question archive Consider the Gibbs energies at 25 ?C
Consider the Gibbs energies at 25 ?C
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Consider the Gibbs energies at 25 ?C.
| SubstanceSubstance | ΔG?f (kJ⋅mol−1)ΔGf? (kJ·mol−1) |
|---|---|
| Ag+(aq)Ag+(aq) | 77.177.1 |
| Cl−(aq)Cl−(aq) | −131.2−131.2 |
| AgCl(s)AgCl(s) | −109.8−109.8 |
| Br−(aq)Br−(aq) | −104.0−104.0 |
| AgBr(s)AgBr(s) | −96.9−96.9 |
(a) Calculate ΔG?rxn for the dissolution of AgCl(s)AgCl(s).
kJ⋅mol−1
(b) Calculate the solubility-product constant of AgCl.
K=
(c) Calculate ΔG?rxnΔGrxn? for the dissolution of AgBr(s)AgBr(s).
kJ⋅mol−1kJ⋅mol−1
(d) Calculate the solubility-product constant of AgBr.
K=K=
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a.
AgCl(s) <---------------> Ag^+(aq) + Cl^-(aq)
ΔG?rxn = ΔG?f products - ΔG?f reactants
= -131.2+77.1 -(-109.8)
= 55.7KJ/mole
b.
ΔG?rxn = 55700J/mole
ΔG?rxn = -RTlnKsp
55700 = -8.314*298lnKsp
lnKsp = 55700/(-8.314*298)
lnKsp = -22.48
Ksp = 1.73*10^-10 >>>>>answer
c.
AgBr(s) <---------------> Ag^+(aq) + Br^-(aq)
ΔG?rxn = ΔG?f products - ΔG?f reactants
= -104+77.1 -(-96.9)
= 70KJ/mole
d.
ΔG?rxn = 70000J/mole
ΔG?rxn = -RTlnKsp
70000 = -8.314*298lnKsp
lnKsp = 70000/(-8.314*298)
lnKsp = -28.25
Ksp = 5.4*10^-13 >>>>>answer
