[A5HLEY]

Ok, this is kinda long, but please bear with me! It took me a while to work on this and I was just curious to see if I could get someone to check my work!

Ok: The following data were measured for the reaction BF3(g) + NH3(g) --> F3BNH3(g)

Experiment       [BF3](M)           [NH3]M     Initial Rate(M/s)
1                       .250                 .250             .2130
2                       .250                 .125             .1065
3                       .200                 .100             .0682
4                       .350                 .100             .1193
5                       .175                 .100             .0596

a). Determine the rate law for this reaction.
    I used the formula m(or n) = [log(R1/R2)]/[log([A]1/[A]2)]
So I used experiments 2 and 3, and calculated the following:
m=[log(.1065/.0682)]/[log(.250/.200)] = 1.997
n=[log(.1065/.0682)]/[log(.125/.100)] = 1.997

So I found that the rate law = k[BF3]^2[NH3]^2

b). What is the overall order of this reaction?
I found it to be four based on my previous answer, by adding up the orders. (2+2)

c). What is the value of the rate constant for this reaction?
I used the formula rate=k[A]^m[B ]^n and found that (.0682)/[(.2)^2][(.1)^2] = 170.5


If someone could check that for me, I would be FOREVER grateful! Thanks!

[Rabn]

Nice to see someone actually post their work in a readable manner. I didn't check the actual math, I figure you've already at least double checked the calcs., but your method looks good. Nice job.

[A5HLEY]

Ok, so when I submitted this to WebAssign, it said the entire problem was wrong. I have two more chances to submit it.

Does anyone see where I could've gone wrong?

[enahs]

The way you typed up how you worked it (the values you used) is wrong.

To compare, the values of one component must be constant.
I.E. If you are solving for M, N must be the same in each case. And if you are solving for N, M must be the same. However, you did not do that in your type up.

http://www.saskschools.ca/curr_content/chem30/modules/module4/lesson3/methodofinitialrates.htm
http://www.chm.davidson.edu/ChemistryApplets/kinetics/MethodOfInitialRates.html

We can check before you hit submit again. Read those and try again.

[A5HLEY]

The way you typed up how you worked it (the values you used) is wrong.

To compare, the values of one component must be constant.
I.E. If you are solving for M, N must be the same in each case. And if you are solving for N, M must be the same. However, you did not do that in your type up.

http://www.saskschools.ca/curr_content/chem30/modules/module4/lesson3/methodofinitialrates.htm
http://www.chm.davidson.edu/ChemistryApplets/kinetics/MethodOfInitialRates.html

We can check before you hit submit again. Read those and try again.

Ok, I get what you're saying. Thanks! I'll work it again and post what I got here.

[A5HLEY]

Ok, so I redid the problem. Here's what I got:


m=[log(.0682/.1193)]/[log(.200/.350)] = .99925 = 1
n=[log(.2130/.1065)]/[log(.250/.125)] = 1

So the rate law is rate=k[BF3][NH3], which is an overall second order reaction.

And when I plugged those numbers into k=(rate)/([A]^m[B ]^n), I got (.1065)/(.250*.125) = 3.408

Does that look better?




By the way, I have a second problem that's similar. The reaction is:
S2O8^(2-)(aq) + 3I^(-)(aq) --> 2SO4^(2-)(aq) + I3(-)(aq)

But the thing is, in that data that I'm given, I could calculate the m because there are two experiments in which [B ] is constant, but there aren't two experiments where [A] is constant, so I can't calculate n. Do I have to do something with the stoichiometry?

Basically, I think I understand the general concept now, but I don't know what to do if there aren't two experiments that have a constant concentration.

[sjb]

By the way, I have a second problem that's similar. The reaction is:
S2O8^(2-)(aq) + 3I^(-)(aq) --> 2SO4^(2-)(aq) + I3(-)(aq)

But the thing is, in that data that I'm given, I could calculate the m because there are two experiments in which [B ] is constant, but there aren't two experiments where [A] is constant, so I can't calculate n. Do I have to do something with the stoichiometry?

Basically, I think I understand the general concept now, but I don't know what to do if there aren't two experiments that have a constant concentration.

If you have something like (using simple figures for ease)
[A]                       [C]                       rate
1                          1                           1
1                          2                           4
2                          3                           18

which is what I think you mean then I think you can use the fact that if [C] is doubled the rate is quadrupled and hence second order to assume a not listed experiment that reads
[A]                       [C]                       rate
1                          3                           9

which you can then combine with the third row of your knowns to get order wrt A.

I'm not 100% sure though, as I think you ought most of the time to be able to get all the info from the data given only, as if you screw up the first set of calculations you may be chasing tails trying to get data to fit for the second.

Also, in other examples the rate determining step (or indeed the mechanism as a whole) may change with concentration changes so again I'm not sure it's a valid approach all of the time.

Order has nothing really to do with stoichiometry, sometimes it appears to be the same but that is just coincidence.

S

[A5HLEY]

By the way, I have a second problem that's similar. The reaction is:
S2O8^(2-)(aq) + 3I^(-)(aq) --> 2SO4^(2-)(aq) + I3(-)(aq)

But the thing is, in that data that I'm given, I could calculate the m because there are two experiments in which [B ] is constant, but there aren't two experiments where [A] is constant, so I can't calculate n. Do I have to do something with the stoichiometry?

Basically, I think I understand the general concept now, but I don't know what to do if there aren't two experiments that have a constant concentration.

If you have something like (using simple figures for ease)
[A]                       [C]                       rate
1                          1                           1
1                          2                           4
2                          3                           18

which is what I think you mean then I think you can use the fact that if [C] is doubled the rate is quadrupled and hence second order to assume a not listed experiment that reads
[A]                       [C]                       rate
1                          3                           9

which you can then combine with the third row of your knowns to get order wrt A.

I'm not 100% sure though, as I think you ought most of the time to be able to get all the info from the data given only, as if you screw up the first set of calculations you may be chasing tails trying to get data to fit for the second.

Also, in other examples the rate determining step (or indeed the mechanism as a whole) may change with concentration changes so again I'm not sure it's a valid approach all of the time.

Order has nothing really to do with stoichiometry, sometimes it appears to be the same but that is just coincidence.

S

Ok, thanks for your advice.

Oh, and does my retry of that first problem look better?

[enahs]

sjb is correct. If you have two unknown reaction orders, and can solve for one; if you assume that it is a well behaved and nice rate function, you can possibly make up your own with the data given.
http://www.chemicalforums.com/index.php?topic=22945.0

Oh, and does my retry of that first problem look better?

You tell me? K is a constant. So if you plug your value of K, with the reaction orders you found, for each condition (Trials) concentration, you should get the rate given in each condition (trial), correct?

Or, another way is, if you use the reaction orders found, and solve each condition (Trials) for K, they should all be the same, right?

Are they? I could tell you. But it is good to think about things like this, so you can check yourself on a test. Or check your data, maybe 4 of 5 agree and one is bad, or not bad but you found a condition where it is not nicely behaved, etc.