42
Chapter 7: Numerical Integration II
From Physical Chemistry, 6th Edition by Peter Atkins:
Problem 4.19
The compound 1,3,5-trichloro-2,4,6-trifluorobenzene is an intermediate in the
conversion of hexachlorobenzene to hexafluorobenzene, and its thermodynamic
properties have been examined by measuring its heat capacity over a wide
temperature range. Calculate the molar enthalpy relative to its value at T = 0 and
the 3rd Law entropy of the compound at these temperatures in kJ/mol and J/K mol,
respectively.
T, K:
C p, J K - 1 mol- 1 :
14.14
9.492
16.33 20.03 31.15 44.08 64.81 100.90
12.70 18.18 32.54 46.86 66.36 95.05
T, K:
C p, J K - 1 mol- 1 :
140.86
121.3
183.59
144.4
225.10
163.7
à
262.99
180.2
298.06
196.4
At constant pressure, the change in the molar enthalpy is dH = Cp dT. The equation can
T2
be otherwise written as H (T2) - H (T1) =
T1
C p â T . Plot Cp vs. T and calculate the
area under the curve from T = 0 to T = 298.06 K.
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Make a list of the t and cp values given in the problem. Using those lists make a new list
of {t, cp} values. Don't forget to transpose.
t = 14.14, 16.33, 20.03, 31.15, 44.08, 64.81, 100.90,
140.86, 183.59, 225.10, 262.99, 298.06 ;
cp = 9.492, 12.70, 18.18, 32.54, 46.86, 66.36, 95.05,
121.3, 144.4, 163.7, 180.2, 196.4 ;
points = t, cp ;
data = Transpose points
14.14 ,
31.15 ,
100.9 ,
225.1 ,
9.492
32.54
95.05
163.7
,
,
,
,
16.33 , 12.7 , 20.03 , 18.18 ,
44.08 , 46.86 , 64.81 , 66.36 ,
140.86 , 121.3 , 183.59 , 144.4 ,
262.99 , 180.2 , 298.06 , 196.4
43
@ 8
Scatter plot the points with ListPlot:
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dataplot = ListPlot data, AxesLabel ® "T,K", "Cp,J K- 1 mol- 1 " ,
PlotLabel ® "Molar Enthalpy", PlotStyle ® PointSize 0.015 ;
Cp ,J K - 1 mol - 1
200
Molar Enthalpy
150
100
50
50
100
150
200
250
300
T,K
We need to extrapolate the data to T = 0. The easiest way to do this is by fitting a
function to the curve. Since we can use the trapezoid rule for all the points between
T=14.14 K and T = 298.06 K, we only need the function to fit the first few data points.
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Make a new list of {t, cp} values using the first four data points:
t4pts = 14.14, 16.33, 20.03, 31.15 ;
cp4pts = 9.492, 12.70, 18.18, 32.54 ;
pts4 = t4pts, cp4pts ;
data4pts = Transpose pts4
14.14 , 9.492 ,
16.33 , 12.7 ,
20.03 , 18.18 ,
31.15 , 32.54
Fit a 2nd order polynomial to the data points and check the fit by plotting the points and
the function together. Remember that if you suppress the graphs with DisplayFunction,
you must unsuppress when you use Show.
fiteqn1 = Fit data4pts, 1, x, x^2 , x
- 14.4242 + 1.83899 x - 0.0106327 x2
fitplot1 = Plot fiteqn1, x, 0, 31.154 ,
DisplayFunction ® Identity ;
Show
dataplot, fitplot1 , DisplayFunction ® $DisplayFunction ;
44
Cp ,J K - 1 mol - 1
200
Molar Enthalpy
150
100
50
20
40
60
80
100
120
T,K
140
Calculate the area as follows:
1) Integrate the area from T = 0 to T = 14.14 K using the fit equation.
2) Find the area from T = 14.14 to T = 298.06 K using the trapezoid rule.
3) Sum up the two areas to find dH in J/mol.
4) Convert dH to kJ/mol and print the result.
