# AQA Chem Unit 4 Order of a Reaction Kinetics Question

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Thread starter 6 years ago
#1
Can anyone explain how to work out the order of A, B and C in question 1a:

Question Paper: http://filestore.aqa.org.uk/subjects...W-QP-JAN07.PDF

Mark Scheme: http://filestore.aqa.org.uk/subjects...W-MS-JAN07.PDF

Thank you very muchh
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6 years ago
#2
(Original post by Mel1312)
Can anyone explain how to work out the order of A, B and C in question 1a:

Question Paper: http://filestore.aqa.org.uk/subjects...W-QP-JAN07.PDF

Mark Scheme: http://filestore.aqa.org.uk/subjects...W-MS-JAN07.PDF

Thank you very muchh
Between reaction 1 and 2

Rate x 4, concentration of B is multiplied by 4 too, and there are no other changes. So the order of the reaction with respect to B is 1

Between reactions 2 and 3, the rate is multiplied by 1/2.

The concentration of B is multiplied by 2, so rate should be multiplied by 2. But it is 1/2. So the change in concentration of A must affect it by 1/4. But the A concentration of A is halved. So the order of the reaction with respect to A must be 2.

between 3 and 4
rate = x 3/8 of the initial
A's concentration doesn't change
B's concentration is x 3/8 of reaction 3, so rate must change to become 3/8th of the original rate of reaction 3.

So far, rate has been x 3/8, which ties up with the change in rate between 3 and 4. So C has no effect, and the order of reaction with respect to C = 0

Rate = k[A]2[B]

If you have any questions, feel free to ask.

Hope this helps
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