Posted by Anonymous on Sunday, February 10, 2008 at 11:53pm.
I don't have time to go through the whole thing; however, here is how you do part of it. I think that will get it for you.
First, look for two experiments in which there is the same concentration of ONE of the reactants. You see #1 has 0.03 for OH and #3 has 0.03 for OH. So let's pick those two to start.
The reaction has this form for the rate.
rate = k(ClO2)^x((OH^-)^y.
We want to determine x and y, the exponents.
rate 1 = 0.0248 = k(0.06)^x(0.03)^y
rate 2 = 0.00276 = k(0.02)^x(0.03)^y
Divide rate 1 by rate 2.
You can do it on paper and follow through to keep me from typing a bunch of stuff.
You see the 0.03^y term top and bottom cancel (that's why we chose one that had the same concentration for one of the reactants---they will cancel).You also see that k cancels (that's the rate constant). And 0.0248/0.00276 = 8.985 so now we have
8.985 = (0.06)^x/(0.02)^x
That is 8.985 = (0.06/0.02)^x
That is 8.985 = (3)^x
Now we take the log of both sides.
log 8.985 = 0.9535
log (3)^x = x*log 3 = x*0.477
So now we have
x*0.477 = 0.9535
Then x = 0.9535/0.477 =1.9989 which rounds to 2. So the exponenet x = 2.
Next you do the same thing but choose experiment 2 and 3 (because in those two reactions, (ClO2) are the same. Go through exactly the same procedure to determine y, the exponent for OH. The k cancels (top and bottom), the ClO2 cancels (top and bottom) leaving just the OH^y part. I determined y to be 1.0 so you will know if you did that part right if you find 1.0.
When you know x and y, then choose ANY experiment you wish, plug in the value from the table of (ClO2) and (OH), plug in your values for x and y, and determine k. Then you will have determined x,y, and k, and those are the only unknowns in the experiment. I hope this helps.