (iii)
Did this help?
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.
April 12, 2009 at 1:51 am
thanks sir.(^_^) its really help.
so,for definition of weak acid, better we used partially dissociate rather than slightly, right?
nway,can i ask u something.?for this 3weeks time before the a level, do u think doing all the past years from 2004-2009 would be better or do the revision for each topic again?(i had revised all topic before the trial)
April 12, 2009 at 5:07 am
its from past year june 2005,unit test 6b(synoptic) question no.1 (b)(ii).
April 12, 2009 at 10:32 am
#86, nurul April 12, 2009 at 1:51 am
Yes, use partially dissociate / partially dissociated for weak acids
April 12, 2009 at 1:45 pm
#87, siti April 12, 2009 at 5:07 am
I’m using abbreviations here…
OK. [S] and [As] both fell by half, and from the graph, the rate fell by a factor of 4. The overall order therefore must be 2.
Possible orders to meet this requirment are:
[S]2 [As]0 (overall order = 2)
[S]1 [As]1 (overall order = 2)
[S]0 [As}2 (overall order = 2)
So how to tell which one it is? Do another experiment. Keep the [ ] of one reactant constant and half the other reactants [ ]. If rate didn’t change, then the order w.r.t. the reactant whose [ ] was halved must be zero and the other must be 2.
If rate dropped by a factor of two, then order w.r.t. the reactant whose [ ] dropped by half must be 1 and the other reactant must therefore also be 1.
If the rate fell by a factor of 4 when the [ ] of reactant was halved, then order w.r.t. that reactant must be 2 and the other must be zero.
Humm, It may be easier for you to not half the [ ] but double it. In which case rate would increase. Most student like to look at increases in conc rather than decreases.
This is one of those times where the fact that mark schemes are made for lecturers and not students, really shines through.
Here’s what the Examiners report said:
“Question 1
(a)(i) The vast majority of candidates scored both marks here.
(a)(ii) Again, it was common to award full marks here. Sometimes the tangent was drawn at the wrong place, and the values for x and y carelessly read.
(b)(i) It was very common to award the ‘slope’ mark here. Better candidates then had little difficulty in explaining that the reaction was second order. Weaker candidates often proposed a fourth order reaction, due to the 4:1 ratio.
(b)(ii) This part was marked consequentially on the candidates’ order given in (b)(i), and it was common to award both ‘rate equation’ marks. Sometimes the rate constant was capitalised, or missing, as was ‘rate’ or even ‘=’. Many candidates were able to explain how to adapt the experiment, but some failed to address the idea of distinguishing between their rate equations.”
April 16, 2009 at 8:17 am
assalamualaikum sir..
here i got 1 question for u..
For thermochemistry practical..
The examples of sources of error is when we use too much water and too little water..
What is the effect on enthalpy when we use too MUCH and too LITTLE water?
Is that because of the heat dispersed slowly in too MUCH water and inversely in too LITTLE water?
I dont get it,Sir..
Help me!!
April 16, 2009 at 5:13 pm
This reply was weird – sorry about that – so I’ve written it in English this time… !
Too much water and the heat will will very ’spread out’ (too highly dispersed), so the temperature rise throughout the solution will be smaller. Therefore, the % error from thermometer readings will have a greater significance upon the temperature range recorded.
If the volume of water was too small, the temperature of the water would be higher, so the rate of heat loss to the surroundings would increase (especially on stirring – note: this doesn’t mean don’t stir!) so the theoretical max temperature may not be achieved in practice. But of course, good insulation (expanded polystyrene cup with a lid with gentle stirring) can minimise this.
I should say, it is rare for the considerations above to be mentioned on any mark schemes.
Common enthalpy errors involve issues of :
1) Heat loss to the surroundings.
2) Accuracy of the thermomenter.
3) Assumptions heat capacity is 4.18 (units)
4) Assumptions about the mass of water present (e.g. if looking at dH(neut), the acid and alkali will make more water. The mass of the extra water produced is usually ignored).
5) Lid not used / solution not stirred.
6) Heat capacity of reactants (and prods) is ignored
7) Heat capacity of the termometer is ignored
The enthalpy Q you are likely to get in 3A or 6A, may contain info (e.g. diagram) which may give a clue about some error which I don’t list above.
Hope that helps.
April 24, 2009 at 7:51 am
Sir..sory for late reply..ok..i understand it..
Sir, i have one more question here…
Actually how to answer the question about Kp and Kc?
I dont understand the mark scheme, especially if I try to compare the older mark scheme and recent years..
Sometime they accept the answer for shift in the equilibrium position..
But recently they dont!!
They rather ask us to answer about the denominator increase or numerator, so on… I dont know how to answer in terms of denominator, numerator or even the ratio..
Sir…help me ya..
April 24, 2009 at 11:07 am
salam sir. could u help me to distinguish between orbital n subshell? i always have problem with the plural n singular usage for these terms. thanks sir.
April 24, 2009 at 2:34 pm
I’ll reply to both Q’s later tonight. I’m just about to leave for home.
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?
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?
April 25, 2009 at 7:56 am
hmm, for example, the ligands split the 3d orbital or orbitals? as i understands b4 this, the orbital is the 1st, 2nd and 3rd…etc.. while the subshells are within the orbital, the spdf….1s2,2s2,2p6,3s2…..etc?
so, for example we can say that magnesium have 3orbitals, and for their 2nd orbital, they have 2subshells of 2s2,2p6. im i right or wrong? plz help me sir..n im sorry if my Q kind of confusing…
thanks sir
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.
