Kinetics Q help! A level

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Bhavika22
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#1
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#1
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Size:  385.9 KB Hey guys, i need help on Q3b(ii) for part b(i) of the same question i calculated that the order of reaction of E is 2. I'm unsure on how to go about calculating the order of reaction for F. I'm more imterested in knowing a way to calculate questions like this thus as well as the answer, please submit some form of method/working out

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LilacFire
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#2
(Original post by Bhavika22)
Name:  image.jpg
Views: 199
Size:  385.9 KB Hey guys, i need help on Q3b(ii) for part b(i) of the same question i calculated that the order of reaction of E is 2. I'm unsure on how to go about calculating the order of reaction for F. I'm more imterested in knowing a way to calculate questions like this thus as well as the answer, please submit some form of method/working out

Thanks
First of all, so you do A-level or AS Maths?
because my description may involve some things that will be easy to understand for maths students but will be more difficult if you don't do maths. So yeah let me know and I'll help you
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Bhavika22
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(Original post by LilacFire)
First of all, so you do A-level or AS Maths?
because my description may involve some things that will be easy to understand for maths students but will be more difficult if you don't do maths. So yeah let me know and I'll help you
Hi, i do A Level Chemistry and Maths
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LilacFire
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(Original post by Bhavika22)
Hi, i do A Level Chemistry and Maths
Great! The reason I asked is because I do this by using algebra. For that particular question, let the order of E be E^2 as you have already calculated, right? Usually they give you a table where only one of the values change at any one time. So for this question use experiment 1 and 2 to find out the order of E. from expt 1 to expt 2, the concentration changes 36/24 times i.e 3/2. Multiply the initial rate by 3/2, which gives 0.012 which is not enough. So square the 3/2. multiply 9/4 by the initial rate and you get 0.018. Eureka!
Okay to do find F we do the same, except now we have to also keep in mind E because there are no experiments where E stays constant. so between expt 1 and 3, E is doubled (x2) but since it is order 2, we multiply by the increase SQUARED i.e 2^2=4. Multiply the rate of expt 1 by this new constant - 0.8x10^-2 x 4=3.2x10^-2. Here we see that it is the rate of experiment 3. So we made it easy, F does not affect the rate of reaction at all. It has order 0.

Okay, I know that is a lot of writing and I'm not sure if that will make any sense to you, but please ask questions if you don't understand. To recap, the steps are:
1. Look at the table to see if there are any experiments in which the concentrations are kept constant. (there usually is at least one pair.)
2. Calculate the increase between the experiment 1 and 2 i.e is the concentration doubled, halved, three times as big etc
3. multiply the initial rate of expt 1 by this increase and check if the answer matches the one in the table.
4. If it doesn't match, square the increase and repeat step 3. The order never goes higher than 2 in our specification so no need to cube it or anything.
5. The value should now match one in the table. The order of the compound is the power you raised the increase to (so in our example it was 2)

Hope this helps!
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Bhavika22
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Thank you so so so much! This helped so much that i had to bookmark it if you don't mind me asking, how do you go about revising Chemistry?
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LilacFire
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(Original post by Bhavika22)
Thank you so so so much! This helped so much that i had to bookmark it if you don't mind me asking, how do you go about revising Chemistry?
Usually, I just do lots of practice questions. I am also lucky to have a Chemistry teacher who has tons of resources so if I don't understand something, he usually has another powerpoint or something that explains it in a different way. I also like to do an exam paper and then mark it myself ( no cheating of course ), as that shows the exact parts needed to get the marks in the question.

It was my pleasure! If you ever need help with any other chemistry (or maths) questions, please ask!
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