Calculating rate law equation by determining initial rate of oxygen produced

[jmarsiglio]

Homework Statement

http://scc.scdsb.edu.on.ca/Students/onlinecourses/Sikora/sch4u/v12/64.pdf

That is the lab I have been working on. I've done the lab but I am answering the questions. I don't understand the Analysis question #1. How do I calculate the reaction rate for oxygen if I don't have its concentration?

Homework Equations

c = n/v
m = m/mr
r = k[h202]^m[NaI]^n

The Attempt at a Solution

Well I know it's completely wrong but I took the volumes recorded of the oxygen produced, converted to moles, divided by the volume of NaI + h202 which was 20ml and divided by 10 seconds (for each trial).

 

[jmarsiglio]

Okay, nevermind, the question specifically says to convert th volume of O2 within the time interval to get the reaction rate, so it can't be that, if so what would be the point of getting the values for the volume.

edit: wouldn't it be the same... what the heck is the point of getting these values though! ARGHHHHHHHH someone please help I'm being torn apart it's 1:50 am but I can't go to sleep pleasezzzzzzzzzz helppppp!

edit2: now that I think of it, since they are different trials, I think they are independent from each other, therefore d[A] and dt don't exist... I believe it is just the mol/L*S of the actual oxygen... hmmmm, do i figure out the moles of oxygen, divide by volume (what would it be... 20ml?) and by the time of the reaction(10 seconds in this case)

please tell me ill hug you

 

[Borek]

Have you read the question? It specifically says to use the stoichiometry of the reaction - oxygen is just a product, and you are interested in the rate of decomposition of the H₂O₂.

 

[jmarsiglio]

Yes, several times.

It says "For each experiment, determine the initial rate of reaction. That is convert the volume, or mass, or increase in pressure of O2 within the time interval to get an initial rate of mol O2/(L*S). Using the stoichiometry of the reaction, convert this initial rate to a rate of decomposition of H202 in mol/(L*S)".

There are 2 questions here. Find the intial rate for O2, then using this rate, stoichiometrically determine the rate of decomposition of H202 which would be double.

I'll try and explain the lab. All trial are independent from each other. That is, they are different concentrations they are left to react for the same time. But it is not one experiment recording at different times, it is 7 different experiments with different concentrations and each 10 seconds. What the lab wants me to do is figure the initial rate for each experiment using the data of volume recorded, with units mol/(L*S). Then I graph the initial rates where H202 is constant and determine what order the reaction is regarding NaI, and vice versa. I can do ALL of this else except I don't know how I am supposed to find the initial rates with the volume of O2 into mol/(L*S). I can get ml/S or mol/S but not mol/(L*S).

Maybe my first attempt was correct when I changed to moles, then to concentration, and divided by time?

The data I have is the volume and concentration of H202 and NaI, and the volume of O2 recorded.

Thanks

to clarify, the concentration was changed manually, as an independent variable to see what affect it would have on the amount of oxygen produced (independent variable).

 

[Borek]

You know how many moles of oxygen were produced, you know what was the volume of the solution, you know in what time, just combine these numbers to get correct units.

There is no such thing as oxygen concentration, as oxygen that you collect is not dissolved, it is a separate phase. However, when you use stoichiometry to find out how much hydrogen peroxide decomposed, you will be able to deal with concentration, as peroxide was in the the solution.

 

[jmarsiglio]

AMAZING! One more thing! The volume of the solution is the catalyst and the hydrogen peroxide? 20 ml in each case? Or just the h202: 10 ml?

 

[jmarsiglio]

I don't think the catalyst counts though, it just speeds up the process. So 10ml, I'll do this.
Correct?

 

[Borek]

No, to calculate concentration you need to know total volume. That is (approximately, in general volumes are not additive) sum of volumes of all samples mixed.

 

[jmarsiglio]

aha... I guess I have to multiply everything by two then -.-

Thanks.

(actually divide)