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OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

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Original post by supercrazyxo
Can't you try again?? Don't you get 3 tries with ISA??


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I did but i didnt do well in it :redface:
sry i dont have it on my computer. I just went to the ocr site and printed the NMR questions and it will take too long to scan it onto laptop :redface:
Hi everyone,

Hows everyone revision going?

I just wanted to ask does anybody know how to do Question 4c)ii) in the OCR June 2014 paper It is a really straightforward question but it seems so tricky to get the initial rate of reaction. For e.g. Experiment 3 the initial rate is 614. How on earth do they get that?
Heres the link:
http://www.ocr.org.uk/Images/243458-question-paper-unit-f325-01-equilibria-energetics-and-elements.pdf
(edited 7 years ago)
Original post by omarali1212
Hi everyone,

Hows everyone revision going?

I just wanted to ask does anybody know how to do Question 4c)ii) in the OCR June 2014 paper It is a really straightforward question but it seems so tricky to get the initial rate of reaction. For e.g. Experiment 3 the initial rate is 614. How on earth do they get that?
Heres the link:
http://www.ocr.org.uk/Images/243458-question-paper-unit-f325-01-equilibria-energetics-and-elements.pdf


They give you the rate question at the beginning of the question. Use that!!

In this question your comparing experiments.
So to find out the initial rate of experiment 3, you compare experiment 1 and 3.

Starting from [I-]
To get from 0.015 to 0.060 you x by 4
We know from the rate equation the [I-] is second order so we square the 4. times 4 squared by 0.60 (initial rate)
And you get X value

[IO3-]

To get from 0.010 to 0.040 to x by 4
We know from the rate equation that
[IO3-] is first order so it 4 to the power of 1 which is 4. 4 times X value to get Y value

[H+]

To get from 0.020 to 0.080 you times by 4. [H+] is second order so thats 4 squared. Y value times 4 squared is Z value which is the I initial rate for experiment 3.

You should get 614.4 and to the 3 SF it's 614.

Don't know of this makes sense.










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Original post by supercrazyxo
They give you the rate question at the beginning of the question. Use that!!

In this question your comparing experiments.
So to find out the initial rate of experiment 3, you compare experiment 1 and 3.

Starting from [I-]

To get from 0.015 to 0.060 you x by 4
We know from the rate equation the [I-] is second order so we square the 4. times 4 squared by 0.60 (initial rate)
And you get X value

[IO3-]

To get from 0.010 to 0.040 to x by 4
We know from the rate equation that
[IO3-] is first order so it 4 to the power of 1 which is 4. 4 times X value to get Y value

[H+]

To get from 0.020 to 0.080 you times by 4. [H+] is second order so thats 4 squared. Y value times 4 squared is Z value which is the I initial rate for experiment 3.

You should get 614.4 and to the 3 SF it's 614.

Don't know of this makes sense.










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Original post by supercrazyxo
They give you the rate question at the beginning of the question. Use that!!

In this question your comparing experiments.
So to find out the initial rate of experiment 3, you compare experiment 1 and 3.

Starting from [I-]
To get from 0.015 to 0.060 you x by 4
We know from the rate equation the [I-] is second order so we square the 4. times 4 squared by 0.60 (initial rate)
And you get X value

[IO3-]

To get from 0.010 to 0.040 to x by 4
We know from the rate equation that
[IO3-] is first order so it 4 to the power of 1 which is 4. 4 times X value to get Y value

[H+]

To get from 0.020 to 0.080 you times by 4. [H+] is second order so thats 4 squared. Y value times 4 squared is Z value which is the I initial rate for experiment 3.

You should get 614.4 and to the 3 SF it's 614.

Don't know if this makes sense.










Posted from TSR Mobile





Posted from TSR Mobile
Original post by supercrazyxo
They give you the rate question at the beginning of the question. Use that!!

In this question your comparing experiments.
So to find out the initial rate of experiment 3, you compare experiment 1 and 3.

Starting from [I-]
To get from 0.015 to 0.060 you times by 4.We know from the rate equation the
[I-] is second order so we square the 4. times 4 squared by 0.60 (initial rate)
And you get X value

[IO3-]

To get from 0.010 to 0.040 you times by 4.We know from the rate equation that
[IO3-] is first order so it 4 to the power of 1 which is 4. 4 times X value to get Y value

[H+]

To get from 0.020 to 0.080 you times by 4. [H+] is second order so thats 4 squared. Y value times 4 squared is Z value which is the I initial rate for experiment 3.

You should get 614.4 and to the 3 SF it's 614.

Don't know of this makes sense.










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How comes your up so early?

I kind of get it

Thanks for the reply.
On question 3)b)ii) on F324 January 2010 (MS here http://www.ocr.org.uk/Images/63346-mark-scheme-january.pdf) says "Ignore bond linkage for all groups" does this mean how you drew the bonds do not matter??
Original post by FutureMedic97
On question 3)b)ii) on F324 January 2010 (MS here http://www.ocr.org.uk/Images/63346-mark-scheme-january.pdf) says "Ignore bond linkage for all groups" does this mean how you drew the bonds do not matter??


I can't see where it says that on the ms but yes, I think that's right. Sometimes they say 'ignore connectivity issues' too, and it just means that it doesn't matter whether you've drawn R---H3N,( implying that the R group is bonded to hydrogen) or drawn it correctly. They seem to say that more often when it's a more complicated drawing, so if it isn't very complicated make very sure the drawing is accurate as possible so you don't lose any marks.
Fe2O3 + 3Cl2 + 10OH- --> 2FeO42- + 6Cl- + 5H2O
Start with two half equations. One for the Fe and one for the Cl2.
First one is right-ish, you have the idea. But you need to go from iron (III) oxide to ferrate (VI) so add the oxygens in and any hydroxide, electrons to balance.

Second one is missing the electrons.
Right, that's the general idea but you must balance it with the species mentioned in the question. You've done the oxidation states but need to incorporate the iron oxide, the ferrate and the hydroxide. Do you want me to post my worked through answer?
OK but you need to prove you could have done it without me having given you the final answer. :wink:
You need a systematic way to build up the half equations.
Do you want me to post the answer to this one then and you can see how I did it. then if you wanted to try a similar example look at Q8e in http://www.ocr.org.uk/Images/243458-question-paper-unit-f325-01-equilibria-energetics-and-elements.pdf
Well stay in your seat and I'll upload it. Deep breathing now! :smile:
Would you know how to balance it if it was in acid?
There is a slightly cheaty way to do it which is to balance as if it was in acid and then add OH- to both sides at the end to cancel out the H+ ions

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