April 26, 2009 at 5:39 am
sir,
dissociation of water will gives pH 6.63, slightly lower than 7…
there’s Q asking about why it is slightly lower. According to the marking scheme its due to [OH] = [H]…I still dont understand if it is equal, then why it is slightly acidic?
April 26, 2009 at 5:41 am
For housekeeping reasons, this question/comment has been moved to
“Past Years Papers etc Questions Here Please” see post comment 48.
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?
April 28, 2009 at 2:33 pm
This Q was moved to “Past Years Papers etc Questions here please.” for reasons of housekeeping.
April 29, 2009 at 2:13 pm
salam sir.
em. a question for you
i dun really get what’s the difference between continous distillation, fractional distillation.
some notes saying the difference is the vertical condenser,
then,if it is vertical condenser,will it be the same as heat under reflux apparatus? hurm..
and how about steam distillation?
.because some past year paper asked us to draw the appartus, n i’m….a bit confusing to draw it..huhu
April 29, 2009 at 2:15 pm
er,sir.. neutral means ?
ph7?
concentration H+ equal to OH-?
is it accepted if we just answer ph7..?
April 29, 2009 at 2:18 pm
owh.i got the view for the answer for my ‘neutral ques’ from your answer above… hee.
April 29, 2009 at 2:59 pm
sir ,
how to sketch a titration curve?
how do we know where is the vertical section?
within what range?
and the level off line as well?
April 29, 2009 at 3:08 pm
sir,
why first ionisation energy for Cl > Na ?
because nuclear charge?
how to compare it? as Cl is -ve charge while Na is +ve charge..huhu
April 29, 2009 at 3:15 pm
#107, siti April 29, 2009 at 2:13 pm
Continuous distillation does have the condenser in an angled position (approx 120 degrees to the vertical) and fractional distillation uses a vertical column.
In fractional distillation the vapours evolved undergo continuous re-boiling. With continuous distillation, there is no condense and re-boil phase.
Fractional distillation is not the same as refluxing because with refluxing, none of the chemical components is given off, unlike fractional distillation in which a component is removed
Steam distillation isn’t on the syllabus is it? Wow. It must be about 14 years since I’ve done a steam distillation. I can’t imagine them asking you to draw fractional distillation apparatus (but as f.d. is on the syllabus I guess there’s the possibility that they might)
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.
April 29, 2009 at 3:39 pm
#111, s April 29, 2009 at 3:08 pm
For gaseous atoms, IE of Cl is > than Na.
Rule is: going to the RHS of a period, the IE generally increases. Same number of inner shells, therefore same shielding and increasing number of protons attracting the outer electrons more strongly therefore energy needed to remove outer (valence e-) is higher: IE is more endothermic.
Exception 1: when a new SUBSHELL is being filled. The previous subshell of he same ENERG SHELL is at a lower energy so closer to the nucleus than the new subshell to be filled, so the lower energy subshell gives a weak shielding effect.
Exception 2: When the first spin pairing occurs in a p subshell. The repulsion of these electrons causes a lowering of the energy needed to remove them.
I doubt they’ll ask about a comparison of IE featuring Cl- and Na+ these species are IONS, but if they they did, it would be harder to remove an electron from a positively charges species. Na+ would have a higher (more endothermic) IE.
May 2, 2009 at 12:25 am
thanks for the explanation sir!! I still wonder how do we know that at which volume we should draw the vertical section… let say we use 20cm3 of weak acid in conical flask and it is titrated with NaOH… why we need to draw vertical section at 40cm3?? and why some questions only draw at 20cm3??
May 2, 2009 at 7:19 am
#116, lucky guy May 2, 2009 at 12:25 am
You may have to calculate the volume at which neutralisation will occur. In the example you gave, you haven’t provided enough data to know the vertical section is at 40 cm3.
so here’s my example:
You are told 25cm3 of 0.15M HCl is titrated against 0,11M NaOH and you need to sketch the pH curve. Where do you draw the vertical section?
Moles of HCl in conical flask = conc x vol (in L) = 0.15 x 25/1000
Moles of NaOH needed to be added from burette = 0.15 x 25/1000 (as balanced chemical eqn. is 1:1)
Conc of NaOH = 0.11M. As volume = moles / conc
therefore 0.034 Litres needed = 34 cm3 of NaOH.
Sometimes you are told how much volume was needed to reach equivalance so it is easier, but to stop it being too easy they may put a ‘twist’ in it. e.g. they may use an old favouraite: ethandioic acid{old name oxalic acid} (a dibasic acid HOOC-COOH) and NaOH. In which case two moles of NaOH are needed to neutralise one mole of ethandoic acid. Bear that x2 factor in mind when doing your calculations.
P.S. Lucky guy… I’ve just thought of another ‘cruel twist’ in respect of this ‘where to draw the vertical portion’ question. They could give you the pKa of the weak acid as used in the titration, and from this, you might be expected to calculate the equivalence point. Do you know how to do this?
May 2, 2009 at 11:07 am
sir…
1 Q here..
After read many kind of books..
