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# Kinetics A2. I AM IN URGENT NEED OF HELP! I HAVE BROKEN TWO BIRO PENS IN ANGER! watch

1. i dont understand. i want to cry.

how do i do this with an extra row? omg please help. im going to cry.....i haaaaaaaaate this
2. (Original post by Mathematics.)
but from row 2 to 3 aren't you halving a and doubling b?
You are, from 0.2 to 0.1 for A, and from 0.4 to 0.8 for B.
3. (Original post by Mathematics.)
did you use an extra row? i dont understand.
no.
just by comparing rows 2 and 3. just think if a and c were constant what effect wud b have on the reaction rate. if b dobles the rate will double. but it doesnt...it quadrouples. thats because a is being doubled. hence a doubling quadrouples the rate. so reaction order 2.
4. thats piss easy
5. (Original post by Mathematics.)
i dont understand. i want to cry.
sorry... maybe just take some time out and come back to the question in the morning or something?
6. (Original post by EoghanM)
thats piss easy
Brilliant input there. I consider myself dazzled by your intellect.
7. i can't believe i dont understand this ok, im going to keep doing this until i get it.
8. Hey OP... If you have the AQA textbook theres a formula on page 11 in the hint box... I used that formula to solve this question and I got it right.. so have a look and you should be fine once youve used it on a few questions !!! good luck!
9. Right so B is pretty clearly 1
From 2 -> 3, double B so rate should double (6.4) but in fact is a quarter of this value. As A has halved as 0.5^2 = 1/4
From 1-> 4, triple B and half A so expect 3/4 the rate [as 1*3 and 2^2 *1 (0.6) so C = 0
10. (Original post by Mathematics.)

I've been trying question 1.a) for one hour and a half...ONE HOUR AND A HALF AND ITS ONLY WORTH 3 ******* MARKS I'm so upset that i can't do this I'm about to cry. I don't know why I can't get it. How can you do a 3 mark question for more than an hour? Am I that stupid, really?

I found the order with respect to [B] = 1. I can't find [A] or [C]. My teacher taught us to add an extra row. please can someone help me/teach me? i'm not the brightest bulb in the box. please can anyone explain how to do this? I will do anything for help. Does anyone know about this extra row?

i know my rep is worth nothing, but i will repay anyone who helps me with this question!
this might be an easy way; if you are good with ratio.

let r = k[A]^a [B]^b [C]^c

then r2/r1 = [0.4/0.1]^b = 3.2/0.8, therefore b = 1

do another one, r3/r1 = (0.1/0.2)^a * (0.8/0.1)^1 * (0.4/0/.4)^c = 1.6/0.8

1 to the power of c = 1, therefore 2 = (1/2)^a * 8 so a = 2

finally, r4/r1 = (1/2)^2 * (3/1)^1 * (2/4)^c = 6/8
that will give you c = 0

therefore r = k[A]^2 [B]^1
11. (Original post by Mathematics.)
..
As general advice with these kinds of questions, I'd recommend checking out each row and seeing how you can easily deduce one of the rates.
For example, you can see from row 1 to 2 there is only a change in B, which means that you can deduce B's order from the given change in rate of the reaction (B is increased fourfold, reaction rate is increased fourfold => FIRST order).

With this information, you can then look at changes in sets of rows where B and one of A and C change (i.e. if only B and C change in concentration, you can deduce A's order). This is the case in the set of row 2 and row 3. You can check out what effect B has on this rate when doubling (= rate doubles from row 2 to row 3). Then, the difference in rates between row 2 and row 3 is 3.20 to 6.40. Yet, it's in fact 3.20 to 1.60, which is then caused by the only thing we haven't taken into account yet: A's effect on the rate. Apparently, A changes the rate in reaction from 6.40 to 1.60 when halving. The change in rate, however, is /4. From this, we can deduce that A's order is 2.

