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HELP needed: RE Decomposition of Hydrogen Peroxide

i need abit of help please, doing my individual investigation on the decomposition of hydrogen peroxide and havent got much of a clue about how to go about doing it.

can someone who has done this investigation please email me or pm me so i can have a chat through msn or something for help?

thanks in advance

yus786

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Reply 1
Probabily a bad choice if you dont have a clue how to go about it!

Start here:

2 H2O2 --> 2 H2O + O2

Google will solve your problems too.
Reply 2
yus786
i need abit of help please, doing my individual investigation on the decomposition of hydrogen peroxide and havent got much of a clue about how to go about doing it.

can someone who has done this investigation please email me or pm me so i can have a chat through msn or something for help?

thanks in advance

yus786


I did this practical some time ago. I am now a science teacher. What help do you require?

You will probably want to look at variables that increase/decrease the rate of decomposition, such as temperature/pH/catalyst/enzymes. Also note that it is stored at a certain pH to stabilise it.

Marcus
Reply 3
right, i kind of know what im doing but need a bit of help in some parts

what im investigating is which catalyst (inorganic or organic) is more efficient when added to hydrogen peroxide

i will need to use the following techniques but dont know how to go about doing them:

titration - what do i need to put in the burette, and what do i need to titre into, i read that i need to put an acid in it but dont know why?

gas syringe method - i kind of know how to do this but im not too sure, do i need to put the hydrogen peroxide in the conical flask or beaker, then add the inorganic or organic catalyst to it and then see how fast the syringe moves to the end?

calorimetry - aint got a clue on how to do this so im not doin it lol

and another question, do i need to work out the activation enthalpy of each or something? and what does the arhyneious (sp?) equation work out as im supposed to be using this to get a good grade?

as i said it would be better if i got an answer via msn messenger as it would be much quicker

thanks for the help in advance
Reply 4
yus786
right, i kind of know what im doing but need a bit of help in some parts

what im investigating is which catalyst (inorganic or organic) is more efficient when added to hydrogen peroxide

i will need to use the following techniques but dont know how to go about doing them:

titration - what do i need to put in the burette, and what do i need to titre into, i read that i need to put an acid in it but dont know why?

gas syringe method - i kind of know how to do this but im not too sure, do i need to put the hydrogen peroxide in the conical flask or beaker, then add the inorganic or organic catalyst to it and then see how fast the syringe moves to the end?

calorimetry - aint got a clue on how to do this so im not doin it lol

and another question, do i need to work out the activation enthalpy of each or something? and what does the arhyneious (sp?) equation work out as im supposed to be using this to get a good grade?

as i said it would be better if i got an answer via msn messenger as it would be much quicker

thanks for the help in advance


Titration - you can use this to determine the concentration of the hydrogen peroxide you are using? I must say I've never used this step as I used stock hydrogen peroxide prepared by the technician and measured the rate of decomposition by measuring the total volume of oxygen produced at certain time intervals.

Gas Syringe - this will accurately measure the amount of gas produced. Since you are measuring the volume of gas produced, you must use a closed system - for example conical flask with delivery tube attached to gas syringe. A beaker will not substitute. Remember rate will be affected by temperature and pH, so unless you are investigating these variables also you will need to keep these constant throughout.

Calorimetry - decomposition of hydrogen peroxide will result in a temperature change - you can use the calorimeter to measure the rate of change in the same way you are measuring the rate of oxygen evolution.

You can use the Arrhenius Equation to work out the activation energy! Hopefully you will get straight line plots of first order reactions from your results. You should be able to rearrange the Arr. Equation to calculate this from your plots - I could post the information here, but it is available on the internet if you google for "Arrhenius Equation" or look it up in your chemistry textbook.

Marcus
Hey

Im doing my chemistry a2 coursework at the moment on the decomposition of hydrogen peroxide - do u know of a more accurate way to measure temperature than use a thermometer?

Thank you!
Reply 6
FayeFayeFaye
Hey

Im doing my chemistry a2 coursework at the moment on the decomposition of hydrogen peroxide - do u know of a more accurate way to measure temperature than use a thermometer?

Thank you!


Nothing that your school is likely to have or will make much difference to your results in this experiment. The main errors will come from you reading the thermometer incorrectly. Thermometers that you use in science lab are pretty accurate.
ok thank you! Also, if you use an inverted burette it has an error of 0.5 and then multiply the value by 2 due to taking 2 readings both inital and final. But in my experiment I used 2 burettes as the gas syringe kept sticking.. so is it right that using 2 inverted burettes has an error of 0.2? Thank you
Reply 8
If you say that the burette error per reading is +/-0.5, then 0.5 x 2 = 1, so the error for one burette is going to be +/-1

Where are you getting 0.2 from? Why two inverted burettes - do you mean you are bubbling the gas into the burette while it is upside down?
Sorry my mistake - I had gone wrong somewhere :s-smilie:

Nope because the reaction took so long - i had to use two inverted burettes

Thank you very much
Sorry the burette scale - 50cm3 progresses every 0.1cm3. Therefore the error = 0.05
0.05 x 2 = 0.1. Then as i have used 2 burettes - 0.01 x 2 = 0.2.