I found that melting point of graphite is greater than diamond..
At first I thought that it will be diamond bcoz it has strong covalent bonds through out its structure..
But it is another way around..
Why graphite has greater MP??
Is it bcoz of the delocalised electrons? Like benzene that makes it more stable??
Help me ya sir!!
May 2, 2009 at 4:21 pm
Graphite is reported to have a higher mpt – like you, I’ll have to trust the books on this one . In doing so, I will deliberately fix my answer around that information. Naughty I know, but I have no choice.
Graphite and diamond are both giant covalent structures, so when discussing melting, we need to discuss weakening of the bonds in those giant covalent structures, between carbon atoms so that they can ‘flow’ over each other.
As graphite has the higher mpt, this means, the total energy of the 3 sigma bonds and the weak london dispersion force (in graphite) must be stronger than the 4 covalent bonds in diamond.
In graphite, the l.d.f. between layers is quite extensive(largely spread) but we know this bond is weak as graphite used in pencil, can easily made to slip over one another onto paper. So we can push this aspect of the structure to the back of our mind (i.e. don’t focus on these particular bonds/forces as you have done)
It suggests the three covalent bonds in graphite are significantly stronger than the covalent bonds in diamond. Why is this? I could offer a few suggestions but they would just be educated guesses and perhaps not that helpful in this case.
Predicted answer sufficient to answer A-level exam Q’s
(It’ll be a U1 question most likely)
“Graphite has a higher mpt than diamond as the total energy needed to weaken the foeces of attraction between C atoms to the point where they can flow over one another is greater for graphite than it is for diamond.”
P.S. people confuse hardness with melting point. It is invalid to do so in this case.
May 3, 2009 at 1:57 am
RE:#116, lucky guy May 2, 2009 at 12:25 am
Does it mean we need to draw vertical section at 34cm3??
Not really… Would you mind to explain it to me?? thanks lot!!
May 3, 2009 at 6:27 am
#119, lucky guy May 3, 2009 at 1:57 am
Yes, the vertical section would be at 34 cm3 IF ‘volume of base added’ was on the x-axis. {Note: They might be really horrible, break with sanity and write: ‘total volume of liquid in conical flask’ on the x-axis instread, so you know, check properly what info is presented to you and what is asked of you in the question.!}
But here, vertical section was at 34 cm3, as that is where the number of moles of base (OH- in the example given) and H+ from the acid are equal. That was the purpose of doing the calculation.
What is a vertical section? It’s where the pH changes enormously for just a tiny addition, [1 drop, (or 1/2 a drop if you can manage it!)] of reagent from the burette. This can only happen very close to the end point (i.e. equivalence point) because there is only a small [H+] in the conical flask…
At the end point, one drop from the burette can produce a typical vertical section o about 4 pH units long. As pH is a log(to the base 10) scale, a pH change of 4, corresponds to a change in [H+] of 10,000 times – all that from just one drop of reagent in the burette!!!
Wow!
If there was a large conc. of H+ in the conical flask, i.e. quite far from the end point, then the change [H+] by adding just one drop would not make a significant difference to the pH so no vertical section.
(of course you could have base in the burettel if so we would need to shuffle the explanation around a bit but the same principles are present.)
If you haven’t already do so Lucky guy, do as many PYP Q’s as you can and you’ll get the necessary exposure to doing these graphs well.
Try looking for some pH graph Q’s from the links I provide.
Hope this helps.
May 3, 2009 at 1:50 pm
Thanks sir…
other question…
what is the function of cryolite ??
to lower the melting point or act as solvent?
May 3, 2009 at 4:52 pm
It acts as a solvent. The mpt of Al2O3 cant vary, it’s a property of the compound. To get mobile ions of pure Al2O3 you must melt it.
But, you can get mobile ions of Al3+ and O2- by dissolving Al2O3 it in a substance/solvent which itself has a lower mpt than Al2O3. Cryolite serves this purpose.
Saying it lowers the mpt of Al2O3 is a common error.
May 4, 2009 at 7:36 am
assalamualaikum sir.
could u give me the important key point for question that ask how a mixture can act as a buffer? coz i do not know how to put in words my answer when came across with this type of Q. maybe coz i dont fully understand this Q? really hope u can help me. thanks sir
May 4, 2009 at 11:58 am
The mark schemes specific to this question are usually good. One went something like this:.
On addition of a small amount of acid, the large reservoir of anion from the salt used in the buffer, will react with the added H+, removing the added, excess H+, from the solution ensuring the pH of the solution does not fall (become more acidic)
A- + (added)H+ –> HA
On addition of small amounts of base, the large reservoir of weak acid used in the buffer, will release a proton to react with the added base preventing presence of excess hydroxide ions hence preventing an increase in the pH (become more basic)
HA + (Added)OH- –> A- + H2O
Mention these things at least :
Large reservoirs.
Prevention of excess H+ / OH-
Give equations
As ever make sure your answer corresponds to any particular requirements in the actual Q.
May 5, 2009 at 6:24 am
Sir,
In colour compounds in transition metal right, what is actually meant by ” it absorbs light in visible region”?
May 5, 2009 at 8:25 am
You can get lots of ‘types’ of light.
Cosmic rays
Gamma rays
X-rays
UV rays
Infra-red rays
Microwaves
Radio waves
All of which cannot be seen.
