This discussion is now closed.

Check out other Related discussions

EdExcel Nuffield Unit 4 CHEMISTRY

well if you look at the end of it...it ends in 0806/6254_01_que_20080612.pdf

if u replace the 6254 with another unit number and the date at the end you can get any of the papers from summer 08, o used 1 from maths to work it out. I wish i could get the mar scheme lol

Reply 41

Mos Def

OP

17

georgeElsworth

http://eiewebvip.edexcel.org.uk/Reports/Confidential%20Documents/0806/6254_01_que_20080612.pdf

i found the paper!!

Lmao you got that off the c4 thread didn't you :P:

Reply 42

effervescence

Anyone fancy doing an "unofficial mark scheme"? I would but I don't have a clue about most of the answers so it would be pretty bare!

Reply 43

georgeElsworth

1a) - Conductivity because I- and H+ ions released
- Titration with sodium thiosulphate (I2)
- Colorimeter, change in colour of iodine
- pH because H+ ions dissociated

b) Add Hydroxide ions which will react with any H+ ions therefore quenching the reaction

c)(i) -Both concentration and rate increase by scale factor 3/2
- therefore 1st order

(ii)zero order
(iii) rate = k[CH3COCH3][H+]
(iv)CH3COCH3 and H+

(V) the hydrogen ions attack the oxygen because of the polarity of the molecule forming a transition state alcohol(LOOK ON PAGE 252 in text book)

2a)(i) 2 double bonds present
(ii) Does not contain OH group
(iii) 1 carbon carbon double bond.
(iv) Must ve a ketone or aldehyde
(v) Must be an aldehyde
(vi) Hard to explain, one carbon at the bottom, one at the top (coming from the two carbons with double bond)
(vii) CH3-CH2-CH2-CH=CH-COH

b(i) step 1 - hydrogen with nickel catalyst
step 2 - LiAlH4

(ii) OH alocohol group

3a (i) PCl3 is a liquid and therefore more energy to move around and therefore increasing the number of random arrangements.
(ii)+215.6
(iii) Yes, because there is an increase in entropy across the system as reactants go from solid phase to liquid and gas phase, which have higher entropy

b) -123.8x1000 / 298 = -415.4

c (i) -415.4 + 215.6 = -199.8

(ii) Eqm will lie very much to the left as Stotal is negative and the reaction is therefore not spontaneous at 298k

d) (i) pPCl3 x pCl2 / pPCl5 units= atm

(ii) From left to right going down.

0.2 0.05 0.05 0.25 2.6 0.86 0.86

(iii) 0.216 x 0.216 / 0.648 = 0.284atm

(iv) A) No change as mole ratio will rematin the same
B) Increase as reaction is endothermic and rise in temperature will favour the forward reaction

4a) (i) fuming sulphuric acid, heat under reflux
sodium hydroxide

(ii) Electrophillic subst. and neutralisation?

b(i) Friedal crafts (alkylation)

(ii) C12H25Cl with AlCl3 catalyst

c) detergent

5a(i) [HCO3-][H+] / [CO2] units = mol dm^3

(ii) pKa = -log Ka

b) A buffer solution resists a change in pH, a substance is added which will react with any acid/base given off, converting it back to its original state, therefore maintaining a constant pH

c) Base, accepts a proton for the backwards reaction (bronsted-lowry def.) (thanks onlysleeping)

d)(i) 7.4 = 6.5 - log (acid/base)

log(acid/base) = -0.9
10^-0.9 = acid/base

acid/base = 0.126

(ii) 0.0028

(iii) hyp 2 because the concentration of CO2 is decreasing which means Hyp 1 is incorrect..

Reply 44

onlysleeping

c) Base (dont no why) i went on about H+ and to neutralise to obtain H2O we must react with a base.

Yeah, because it was accepting a proton, which is the Bronsted-Lowry definition of a base :smile:

Reply 45

sahl

for 4c) can't it be used to makes dyes and drugs as well? due to the sulphonic acid group.
And i was running out of ideas so I'm probably completely wrong, but for 1b) they were hinting at the reaction being really slow at room temperature therefore couldn't you quench using ice?
Also why is 5ai) over [c02] i put [HCO3-][H+]/[H2CO3]

Reply 46

georgeElsworth

yeh looking again i think you could use ice aswell,
well my chem teacher always said when water is there you can leave it out so i did lol

Reply 47

sahl

you can ignore water if it is in excess and everything else is aqueous and thats with Kc, but this is an acid disassociation and from biology you know that within a red blood cell the carbon dioxide combines with the water to produced carbonic acid which disassociates to give you the HCO3- and H+, but then i guess this is chemistry not biology, urg i puzzled over that question for so long!

Reply 48

georgeElsworth

i have a feeling your answer may be the right one and then mine will be in the accepted answer bit

Reply 49

effervescence

Anyone got any ideas what percentages or raw marks you need for each grade. I think I've got about 65% or 39/60, is that a D?

Reply 50

Mos Def

OP

17

Damn i failed it seems

Reply 51

georgeElsworth

effervescence

Anyone got any ideas what percentages or raw marks you need for each grade. I think I've got about 65% or 39/60, is that a D?

could be just under an A, sometimes for an A it can be as low as 40/60

Reply 52

clareyfairy

gwra

for 4c) can't it be used to makes dyes and drugs as well? due to the sulphonic acid group.
And i was running out of ideas so I'm probably completely wrong, but for 1b) they were hinting at the reaction being really slow at room temperature therefore couldn't you quench using ice?
Also why is 5ai) over [c02] i put [HCO3-][H+]/[H2CO3]

I think what the question means is that it WILL be slow at room temp if you remove the H+ ions, i.e. with a base. I think this is one of those rare occasions when quenching with ice won't work.

Reply 53

JayG99

ahezhara

I supported Hypothesis I. Did anyone else?

I think my reasoning was that because the ratio of acid to base was higher, one would expect a decrease in pH... and there was. Therefore the CO2 was making it more acidic, hence hypothesis I, blah blah blah.

Not sure about it though, I may have argued it circularly.

That was exactly my reasoning!
I thought it was far too easy to pick Hypothesis II and there must be a catch. I'm pretty sure it was hypothesis 1, because even though the Co2 concentration decreased, RELATIVE to the base, it was higher. Thus the ratio of acid to base was higher so when you plug that in...your overall pH should be lower (more acidic). I'm not saying that's 100% correct..just my answer. Page 346 of the textbook (near the top) if you wanted to know...