Chem 301 Problem Day #9
Fall 2008
1. Calculating the thermodynamic values for ionic reactions is similar to how we did it for neutral
reactions. For example ΔGrxn=Σ νΔGf(products) – Σ νΔGf(reactants). For ionic reactions, we will
need to make sure that we are dealing with net ionic reactions.
The energy and entropy of formation values, however, are quite difficult to measure for solvated
species. Therefore, an arbitrary choice has been made that assigns a value of zero to the Gibbs energy
change of formation, the enthalpy change of formation, and the molar entropy of the hydrogen ion in
its standard state in aqueous solution (H+(aq)). All other values of aqueous ions are made consistent
with this convention giving them the name “conventional formation enthalpies”, etc.
Table 10.1 lists values of ΔHof , ΔGof, and So for common solvated ions. They are also included in
Table 4.1 in Appendix A.
a. Calculate the ΔGrxn , ΔHrxn , ΔSrxn for the dissolution of CaF2(s) if the values of ΔHof and ΔGof for
CaF2(s) are -1225.9 kJ/mole and -1170.2 kJ/mole, respectively.
CaF2(s) ⇄ Ca2+(aq) + 2 F-(aq)
ΔHrxn = -542.8 + 2(-332.6) – (-1225.9) = 17.9 kJ/mole
ΔGrxn = -553.6 + 2(-278.8) – (-1170.2) = 59.0 kJ/mole
ΔSrxn = -(ΔGrxn - ΔHrxn)/T = -(59.0-17.9)/298 = -137.9 J/Kmole
b. Calculate Ksp using the ΔGrxn value found above.
Ksp = exp(-ΔGrxn/(RT)) = exp(-59000/(8.3145*298)) = 4.5 x 10-11
2. Determine the mean activity coefficient and the mean activity of the electrolyte in a 0.250 m solution
of ZnSO4
a. by assuming the Debye-Hückel model ZnSO4 Zn2+ + SO42-
I = (1/2) [(0.250)(2)2 + (0.250)(2)2 ] = 1.00
ln γ± = -1.173(z+)(z-)(I)1/2 = -1.173(2)(2)(1.00)1/2 = -4.692 ; γ± = 0.00917
a± = (m+m-(γ±)2)1/2 = (0.250)(0.250)(0.00917)1/2 = 0.0023
b. by using Table 10.3
From the table: γ±=0.140 for m=0.20 and γ± = 0.0835 for m= 0.30. So, take the average.
Therefore, γ± = 0.112
a± = (m+m-(γ±)2)1/2 = (0.250)(0.250)(0.112)1/2 = 0.028
