## 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 MgF2g/L 1 CaF2g/L 2 SrF2g/L 3 BaF2g/L 4 PbCl2g/L 5 PbBr2g/L 6 PbI2g/L 7 Hg2Cl2g/L 8 Hg2Br2g/L 9 Hg2I2g/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

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.