Question

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=

Answer 1

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