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ocr a f325 revision thread

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Original post by k1rby
The question said molar mass was 214.7 and it was [Ni(CN)6]3-. Potassium and nickel both form positive ions, they wouldn't form a complex together :smile:



For "Explain why H2O is zero order (1 mark)" another answer is that H2O isn't in the rate-determining step.

For the table about H2 concentration increase affect on equilibrium values, I googled it and I'm pretty sure the H2 would stay slightly greater, I2 would get smaller (because some reacted to decrease H2 concentration) and HI would get greater (because more products formed). :smile:


I agree!
Original post by otrivine
I put for the box


greater,smaller,greater

makes sense because there was more h2 and more HI as there was a squared on HI


same
Original post by keepontrying
I agree!


for the H2O if I said not in the rate equation as specie is not involved?
Reply 4443
I don't understand why the concentration of all reactants and products in the equilibrium wouldn't increase.

You have more H2, so the equilibrium shifts to the right. This decreases the concentration of both H2 and I2. HI is increased. However it is increased to a point where it is greater than both H2 and I2 - the equilibrium will thus shift to the left. This fluctuation keeps repeating until Kc is re-established. For it to be the same value, both the numerator and denomenator must increase - by the same proportions. That's what I'd thought at least.. but looking at this graph from google..

image014.gif

..where I can only assume A + B = C + D, it would seem in our example H2 and HI increase but I2 decreases. That question was so annoying..

15-10.jpg
(edited 10 years ago)
Reply 4444
i think is said H2 stays the same, I2 decreases and HI increases, i hope it s right :/
Original post by Tania2k9
i think is said H2 stays the same, I2 decreases and HI increases, i hope it s right :/


I did this, but not sure if it's right or not. My thinking was that the system wants to reverse the change. Reversing the change means restoring the previous value of H2, so that stays the same.
In order to do this, the concentration of HI needs to increase and the concentration of I2 needs to decrease for Kc to stay the same. I didn't think about it for long, so I'm probably not right. Now I think about it, if it did shift all the way to the right, then I2 will have decreased by too much, so it will shift back a bit and form a compromise between too much H2 and too little I2. This compromise leads to [H2] still being higher, [I2] being lower and [HI] being the same. So, I literally have no idea on this question. Oh well, it's only two marks :smile:
Reply 4446
Original post by Holz888
I did this, but not sure if it's right or not. My thinking was that the system wants to reverse the change. Reversing the change means restoring the previous value of H2, so that stays the same.
In order to do this, the concentration of HI needs to increase and the concentration of I2 needs to decrease for Kc to stay the same. I didn't think about it for long, so I'm probably not right. Now I think about it, if it did shift all the way to the right, then I2 will have decreased by too much, so it will shift back a bit and form a compromise between too much H2 and too little I2. This compromise leads to [H2] still being higher, [I2] being lower and [HI] being the same. So, I literally have no idea on this question. Oh well, it's only two marks :smile:


I understand the logic there, and for another for the entropy / enthalpy question is stupidly made a grave mistake by using delt as as my calculated standard entropy of glucose, and by using that i obtained a positive value and using that i concluded that delta S is more negative than delta H and delta G>0. I hope i have some marks , how much do u think?
Reply 4447
first dentecne i mean t delta S and dint use -256
Original post by Tania2k9
I understand the logic there, and for another for the entropy / enthalpy question is stupidly made a grave mistake by using delt as as my calculated standard entropy of glucose, and by using that i obtained a positive value and using that i concluded that delta S is more negative than delta H and delta G>0. I hope i have some marks , how much do u think?


If you wrote delta S on the answer line for S for glucose, I think you'd get ECF for the whole of the next Q and not lose any marks for that, possibly. Not sure about the last bit though
What do you think the grade boundaries for the chemistry coursework will be? and are they same every year?
Original post by A-New-Start
What do you think the grade boundaries for the chemistry coursework will be? and are they same every year?


Hoping super low lol. I got 31 and am praying that's enough for 40 UMS.

Coursework boundaries are generally VERY high because most students get told the answers by the teachers and therefore pretty much everyone does well on it.

I doubt I'll get less than 35 UMS, meaning I need a B in this exam to get my A overall.
i need at least 50 ums to get an A* , surely it's only AS grade boundaries that are high and A2 cwk's are harder to do well in?
Reply 4452
Someone should make a facebook complain page about how hard his paper was!!!
Stupid OCR dont want us to go to Uni!!!!
Reply 4453
Original post by emah123
Someone should make a facebook complain page about how hard his paper was!!!
Stupid OCR dont want us to go to Uni!!!!


I did OCR B chemistry F335 yesterday and it was a horrible paper. They've just decided to throw in some super nasty questions :frown:
I guess you can make a page, there's this complain page on Facebook on Edexcel C3 lol


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Reply 4454
Original post by emah123
Someone should make a facebook complain page about how hard his paper was!!!
Stupid OCR dont want us to go to Uni!!!!


It was hard but I think it was fair.

The paper involved more critical thinking rather than just memorizing information. If we just keep expecting easy papers which just test our memories, then just imagine how high the grade boundaries will go.

They have to try and differentiate between people who put the hard work in and who didn't, but they also have to see if people can think on their feet and tackle ambiguous data.
can anyone remember whether these were the figures used to find out the entropy of glucose (1230+x)-(1704)=-256 and then rearrange equation? do the numbers look familiar? x is to represent the glucose
Original post by keepontrying
can anyone remember whether these were the figures used to find out the entropy of glucose (1230+x)-(1704)=-256 and then rearrange equation? do the numbers look familiar? x is to represent the glucose


the entropy should have been positive not negative if I remember correctly
Original post by otrivine
the entropy should have been positive not negative if I remember correctly

nope it was negative that's why the reaction would never be feasible
the entropy of glucose was positive I got +218 im just trying to remember the equation I got it from the "-256" is the entropy change not the entropy of glucose
(edited 10 years ago)
Original post by keepontrying
nope it was negative that's why the reaction would never be feasible
the entropy of glucose was positive I got +218 im just trying to remember the equation I got it from


yes I got +218 for the entropy of glucose? which question are u talking about sorry the reasoning
Original post by otrivine
yes I got +218 for the entropy of glucose? which question are u talking about sorry the reasoning
The method I got the entropy change of glucose I basically did the equation of sum of the entropy of products- sum of entropy of reactants= =-256 and then rearranged to make x the subject which was the entropy of glucose I was just wondering if anyone used the same method, :colondollar:

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