[jenri619june]

Hi,first of all thank you for taking time to read this. I am a student currently working on a project, that requires me to solve a few equations, i am currently v confused with what i have now and wonder if you could spare me some of your time to enlighten me.

I am working with the decomposition of hydrogen peroxide.
2H2O2 --> 2H2O + O2

i need the amount of O2 produced to be at least 8litres per min over a period of time, thus i am trying to find out the concentration of hydrogen peroxide to use. I will need to be able to draw a graph that shows the amount of O2 produced per min at the end of the day.

I actually have two eq.

1.The Arrhenius equation of decomposition of hydrogen peroxide:
k=Ae^-(Ea/(RT)),
Ea to be 75kJ/mol
R=8.314Jmol^-1K^-1
T= 298K
A= 10^11 s^-1


So what i got after i sub the value in is k to be 0.0071322 units??
2.Assumed kinetic equation: (suggested)

dO2 over dt = 1/2[H2O2]k     ,k from the above equation
How will I be able to get a graph that shows the amount of O2 produced per min at the end of the day?

I tried the equation : (got from internet (first order))

-d[H2O2] over dt =[H2O2]k

d[H2O2] over [H2O2] = -kdt
and after integration : ln[H2O2]= -kt+c
then i am stucked...again.

I am very confused, please help me.

Thanks.

[tamim83]

So what i got after i sub the value in is k to be 0.0071322 units??

Units for k (first order) = s^-1

i am trying to find out the concentration of hydrogen peroxide to use

You should be able to use the rate law, since you know k and the rate. 

rate = k[H₂0₂]

I don't think you need to integrate for an initial concentration.  Just plug in your rate and k and rearrange to solve for the concentration (in litres). 

Hope that helps some. 

[Abstractineum]

If you want to know the amount of O₂ produced, you should multiply the change in concentration with the volume of the system (n = c * V). The change of the amount of O₂ produced thus equals (dO₂/dt) * V.

Now you need a way to describe dO₂/dt in terms of the rate equation. For an elementary reaction, I think that would be dO₂/dt = [H₂O₂]²k₁.
 (I have disregarded the possibility of the reaction going backwards, so I have emitted the k_-1 konstant)

You have already calculated k, so now you can substitute that in the rate equation. The change in the concentration of oxygen would then depend on H₂O₂. Now you can plot (dO₂/dt) * V against H₂O₂. I think.

Please correct me if I am wrong someone

EDIT: What tanim83 said Be aware that the rate changes with the concentration of Hydrogen Peroxide, so if you want the amount of oxygen produced to be at least 8L/min over a period of time, the longer the time the higher concentration of [H₂O₂]₀ you need.

[Mitch]

8 liters per minute!? Making rockets or something else?

[jenri619june]

If you want to know the amount of O₂ produced, you should multiply the change in concentration with the volume of the system (n = c * V). The change of the amount of O₂ produced thus equals (dO₂/dt) * V.

Now you need a way to describe dO₂/dt in terms of the rate equation. For an elementary reaction, I think that would be dO₂/dt = [H₂O₂]²k₁.
 (I have disregarded the possibility of the reaction going backwards, so I have emitted the k_-1 konstant)

You have already calculated k, so now you can substitute that in the rate equation. The change in the concentration of oxygen would then depend on H₂O₂. Now you can plot (dO₂/dt) * V against H₂O₂. I think.
Please correct me if I am wrong someone

EDIT: What tanim83 said Be aware that the rate changes with the concentration of Hydrogen Peroxide, so if you want the amount of oxygen produced to be at least 8L/min over a period of time, the longer the time the higher concentration of [H₂O₂]₀ you need.

if equation is dO₂/dt = [H₂O₂]²k₁.


how do u get to plot  (dO₂/dt) * V against H₂O₂.

[Abstractineum]

What I meant (without specifying the relationship) was that you could make a plot containing the values of  (dO₂/dt) * V and [H₂O₂]. You are right that the x-axis would be measured in [H₂O₂]².