I hope thats right :s
FayeFayeFaye
Nope because the reaction took so long - i had to use two inverted burettes


That's why the way to go is measure the volume of gas is produced in a specific short period of time - just a few minutes, not time taken to produce a certain specific volume.

FayeFayeFaye
Sorry the burette scale - 50cm3 progresses every 0.1cm3. Therefore the error = 0.05
0.05 x 2 = 0.1. Then as i have used 2 burettes - 0.01 x 2 = 0.2.

I hope thats right :s


0.1 x 2 = 0.2.

The main problem is that you haven't greased your gas syringes properly. Whether that's up to you or the tech, I don't know. They shouldn't stick. I've never had this problem with them.
Thank you! Well i had to get to the final volume of oxygen produced in order to work out V final - V t to calculate the half life of each mass change in catalyst! However my half lives seem to be like 436, 530 and then 400 :s and others just seem to decrease so i cant say whether the reaction is first order or not! And it should be! So i dont know what to do about that!

Yeap well the tech guy in our school said it was fine, and it was at the first few cm3 etc but around 40 it just seemed to stick! Thanks again for your help!
FayeFayeFaye
Thank you! Well i had to get to the final volume of oxygen produced in order to work out V final - V t to calculate the half life of each mass change in catalyst! However my half lives seem to be like 436, 530 and then 400 :s and others just seem to decrease so i cant say whether the reaction is first order or not! And it should be! So i dont know what to do about that!

Yeap well the tech guy in our school said it was fine, and it was at the first few cm3 etc but around 40 it just seemed to stick! Thanks again for your help!


Why V final? If you are looking for half life, which you didn't specify originally, you are only going to calculate how long it takes for half of the hydrogen peroxide to decompose using each catalyst, you don't say which - I presume you were comparing biological enzyme with manganese dioxide?

And then 50% of the remainder and after, 50% of the remainder of that, and continue - surely you don't have to go to the end? In fact, if you do, then technically by the strict definition of half life, you could be waiting for ever for the last molecule to decompose :confused:

The question is, how do we know that half of the hydrogen peroxide has decomposed? We know that oxygen gas is produced when the hydrogen peroxide decomposes. Using stoichiometry and the gas laws, it is possible to predict how much oxygen gas should be produced when a given amount of hydrogen peroxide decomposes. Then, all we have to do is measure how long it takes for the decomposition of hydrogen peroxide to produce half of the total oxygen gas the reaction is capable of producing.

Since you said you needed two burettes, it would have been better to use a volume of peroxide predicted to decompose into a volume that would fill the burette, or otherwise a large graduated cylinder. Whatever accuracy you are gaining by using the greater precision of the measurement units of the burette, you are going to be losing when you switch over burettes.

If you were just looking at rates, then just measure the volume every 10-20 or 30 seconds, or the time taken to produce 5 ml until the reaction gets so slow that it is taking <60sec for 5ml, stop, and then you can get first/second/third order from that.
FayeFayeFaye
Hi

I used manganese (IV) oxide and lead (IV) dioxide - I do not know what to do with my half life data! As wen i plotted the graphs they do not show anything!
I used 2.0 volume of hydrogen peroxide, which should produce 40.00cm3 of oxygen. You said: Then, all we have to do is measure how long it takes for the decomposition of hydrogen peroxide to produce half of the total oxygen gas the reaction is capable of producing. So do i just halve 40 three times? and do it that way! I dont know what to do - I have over 500 burette readings and doing a percentage error on each :s-smilie:
Also i did the correct thing about changing the concentration of hydrogen peroxide however, lead iv dioxide shows like zero order and manganese is like first - ignoring one data point.

Thank you very much for your help!


Say you have 200 molecules. The first half life is the time taken for 100 of these to be converted.

The second half life is the time for 50 of the remaining 100 to be converted.

And then you'll have 50 left, the time taken for half of these to be converted is the third half life.

When you are getting down to small numbers, your results will be less accurate.

So if you calculate that you will produce 40 cubic cm of oxygen, the half life will be the time taken to produce the first 20 cubic cm. The second half life will be the time for the next 10 cubic cm and half of the remainder of that, 5 cubic cm and half again and half again... you get the idea.