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area0to14 = Integrate fiteqn1, x, 0, 14.14
- 30.1355
There are 12 data points and 11 trapezoids, so i will range from 1 to 11.
area14to298 =
1
* Sum cp i + 1
2
+ cp
i
*
t
i+1
- t
i
, i, 1, 11
34475.8
totalAreaJ = area0to14 + area14to298;
totalAreakJ = totalAreaJ 1000;
Print "dH = ", totalAreakJ, " kJ mol"
dH = 34.4456
kJ mol
Since dH = H (298.06) - H (0) and H (0) = 0, then H (298.06) = dH.
Print "Molar enthaply relative to 0 K = ", totalAreakJ,
" kJ mol"
Molar enthaply relative to 0 K = 34.4456 kJ mol
45
à
Cp
At constant pressure, the change in the molar entropy is dS = T
T2
be otherwise written as S (T2) - S (T1) =
T1
dT. The equation can
Cp
Cp
â T . Plot
T vs. T and calculate
T
the area under the curve from T = 0 to T = 298.06 K.
The list of t and cp values was defined previously. Make a new list of cp/t values and use
it to make a list of {t, cp/t} values. Make sure to name your definitions with different
names.
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cpT = cp t;
points2 = t, cpT ;
data2 = Transpose points2
14.14 ,
31.15 ,
100.9 ,
225.1 ,
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0.671287 , 16.33 , 0.77771 , 20.03 , 0.907639 ,
1.04462 , 44.08 , 1.06307 , 64.81 , 1.02392 ,
0.942022 , 140.86 , 0.861139 , 183.59 , 0.786535
0.727232 , 262.99 , 0.685197 , 298.06 , 0.658928
Scatter plot the points with ListPlot:
,
dataplot2 =
ListPlot data2, AxesLabel ® "T,K", "Cp T,J K- 2 mol- 1 " ,
PlotLabel ® "3rd Law Entropy", PlotStyle ® PointSize 0.015

Cp T,J K - 2 mol - 1
50
3rd Law Entropy
100
150
200
250
300
T,K
0.9
0.8
0.7
Similar to molar enthalpy, extrapolate the data to T = 0 by fitting a function to the first
few data points.
Given the shape of the curve, it's best to fit an equation to only three points, instead of
four. Make a list of the first three t and cp values. Then define a list of cp/t values to
make a new list of {t, cp/t} values.
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t3pts = 14.14, 16.33, 20.03 ;
cp3pts = 9.492, 12.70, 18.18 ;
cpT3pts = cp3pts t3pts;
points3 = t3pts, cpT3pts ;
data3pts = Transpose points3
14.14 , 0.671287
,
16.33 , 0.77771
,
20.03 , 0.907639
Fit a 2nd order polynomial to the data points and check the fit by plotting the points and
the function together. Remember that if you suppress the graphs with DisplayFunction,
you must unsuppress when you use Show.
fiteqn2 = Fit data3pts, 1, x, x^2 , x
- 0.544258 + 0.118324 x - 0.00228844 x2
fitplot2 = Plot fiteqn2, x, 0, 20.03 ,
DisplayFunction ® Identity ;
Show dataplot2, fitplot2 ,
DisplayFunction ® $DisplayFunction ;

Cp T,J K - 2 mol - 1
3rd Law Entropy
1
0.75
0.5
0.25
25
50
75
100
125
150
T,K
-0.25
-0.5
Calculate the area as follows:
1) Integrate the area from T = 0 to T = 14.14 K using the fit equation.
2) Find the area from T = 14.14 to T = 298.06 K using the trapezoid rule.
3) Sum up the two areas to find dS in J/K mol.
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Sarea0to14 = Integrate fiteqn2, x, 0, 14.14
1.97639
There are 12 data points and 11 trapezoids, so i will range from 1 to 11.
Sarea14to298 =
1
* Sum cpT i + 1
2
i, 1, 11
239.268
+ cpT
i
*
t
i+1
- t
i
,
totalSAreaJ = Sarea0to14 + Sarea14to298;
Print "dS = ", totalSAreaJ, " J K mol"
dS = 241.244 J K mol
Since dS = S (298.06) - S (0) and S (0) = 0, then S (298.06) = dS.