[jenri619june]

What I meant (without specifying the relationship) was that you could make a plot containing the values of  (dO₂/dt) * V and [H₂O₂]. You are right that the x-axis would be measured in [H₂O₂]².

i am sorry i dont quite get it...can u please rephrase? i really need your help

[Abstractineum]

The relationship between d(n_O2)/dt (the amount of oxygen produced per time unit, which is the same as V * d[O₂]/dt) and [H₂O₂] can be described by:

d(n_O2)/dt = V * [H₂O₂]²k

So you can plot d(n_O2)/dt against (V * [H₂O₂]²k). With a konstant value of k and V, the variable is [H₂O₂]². Hope that clarified it a bit

EDIT: I realized that I lacked a V on one side of the equation earlier. It should be on both sides, and the reason for multiplying at all is that it is V * d[O₂]/dt you are interested in. (Why can't I modify my earlier posts?  )

[jenri619june]

The relationship between d(n_O2)/dt (the amount of oxygen produced per time unit, which is the same as V * d[O₂]/dt) and [H₂O₂] can be described by:

d(n_O2)/dt = V * [H₂O₂]²k

So you can plot d(n_O2)/dt against (V * [H₂O₂]²k). With a konstant value of k and V, the variable is [H₂O₂]². Hope that clarified it a bit

EDIT: I realized that I lacked a V on one side of the equation earlier. It should be on both sides, and the reason for multiplying at all is that it is V * d[O₂]/dt you are interested in. (Why can't I modify my earlier posts?  )

eh sorry...i have never worked with this before. may i know whats the unit for d(nO2) over dt? ? mol/s???

and the v[H2O2]k

[Abstractineum]

Yeah, the units for d(n_O2)/dt are mol s^-1. The units for the parts in V * [H₂O₂]²k are the following (SI units are written parenthetical):

V is measured in dm³ (m³)
[H₂O₂]² is measured in mol² dm^-6 (mol² m^-6)
k are for a second order reaction measured in dm³ mol^-1 s^-1 (m³ mol^-1 s^-1)

Put together the units become the same as for d(n_O2)/dt, which is mol s^-1.

[jenri619june]

Yeah, the units for d(n_O2)/dt are mol s^-1. The units for the parts in V * [H₂O₂]²k are the following (SI units are written parenthetical):

V is measured in dm³ (m³)
[H₂O₂]² is measured in mol² dm^-6 (mol² m^-6)
k are for a second order reaction measured in dm³ mol^-1 s^-1 (m³ mol^-1 s^-1)

Put together the units become the same as for d(n_O2)/dt, which is mol s^-1.

Thanks erm...i dont mean to bug you but could you explain what's zero, first and second order? before i post this qn, i actually found many differnent websites saying decompositon of hydrogen peroxide is a first order reaction. while others mention 2nd.

[Abstractineum]

I'll try to answer your question without being to lengthy:

The overall reaction order for an elementary(!) reaction is equal to the sum of the coefficients of the reactants. Some examples:

2 H   H₂ has a reaction order of 2
H + Cl   HCl has a reaction order of 2 (1 + 1)
aA + bB   A_aB_b has a reaction order of (a +b)

You can also talk about the reaction order of a specific reactant in a reaction. This is equal to the coefficient of that particular reactant.

H + Cl   HCl has a reaction order of one with respect to H

 An elementary reaction is a simple reaction, where the whole reaction takes place in one step. Generally, the overall reaction order of an elementary reaction is no higher than two. But how do you know that a reaction is an elementary one? Well, you often can't say by simply looking at the reaction formula.
  Most reactions are complex, and complex reaction can be divided into elementary steps. Complex reactions does not have to have to reaction orders implied by the stoichiometry. I simply assumed that the reaction of Hydrogen Peroxide was elementary, which may be completely wrong. Reactions can be described by reaction paths, which is the way the reaction is assumed to happen (for H₂O₂, it could be that one molecule dissociate to water and an oxygen atom, and that two oxygen atoms then form an oxygen molecule, or it could be that two H₂O₂ molecules collide, and form the products.) When you have an assumed reaction path, you can work out what order of reaction of the reactants that path would lead to. You can then compare to measured results (you can for example plot the rate of the reaction to the concentration of the reactant) to those that you predicted, and see if they fits.
  You can do this because for a reaction, complex ones too, the reaction order of a reactant corresponds to the relationship between the concentration of that reactant and the rate of reaction.

So to sum it up: A zero order reaction with respect to a reactant is one where the concentration of the reactant doesn't matter, a first order is where the concentration is proportional to the rate of reaction and a second order is one where the relationship between rate of reaction and concentration is rate=constant * concentration².
  If you want a more substantial knowledge of reaction orders etc you should find a general chemistry textbook at a library or search the net. http://en.wikipedia.org/wiki/Reaction_rate for example
  

[jenri619june]

Oh I see..Thank you for clearing my doubts. =)