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CHM 112
Section 1

Homework Assignment #27
Due April 19, 2006 by 8:05 a.m.

Practice (not to be turned in): 17.16, 17.52, 17.53, 17.54, 17.61, 17.62, 17.79, 17.84, 17.85

Choose the data set corresponding to the last digit of your Student ID Number.

The data given is the solubility in g/L for the indicated salt at each temperature. Use this data to find the Ho and So for the solubilization of the salt by using a van't Hoff plot. Find the slope and intercept graphically, using all of the data. Using the data from the Table of Thermodynamic Quantities, compare the calculated values for Ho and So to those found from your graph. Is the agreement good, bad, or indifferent?

Hints:

Write the reaction. This is critical for solving this problem correctly.

Note units of the solubility. You will need to find the equilibrium constant from the solubility, which we did not do in class. Recall that mass action expressions are expressed in terms of molarity and use the principals described earlier.

For finding the slope and intercept, use the techniques we learned earlier for Arrhenius plots of kinetic data.

To find Ho and So from data in the Table of Thermodynamic Quantities, follow the procedures in Homework 24 and Homework 25.

If you are assigned one of the mercury salts, note that Hg22+ is a single ion that contains two atoms. The mercury does not break apart into 2 Hg+ ions.

This is a long homework assignment and integrates ideas from several parts of the course, so will be worth 20 points.



Set:

T (oC)

0

MgF2
g/L

1

CaF2
g/L

2

SrF2
g/L

3

BaF2
g/L

4

PbCl2
g/L

5

PbBr2
g/L

6

PbI2
g/L

7

Hg2Cl2
g/L

8

Hg2Br2
g/L

9

Hg2I2
g/L

5
0.0167
0.0202
0.115
0.459
3.93
2.17
0.314
0.000134
2.86×10–6
3.48×10–8
15
0.0157
0.0215
0.128
0.477
4.35
2.70
0.450
0.000193
5.79×10–6
7.57×10–8
25
0.0155
0.0227
0.124
0.483
4.82
3.13
0.601
0.000317
9.81×10–6
1.53×10–7
35
0.0145
0.0241
0.131
0.511
5.44
3.63
0.812
0.000534
1.47×10–5
3.81×10–7
45
0.0143
0.0254
0.137
0.503
6.00
4.26
0.953
0.000806
3.63×10–5
6.58×10–7
55
0.0138
0.0264
0.133
0.527
6.43
4.76
1.27
0.00139
5.87×10–5
9.96×10–7
65
0.0136
0.0278
0.151
0.542
7.07
5.47
1.56
0.00180
6.65×10–5
1.89×10–6
75
0.0130
0.0287
0.141
0.556
7.67
5.72
1.94
0.00222
1.23×10–4
6.54×10–6
85
0.0127
0.0295
0.153
0.578
8.16
7.24
2.58
0.00425
1.89×10–4
8.48×10–6
95
0.0123
0.0305
0.151
0.576
8.80
7.77
3.28
0.00461
3.89×10–4
1.27×10–5


Answers:

MgF2

The reaction is: MgF2(s) Mg2+(aq) + 2 F(aq)

The molar solubilities of magnesium fluoride are found by dividing the given mass solubilites (g/L) by the molar mass, 62.3 g/mol. Since this is a 1:2 salt, Ksp = [Mg2+]e[F]e2 = 4s3, where s is the molar solubility.

T (oC)
T (K)
1/T (K–1)
solubility (g/L)
solubility (M)
Ksp = 4s3
ln Ksp
5
278
3.60×10–3
0.0167
2.68×10–4
7.64×10–11
–23.295
15
288
3.47×10–3
0.0157
2.52×10–4
6.40×10–11
–23.473
25
298
3.36×10–3
0.0155
2.49×10–4
6.12×10–11
–23.517
35
308
3.25×10–3
0.0145
2.33×10–4
5.07×10–11
–23.704
45
318
3.14×10–3
0.0143
2.30×10–4
4.84×10–11
–23.752
55
328
3.05×10–3
0.0138
2.22×10–4
4.33×10–11
–23.863
65
338
2.96×10–3
0.0136
2.18×10–4
4.19×10–11
–23.896
75
348
2.87×10–3
0.0130
2.09×10–4
3.63×10–11
–24.040
85
358
2.79×10–3
0.0123
1.97×10–4
3.10×10–11
–24.197
95
368
2.72×10–3
0.0127
2.04×10–4
3.43×10–11
–24.097

The highlighted cells are plotted, as shown below:

The slope of the plot = 984 and the intercept is –26.9.