In between UV and IR, we get light which we can see, so it’s called visible light.
Each ‘type’ of light has its own range of energy. X-rays are high energy, Radio waves are low energy. Visible light is pretty low energy too [Good or else when we turned on a light bulb it would 'hurt' us!]
When you shine white light (visible light – we can see the white colour!) which is a mixture of all visible colours, being: Red, Orange ,Yellow, Blue, Indigo and Violet, then only SOME of those colours are absorbed.
Why?
The energy of SOME of the colours corresponded to the energy gap between the split d-subshell. So it may be possible for an e- in a split d-subshell to absorb that specific frequency of light (called quanta).
Note the e- needs a vacancy in the upper set of orbitals from the split 3d for this promotion to happen.
If say, a RED colour was absorbed, your eyes will see the remaining UNABSORBED colours (at a guess it might look green), all in the visible part of the electromagnetic spectrum
May 7, 2009 at 1:22 am
mr allen can u post the practical hint sheet for june 2009 3A and 6A practical? thanks
May 7, 2009 at 3:09 am
salam sir..regarding on your #116 post..u gave a Q on ur PS..n i believe u havent answered that one yet..(or did i miss a post?)
but umm..i thought of giving it a shot but got stuck in the middle..
we can find the ka of the acid first by 10^(-pka). as its a weak acid, ka=[H+] at the half of the equivalent point since [A-]=[HA] at this point..rite?n then…???yep got stuck here. . .
May 7, 2009 at 3:17 am
Sir,
can you give us some advice on the coming practical papers?
May 7, 2009 at 3:43 am
#129, “A” May 7, 2009 at 1:22 am
I don’t actually have the ‘hints/advice’ as I’m not involved with the running of the final exams this time around.
I’ve just asked for a copy. If I get it i’ll post it here ASAP.
May 7, 2009 at 4:17 am
#130, lolipop May 7, 2009 at 3:09 am
I didn’t answer the “P.S. in #116 ‘cos it was a Q for ‘lucky guy’. But the answer was really just to do the reverse procedure.
the pKa will = the pH at half neutralisation. Now they would have to give you some more info at this stage for you to calculate the volume of acid that had been consumed at that pH. Then you would have to double this volume to get the end point.
I don’t have an actual example to give you as I was asking a general Q, and not a specific Q for ‘lucky’ guy’ to do, so don’t worry if you got ’stuck’ in actually trying to do it. What I am saying really is, “be able to think about the reverse procedure too.”
‘Twists’ in questions are sometimes just getting students to work in the opposite way. they may do this ‘cos it requires the same knowledge requirement (i.e. same syllabus!)but simply a different approach to answer the Q.
The early syllabus naturally works in the traditional direction but eventually, to make the Q’s appear ‘new’, they may give the students the answer and then ask them to work in the reverse direction.
I must say, what can be loosely described as “working in ‘reverse” appearing on exam papers is NOT likely to be the way most of the paper is constructed. Hopefully you will have come accross these things in you revision of past years papers.
May 7, 2009 at 4:18 am
Sir, for our upcoming 3A and 6A practical exams , we can bring anything we want into the exam halls? all the books.the past year question with answer, everything except planning excercise?
May 7, 2009 at 4:24 am
#131, lucky guy May 7, 2009 at 3:17 am
Were you refering to anything in particular?
Naturally I can’t say anything specific about the pracs, not only because at this stage I know absolutely nothing about them, but also becasue it would be unethical. All I can do really, is repeat the same info that I gave January’s students.
http://intechemistry.wordpress.com/2009/01/02/unit-3a-and-6a-practical-exam-notice/
If you’ve got a particular query then ask away.
May 7, 2009 at 4:36 am
#135, desperate guy May 7, 2009 at 4:18 am
YIP.
You can bring in any ‘written material, apart from anything to do with planning exercises – either hand written or as part of your past practicals.
If you’re not sure, don’t bring it into the exam room.
The chance of you needing anything other than data relating to ‘tests/analysis’ is very small. Edexcel want to test your practical skills, not your ‘luck’ at having brought a certain book into the room. Students often lose sight of the face it’s a practical assessment There should be nothing that essentially you haven’t done before.
They simply want to test the lab skills you’ve learned over the last two years and whether you can make a scientifically valid experiment as well as to check you understand what problems (e.g. sources of errors in enthalpy expts and how to minimise them) may occur in an expt.
Most stu’s do very well in practicals. I’m sure you will do too
May 7, 2009 at 4:41 am
ic…Thanks Sir, btw ..sir do you have a copy of jan 2009 practical question and ms? if you have could you give me?
May 7, 2009 at 5:36 am
sorry sir, if my question led you misunderstood!! The “advice” i meant is the general thing we should or shouldn’t do during the practical…
Eg: 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?
May 7, 2009 at 5:39 am
yea desperate guy!! I was really desperate to find the question papers for the practical for Jan 2009…Anybody got it?
May 7, 2009 at 6:56 am
sir, can you give more details regarding to the significant figures and decimal places?
May 7, 2009 at 8:19 am
#138, desperate guy May 7, 2009 at 4:41 am
I HAD a copy of Jan 09 U3a and U6a, but I’m pretty sure ONE of them (3a or 6a – I can’t remember) went missing.
