GChem: Creating 4th experimental trial (rate laws)

[snaggletoof]

---

Members don't see this ad.

Hi All,
One thing that I have been having serious problems with is creating an extra trial in an experimental data set. I had this question pop up on the DAT for me once, and I feel like I totally missed it because I had no idea how to do it/understand the method. Here is an example if someone could explain it nicely.



So in the trial, B needs to be held constant.
I first convert the data from scientific notation to whole numbers:

Trial [A] Rate
1 1 1 1
2 1 2 4
3 2 4 16

Now, how do you make a 4th trial? Any help would be greatly appreciated!!!

 

[DQLEUCD]

Hi All,
One thing that I have been having serious problems with is creating an extra trial in an experimental data set. I had this question pop up on the DAT for me once, and I feel like I totally missed it because I had no idea how to do it/understand the method. Here is an example if someone could explain it nicely.



So in the trial, B needs to be held constant.
I first convert the data from scientific notation to whole numbers:

Trial [A] Rate
1 1 1 1
2 1 2 4
3 2 4 16

Now, how do you make a 4th trial? Any help would be greatly appreciated!!!

You don't need a 4th trial, Just solve for B and divide by B's factor to get the change due to A.

k^2 = rate law

Since when you double B, the rate goes up by 4.

Now trials 2 and 3 for A, since you doubled B, you divide the rate by 4 and get get the same rate therefore A = 0 order

 

[rmm30]

what if you wanted to use trials 1 and 3 to find order A? how does that work?

 

[DQLEUCD]

what if you wanted to use trials 1 and 3 to find order A? how does that work?

Same method. Once you have found the order for B. You divide the rates for 3 over rates for 1. You get an increase of 16. Now Divide B's concentration from 3 and 1. You get an increase of 4. Since you know that the order for B is 2. 4^2 = 16. If you double A, nothing happens so the order for A is 0.

again the rate law is k^2

 

[rmm30]

thanks a lot DQ. so mathematically using eq 2 and 3 it looks like this?

rate 3
-----= 4 = 2^m x 4 ; solve for m; m=0
rate 2