By the same approach, we can deduce the effect of C on the rate of the reactions.
Compare row 1 and 4 for example. A is halved, B is multiplied by 3. The effect these two have on the rate of reaction is known: A does (rate*1/4) and B does (rate*3) => rate*3/4. Now we compare the rates of reactions to see what the difference of [C] has on the rate:
rate 1 = 0.8
rate 4 = 0.6
This is the same as rate*3/4. Apparently, [C] has no effect on the rate of the reaction. => 0th order.

edit:
Seriously, check my post out thoroughly. I think it's pretty clear, even though you might not get it at first reading it.

Ok so B is 1, A is 2 and C is zero.

I think you already know how to find B..

Then for A, in rows 1 and 3, the rate multiplies by 8 to account for B to give 6.4, but then when divided by 1.6 = 4. A has been halved, and 2^2 = 4, so the order is 2.

And then for C, in rows 1 and 4, the rate multiplies by 3 to account for B to give 2.4, then divides by 4 to account for A, to give 0.6, and this is the actual rate of equation 4, hence the order with respect to C must be zero as it hasn't affected it.

Sorry if i'm just repeating someone.
13. I looked at that question and Im now praying this has nothing at all to do with chemistry.
14. Thanks everyone, I still don't understand. I'm taking a break and coming back to it. thanks
15. Still if anyone can explain using the "extra row" technique? or if anyone can explain in a super broken down way?
16. (Original post by Mathematics.)
Still if anyone can explain using the "extra row" technique? or if anyone can explain in a super broken down way?
there are often more than one ways to tackle these type of questions. if you are having difficulty with the one your teacher hinted, perhaps it is time you move on to the next best one you think you can always perform under exam conditions.
17. this is mind-boggling. I'm baffled.. flabbergasted.

WORDS LACK TO DESCRIBE MY AMAZEMENT
18. (Original post by phen)
As general advice with these kinds of questions, I'd recommend checking out each row and seeing how you can easily deduce one of the rates.
For example, you can see from row 1 to 2 there is only a change in B, which means that you can deduce B's order from the given change in rate of the reaction (B is increased fourfold, reaction rate is increased fourfold => FIRST order).

With this information, you can then look at changes in sets of rows where B and one of A and C change (i.e. if only B and C change in concentration, you can deduce A's order). This is the case in the set of row 2 and row 3. You can check out what effect B has on this rate when doubling (= rate doubles from row 2 to row 3). Then, the difference in rates between row 2 and row 3 is 3.20 to 6.40. Yet, it's in fact 3.20 to 1.60, which is then caused by the only thing we haven't taken into account yet: A's effect on the rate. Apparently, A changes the rate in reaction from 6.40 to 1.60 when halving. The change in rate, however, is /4. From this, we can deduce that A's order is 2.

By the same approach, we can deduce the effect of C on the rate of the reactions.
Compare row 1 and 4 for example. A is halved, B is multiplied by 3. The effect these two have on the rate of reaction is known: A does (rate*1/4) and B does (rate*3) => rate*3/4. Now we compare the rates of reactions to see what the difference of [C] has on the rate:
rate 1 = 0.8
rate 4 = 0.6
This is the same as rate*3/4. Apparently, [C] has no effect on the rate of the reaction. => 0th order.

edit:
Seriously, check my post out thoroughly. I think it's pretty clear, even though you might not get it at first reading it.
Hey, thanks for your post, (even though I still think you're kinda mean... ) your post helped me

So I read through everyones posts again and I kind of understand it. Like 80% I need to look for some more examples though! Thanks everyone!!!!!!! Exam's not 'til the 27th so I've got a bit of time to perfect everything
19. (Original post by phen)
this is mind-boggling. I'm baffled.. flabbergasted.

WORDS LACK TO DESCRIBE MY AMAZEMENT
20. (Original post by shengoc)
there are often more than one ways to tackle these type of questions. if you are having difficulty with the one your teacher hinted, perhaps it is time you move on to the next best one you think you can always perform under exam conditions.
lol I only know one method. I'm a very stupid person.

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Updated: January 15, 2010
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