I’m in the main campus at the moment, so perhaps I give it to you tomorrow?
I don’t have mark schemes for ANY Jan 09. Sorry. I’m sure they are on the net somewhere. If you did find papers and mark schemes, I woudln’t mind if you posted the info here.
It’s a paper copy, not an e-copy.
May 7, 2009 at 8:22 am
alright..when are u free tmrw sir? and where can i meet you? where’s ur office etc?…
May 7, 2009 at 8:32 am
#143, desperate guy, May 7, 2009 at 8:22 am | edit
I may be free at about 10:45. Sms me at that time, and I’ll let you know more clearly when I can meet you.
If you don’t know my HP, you will need to ask your fellow students. I’m not giving it out over the internet.
You will have to return it to me after a few minutes!
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.?
May 7, 2009 at 10:05 am
s.f for concentration of a solution? We must put three s.f ??
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.
May 8, 2009 at 6:21 am
salam sir..
hw do we describe bout the trend of increasing in mpt across the period 3?is it bcoz of increasing the metallic character ?especially when we compare btwn sodium and aluminium..
help me sir..thnx!
May 8, 2009 at 11:33 am
Na Mg and Al are metals so you talk about the increasing metallic bond strength between them. Then Si which is (pretty much) unique in period 3 as being the only element to there to have a giant covalent structure. Lots of strong bonds per atom, stronger than the non directional metallic bonds. mpt jumps.
Then P4, S8 and Cl2. Each classified as ’simple molecular’ forces between molecules are weak and easily overcome compared to previous 4 stronger bonds. No of e- in the molecule basically determines the mpt here (note I’m not mentioning red P).
Cl2 << P4 < S8. Ar is simple atomic weak temporary dispersion forces here so lowest mpt of all.
That forms the basis of these kind of Q’s
May 8, 2009 at 1:18 pm
All for the purposes of A-level, Yes.
It’s been a while since I’ve looked into the finer details of mechanistic profiles, so whether it’s ‘universal’ I’m not too sure but it would be my default position to think it was the case.
May 8, 2009 at 3:28 pm
salam, i wanna ask about the test for halogen atoms in halogenoalkanes, besides the aqueous silver nitrate that we added, what other reagents should we use? i know that most of us is familiar with the steps
1. warm the halogenoalkane with the NaOH
2.acidify with HNO3
3. add silver nitrate solution..
but, some mark scheme from pastpapers use ethanol as the solvent, add silver nitrate solution without the use of NaOH and HNO3. they also stated that the students won’t get full mark if we use the former steps…help me, i’m so confuse……
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.
May 9, 2009 at 1:04 am
ohh.ok… many past year questions use the ethanol and silver nitrate and not NaOH… like in JUNE 2006 U3B question nmber 6 and JAN 05 U3A question nmber 4, they use hydrolysis process without the use of any base….
May 9, 2009 at 9:54 am
#156, “sylar” May 9, 2009 at 1:04 am
June 06 U3b Q6 sees ethanol being used to dissolve the halogenoalkane. It’s the water in the AgNO3 solution that acts as a nucleophile. Usually we use OH- from NaOH as the nucleophile, ‘cos it’s a better nucleophile.
Mentioning NaOH in part (c) is penalised because you were told to use the experiment in (a) which never mentioned NaOH(aq). so it’s not that NaOH(aq) is a wrong reagent to use in identifying halogenoalkanes.
If you had a “fresh” question, then it would be perfectly valid to use NaOH, and in fact it’s better.
It’s questions like this that causes students to lose marks. They wrote ‘in bold type’ the bit about using expt (a) to stress ‘don’t do anything new’, trying to steer you away from making that mistake. Jan 05 U3A Q4 uses ethanol and aq NaNO3 also simply because it means the students don’t have to go about doing 3 extra things. 1) Add NaOH(aq) 2) acidify with HNO3 after warming. 3) checking to make sure the solution is acidic before addn of AgNO3.
So the ethanol + aqueous AgNO3 is simply a short cut in terms of giving directions in an experiment, as either a Q for the stu’s to read or an A for the stu’s to write.
Good Question. Thanks for bringing it up.
May 11, 2009 at 1:44 am
sir for the halogen displacement test, the upper element will displace the lower one right? Is it possible for them to ask us to use fluorine compounds? why somebody said the Cl is more reactive than F?
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
May 13, 2009 at 1:22 am
sir, does2,4-DNP only give positve result to carbonyl group? will it give positive result to functional group such as carboxylic acid?
May 13, 2009 at 11:33 am
2,4-dnp will not give a positive test result with carboxylic acids.
For A-level, if a orange solution of 2,4-dnp gives a yellow/orange ppte, then you had a carbonyl present.
The OH group and the =O on the same C, affect each others properties so much so that it becomes a new functional group with specific characteristics.
They are acids (alcohols are so poorly acidic that at A-level we almost never refer to the, being acidic). They tend to undergo nucleophilic substitution (e.g. esterification) rather than nucleophilic addition as carbonyls do.
May 16, 2009 at 3:31 pm
salam..sir, can ester be hydrolised by any reducing reagent?? e.g. LiAlH4…..I thought the reaction involving esters is just hydrolysis…
May 16, 2009 at 6:23 pm
#165 sylar
I’ve updated this as I was half asleep last night when I write it.
