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Consider the following Gibbs energies at 25 °C
Consider the following Gibbs energies at 25 °C.
Substance Ag (aq) Cl–(aq) AgCl(s) Br–(aq) AgBr(s)
(a) Calculate ?G°rxn for the dissolution of AgCl(s).
(b) Calculate the solubility-product constant of AgCl.
(c) Calculate ?G°rxn for the dissolution of AgBr(s).
(d) Calculate the solubility-product constant of AgBr.
| Substance |  Gf |
| Ag+ | 77.1 |
| Cl- | -131.2 |
| AgCl | -109.8 |
| Br- | -104 |
| AgBr | -96.9 |
Expert Solution
(a) AgCl(s) -----> Ag+(aq) + Cl-(aq)
ΔG°rxn =[ total (G°f products)] - [ (G°f reactants)]
= [(1xG°f ( Ag+)(aq) + (1xG°f ( Cl-)(aq)] - [ 1xG°f ( AgC)(s)]
=[(77.1 + (-131.2)] - (-109.8 ) = +55.7 kJ/mole
(b) The solubulity product constant is Ksp for the reaction in (a).
ΔG° = -RTlnK
K = e(-ΔG°/RT) = e(-55700 J) /(8.31 J/mole K)(298 K) = e-22.5 = 1.7 x 10-10
(c)AgBr(s) -----> Ag+(aq) + Br-(aq)
ΔG°rxn = [(1xG°f ( Ag+)(aq) + (1xG°f ( Br-)(aq)] - [ 1xG°f ( AgBr)(s)]
ΔG°rxn = [77.1+ (-104) ] - (-96.9)
ΔG°rxn = +70.0 kJ
d) K = e(-ΔG°/RT) = e(-70000 J) /(8.31 J/mole K)(298 K)
Ksp AgBr = 5.3 x 10-13
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