Ho = –R×slope = –8.314×(984) = 8180 J = –8.2 kJ

So = R×intercept = 8.314×(–26.9) = –224 J/K

To find the values from the Table of Thermodynamic Quantities :

Hfo(Mg2+(aq)) = –466.9 kJ/mol

Hfo(F(aq)) = –332.6 kJ/mol

Hfo(MgF2(s)) = –1124 kJ/mol

So(Mg2+(aq)) = –138.1 J/mol•K

So(F(aq)) = –13.8 J/mol•K

So(MgF2(s)) = 57.24 J/mol•K

Ho = [–466.9 + 2(–332.6)] – [–1124] = –8. kJ/mol

So = [–138.1 + 2(–13.8)] – [57.24] = –222.9 J/mol•K

The agreement is good for the enthalpy change and better than expected for the entropy change.



CaF2

The reaction is: CaF2(s) Ca2+(aq) + 2 F(aq)

The molar solubilities of calcium fluoride are found by dividing the given mass solubilites (g/L) by the molar mass, 78.1 g/mol. Since this is a 1:2 salt, Ksp = [Ca2+]e[F]e2 = 4s3, where s is the molar solubility.

T (oC)
T (K)
1/T (K–1)
solubility (g/L)
solubility (M)
Ksp = 4s3
ln Ksp
5
278
3.60×10–3
0.0202
2.59×10–4
6.92×10–11
–23.395
15
288
3.47×10–3
0.0215
2.75×10–4
8.37×10–11
–23.204
25
298
3.36×10–3
0.0227
2.91×10–4
9.82×10–11
–23.044
35
308
3.25×10–3
0.0240
3.07×10–4
1.17×10–10
–22.871
45
318
3.14×10–3
0.0254
3.25×10–4
1.38×10–10
–22.706
55
328
3.05×10–3
0.0264
3.38×10–4
1.55×10–10
–22.588
65
338
2.96×10–3
0.0278
3.56×10–4
1.81×10–10
–22.434
75
348
2.87×10–3
0.0287
3.67×10–4
1.98×10–10
–22.343
85
358
2.79×10–3
0.0295
3.78×10–4
2.15×10–10
–22.262
95
368
2.72×10–3
0.0305
3.91×10–4
2.38×10–10
–22.158

The highlighted cells are plotted, as shown below:

The slope of the plot = –1417 and the intercept is –18.3.

Ho = –R×slope = –8.314×(–1417) = 11780 J = 11.8 kJ

So = R×intercept = 8.314×(–18.3) = –152 J/K

To find the values from the Table of Thermodynamic Quantities :

Hfo(Ca2+(aq)) = –542.96 kJ/mol

Hfo(F(aq)) = –332.6 kJ/mol

Hfo(CaF2(s)) = –1220 kJ/mol

So(Ca2+(aq)) = –55.2 J/mol•K

So(F(aq)) = –13.8 J/mol•K

So(CaF2(s)) = 68.87 J/mol•K

Ho = [–542.96 + 2(–332.6)] – [–1220.] = 12. kJ/mol

So = [–55.2 + 2(–13.8)] – [68.87] = –151.7 J/mol•K

The agreement is good for the enthalpy change and better than expected for the entropy change.



SrF2

The reaction is: SrF2(s) Sr2+(aq) + 2 F(aq)

The molar solubilities of strontium fluoride are found by dividing the given mass solubilites (g/L) by the molar mass, 125.6 g/mol. Since this is a 1:2 salt, Ksp = [Sr2+]e[F]e2 = 4s3, where s is the molar solubility.

T (oC)
T (K)
1/T (K–1)
solubility (g/L)
solubility (M)
Ksp = 4s3
ln Ksp
5
278
3.60×10–3
0.115
9.16×10–4
3.05×10–9
–19.609
15
288
3.47×10–3
0.128
1.02×10–3
4.23×10–9
–19.281
25
298
3.36×10–3
0.124
9.87×10–4
3.86×10–9
–19.373
35
308
3.25×10–3
0.131
1.04×10–3
4.53×10–9
–19.213
45
318
3.14×10–3
0.137
1.09×10–3
5.16×10–9
–19.083
55
328
3.05×10–3
0.133
1.06×10–3
4.78×10–9
–19.160
65
338
2.96×10–3
0.151
1.20×10–3
6.93×10–9
–18.787
75
348
2.87×10–3
0.141
1.12×10–3
5.64×10–9
–18.994
85
358
2.79×10–3
0.153
1.22×10–3
7.27×10–9
–18.739
95
358
2.72×10–3
0.151
1.20×10–3
6.91×10–9
–18.790

The highlighted cells are plotted, as shown below:

The slope of the plot = –656.5 and the intercept is –17.1.