Obviously you need to take into account that you have excess catalyst and very much later half lives will not be as accurate as the first ones because water, which is also a product, is diluting the reaction and will slow down the collisions between peroxide molecules and catalyst.

If the half lives remain approximately the same each time, then the reaction is first order. If the half life doubles as the concentration halves, then it is second order.

I know that this is a first order reaction when you use manganese dioxide, so the half lives should be approximately the same. With catalase enzyme, the reaction is zero order. At low concentrations, Michaelis-Menten kinetics predict that the rate will become first order. I don't know what it should be with lead dioxide.

I was unaware that you were using these specific catalysts as you didn't post your objectives.
FayeFayeFaye
marcusfox
FayeFayeFaye
Hi

I used manganese (IV) oxide and lead (IV) dioxide - I do not know what to do with my half life data! As wen i plotted the graphs they do not show anything!
I used 2.0 volume of hydrogen peroxide, which should produce 40.00cm3 of oxygen. You said: Then, all we have to do is measure how long it takes for the decomposition of hydrogen peroxide to produce half of the total oxygen gas the reaction is capable of producing. So do i just halve 40 three times? and do it that way! I dont know what to do - I have over 500 burette readings and doing a percentage error on each :s-smilie:
Also i did the correct thing about changing the concentration of hydrogen peroxide however, lead iv dioxide shows like zero order and manganese is like first - ignoring one data point.

Thank you very much for your help!


I just added more to the thread, hope it helps.


Hii

Thank you very much - that really helps! Although, because ive got my results and graphs drawn I think it would take up too much time to process them again! I have already worked out 3 half lives for manganese (IV) oxide and the values go like this:

408, 480 and 288 - 60g

I have just realized something, I had changed the mass of the catalyst 4times - 0.20-0.80g to see what order the reaction is but surely i dont need them. As changing the mass of the catalyst willl produce the sme amount of oxygen as the concentration and volume of hydrogen peroxide was kept constant. Therefore each should have the same order etc. Therefore i only need one graph to show the half life as shown above! I hope that makes sense... if i am right in doing that, that would help alot! Also thank you for your notes - I think they will be very good to include in my analysis.

Thank u again! Btw I noticed youve done a Bsc in chemistry at uni...thats what ive applied to do aswell... but i obviously cant do chemistry!


OK, so you have measured amounts of catalyst. If you change the amount of catalyst to match the concentration after each half life, doesn't that mean that successive half lives won't be as expected if you just let the reaction proceed?

What does your teacher say?
My teachers arent replying to their e-mails and i dont think anybody knows.
Basically i just used 20cm3 of h202 and then changed the mass of catalyst 0.20 - measured the vol of o2 prod and worked out vfinal - vt then i repeated the experiment and used 0.40g etc

To see if the reaction is first order - i just used 0.20g.

What do u mean by:

If you change the amount of catalyst to match the concentration after each half life

I just measured the burette reading, then worked out vol of o2 produced and then worked out vfinal - vinital and then did the half lives from that. Surely i dont need the results for 0.60g etc or do i?
Im very confused! Thanks again
FayeFayeFaye
My teachers arent replying to their e-mails and i dont think anybody knows.
Basically i just used 20cm3 of h202 and then changed the mass of catalyst 0.20 - measured the vol of o2 prod and worked out vfinal - vt then i repeated the experiment and used 0.40g etc

To see if the reaction is first order - i just used 0.20g.

What do u mean by:

If you change the amount of catalyst to match the concentration after each half life

I just measured the burette reading, then worked out vol of o2 produced and then worked out vfinal - vinital and then did the half lives from that. Surely i dont need the results for 0.60g etc or do i?
Im very confused! Thanks again


If the half lives remain approximately the same each time, then the reaction is first order. If the half life doubles as the concentration halves, then it is second order. If you are artificially changing the concentration, this will no longer hold.
Sorry im still confused :s-smilie::s-smilie: Not of my results show first or second order. Do i need to do this for all 4 masses of catalyst? As i was told that as long as u keep the amount and concentration of h202 constant, then changing the mass of the catalyst will not affect it - therefore u will always get the same amount of o2 produced.

Could i just not do 0.20g show the graph and work out the half life and thats it? rather than repeating this 5 times and the same with lead dioxide

thanks again
I used Nuffield Chemistry : which mentions this

You can either follow the reaction by measuring the volumes of oxygen given off at set time intervals

You can subtract each volume from the final reading and then by plotting V final - V t against time you can draw a curve of best and measure the half lifes. So surely just doing the experiment using 0.20g of catalyst will be sufficent.

I think what i wanted to do was see if increasing the mass of catalyst affected the rate of the reaction - but adding more of a catalyst wont. Its about changing the surface area such as crushing it into a powder and i did not do this.
Is what I have said above wrong?