NaBH4 is too weak a reducing agent to reduce the ester. LiAlH4 is strong enough.
Esters will react with LiAlH4 in a reduction reaction – it is not classisified as a hydrolysis, even though the last step (adding aqueous acid) is a hydrolysis of the “intermediate” alkoxide salt formed You will get these following products: From the original ‘carboxyilc part’ of the ester, you will get a primary alcohol. From the original ‘alcohol part’ in the ester, it will reforms the initial alcohol.
e.g. ethyl propanoate is added to LiAlH4 in dry ether. It will produce alkoxides of alcohols (alkoxide = RO- where R is an alkyl group,{similar to the product of ethanol with sodium}). When complete, (we know it’s compelte as no more heat change occurs) then we slowly add water. This is the hydrolysis step, but the reduction part wins in terms of ‘classifying the reaction. Propan-1-ol and ethanol are formed.
(Normal) Hydrolysis of ester is different. We get the carboxylate anion if alkaline hydrolysis is used. If we used acid conditions, we would get the actual carboxylic acid itself. The oxidation number of the C in the ester gp (the one carrying the =O and -O) does NOT change.
There are quite a few reactions involving esters, but not many are mentioned in the A-level syllabus. They are a ‘bit’ like amines and amides – resembling ‘end of the road’ functional groups. {Note this is only in the A-level sence} Reactions of esters would start to be somewhat ’specilist’ knowledge and getting too deep into organic. I’d like that, but A-level is quite general.
A similar case exists for a compound of Lead (still cant remembe the specific compoind. PbCl2 I thin). It may look like an acid/base reaction or, w.r.t. the lead compound, a redox reaction. It is not classified as an acid/base reaction but as a Redox reaction instead. (see your notes).
May 23, 2009 at 6:35 am
It has two main uses in A-level.
One is as a mild oxidising agent
Reduction potential, E(std), = +0.80V
which can oxidise alheydes.
(I think it may also may oxidise ethanol but I can’t remember the oxidation potential for ethanol).
Secondly, it could be used as a reagent to distinguish solutions containing Cl- and Br- if there wasn’t excess NH3 present.
[Ag(NH3)2]+ + Cl- (aq) –> no ppte produced
[Ag(NH3)2]+ + Br- (aq) –> cream ppte forms,
but in this form, it’s not how Ag+ is usually employed to test for halide ions or halogenoalkens, but Edexcel do sometimes ‘twist’ things around in ways similar to this
The amine ligands can ‘protect’ the Ag+ from reacting with OH- when basic conditions are needed, a bit like the tartrate ligands in Fehlings reagent which protect Cu2+ from OH- ions.
The nitrate in silver nitrate is just to ensure you get aqueous Ag+ ions because all nitrates are soluble. If you took silver carbonate and tried to dissolve it to get aq. Ag+ it wouldn’t work.
what did you come across that made you ask this question?
May 23, 2009 at 12:01 pm
it is from chemistry unit 5 pass year( summer 2002) question 6d. I don’t really understand the answer given.
May 23, 2009 at 1:04 pm
sir how an order of reaction can suggests about the mechanism of the reaction? let take the first order of N2O5..
May 23, 2009 at 1:58 pm
Sorry, rushing right now. will try & reply in a couple of hours.
May 23, 2009 at 5:29 pm
June 2002 U5 Q6d is using ammoniacal silver nitrate as an oxidising agent (really, it’s the reagent behind Tollens silver mirror test). The reactivity with 2,4-DNP tells you B is a carbonyl and the lack fo reactivity with [Ag(NH3)2]+ tells you B is a ketone. Ketones (for the scope of A-level) cannot be oxidised.
In answer to comment 170:
The order of reaction, with respect to each individual reactant, tells you how many molecules (yes, molecules, not moles! as mechanisms involve molecular considerations) of it were used “up to and including the rate determining step” (rds). The rds being the slowest step of the reaction.
The overall order is the total number of molecules of all reactants up to and including the rds.
So if order w.r.t. N2O5 was one. then a plausible mechanism could be something like
N2O5 –> 2NO2 + 1/2 O2 (slow)
May 23, 2009 at 5:57 pm
then how do we know the least number of step involved?
other question:
how to distinguish between primary and secondary alcohol? is it the way we differentiate aldehyde and ketone?
May 24, 2009 at 7:16 am
A mechanism can have any number of steps involved, but from the point of view of probability, the fewer the steps a mechanism has, the more likely it is to happen. So when we propose a mechanism, we try and do it in as few steps as possible.
(note: the proposal may be wrong and needs evidence to support it before we can be confident it may be ‘true’)
When it comes to porposing mechanism, we can do experiments which can be used to construct a rate equation, and this rate equation tells us about the SLOWEST step of a reaction. It reveals the maximum number of reactant molecules used up to and including the slow step.
So when proposing mechanisms, we try to keep the number of proposd steps to a minimum (and hypothesise bimolecular processes – again for reasons of greater likelhood). The minimum number of steps will the the overall order order divided by two {divide by 2 because at best, two reactant molecules will be used in each step}
So if rate = k [A]^p [B]^q [C]^r
and p+q+r (the ‘overall order’) = 6 then we could propose a mechanism that would have 6/2 steps, i.e. 3 steps, before the rds step will occur.
So it’s based on the overall order.
