rate eqation exam question Watch

usernamenew
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I don't know how to work out the order when both reactants are changing concentrations, is there a method to workit out? once ive found A is 2nd order, how do I work [B] which experiments do I look at? 1and2, 1and3? how do I decide what to do next
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usernamenew
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(Original post by usernamenew)
I don't know how to work out the order when both reactants are changing concentrations, is there a method to workit out? once ive found A is 2nd order, how do I work [B] which experiments do I look at? 1and2, 1and3? how do I decide what to do next
heres the quesiton
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usermeme1112
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(Original post by usernamenew)
I don't know how to work out the order when both reactants are changing concentrations, is there a method to work out? once I've found A is 2nd order, how do I work [B] which experiments do I look at? 1and2, 1and3? how do I decide what to do next
What is the Question? If you have worked out A then work out B by looking at 1 and 2 or 1 and 3 whichever one of them is the easiest e.g 1 and 2 may be x1.5 and 1 and 3 may be x3 which is easier to work out is your going to use interim rate as then you can look at the rate of 1 and 3 if A changes by x2 then you would times the rate of 1 by 2squared as A is 2nd order then times by 3 as 1 and 3 of B x3
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I'mComingOxford
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answer is a+b=c
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usernamenew
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(Original post by usermeme1112)
What is the Question? If you have worked out A then work out B by looking at 1 and 2 or 1 and 3 whichever one of them is the easiest e.g 1 and 2 may be x1.5 and 1 and 3 may be x3 which is easier to work out is your going to use interim rate as then you can look at the rate of 1 and 3 if A changes by x2 then you would times the rate of 1 by 2squared as A is 2nd order then times by 3 as 1 and 3 of B x3
i posted the question above, whats the interim rate , never heard of that
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charco
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(Original post by usernamenew)
heres the quesiton
Look at expt 1 & 2: [B ] is constant, so any change in rate is due to change in [A]

[A] triples and the rate increases by a factor of 18.9/2.1 = 9

Hence there is a squared dependency between [A] and the rate. In other words the order with respect to [A] is 2.

Now you can use this fact to find out how the rate should alter between experiments 2 and 3 due to change in [A].

[A] halves from expt 3 to expt 2 so you would expect the rate to decrease by a factor of 4. In fact it only decreases by a factor of 2. Hence the change in [B ] is causing a doubling of the rate. [B ] doubles from 3 to 2, so the change is commensurate with an order of 1.

Hence the rate equation is:

Rate = k[A]2[B ]
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usermeme1112
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(Original post by usernamenew)
I don't know how to work out the order when both reactants are changing concentrations, is there a method to workit out? once ive found A is 2nd order, how do I work [B] which experiments do I look at? 1and2, 1and3? how do I decide what to do next
There you go that's one method
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usernamenew
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(Original post by usermeme1112)
There you go that's one method
Thank u! So when u have to find B, u still have to look at the effect A also had and find the rate and then divide that new rate answer by 2 bc of B? How does this make it 1st order
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usernamenew
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(Original post by charco)
Look at expt 1 & 2: [B ] is constant, so any change in rate is due to change in [A]

[A] triples and the rate increases by a factor of 18.9/2.1 = 9

Hence there is a squared dependency between [A] and the rate. In other words the order with respect to [A] is 2.

Now you can use this fact to find out how the rate should alter between experiments 2 and 3 due to change in [A].

[A] halves from expt 3 to expt 2 so you would expect the rate to decrease by a factor of 4. In fact it only decreases by a factor of 2. Hence the change in [B ] is causing a doubling of the rate. [B ] doubles from 3 to 2, so the change is commensurate with an order of 1.

Hence the rate equation is:

Rate = k[A]2[B ]
Thank u
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usermeme1112
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(Original post by usernamenew)
Thank u! So when u have to find B, u still have to look at the effect A also had and find the rate and then divide that new rate answer by 2 bc of B? How does this make it 1st order
Yes!! you always have to look if [A] has an effect, well its First order because you divided it by 2 which is proportional to B as it goes from 24 to 12 which is divided by 2 if you divided the rate by 2 squared and got the same rate then it would be 2nd order reactant with respect to B.
If [A] effects the rate when your working out B you always have to times or divide the rate by whatever A is proportional to so if A is 2nd order and if it goes from 10 in experiment 1 to 30 in experiment 3, and B goes from 10 in experiment 1 to 20 in experiment 3 then it would be x2, the rate would first times by 3 squared as A is 2nd order and then after you got the answer for that then times it by B experiment 1-3 which is times by 2 which would be 1st order reactant.
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usernamenew
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(Original post by usermeme1112)
Yes!! you always have to look if [A] has an effect, well its First order because you divided it by 2 which is proportional to B as it goes from 24 to 12 which is divided by 2 if you divided the rate by 2 squared and got the same rate then it would be 2nd order reactant with respect to B.
If [A] effects the rate when your working out B you always have to times or divide the rate by whatever A is proportional to so if A is 2nd order and if it goes from 10 in experiment 1 to 30 in experiment 3, and B goes from 10 in experiment 1 to 20 in experiment 3 then it would be x2, the rate would first times by 3 squared as A is 2nd order and then after you got the answer for that then times it by B experiment 1-3 which is times by 2 which would be 1st order reactant.
thank u so much!!
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