Ho = –R×slope = –8.314×(–656.5) = 5458 J = 5.5 kJ

So = R×intercept = 8.314×(–17.1) = –142 J/K

To find the values from the Table of Thermodynamic Quantities :

Hfo(Sr2+(aq)) = –545.8 kJ/mol

Hfo(F(aq)) = –332.6 kJ/mol

Hfo(SrF2(s)) = –1216.3 kJ/mol

So(Sr2+(aq)) = –32.6 J/mol•K

So(F(aq)) = –13.8 J/mol•K

So(SrF2(s)) = 82.1 J/mol•K

Ho = [–545.8 + 2(–332.6)] – [–1216.3] = 5.3 kJ/mol

So = [–32.6 + 2(–13.8)] – [82.1] = –142.3 J/mol•K

The agreement is good for the enthalpy change and better than expected for the entropy change.



BaF2

The reaction is: BaF2(s) Ba2+(aq) + 2 F(aq)

The molar solubilities of barium fluoride are found by dividing the given mass solubilites (g/L) by the molar mass, 175.3 g/mol. Since this is a 1:2 salt, Ksp = [Ba2+]e[F]e2 = 4s3, where s is the molar solubility.

T (oC)
T (K)
1/T (K–1)
solubility (g/L)
solubility (M)
Ksp = 4s3
ln Ksp
5
278
3.60×10–3
0.459
2.62×10–3
7.19×10–8
–16.448
15
288
3.47×10–3
0.477
2.72×10–3
8.05×10–8
–16.335
25
298
3.36×10–3
0.483
2.76×10–3
8.37×10–8
–16.297
35
308
3.25×10–3
0.511
2.92×10–3
9.88×10–8
–16.130
45
318
3.14×10–3
0.503
2.87×10–3
9.47×10–8
–16.172
55
328
3.05×10–3
0.527
3.01×10–3
1.09×10–7
–16.033
65
338
2.96×10–3
0.542
3.09×10–3
1.18×10–7
–15.952
75
348
2.87×10–3
0.556
3.17×10–3
1.27×10–7
–15.876
85
358
2.79×10–3
0.578
3.30×10–3
1.44×10–7
–15.756
95
368
2.72×10–3
0.576
3.29×10–3
1.42×10–7
–15.767

The highlighted cells are plotted, as shown below:

The slope of the plot = –804.2 and the intercept is –13.6.

Ho = –R×slope = –8.314×(–804.2) = 6686 J = 6.7 kJ

So = R×intercept = 8.314×(–13.6) = –113 J/K

To find the values from the Table of Thermodynamic Quantities :

Hfo(Ba2+(aq)) = –537.6 kJ/mol

Hfo(F(aq)) = –332.6 kJ/mol

Hfo(BaF2(s)) = –1209 kJ/mol

So(Ba2+(aq)) = 9.6 J/mol•K

So(F(aq)) = –13.8 J/mol•K

So(BaF2(s)) = 96.40 J/mol•K

Ho = [–537.6 + 2(–332.6)] – [–1209] = 6. kJ/mol

So = [9.6 + 2(–13.8)] – [96.40] = –114.4 J/mol•K

The agreement is reasonable for the enthalpy change and better than expected for the entropy change.



PbCl2

The reaction is: PbCl2(s) Pb2+(aq) + 2 Cl(aq)

The molar solubilities of lead(II) chloride are found by dividing the given mass solubilites (g/L) by the molar mass, 278.2 g/mol. Since this is a 1:2 salt, Ksp = [Pb2+]e[Cl]e2 = 4s3, where s is the molar solubility.

T (oC)
T (K)
1/T (K–1)
solubility (g/L)
solubility (M)
Ksp = 4s3
ln Ksp
5
278
3.60×10–3
3.93
1.41×10–2
1.13×10–5
–11.389
15
288
3.47×10–3
4.35
1.56×10–2
1.53×10–5
–11.088
25
298
3.36×10–3
4.81
1.73×10–2
2.08×10–5
–10.782
35
308
3.25×10–3
5.44
1.96×10–2
3.00×10–5
–10.416
45
318
3.14×10–3
6.00
2.16×10–2
4.02×10–5
–10.122
55
328
3.05×10–3
6.43
2.31×10–2
4.94×10–5
–9.915
65
338
2.96×10–3
7.07
2.54×10–2
6.56×10–5
–9.631
75
348
2.87×10–3
7.67
2.76×10–2
8.39×10–5
–9.386
85
358
2.79×10–3
8.16
2.93×10–2
1.01×10–4
–9.200
95
368
2.72×10–3
8.80
3.16×10–2
1.27×10–4
–8.974

The highlighted cells are plotted, as shown below:

The slope of the plot = –2774 and the intercept is –1.43.