OTHER Q:
You can use the Lucas reagent (ZnCl2 in conc HCl) to DIRECTLY distunguish between 1o, 2o and 3o alcohols. 3o alcohols produce a ‘cloudy’ looking solution instantly, 2o alcohols give a cloudy looking solution after about a minute or so, 1o alcohols do not give a cloudy solution. But the Lucas test isn’t in the syllabus which is prob. why might not have heard of it, and alcohols with >6 C atoms don’t really dissolve in c.HCl so the test is limited to short chain alcohols.
Within the syllabus there are a few things you could do to distinguish 1o from 2o alcools. You could use a solution of potassium chromate acidified with dil.H2SO4 (to produce ‘acidified orange dischromate’ solution) and continuopusly distill (so that the carbox acid won’t be formed in the case of the 1o alcogol) then test the distillate with Tollens reagent (alkaline ‘diammine silver (I)’ solution). A silver mirror would indicate the initial alcohol was the 1o alcohol. No mirror indicates the initial substance was the 2o alcohol.
Note: distinguishing between 1o and 2o alcohols is not the same as identifying a 1o or a 2o alcohols from a completely unknown substance. You would have to do more tests in such a case.
June 2, 2009 at 6:22 am
‘why the first ionisation energy of aluminium is less than magnesium’ The answer said in al,the outer electron is in the 3p orbital whereas in mg it is in the 3s orbital so outer electron is in a higher energy level. Is it correct the word ‘orbital’? I thought the outer electron removed is in 3p subshell.. help me sir. Thanks..
June 2, 2009 at 8:35 am
Before I talk about orbital and subshell, your answer might be wrong as Aluminium has a less endothermic value of IE than Magnesium.
1st IE data:
Mg = +738 kJ/mol
Al = +578 kJ/mol
So = +789 kJ/mol
p.s. I seriously advise you to be able to sketch the relative IE’s without data for the 1st 20 elements.
So that answer is not good (or maybe I understood it wrongly, in which case, it may be not good ‘cos it’s not so clear)
On average, (3p can accommodatite 6e-, while 3s can accommodate only two) the 3p subshell is of a higher energy than 3s. But when you look at the specific electron which is being removed in these cases, the 3s e- in Mg is of a lower energy than the 3p e- in Al. That is why it takes more energy to remove the e- from 3s than 3p IN THIS CASE.
The answer I would give is:
Compared to Mg, Al’s outer electron is shielded by the inner 3s subshell. This does not happen in the case for Mg. The increased shielding (which lowers the IE value) in Al is greater than the increased attraction from the extra proton in the Al nucleus which would increase IE.
This only happens for the 2p1 and 3p1 electrons. After that, the shielding the inner subshell is no longer effective, something you can see from Si which has a greater IE than Mg.
Getting back to subshell/orbital, in this case either word is OK because we are only removing one e- from each atom (Mg and Al) each e- must be in an orbital, so, from what you have said in your question you could use either word.
If we were removing say three e- from the atoms, or comparing different elements then maybe – depending on the question, we may have to talk more about subshells.
Does that help?
June 2, 2009 at 8:40 am
Actually, this site: http://www.creative-chemistry.org.uk/alevel/module1/trends6.htm
gives a similar answer to what you gave, but it’s still a bad answer as you can’t apply that very same arguement if you wanted to compare Mg with Si (or the remainder of the Per3 elements)
June 4, 2009 at 2:10 am
sir in the exam the paper stated “do not use pencil” … can we use pencil actually for drawing?
June 4, 2009 at 12:32 pm
Damian Riddle{Damian is representing Edexcel}
20/04/2009 01.44 PM
Dear Sir,
Students have a limited time to complete the paper and are strongly recommended to answer only two questions in Section B. If they attempt all three, they are likely to spend too little time on each one and will produce three average answers, rather than two good ones. However, we would mark all three and credit the best two.
Students should use pencil only for graphs and diagrams. Mechanisms can be done in pen.
Damian Riddle
June 9, 2009 at 7:26 pm
I addressed this point further up. See #174, intechemistry,
May 24, 2009 at 7:16 am (sorry for the typos)
June 13, 2009 at 3:47 pm
sir what is the function of NaOH in reduction of C6H5NO2 to phenylamine? what is the reaction taking place?
June 13, 2009 at 5:28 pm
Sorry for the late reply, I’ve been terrifying myself watching some thriller and horror movies tonight.
As for your Q: I can’t remember it’s exact function to be honest! (I can’t recall the exact mechanism)
I have a very strong feeling however that it’s to ensure the aromatic amine is produced and NOT the salt (-NH3+Cl-).
I think this because one way of doing it is with Fe and HCl. After that reaction, the OH- is added in a ’step 2′ fashion. In fact I’d be really really surprised if that educated ‘guess’ was wrong.
June 14, 2009 at 7:08 am
it was one of the unit 5 questions and i screwed up that question
June 14, 2009 at 7:32 am
Hummm. I think this U4 and U5 will become legendary as ‘nightmare papers’.
Everyone seems ‘in the same boat’ however, so it won’t be as bad as it may seem.
June 16, 2009 at 1:39 am
sir what is quenching? is it same with titrating? how is it be done?
