Solving Chemical Reactions in Greenburg's Advanced Eng. Mathematics #8

In summary: 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 = 0z’= βyx + y + z = γ solve for x(t), y(t), and Z(t) subject to intitial conditionsz(0)=z'(0)=0for these cases: α≠β,α=βThe solution is x=γ and y=0.
  • #1
LabRat
3
0
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
z(0)=z'(0)=0
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:
 
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  • #2
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".

Daniel.
 
  • #3
LabRat said:
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
z(0)=z'(0)=0
for these cases: α≠β,α=β

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.
 
  • #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.
 
  • #5
LabRat said:
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.

Well, I don't know. You?

[tex](CH_3)_2N(NH_2)+N_20_4\to?[/tex]

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.
 
  • #6
Lab, where you 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 textbook. Does the text give values for the constants?

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

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  • #7
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]x+y+z=\gamma[/tex]

with:

[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:

[tex]\alpha=\beta[/tex]

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:

[tex]c=-\gamma[/tex]
[tex]K=\gamma[/tex]

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 . . .
 
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  • #8
Saltydog,

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 !
 
  • #9
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]

[tex]K=\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]
 

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  • #10
FYI - (CH3)2NNH2 (l) + N2O4 (g) → 3 N2 (g) + 4 H2O(g) + 2 CO2(g) + heat
 

1. What is the purpose of solving chemical reactions in Greenburg's Advanced Eng. Mathematics #8?

The purpose of solving chemical reactions in Greenburg's Advanced Eng. Mathematics #8 is to apply mathematical principles and equations to analyze and predict the outcomes of chemical reactions. This allows for a better understanding of the underlying principles behind chemical reactions and can aid in the development of new technologies and products.

2. What types of chemical reactions can be solved using Greenburg's Advanced Eng. Mathematics #8?

Greenburg's Advanced Eng. Mathematics #8 can be used to solve a wide range of chemical reactions, including but not limited to acid-base reactions, redox reactions, and precipitation reactions. It can also be applied to more complex reactions such as combustion and organic reactions.

3. How does Greenburg's Advanced Eng. Mathematics #8 simplify the process of solving chemical reactions?

Greenburg's Advanced Eng. Mathematics #8 provides a systematic approach to solving chemical reactions by using mathematical equations and principles. This simplifies the process by breaking down the reaction into smaller, more manageable steps and allows for a more accurate analysis of the reaction.

4. Can Greenburg's Advanced Eng. Mathematics #8 be used for all chemical reactions?

No, Greenburg's Advanced Eng. Mathematics #8 may not be applicable to all chemical reactions. Some reactions may be too complex or may require specialized mathematical techniques that are not covered in this particular resource. It is important to consult other sources and experts when solving chemical reactions.

5. How can solving chemical reactions in Greenburg's Advanced Eng. Mathematics #8 benefit the scientific community?

Solving chemical reactions using Greenburg's Advanced Eng. Mathematics #8 can benefit the scientific community in several ways. It can lead to a deeper understanding of chemical processes, aid in the development of new technologies and products, and contribute to the advancement of various fields such as medicine, materials science, and environmental science.

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