2ClO2 + 2OH- --> ClO2- + H2O

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 through this one I would appreciate it. Thanks.

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.

To determine the reaction order of [ClO2], we need to compare the rates of the reaction under different concentrations of [ClO2] while keeping the concentration of OH- constant.

Let's compare Experiment 1 and Experiment 2 first. In these experiments, the concentration of OH- is the same (.030 M), but the concentration of [ClO2] is different.

Experiment 1:
M of [ClO2] = 0.060 M
Rate of disappearance of ClO2 = 0.0248 M/s

Experiment 2:
M of [ClO2] = 0.020 M
Rate of disappearance of ClO2 = 0.00276 M/s

To determine the reaction order with respect to [ClO2], we can use the rate equation:

Rate = k * [ClO2]^x * [OH-]^y

Since the concentration of OH- is constant and the same in both experiments, we can ignore it for now. Let's compare the rates of the two experiments:

Rate1 / Rate2 = (k * [ClO2]1^x * [OH-]^y) / (k * [ClO2]2^x * [OH-]^y)

Simplifying:

Rate1 / Rate2 = ([ClO2]1^x) / ([ClO2]2^x)

Plugging in the values from the experiments:

0.0248 M/s / 0.00276 M/s = (0.060 M)^x / (0.020 M)^x

Simplifying further:

(x = reaction order of [ClO2])

8.9855 = (3)^x

To solve for x, we can take the logarithm of both sides:

log(8.9855) = log(3^x)

Using logarithmic properties:

log(8.9855) = x * log(3)

Solving for x:

x = log(8.9855) / log(3)
x ≈ 1.55

So, the reaction order of [ClO2] is approximately 1.55.

To determine the reaction order of [ClO2], we can use the rate data from the experiments provided. The reaction order of [ClO2] can be determined by comparing the rates of reactions when the concentration of [ClO2] is changed while keeping the concentration of [OH-] constant. The rate equation for this reaction can be written as:

Rate = k[ClO2]^a[OH-]^b

Where:
- Rate is the rate of disappearance of ClO2 (M/s) as given in the experiments.
- [ClO2] is the concentration of ClO2.
- [OH-] is the concentration of OH-.
- k is the rate constant.
- a and b are the reaction orders with respect to [ClO2] and [OH-], respectively.

To determine the reaction order of [ClO2], we need to compare the rates of ClO2 disappearance in experiments 1 and 2, while keeping [OH-] constant.

For Experiment 1:
[ClO2] = 0.060 M
Rate = 0.0248 M/s

For Experiment 2:
[ClO2] = 0.020 M
Rate = 0.00276 M/s

By comparing these two experiments, we can set up a ratio to determine the reaction order of [ClO2]:

Rate1 / Rate2 = ([ClO2]1 / [ClO2]2)^a

Substituting the given values:
0.0248 / 0.00276 = (0.060 / 0.020)^a

Simplifying the ratio gives:
8.99 = 3^a

To solve for 'a', take the logarithm (base 3) of both sides:
log3(8.99) = a

Using a calculator, we find:
a ≈ 1.5849

Therefore, the reaction order of [ClO2] is approximately 1.585.

Explanation:
In this case, we determined the reaction order of [ClO2] by comparing the rates of reaction for two different concentrations of ClO2 while keeping the concentration of OH- constant. The reaction order, 'a', was determined by setting up a ratio of the rates and solving for 'a' using logarithms.