Ho = –R×slope = –8.314×(–2774) = 23060 J = –23.1 kJ

So = R×intercept = 8.314×(–1.43) = –11.9 J/K

To find the values from the Table of Thermodynamic Quantities :

Hfo(Pb2+(aq)) = –1.7 kJ/mol

Hfo(Cl(aq)) = –167.2 kJ/mol

Hfo(PbCl2(s)) = –359 kJ/mol

So(Pb2+(aq)) = 10.5 J/mol•K

So(Cl(aq)) = 56.5 J/mol•K

So(PbCl2(s)) = 136 J/mol•K

Ho = [–1.7 + 2(–167.2)] – [–359] = 23. kJ

So = [10.5 + 2(56.5)] – [136] = –13 J/mol•K

The agreement is good for the enthalpy change and better than expected for the entropy change.



PbBr2

The reaction is: PbBr2(s) Pb2+(aq) + 2 Br(aq)

The molar solubilities of lead(II) bromide are found by dividing the given mass solubilites (g/L) by the molar mass, 367.0 g/mol. Since this is a 1:2 salt, Ksp = [Pb2+]e[Br]e2 = 4s3, where s is the molar solubility.

T (oC)
T (K)
1/T (K–1)
solubility (g/L)
solubility (M)
Ksp = 4s3
ln Ksp
5
278
3.60×10–3
2.17
5.91×10–3
8.26×10–7
–14.007
15
288
3.47×10–3
2.70
7.36×10–3
1.59×10–6
–13.351
25
298
3.36×10–3
3.13
8.53×10–3
2.48×10–6
–12.907
35
308
3.25×10–3
3.63
9.89×10–3
3.88×10–6
–12.459
45
318
3.14×10–3
4.26
1.16×10–2
6.27×10–6
–11.780
55
328
3.05×10–3
4.76
1.30×10–2
8.73×10–6
–11.648
65
338
2.96×10–3
5.47
1.49×10–2
1.32×10–5
–11.232
75
348
2.87×10–3
5.72
1.56×10–2
1.51×10–5
–11.098
85
358
2.79×10–3
7.24
1.97×10–2
3.07×10–5
–10.392
95
368
2.72×10–3
7.77
2.12×10–2
3.80×10–5
–10.178

The highlighted cells are plotted, as shown below:

The slope of the plot = –4252 and the intercept is 1.34.

Ho = –R×slope = –8.314×(–4252) = 35350 J = 35.4 kJ/mol

So = R×intercept = 8.314×(1.34) = 11.1 J/K

To find the values from the Table of Thermodynamic Quantities :

Hfo(Pb2+(aq)) = –1.7 kJ/mol

Hfo(Br(aq)) = –120.9 kJ/mol

Hfo(PbBr2(s)) = –278.7 kJ/mol

So(Pb2+(aq)) = 10.5 J/mol•K

So(Br(aq)) = 80.71 J/mol•K

So(PbBr2(s)) = 161.5 J/mol•K

Ho = [–1.7 + 2(–120.9)] – [–278.7] = 35.2 kJ/mol

So = [10.5 + 2(80.71)] – [161.5] = 10.4 J/mol•K

The agreement is good for the enthalpy change and better than expected for the entropy change.



PbI2

The reaction is: PbI2(s) Pb2+(aq) + 2 I(aq)

The molar solubilities of lead(II) iodide are found by dividing the given mass solubilites (g/L) by the molar mass, 461.0 g/mol. Since this is a 1:2 salt, Ksp = [Pb2+]e[I]e2 = 4s3, where s is the molar solubility.

