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

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Hey i'm doing a rate of reaction investigation with hydrogen peroxide and catalase.

I'm really confused about something. I've spent 4 hours writing out 30 graphs on substrate conc, enzyme conc, temperature and PH..

I've worked out the gradient of each concentration of enzyme and substrate..

And now my teacher says I should change my answer, from tracking oxygen per second/time to hydrogen peroxide used up.

I have already done all of the graphs with oxygen increase/s and NOT hydrogen peroxide decrease/s... apparently I can use the gradient i have calculated to convert rate in cm3s-1 to rate in mol dm-3s-1... how on earth do I go about doing this with a gradient?

Thanks!

I'm really confused about something. I've spent 4 hours writing out 30 graphs on substrate conc, enzyme conc, temperature and PH..

I've worked out the gradient of each concentration of enzyme and substrate..

And now my teacher says I should change my answer, from tracking oxygen per second/time to hydrogen peroxide used up.

I have already done all of the graphs with oxygen increase/s and NOT hydrogen peroxide decrease/s... apparently I can use the gradient i have calculated to convert rate in cm3s-1 to rate in mol dm-3s-1... how on earth do I go about doing this with a gradient?

Thanks!

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#2

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?

Does that help at all?

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(Original post by

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?

**Pendulum**)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?

__rate__of oxygen produced to

__rate__of hydrogen peroxide used.

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#5

(Original post by

I don't want to convert my volume of oxygen produced I want to convert my

**Cinamon**)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.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

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.

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(Original post by

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)

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

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.

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.

**Pendulum**)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 second2. 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.

I can't go through that again

lol XD stupid chemistry =[

thanks for your help though +

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#7

(Original post by

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

[SIZE="4"]I can't go through that again[/SIZE]

lol XD stupid chemistry =[

thanks for your help though +

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

[SIZE="4"]I can't go through that again[/SIZE]

lol XD stupid chemistry =[

thanks for your help though +

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#8

im doing this right now at 4am and ive realised that ive made the same mistake as Cinamon, makes me want to give up on everything >.<

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