HELP needed: RE Decomposition of Hydrogen Peroxide

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Reply 20

FayeFayeFaye

http://www.chemistry-react.org/go/Tutorial/Tutorial_4643.html

Reply 21

marcusfox

FayeFayeFaye

http://www.chemistry-react.org/go/Tutorial/Tutorial_4643.html

Yes, I saw that one - I take your point about half lives, but at that point I wasn't exactly sure what your objectives were.

Reply 22

FayeFayeFaye

marcusfox

Yes, I saw that one - I take your point about half lives, but at that point I wasn't exactly sure what your objectives were.

Okay thank you!

So should i just use the data i have for the 0.20g of catalyst to work out the half life? And ignore/leave the other bits of data i got!

As it was a waste to calculate the volume of o2 for 0.40, 0.60 0.80g - on my starter sheet i was given it said u cud change the amount of catalyst to see what affect it had on the rate of the reaction, so cud i plot a graph using the data i got above - but only do the volume of o2 produced for each one in 5 minutes and calculate the gradient - as this is just something extra to put in.

Reply 23

marcusfox

FayeFayeFaye

Okay thank you!

So should i just use the data i have for the 0.20g of catalyst to work out the half life? And ignore/leave the other bits of data i got!

Haven't seen your data. Why don't you look at all of it, then compare it to just that for 0.20g catalyst.

Reply 24

FayeFayeFaye

Homm ok!

I just dont see why i can't just measure the half live using 0.20g and thats it :s-smilie:

Sorrrry to keep asking all these questions

Reply 25

FayeFayeFaye

Sorry - I have now decided to:

1. Measure the half life of the reaction using 0.20g of manganese (IV) oxide - I kept this amount of catalyst constant wen changing h202 and temperature.
2. Measure the half life of the reaction using 0.02g of lead (IV) dioxide - Same reasons as above.

3. Then because i have data for 0.40, 0.60, 0.80g, I will see how changing the amount of catalyst - affects the rate of the reaction - so just measure the volume of oxygen produced for 10 times. Draw a graph and then calculate the rate, and i shud find that as I increase the amount of catalyst, the rate increases, possibly due to the fact there a there is more catalyst, more reactants should become adsorbed.

Will that be right? None of the teachers are answering their e-mails!
Thank you if the above is right, then I have no more questions! You have been a great help

Reply 26

FayeFayeFaye

I have just plotted the graph for changing the catalyst - vol of 02 produced for 10 minutes against time, and the rate increases by alot especially for lead (IV) dioxide. :smile:

Reply 27

FayeFayeFaye

and the half life for 0.20g is not constant and it does not halve either - so it shows that it must be zero?
and thats the same for lead (IV) dioxide
Thanks again!

Reply 28

Yellow_Paper

Original post by FayeFayeFaye

and the half life for 0.20g is not constant and it does not halve either - so it shows that it must be zero?
and thats the same for lead (IV) dioxide
Thanks again!

Hii, I know I'm a couple years late- but I was researching this topic for my own chem coursework and I was wondering what grade you got at the end