T (oC)
T (K)
1/T (K–1)
solubility (g/L)
solubility (M)
Ksp = 4s3
ln Ksp
5
278
3.60×10–3
0.314
6.81×10–4
1.27×10–9
–20.486
15
288
3.47×10–3
0.450
9.76×10–4
3.72×10–9
–19.408
25
298
3.36×10–3
0.601
1.30×10–3
8.85×10–9
–18.543
35
308
3.25×10–3
0.812
1.76×10–3
2.18×10–8
–17.640
45
318
3.14×10–3
0.953
2.07×10–3
3.53×10–8
–17.159
55
328
3.05×10–3
1.27
2.75×10–3
8.35×10–8
–16.298
65
338
2.96×10–3
1.56
3.38×10–3
1.54×10–7
–15.686
75
348
2.87×10–3
1.94
4.21×10–3
2.98×10–7
–15.025
85
358
2.79×10–3
2.58
5.60×10–3
7.03×10–7
–14.168
95
368
2.72×10–3
3.28
7.11×10–3
1.44×10–6
–13.449

The highlighted cells are plotted, as shown below:

The slope of the plot = –7731 and the intercept is 7.34.

Ho = –R×slope = –8.314×(–7731) = 64300 J = 64.3 kJ

So = R×intercept = 8.314×(7.34) = 61.0 J/K

To find the values from the Table of Thermodynamic Quantities :

Hfo(Pb2+(aq)) = –1.7 kJ/mol

Hfo(I(aq)) = –55.19 kJ/mol

Hfo(PbI2(s)) = –175.5 kJ/mol

So(Pb2+(aq)) = 10.5 J/mol•K

So(I(aq)) = 111.3 J/mol•K

So(PbI2(s)) = 174.8 J/mol•K

Ho = [–1.7 + 2(–55.19)] – [–175.5] = 63.4 kJ/mol

So = [10.5 + 2(111.3)] – [174.8] = 58.3 J/mol•K

The agreement is good for the enthalpy change and better than expected for the entropy change.



Hg2Cl2

The reaction is: Hg2Cl2(s)Hg22+(aq) + 2 Cl(aq)

The molar solubilities of mercury(I) chloride are found by dividing the given mass solubilites (g/L) by the molar mass, 472.2 g/mol. Since this is a 1:2 salt, Ksp = [Hg22+]e[Cl]e2 = 4s3, where s is the molar solubility.

T (oC)
T (K)
1/T (K–1)
solubility (g/L)
solubility (M)
Ksp = 4s3
ln Ksp
5
278
3.60×10–3
0.000134
2.84×10–7
9.20×10–20
–43.832
15
288
3.47×10–3
0.000193
4.09×10–7
2.74×10–19
–42.740
25
298
3.36×10–3
0.000317
6.71×10–7
1.21×10–18
–41.259
35
308
3.25×10–3
0.000534
1.13×10–6
5.78×10–18
–39.692
45
318
3.14×10–3
0.000806
1.71×10–6
1.99×10–17
–38.455
55
328
3.05×10–3
0.00139
2.94×10–6
1.02×10–16
–36.818
65
338
2.96×10–3
0.00180
3.81×10–6
2.22×10–16
–36.044
75
348
2.87×10–3
0.00222
4.70×10–6
4.15×10–16
–35.419
85
358
2.79×10–3
0.00425
9.00×10–6
2.92×10–15
–33.467
95
368
2.72×10–3
0.00461
9.76×10–6
3.71×10–15
–33.227

The highlighted cells are plotted, as shown below:

The slope of the plot = –12960 and the intercept is 1.22.

Ho = –R×slope = –8.314×(–12960) = 107700 J = 107.7 kJ

So = R×intercept = 8.314×(1.22) = 10.1 J/K

To find the values from the Table of Thermodynamic Quantities :

Hfo(Hg22+(aq)) = 172.4 kJ/mol

Hfo(Cl(aq)) = –167.2 kJ/mol

Hfo(Hg2Cl2(s)) = –265.4 kJ/mol

So(Hg22+(aq)) = 84.5 J/mol•K

So(Cl(aq)) = 56.5 J/mol•K

So(Hg2Cl2(s)) = 191.6 J/mol•K

Ho = [172.4 + 2(–167.2)] – [–265.4] = 103.4 kJ/mol

So = [84.5 + 2(56.5)] – [191.6] = 5.9 J/mol•K

The agreement is modest for both the enthalpy change and the entropy change.



Hg2Br2

The reaction is: Hg2Br2(s)Hg22+(aq) + 2 Br(aq)

The molar solubilities of mercury(I) bromide are found by dividing the given mass solubilites (g/L) by the molar mass, 561.0 g/mol. Since this is a 1:2 salt, Ksp = [Hg22+]e[Br]e2 = 4s3, where s is the molar solubility.

