# High school chemistry EEI

1. May 25, 2012

### thestudent101

I'm doing an EEI on enzymes. I'm using catalase and hydrogen peroxide to produce oxygen. The oxygen is then collected into a measuring cylinder. My three variables are the concentration of catalase, concentration of hydrogen peroxide and the temperature of both. Increasing the concentration of catalase is supposed to speed up the reaction, which it has, but it is meant keep the amount of oxygen produced the same, which it also has. I have calculated the theoretical yield and our actual yield is really close. My question is, how do you calculate the theoretical time the reaction should occur for. I asked my teacher if there is a way to do it and he said that there is however it is beyond a highschool kid. I get A+'s in chemistry and top the class, so I would really like to impress him by being able to calculate the theoretical time of each reaction.
Catalase concentrations are 5%, 10%.....30% with 1% hydrogen peroxide, both kept at 40°C.

Last edited: May 25, 2012
2. May 26, 2012

### Staff: Mentor

Really that high?

In general you should start from rate laws (http://en.wikipedia.org/wiki/Rate_law), integrate them to find out amount of hydrogen peroxide consumed as a function of time, then calculate how long it will take consume all hydrogen peroxide present.

You will find that this time theoretically equals infinity. In reality observable reaction goes till the concentration of the hydrogen peroxide falls below some level, but what "some" means in this context depends on many things like concentration, oxygen solubility in the solution, mixing and so on.

It will be much easier to calculate time required to reach 99% or even 99.9%.

3. May 28, 2012

### thestudent101

Thankyou for replying and sorry for not replying back quickly. Anyway about the catalase concentrations. Do you think that they are too high? They have been mixed with 1% hydrogen peroxide. And the mean time for each set of experiments has decreased in a fairly linear fashion, starting from 11.3s with 5% and going down to 3.3s with 30%.

I've had a look at the wikipedia site you suggested. I sort of understand it. From what I can gather this is a second order reaction? But I'm really not sure how to actually calculate anything.

4. May 28, 2012

### Staff: Mentor

I am sure they are. What you have listed is mots likely result of dilution calculation that assumes you used a pure catalase. That's not the case. You have not used a pure catalase, but an already diluted solution. To calculate the real concentration you should know initial concentration and calculate its concentration after dilution.

Yes, this is a second order reaction. Unfortunately, to calculate anything you would need at least some basic knowledge of calculus (unless you will be able to find the integrated forms of equations).

5. May 28, 2012

### thestudent101

I watched a youtube video on rate laws, it helped a fair bit. What I have done so at the moment is calculated the molarity of hydrogen peroxide used with each set of tests. The first with 1.66mL (1%) is 4.18x10^-3 mol, the second using 3.33mL (2%) is 8.4x10^-3 and so on...But I'm unsure on how to calculate the catalase. My teacher said that it was a bottle of pure catalase. Why do you think its been pre-diluted? Anyway, if it was originally 100% concentration, and I calculated the molarity of the catalase, where does that come into the equation, because it acts as a catalyst only?
For the equation r=k(A)^m*(b)^n
Is the b value the molarity of catalase?
And I'm also unsure on how to calculate the r value. For 1% hydrogen peroxide it took on average 11.3s before the reaction appeared to have stopped. Also I do have basic knowledge with calculus, I also study maths B and C. Thanks again.