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Chemistry Exams 19/06/06

Is it just me or is it really confusing that people are discussing (at least) two different exams on the same thread :s
The coordination number was a one mark q, the drawing was for 2, I think.

Reply 41

uk6458

shinyhappy

Is it just me or is it really confusing that people are discussing (at least) two different exams on the same thread :s
The coordination number was a one mark q, the drawing was for 2, I think.

yeah i keep thinking, " we didn't have a question like that, did we?" maybe we should have a separate thread for salters?

Reply 42

Prometheus

For the record the hydrazone derivitive was

C6H3(NO2)2--NH--N==C(OH)-CH=CH2

Reply 43

Prometheus

And for the Kc on Unit4 i got something like 6.8x10-4 moldm (cant remeber exactly or what units were)

Reply 44

littleone

Prometheus

And for the Kc on Unit4 i got something like 6.8x10-4 moldm (cant remeber exactly or what units were)

did u get 8.75*10-3 for ka?

Reply 45

gfdsa

didnt anyone do the mechanisme in the rate equation like this :

Step1 A+2B--> AB2
Step 2 AB2 + B --> AB3
overall A+3B--> AB3
Step 1 is the rate determining step

i remember my lecturer couple of months back taught us to do this way for this type of question.. Anyone with me on this ?

Reply 46

tazzalc

why didn't the question with ethanoyl chloride need a catalyst??

Reply 47

Prometheus

did u get 8.75*10-3 for ka?

Mebe, i cant remeber, but it seems about right, i got a small number like that and i think it was something like that.

Reply 48

littleone

carboxylic acids need h2s04+refluc
ethanoyle chloride doesnt

Reply 49

Prometheus

Step1 A+2B--> AB2
Step 2 AB2 + B --> AB3
overall A+3B--> AB3
Step 1 is the rate determining step

Yeh thats what i mean exactly, except thats simpler than mine :-p i had Ds and Es as intermedietary, but it still works the same so should be ok, markschemes tend to say, "any sensible and working mechanism".

Reply 50

tazzalc

gfdsa

didnt anyone do the mechanisme in the rate equation like this :

Step1 A+2B--> AB2
Step 2 AB2 + B --> AB3
overall A+3B--> AB3
Step 1 is the rate determining step

i remember my lecturer couple of months back taught us to do this way for this type of question.. Anyone with me on this ?

yeah i did something similar but i drew it out.

Reply 51

littleone

Prometheus

Mebe, i cant remeber, but it seems about right, i got a small number like that and i think it was something like that.

yay!!
also did u get moles of the 1st reactant to be0.35/20 at eqm?

Reply 52

Clearvision

tazzalc

why didn't the question with ethanoyl chloride need a catalyst??

electronegative O and CL make C +ve, this makes it very susceptible to nuclephilic attack so low Ea, no catalyst/heat needed.

But i wrote COCl is polar (haha) and so it is more susceptible to nucleophilic attack and so happens at RT without catalyst.

Maybe 1/2 or 2/3 for me whichever it was out of.

Reply 53

littleone

Prometheus

And for the Kc on Unit4 i got something like 6.8x10-4 moldm (cant remeber exactly or what units were)

i got 50.....or sumthing oooooooooooooopppppps!

Reply 54

Prometheus

also did u get moles of the 1st reactant to be0.35/20 at eqm

Yes that i did get, and the others were 0.15/20 and 0.075/20 i believe

Reply 55

gfdsa

tazzalc

why didn't the question with ethanoyl chloride need a catalyst??

-the compund is trigonal planar
- nucleophile is able to attack from either side of the molecule .
or
-large (+) on carbon atom as attached to 2 highly electronegative atoms . -attracts nucleophile effectively

i think this is it . this is actually part of the unit 4 syllabus i think.

Reply 56

littleone

Prometheus

Yes that i did get, and the others were 0.15/20 and 0.075/20 i believe

i gt the rest wrong :frown:

Reply 57

tazzalc

gfdsa

-the compund is trigonal planar
- nucleophile is able to attack from either side of the molecule .
or
-large (+) on carbon atom as attached to 2 highly electronegative atoms . -attracts nucleophile effectively

i think this is it . this is actually part of the unit 4 syllabus i think.