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OCR B F335 - Chemistry by Design - 13th June 2012

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Original post by master_blaster66
january 2006 bro


I'll do the paper tonight, where is the grade boundaries for it? I can't find them. I took a quick glance at it and it was trickier than normal, the grade boundaries should be low.
Original post by Bi0logical
What type of bonding would occur between the Sudan and Triglyceride then bro?


if it is to be soluble, ion-dipole bonds should form.
Original post by Ilyas
I'll do the paper tonight, where is the grade boundaries for it? I can't find them. I took a quick glance at it and it was trickier than normal, the grade boundaries should be low.


thats what i thought but the boundaries are not low...85/120 for an A :eek:
Original post by Ilyas
if it is to be soluble, ion-dipole bonds should form.


Ah right, I need to take a look at the question on the paper and see the molecule so I'll understand it.
Does anyone have any predictions on this exam? Even though they tend to ask similar stuff.
Original post by Bi0logical
Ah right, I need to take a look at the question on the paper and see the molecule so I'll understand it.
Does anyone have any predictions on this exam? Even though they tend to ask similar stuff.


well im just hoping that a paper like the jan f334 paper doesnt come up, and for f335 it will be even harder.:s-smilie:
Original post by Bi0logical
Why some molecules cannot bond with non-polar substances such as petrol;
The ion-dipole bonds would form between the non-polar substances, these would be very weak ion/dipole bonds;
this is because its a polar substance (e.g. alkane) so bonds formed are weak.
More energy would therefore be needed to break the actual molecule lattice than would be gained when the individuals ions undergo solvartion; this means the process would be endothermic.


huh how do ion dipole bonds form between non polarsubstances, i swear ion dipoles only form when the ion lattice breaks up and then then water molecules surround the ions??
i think this site is very helpful if someone needs some help with intermolecular bonds
http://www.elmhurst.edu/~chm/vchembook/170solutions.html
Reply 247
Avatar for nosh1994
nosh1994
guys, for january 2011, question 4c) iii) what on earth? i don't get how were meant to draw an energy diagrams nd where that is in c.i? help is much appreciated :smile:
Original post by nosh1994
guys, for january 2011, question 4c) iii) what on earth? i don't get how were meant to draw an energy diagrams nd where that is in c.i? help is much appreciated :smile:


you gotta use your f332 knwoledge here, this is the stuff where you learnt about catalyst lowering activation enthalpy and so less energy is needed for a successful reaction to occur. in the CI it should be near the collision theory stuff
Original post by master_blaster66
huh how do ion dipole bonds form between non polarsubstances, i swear ion dipoles only form when the ion lattice breaks up and then then water molecules surround the ions??

Yeah they cant, but when they do their really weak, I think?
Original post by Bi0logical
Yeah they cant, but when they do their really weak, I think?


hmm yeah ive seen that in the markschemes too, its soo confusing man..F335 is frustrating my life :frown:
'like dissolves like"
substances that are mostly polar (like glucose) dissolve in polar solvents like water but not in a nonpolar solvent like hexane, presumably since the dipole forces (specifically H-bonds) holding polar solute molecules together can be replaced with polar solute-solvent interactions (H-bonds).
substances that are mostly nonpolar dissolve ion nonpolar solvents like hexane but not in polar solvents like water, presumably since the id-id forces holding nonpolar solute molecules together can be replaced with nonpolar solute-solvent interactions.
Molecules that are both about equal amount of polar and nonpolar parts may or maynot dissolve readily in different solvents.
Reply 252
Avatar for nosh1994
nosh1994
Original post by master_blaster66
you gotta use your f332 knwoledge here, this is the stuff where you learnt about catalyst lowering activation enthalpy and so less energy is needed for a successful reaction to occur. in the CI it should be near the collision theory stuff


oh, i haven't looked at f332 in ages, since i only retook f331, just found the page in c.i, thanks a lot! :smile:

p.s- do we need to know how to do all the experiments which we learnt in f334? (titration, chromatography, recrystallisation)
Original post by nosh1994
oh, i haven't looked at f332 in ages, since i only retook f331, just found the page in c.i, thanks a lot! :smile:

p.s- do we need to know how to do all the experiments which we learnt in f334? (titration, chromatography, recrystallisation)


ive never seen them coming up in f335 so i hope not
Reply 254
Avatar for examtalk
examtalk
can anyone help me with this question...

At 500 K the value of ΔStot for the forward reaction is –1784.
Calculate the value of ΔStot at 1000 K.
Assume that ΔSsys does not change with temperature (the value is +216)

this is the correct workings according to the mark scheme...
–1784 = +216 – ΔH/500 , thus ΔH = +1 x 106 J mol–1
ΔS = +216 – 1 x 106 /1000 = –784 J K–1 mol–1

the only thing i'm not sure on is how they got that ΔH = +1 x 106 J mol–1 and I don't see how 216 - +1x106/500 = –1784 (my calculator gets the answer 215.788)

thanks for the help in advance and yeh i'm probably just being silly and missing out something obvious :smile:

i should also say that the equation is...
CH4(g) + H2O(g) ---- CO(g) + 3H2(g)
(edited 11 years ago)
If you have the OCR Salters revision guide, there's a page with a list of synoptic stuff u need to know.
It's mainly the basics; electron config, different types of bonding, shapes, affects of catalyst/rate, etc etc
Original post by examtalk
can anyone help me with this question...

At 500 K the value of ΔStot for the forward reaction is –1784.
Calculate the value of ΔStot at 1000 K.
Assume that ΔSsys does not change with temperature (the value is +216)

this is the correct workings according to the mark scheme...
–1784 = +216 – ΔH/500 , thus ΔH = +1 x 106 J mol–1
ΔS = +216 – 1 x 106 /1000 = –784 J K–1 mol–1

the only thing i'm not sure on is how they got that ΔH = +1 x 106 J mol–1 and I don't see how 216 - +1x106/500 = –1784 (my calculator gets the answer 215.788)

thanks for the help in advance and yeh i'm probably just being silly and missing out something obvious :smile:

i should also say that the equation is...
CH4(g) + H2O(g) ---- CO(g) + 3H2(g)


I think they meant +1x10^6, not +1x106. I did this question yesterday and got it right, it must be some typo in the MS.

Stot = 216 + -(1x10^6)/1000 = -784
Two assumptions made in buffer calculations:
- All the anions (A-) have come from the salt, so the contribution from the acid is negligible
- The concentration of the acid in the solution [HA(aq)] is the same as the amount of acid put into the solution - in other words ignore any dissociation
Original post by nosh1994
oh, i haven't looked at f332 in ages, since i only retook f331, just found the page in c.i, thanks a lot! :smile:

p.s- do we need to know how to do all the experiments which we learnt in f334? (titration, chromatography, recrystallisation)


I think we only have to know about GLC and the NMR stuff... but its a synoptic paper so you never know... But I highly doubt that it would
Reply 259
Avatar for examtalk
examtalk
Original post by Ilyas
I think they meant +1x10^6, not +1x106. I did this question yesterday and got it right, it must be some typo in the MS.

Stot = 216 + -(1x10^6)/1000 = -784


ohh haha silly me, thank you that makes much more sense :smile:

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