June 16, 2009 at 4:57 am
Quenching is to stop a reaction (or slow it down so much that in effect we can assume the reaction has essentailly stopped)
It is done by withdrawing a known volume of qolution, using a pipette, fromt the reaction mixture, then draining the liquid into an accurately volume of ICE cold water. This is a technique used in kinetics measurements. Your stopwatch should be running from the time the reactants were added together and when you state to drain the reaction mixture into the water you STOP the stopwatch. (note there will be a small error here as reaction proceeds in the pipette as it drains out just before it is quenched. Perhaps stopping the stopwatch when the pipette is half drained will give a better time measurement)
1) By putting the reation mixture in a COLD solvent – usually ICE cold water, the rate of rxn will fall as Temp will fall.
2) By diluting reacting mixture conc of reactants will fall therefore reaction rate will fall again.
** the two factors above combined, the rate should be low enough to assume the rxn has stopped **
Sometimes (but v. rarely at A-level) you use a non water based ice cold solvent – if say the water was to react with one of the reactants or products that you were trying to measure.
Quenching is done before titration, and the titrations are done using standard solutions (solutions whose concn is accurately known)
August 24, 2009 at 8:15 pm
4 d case of MgSO4 n BaSO4, d decrease in hydration enthalpy {less exo} > decrease in LE {also becomes less exo} so is it becmg less soluble? Why?
Im gttg cnfused. sorry.
August 24, 2009 at 9:32 pm
Dowing down gp II, sulphate solubility decreases.
solubility = balance between dH(hyd) and dH(LE)
dH(solution)=dH(hyd)-dH(LE)
Lets use some numbers (taken from G. Facer p45) and from various places on the net. If you work through the Q’s below you will hopefully see the answer to your enquiry. If not, read the explanations that follow (again taken from the new). If that doesn’t help, then tell me what’s causing you the problem.
HAve you read your book about it?
Note: data values show some relatively insignificant variation according to what source you look up. All units kJ mol-1
data:
dH(hyd of Mg2+) = – 1920
dH(hyd of SO4 2-) = -1004
dH(LE of MgSO4) = – 2833
dH(hyd of Ba2+) = – 1360
dH(hyd of SO4 2-) = -1004
dH(LE of MgSO4) = – 2474
Calc dH(hyd) MgSO4 and BaSO4
going down gp II, what happened to the magnitude of dH(hyd) for each compound?
Comment on dH(LE) going down gp II
Calculate dH(solution) for MgSO4 and BaSO4
ans: should be approximately about – 90 for MgSO4 and approx about +20 for BaSO4 {of that kind of order}
Which becomes less exo dH(hyd's) or dH(LE's)
How will the change in dH(hyd) and dH(LE) affect the solubility of each compound?
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Dr Bateman says:
Explaining solubilities of hydroxides and sulphates of group 2
Before a compound dissolves, its lattice has to be broken down. A high lattice enthalpy tends to leave an unbroken lattice which will mean an insoluble compound. The energy needed to break the lattice is taken from the hydration process.
A high hydration enthalpy for an ion tends to give a soluble compound. Hydration enthalpy increases as the size of an ion decreases and as the charge on an ion increases. This is because both of these factors create a high charge density and cause the ion to attract more water molecules making hydration more exothermic.
As we go down group 2, the increase in cationic size decreases the hydration enthalpy, tending to make all compounds less soluble. Also as we descend the group the same increasing cationic size reduces the lattice enthalpy, tending to make compounds more soluble.
The solubility of the sulphates decreases down the group because the large sulphate ion makes the lattice enthalpy changes less than hydration enthalpy. The hydration enthalpy is therefore dominant.
The solubility of the hydroxides increases down the group because the small hydroxide ion makes the lattice enthalpy change more than hydration enthalpy. The lattice energy is therefore dominant.
http://www.drbateman.net/asa2sums/sum4.1/sum4.1.htm
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Something I found in http://www.thestudentroom.co.uk
The solubility of group 2 sulphates depends on the lattice enthalpy and the hydration enthalpy of the resulting cation. The sulphate ion is large, and therefore changing the size of the cation has only a small effect on the lattice enthalpy, which is in part a function of the sum of the radii of the compound. Therefore, as the group is descended, there is only a small change in lattice enthalpy, but the decreasing charge density of the cations means that the hydration enthalpy yields more energy upon solvation. Because the solubility is favoured by compounds with high hydration enthalpies and low lattice enthalpies (the energy required to break the lattice is compensated for by the energy released from the formation of dative bonds from the water molecules), the solubility increases as the group is ascended because they group two sulphates will have similar lattice enthalpies, however the energy yielded by hydration of smaller group two cations means they are more soluble. There, Magnesium, which is smaller than Barium and has a higher charge density, will release more energy on hydration with water than the energy required to break the lattice and so is more soluble.
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Also found this:
The solubility of the sulphate falls as the group is descended; BaSO4 is some 105 times less soluble than MgSO4. The large sulphate ion dominates the interionic distance so that the lattice enthalpies do not change much with increasing cation size. However the hydration enthalpy of the cation is now the dominant factor, and as this becomes less exothermic with increasing cation size, so the solubility becomes less.
November 4, 2008 at 6:39 am
(original: November 3rd, 2008 at 5:49 am)
dear intechem,
i’ve a bit confusion here.why is that NaCl and NaI hv very high bp?
_thnx_