Print "3rd law entropy relative to 0 K = ", totalSAreaJ,
" J K mol"
3rd law entropy
relative
to 0 K = 241.244
J K mol
From Physical Chemistry, 6th Edition by Peter Atkins:
Problem 4.18
The heat capacity of anhydrous potassium hexacyanoferrate(II) varies with
temperature. Calculate the molar enthalpy relative to its value at T = 0 and the 3rd
Law entropy of the compound at these temperatures in kJ/mol and J K-1 mol-1,
respectively.
T, K:
C p, J K - 1 mol- 1 :
T, K:
C p, J K - 1 mol- 1 :
10
20
30
40
50
60
70
80
90
2.09 14.43 36.44 62.55 87.03 111.0 131.4 149.4 165.3
100
110
150
160
170
180
190
200
179.6 192.8 237.6 247.3 256.5
265.1
273.0 280.3
Ù
à
At constant pressure, the change in the molar enthalpy is dH = Cp dT. The equation can
T
be otherwise written as H (T2) - H (T1) = T12 C p â T . Since H (0) = 0, dH = H (200 K).
Plot Cp vs. T and calculate the area under the curve from T = 0 to T = 200 K.
Cp
At constant pressure, the change in the molar entropy is dS = T dT. The equation can
T2 Cp
â T . Since S (0) = 0, dS = S (200 K).
be otherwise written as S (T2) - S (T1) =
T1 T
Ù
T
Plot T12 C p â T vs. T and calculate the area under the curve from T = 0 to T = 200 K.
48
Answer: Problem 4.18
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Make a list of the t and cp values given in the problem. Using those lists make a new list
of {t, cp} values. Don't forget to transpose.
t = 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 150, 160,
170, 180, 190, 200 ;
cp = 2.09, 14.43, 36.44, 62.55, 87.03, 111.0, 131.4,
149.4, 165.3, 179.6, 192.8, 237.6, 247.3, 256.5, 265.1,
273.0, 280.3 ;
points = t, cp ;
data = Transpose points
10 , 2.09 , 20 , 14.43 ,
60 , 111. , 70 , 131.4 ,
100 , 179.6 , 110 , 192.8
170 , 256.5 , 180 , 265.1
30 , 36.44 , 40 , 62.55 , 50 , 87.03 ,
80 , 149.4 , 90 , 165.3 ,
, 150 , 237.6 , 160 , 247.3 ,
, 190 , 273. , 200 , 280.3
Scatter plot the points with ListPlot:
dataplot = ListPlot data, AxesLabel ® "T,K", "Cp,J K- 1 mol- 1 " ,
PlotLabel ® "Molar Enthalpy", PlotStyle ® PointSize 0.015 ;
Cp ,J K - 1 mol - 1
Molar Enthalpy
250
200
150
100
50
50
100
150
200
T,K
We need to extrapolate the data to T = 0. The easiest way to do this is by fitting a
function to the curve. Since we can use the trapezoid rule for all the points between
T=14.4 K and T = 298.06 K, we only need the function to fit the first few data points.
Given the shape of the curve, three points might be best.
49
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Make a new list of {t, cp} values using the first three data points:
t3pts = 10, 20, 30 ;
cp3pts = 2.09, 14.43, 36.44 ;
pts3 = t3pts, cp3pts ;
data3pts = Transpose pts3
10, 2.09 ,
20, 14.43 ,
30, 36.44
Fit a 2nd order polynomial to the data points and check the fit by plotting the points and
the function together. Remember that if you suppress the graphs with DisplayFunction,
you must unsuppress when you use Show.
fiteqn1 = Fit data3pts, 1, x, x^2 , x
- 0.58 - 0.2165 x + 0.04835 x2
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fitplot1 = Plot fiteqn1, x, 0, 30 , DisplayFunction ® Identity ;
Show dataplot, fitplot1 , DisplayFunction ® $DisplayFunction ;
Cp ,J K - 1 mol - 1
Molar Enthalpy
250
200
150
100
50
20
40
60
80
100
120
T,K
Calculate the area as follows:
1) Integrate the area from T = 0 to T = 10 K using the fit equation.
