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Chemical Reactions

  1. Jun 4, 2005 #1
    Working a problem in Greenburgs "Advanced Eng. Mathematics", #8 page 170. Two substances w/ concentration x(t) and y(t), react to form a 3rd substance w/ concentration z(t). The reaction is governed by the system
    x’+ αx = 0
    z’= βy
    x + y + z = γ
    solve for x(t), y(t), and Z(t) subject to intitial conditions
    for these cases: α≠β,α=β

    My approach…I would like set the equation such that:
    x’+ αx = 0
    z’ – βy = 0
    and set
    L1= 1, L2=α,
    L3= 1, L4= -β
    (Where α≠β, α=β for the 2 cases) and use Cramer’s rule to solve…but I am missing the relationship between all three: x + y + z = γ
    What I am missing? I have never done a problem with 3 variables. :yuck:
  2. jcsd
  3. Jun 4, 2005 #2


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    You can integrate the first directly and and then plug 'y' as a function of "z"-s derivative in the last.U'll have a linear nonhomogenous equation in "z".

  4. Jun 4, 2005 #3


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    Alright, what's big-y? Is that a typo and you mean y'? And by the way, once you get the solution, is this a real model of chemical kinetics? What's an example of two reagents reacting this way? Why do they do so that way? Suppose you had to go into a lab and verify it experimentally, how to do so? It's just me. Do as you wish.
  5. Jun 4, 2005 #4
    Big Y?

    I assuming you're talking about the third equation, but excuse my limitation on expressing the equation; its actually

    x+y+z= gamma (i.e. the symbol appears like a "y" in my original post). Where in all the equations; alpha, beta, and gamma-are all constants.

    As far as a "real" model, I doubt it. Maybe a hypergolic reaction, say between UDMH and N2O4. But then again to verify in the lab, you would have to also measure the energy generated. I think this is just an oversimplified chem kinectics problem made into a Diff Eq problem-to frustrate chemistry majors.
  6. Jun 4, 2005 #5


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    Well, I don't know. You?


    And what is the kinetics of this anyway? Bomb-calorimeter first comes to mind. Kinda' dangerous though. Surely someone's done it already. I know, it's off the subject. Whatever.
  7. Jun 4, 2005 #6


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    Lab, where ya at with this? Got it already? I have the solutions for [itex]\alpha=\beta[/itex]. I'm pretty sure it's correct by back-substitution into the third equation. You got that or need help? I've attached a plot with all constants set equal to 1. Note that y(t) is not reasonable for a chemical reaction since it should be steadly decreasing (I guess as I don't have the balanced equation). I know it's just an example in a text book. Does the text give values for the constants?

    Edit: I mean [itex]\alpha,\beta,\gamma=1[/itex]

    Attached Files:

    Last edited: Jun 4, 2005
  8. Jun 5, 2005 #7


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    Lab, I'm assuming you've done all this already as I'd like to see this one through to completion and don't wish to do the work for you in your best interest.

    So we have:

    [tex]x^{'}=-\alpha x[/tex]

    [tex]z^{'}=\beta y[/tex]



    [tex]z(0)=0 \quad \text{and} \quad z^{'}(0)=0[/tex]

    So the first one is solvable by inspection. It's:

    [tex]x(t)=Ke^{-\alpha t}[/tex]

    Substituting this and the second one into the third gives:

    [tex]z^{'}+\beta z=\beta \gamma- \beta Ke^{-\alpha t}[/tex]

    Treating this as a standard first order ODE solvable by use of an integrating factor, gives:

    [tex]d[e^{\beta t}z]=\beta \gamma e^{\beta t}-\beta Ke^{t(\beta-\alpha)}[/tex]

    Now, at this point we consider two cases:

    Case 1:


    Thus we have:

    [tex]d[e^{\beta t}z]=\beta \gamma e^{\beta t}-\beta K[/tex]

    Solving for z(t) gives:

    [tex]z(t)=\gamma-\beta K te^{-\beta t}+Ce^{-\beta t}[/tex]

    Taking the derivative of z(t) and using the initial conditions given, we can then determine the values of K and c.

    We obtain:


    Thus we have:

    [tex]x(t)=\gamma e^{-\alpha t}[/tex]

    [tex]y(t)=\beta \gamma te^{-\beta t}[/tex]

    [tex]z(t)=\gamma-\beta \gamma t e^{-\beta t}-\gamma e^{-\beta t}[/tex]

    Using [itex]\alpha=1 \qquad \beta=1 \qquad \gamma=1[/itex] yields the plots above. I'm a bit dissapointed in them. I suppose if I start with a beaker full of x, then at time 0, start adding y, they react kinda slowly making z, then the plots resemble what I think the chemistry should be but I'd need chemically quantifiable data to be convinced.

    Suppose I should be happy I don't have access to a chem lab or else I'd be stuffing a bomb-calorimeter full of rocket fuel right about now and end up . . . well just a gram or two shouldn't do anything . . .
    Last edited: Jun 5, 2005
  9. Jun 5, 2005 #8


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    The only real situation that will fit the equations (in my opinion) is a reaction that is first order in X and Y, where Y is being added to the reaction mixture at a uniform rate - such as in a titration !
  10. Jun 5, 2005 #9


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    Thanks Gokul. I was just joking about the rocket fuel. I'm actually a very cautious person:

    Here's the case two. I'll leave out the steps for others to do if they wish.

    Taking the case [itex]\alpha \neq \beta [/itex] and solving for z(x) we obtain:

    [tex]z(x)=\gamma-\frac{\beta K}{\beta-\alpha}e^{-\alpha t}+C e^{-\beta t}[/tex]

    Taking the derivative of z(x) and solving for K and c we obtain:

    [tex]C=\frac{\beta K}{\beta-\alpha}-\gamma[/tex]


    This leaves us with:

    [tex]x(t)=\gamma e^{-\alpha t}[/tex]

    [tex]y(t)=\frac{\alpha \gamma}{\beta-\alpha}e^{-\alpha t}-(\frac{\beta \gamma}{\beta-\alpha}-\gamma)e^{-\beta t}[/tex]

    [tex]z(t)=\gamma-\frac{\beta \gamma}{\beta-\alpha}e^{-\alpha t}+(\frac{\beta \gamma}{\beta-\alpha}-\gamma)e^{-\beta t}[/tex]

    A plot is attached for [itex]\alpha=1\quad\beta=0.2\quad\gamma=1[/tex]

    Attached Files:

  11. Jun 12, 2005 #10
    FYI - (CH3)2NNH2 (l) + N2O4 (g) → 3 N2 (g) + 4 H2O(g) + 2 CO2(g) + heat
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