T (oC)
T (K)
1/T (K–1)
solubility (g/L)
solubility (M)
Ksp = 4s3
ln Ksp
5
278
3.60×10–3
2.86×10–6
5.10×10–9
5.35×10–25
–55.888
15
288
3.47×10–3
5.79×10–6
1.03×10–8
4.45×10–24
–53.769
25
298
3.36×10–3
9.81×10–6
1.75×10–8
2.16×10–23
–52.189
35
308
3.25×10–3
1.49×10–5
2.66×10–8
7.25×10–23
–50.979
45
318
3.14×10–3
3.63×10–5
6.47×10–8
1.09×10–21
–48.267
55
328
3.05×10–3
5.87×10–5
1.05×10–7
4.63×10–21
–46.823
65
338
2.96×10–3
6.65×10–5
1.19×10–7
6.75×10–21
–46.445
75
348
2.87×10–3
1.23×10–4
2.19×10–7
4.30×10–20
–44.593
85
358
2.79×10–3
1.89×10–4
3.37×10–7
1.54×10–19
–43.315
95
368
2.72×10–3
3.89×10–4
6.93×10–7
1.35×10–18
–41.146

The highlighted cells are plotted, as shown below:

The slope of the plot = –16095 and the intercept is 1.89.

Ho = –R×slope = –8.314×(–16095) = 133800 J = 133.8 kJ

So = R×intercept = 8.314×(1.89) = 15.7 J/K

To find the values from the Table of Thermodynamic Quantities :

Hfo(Hg22+(aq)) = 172.4 kJ/mol

Hfo(Br(aq)) = –120.9 kJ/mol

Hfo(Hg2Br2(s)) = –206.9 kJ/mol

So(Hg22+(aq)) = 84.5 J/mol•K

So(Br(aq)) = 80.71 J/mol•K

So(Hg2Br2(s)) = 218.0 J/mol•K

Ho = [172.4 + 2(–120.9)] – [–206.9] = 137.5 kJ/mol

So = [84.5 + 2(80.71)] – [218.0] = 27.9 J/mol•K

The agreement is moderate for the enthalpy change and poor for the entropy change.



Hg2I2

The reaction is: Hg2I2(s)Hg22+(aq) + 2 I(aq)

The molar solubilities of mercury(I) iodide are found by dividing the given mass solubilites (g/L) by the molar mass, 655.0 g/mol. Since this is a 1:2 salt, Ksp = [Hg22+]e[I]e2 = 4s3, where s is the molar solubility.

T (oC)
T (K)
1/T (K–1)
solubility (g/L)
solubility (M)
Ksp = 4s3
ln Ksp
5
278
3.60×10–3
3.49×10–8
5.33×10–11
6.07×10–31
–69.577
15
288
3.47×10–3
7.57×10–8
1.16×10–10
6.17×10–30
–67.258
25
298
3.36×10–3
1.53×10–7
2.34×10–10
5.10×10–29
–65.147
35
308
3.25×10–3
3.81×10–7
5.82×10–10
7.88×10–28
–62.408
45
318
3.14×10–3
6.58×10–7
1.00×10–9
4.06×10–27
–60.768
55
328
3.05×10–3
9.96×10–7
1.52×10–9
1.40×10–26
–59.528
65
338
2.96×10–3
1.89×10–6
2.89×10–9
9.64×10–26
–57.602
75
348
2.87×10–3
6.54×10–6
9.98×10–9
3.98×10–24
–53.882
85
358
2.79×10–3
8.48×10–6
1.29×10–8
8.67×10–24
–53.103
95
368
2.72×10–3
1.27×10–5
1.94×10–8
2.92×10–23
–51.887

The highlighted cells are plotted, as shown below:

The slope of the plot = –20530 and the intercept is 3.96.

Ho = –R×slope = –8.314×(–20530) = 170700 J = 170.7 kJ

So = R×intercept = 8.314×(3.96) = 32.9 J/K

To find the values from the Table of Thermodynamic Quantities :

Hfo(Hg22+(aq)) = 172.4 kJ/mol

Hfo(I(aq)) = –55.19 kJ/mol

Hfo(Hg2I2(s)) = –121.3 kJ/mol

So(Hg22+(aq)) = 84.5 J/mol•K

So(I(aq)) = 111.3 J/mol•K

So(Hg2I2(s)) = 233.5 J/mol•K

Ho = [172.4 + 2(–55.19)] – [–121.3] = 183.3 kJ

So = [84.5 + 2(111.3)] – [233.5] = 73.6 J/mol•K

The agreement is modest for the enthalpy change and poor for the entropy change.