2) Find the area from T = 10 to T = 200 K using the trapezoid rule.
3) Sum up the two areas to find dH in J/mol.
4) Convert dH to kJ/mol and print the result.
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area0to10 = Integrate fiteqn1, x, 0, 10
- 0.508333
There are 17 data points and 16 trapezoids, so i will range from 1 to 16.
area10to200 =
1
* Sum cp i + 1
2
31962.5
+ cp
i
*
t
i+1
- t
i
, i, 1, 16
totalAreaJ = area0to10 + area10to200;
totalAreakJ = totalAreaJ 1000;
Print "dH = ", totalAreakJ, " kJ mol"
dH = 31.9619
kJ mol
Since dH = H (298.06) - H (0) and H (0) = 0, then H (298.06) = dH.
Print "Molar enthaply relative to 0 K = ", totalAreakJ,
" kJ mol"
Molar enthaply
relative
to 0 K = 31.9619
kJ mol
The list of t and cp values is already defined. To calculate dS, make a new list of cp/t
values and use it to make a list of {t, cp/t} values. Make sure to name your definitions
with different names.
cpT = cp t;
points2 = t, cpT ;
data2 = Transpose points2
10 , 0.209 , 20 , 0.7215 , 30 , 1.21467 ,
40 , 1.56375 , 50 , 1.7406 , 60 , 1.85 , 70 , 1.87714 ,
80 , 1.8675 , 90 , 1.83667 , 100 , 1.796 , 110 , 1.75273
150 , 1.584 , 160 , 1.54563 , 170 , 1.50882 ,
180 , 1.47278 , 190 , 1.43684 , 200 , 1.4015
,
51
Scatter plot the points with ListPlot:
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dataplot2 =
ListPlot data2, AxesLabel ® "T,K", "Cp T,J K- 2 mol- 1 " ,
PlotLabel ® "3rd Law Entropy", PlotStyle ® PointSize 0.015

Cp T,J K - 2 mol - 1
;
3rd Law Entropy
1.75
1.5
1.25
1
0.75
0.5
0.25
50
100
150
200
T,K
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Similar to molar enthalpy, extrapolate the data to T = 0 by fitting a function to the first
three data points. You can use the lists already defined for the first 3 points and just
make a new list of cp/t values. Using those lists, make a new list of {t, cp/t} values.
cpT3pts = cp3pts t3pts;
points3 = t3pts, cpT3pts ;
dataS = Transpose points3
10, 0.209 ,
20, 0.7215 ,
30, 1.21467
Fit a 2nd order polynomial to the data points and check the fit by plotting the points and
the function together. Remember that if you suppress the graphs with DisplayFunction,
you must unsuppress when you use Show.
fiteqn2 = Fit dataS, 1, x, x^2 , x
D
- 0.322833 + 0.05415 x - 0.0000966667 x2
fitplot2 = Plot fiteqn2, x, 0, 30 , DisplayFunction ® Identity ;
Show dataplot2, fitplot2 ,
DisplayFunction ® $DisplayFunction ;

52
Cp T,J K - 2 mol - 1
3rd Law Entropy
1.5
1
0.5
50
100
150
200
T,K
Calculate the area as follows:
1) Integrate the area from T = 0 to T = 10 K using the fit equation.
2) Find the area from T = 10 to T = 200 K using the trapezoid rule.
3) Sum up the two areas to find dS in J/K mol.
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Sarea0to10 = Integrate fiteqn2, x, 0, 10
- 0.553056
There are 17 data points and 16 trapezoids, so i will range from 1 to 16.
Sarea10to200 =
1
* Sum cpT i + 1
2
i, 1, 16
295.79
+ cpT
i
*
t
i+1
- t
i
,
totalSAreaJ = Sarea0to10 + Sarea10to200;
Print "dS = ", totalSAreaJ, " J K mol"
dS = 295.237
J K mol
Since dS = S (200) - S (0) and S (0) = 0, then S (200) = dS.
Print "3rd law entropy relative to 0 K = ", totalSAreaJ,
" J K mol"
3rd law entropy
relative
to 0 K = 295.237
J K mol