converting rate in cm3s-1 to rate in mol dm-3s-1

So what are you measuring? I'm assuming you're measuring the volume of oxygen produced? What you would need to do is assume that oxygen has the same molar volume as an ideal gas (24.465 L/mol at 25 °C). Using this you can then convert your volume of oxygen produced to molar which lets you plot moldm-3s-1 for your oxygen production, and therefore your consumption of peroxide.

Does that help at all?

I don't want to convert my volume of oxygen produced I want to convert my rate of oxygen produced to rate of hydrogen peroxide used.

Yes but the only information you have is the rate of production of oxygen by volume is what I'm assuming?

Ok, so best way to do this is (to get marks on coursework at least)

1. Calculate your concentration of oxygen produced.

Calculate the molar volume of O2 gas for the approximate temperature of your reaction (298K) which will be about 24.77dm3/mol.

The inverse of this 1/molar volume is moles per unit volume, in this case mol/dm3 which is ~0.04moldm-3 for oxygen at 298K

2. Calculate moles of peroxide consumed per second

Multiply this with the rate of change of volume (in dm3s-1) to get moles of oxygen produced per second.

Balance stoichiometry (2peroxide per oxygen) to get the moles of hydrogen peroxide consumer per second.

3. Calculate the remaining concentration of peroxide from your initial value for each of your data points

You know your volume and initial concentration of hydrogen peroxide solution, so you can calculate your rate of consumption from there.

For example your starting concentration was 0.01M peroxide and 500ml volume, and your rate of production of o2 was 0.001dm3s-1 you would be consuming 0.00008 moles of peroxide per second. so at t=10seconds your moles of peroxide left would be (0.005moles - 10x0.00008) = 0.0042 moles. Divide that by your volume to get your new concentration (0.0084M) then you can plot concentration of peroxide vs time and therefore calculate the rate using your slope.

Sorry if this is a confusing way but I haven't done anything like that in nearly 7 years. If anything is unclear ask me and I'll try and be clearer.

Yep totally get you, it's just... i've done like 30 graphs... it took me 4 hours...


I can't go through that again

lol XD stupid chemistry =[

thanks for your help though +