Tuesday

September 30, 2014

September 30, 2014

Posted by **Anonymous** on Sunday, February 10, 2008 at 11:53pm.

Experiment 1, M of [ClO2] = .060

M of [OH-] = .030

Rate of disappearance of ClO2 (M/s) = .0248

Experiment 2, M of [ClO2] = .020

M of [OH-] = .030

Rate of disappearance of ClO2 (M/s) = .00276

Experiment 3, M of [ClO2] = .020

M of [OH-] = .090

Rate of disappearance of ClO2 (M/s) = .00828

What is the reaction order of [ClO2]? Expain.

I have a bunch of these questions so if someone could walk me thru this one I would appreciate it. Thanks.

- Chem -
**DrBob222**, Monday, February 11, 2008 at 12:59amI 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.

**Answer this Question**

**Related Questions**

Chemistry - i need help with a rate law question: The reaction is 2ClO2(aq) + ...

Chem Help .plz - pre-lab questin.. part a was to determine the rate law from the...

CHEMISTRY - i need help with this pre lab question:| plse help me! teh following...

chemistry - If HClO2 is a stronger acid than HF, which is stronger than HOCl, ...

chemistry - Given the table of Ka values on the right below, arrange the ...

Chem 1046 - An electrochemical cell is based on these two half-reactions: Ox: Sn...

chem 1046 - An electrochemical cell is based on these two half-reactions: Ox: Sn...

chemistry - Chlorine dioxide, ClO2, has been used as a disinfectant in air ...

chemistry - Chlorine dioxide, ClO2, has been used as a disinfectant in air ...

science - How would you balance this reaction(it's a half reaction problem): ...