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Douglas Tobias
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University of California - Irvine

Chemistry

CHEM 5

Douglas Tobias

Fall

Description

LESSON #7
FINDING ROOTS OF POLYNOMIAL EQUATIONS
- Finding the roots of equations and systems of equations, such as type that arise in chemi-
cal equilibrium and kinetics problems.
- There are several commands available for this purpose.
The Solve command:
Example # 1:
In[1]:[email protected] + 1 ã 0, xD
Out[1]=x Ø -1<<
Example # 2
Applying solve to a generic quadratic polynomial gives the familar quadratic formula:
In[2]:=veAa * x + b * x + c ã 0, xE
2 2
-b - b - 4 a c -b + b - 4 a c
Out[2]=x Ø >, :x Ø >>
2 a 2 a
- You can use the replacement rules to evaluate the roots for a particular choice ofthe coefficients:
In[3]:=ts = SolveAa * x + b * x + c ã 0, xE;
roots ê. 8a Ø 1, b Ø 3, c Ø 1<
1 1
Out[4]=x Ø I-3 - 5 M>, :x Ø I-3 + 5 M>>
2 2
In[5]:=D
Out[5]=x Ø -2.61803, :x Ø I-1 + 13 M>>
2 2
In[12]:=D
Out[12]=x Ø -2.30278>
2 µ 21ê3
In[16]:=%D
Out[16]=8x Ø 0., :x Ø -2 ProductLogB F>>
2 2
These roots are not very useful
In[18]:=ProductLog
w
[email protected] gives the principal solution for w in z ▯ we .
[email protected], zD gives the khsolution. à
Printed by Wolfram Mathematica Student Edition 4 Untitled-1
x 2
In[19]:=olveAE ã x , xE
NSolve::ifun : Inverse functions are being used by NSolve, so
some solutions may not be found; use Reduce for complete solution information. à
Out[19]=8x Ø -0.703467, :x Ø -H-1L ProductLogB-1, - 6F >>
3 ‰ 3 ‰
3
In[25]:[email protected] ã x - 2, xE
NSolve::ifun : Inverse functions are being used by NSolve, so
some solutions may not be found; use Reduce for complete solution information. à
Out[25]=8x Ø 0.135674 + 0. Â, :x Ø I-1 + 33 M, y Ø I3 + 33 M>>
2 2 2 2
In[35]:=%D
Out[35]=8x Ø -3.37228, y Ø -1.37228>
b d - a e b d - a e
CHEMICAL EXAMPLES
The solutions of polynomial equations is routinely encountered int eh analysis of chemical equilibra.
Below we consider several examples
Example #1:
EQUILIBRIUM
Consider the steam reforming of methane reaction, which is an industrial proess used to produce
hydrogen gas:
CH 4HgL+ H 2 HgLFCO HgL+ 3H 2HgL
-7
The equilibrium constatn for this reaction is 1.8 x 10 at 600K.
Gaseous methane, water, and carbon monoxide are introduced at partial pressures of 1.40 atm, 2.30
atm, respectively, into a reaction vessel at 600K.
Determine the partial pressures fo all species at equilibrium.
The law of mass action for this reaction is:
3
K = PCOP H2
PCH4PH2O
The Following table is used to derive an equaiton that can be solved to determine the equilibrium partial
pressures of the species present:
Printed by Wolfram Mathematica Student Edition Untitled-1 7
In[39]:=ble = 881.40, 2.30, 1.6, 0> k
a w
kw= @H [email protected] D = 1.0*10 -14
We therefore draw up the follwing table
In[57]:[email protected], tableD;
table = 881.000, 0, 0

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