# Chem

posted by .

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.

• Chem -

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.

## Similar Questions

1. ### chemistry

Chlorine dioxide, ClO2, has been used as a disinfectant in air conditioning systems. It reacts with water according to the following equation. 6ClO2 + 3H2O --> 5HClO3 + HCl If 142.0 g of ClO2 is mixed with 33.0 g of H2O, how many …
2. ### Chem Help .plz

pre-lab questin.. part a was to determine the rate law from the given table... finally i was able to calculate ..rate law which was k[ClO2]2[OH-] since i got m =2 n=1 ..overall order was 3 ... the equation was .. 2ClO2(aq) + 2OH-(aq)---ClO3-(aq) …
3. ### CHEMISTRY

i need help with this pre lab question:| plse help me! teh following data were obained for the reaction 2ClO2(aq)+ 2OH^-(aq) -> ClO3^-(aq) + ClO2^-(aq) +H2O (l) Where rate = delta[ClO2]/ delta t (mol/L) (mol/L) (mol/L*s) [ClO2]0 …
4. ### Chemistry

i need help with a rate law question: The reaction is 2ClO2(aq) + 2OH -(aq)---> ClO3-(aq)+ ClO2-(aq) + H2O(l) Experiment 1: [ClO2]0= 0.0500 [OH-]0= 0.100 INITIAL RATE= 5.75*10^-2 Experiment 2: [ClO2]0= 0.100 [OH-]0= 0.100 INITIAL …
5. ### chemistry

Chlorine dioxide, ClO2, has been used as a disinfectant in air conditioning systems. It reacts with water according to the following equation. ... HCl If 142.0 g of ClO2 is mixed with 33.0 g of H2O, how many grams of which reactant …
6. ### chemistry

Given the table of Ka values on the right below, arrange the conjugate bases in order from strongest to weakest. Acid Ka HClO 3.5 e-8 HClO2 1.2 e-2 HCN 6.2 e-10 H2PO4- 6.2 e-8 A. ClO2-, ClO¬-, HPO42-, CN- B. ClO2-, HPO42-, ClO¬-, …
7. ### chemistry

If HClO2 is a stronger acid than HF, which is stronger than HOCl, then the order of strengths of the conjugate bases of these acids is ______ < _____ < _____ a) ClO- < F- < ClO2- b) ClO2- < ClO- < F- e) F- < ClO2- …
8. ### chem 1046

An electrochemical cell is based on these two half-reactions: Ox: Sn(s) → Sn2+(aq, 1.74 M) + 2 e- Red: ClO2(g, 0.120 atm) + e- → ClO2-(aq, 1.44 M) Calculate the cell potential at 25°C E Sn>Sn2+ = -.14 E ClO2 = .95
9. ### Chem 1046

An electrochemical cell is based on these two half-reactions: Ox: Sn(s) → Sn2+(aq, 1.74 M) + 2 e- Red: ClO2(g, 0.120 atm) + e- → ClO2-(aq, 1.44 M) Calculate the cell potential at 25°C E Sn>Sn2+ = -.14 E ClO2 = .95