Answer:
A)ΔGrxn∘=55.7kJ/mol
B)Ksp=1.75×10^−10
C)ΔGrxn∘=70.0kJ/mol
D)Ksp=5.45×10^−13
Explanation:
ΔGrxn∘=ΔGf,products∘−ΔGf,reactants∘
To calculate for the
Ksp
of the dissolution reaction can be claculated
ΔGrxn∘=−RTlnKsp
where R is the proportionality constant equal to 8.3145 J/molK.
A)
ΔGrxn∘=[ΔGf,Ag(aq)+∘+ΔGf,Cl(aq)−∘]−ΔGf,
AgCl(s)⇌Ag(aq)++Cl(aq)−
ΔG∘rxn=[77.1kJ/mol+(−131.2kJ/mol)]−(−109.8kJ/mol)
ΔGrxn∘=55.7kJ/mol
b) Calculate the solubility-product constant of AgCI.
ΔGrxn∘=−RTlnKsp
55.7kJ/mol=−(8.3145×10−3J/molK)(298.15K)InKsp
Ksp=1.75×10^−10
c) Calculate
To calculate ΔG°rxn
for the dissolution of AgBr(s).
ΔGrxn∘=[ΔGf,Ag(aq)+∘+ΔGf,Br(aq)−∘]−ΔGf,
ΔGrxn∘=[77.1kJ/mol+(−104.0kJ/mol)]−(−96.90kj/mol
ΔGrxn∘=70.0kJ/mol
d)To Calculate the solubility-product constant of AgBr.
ΔGrxn∘=−RTlnKsp
70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K)lnKsp
70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K
Ksp=5.45×10^−13
The free energy change for the dissolution of AgCl is 55.7 kJ⋅mol−1 and the solubility product of silver chloride is 1.7 × 10^-10.
We need to obtain the ΔG∘rxn for the dissolution of each specie and the equilibrium constant from the following;
a) AgCl(s) ⇄ Ag+(aq) + Cl−(aq)
ΔG∘rxn = 77.1 kJ⋅mol−1 + (−131.2 kJ⋅mol−1) - ( −109.8 kJ⋅mol−1)
ΔG∘rxn = 55.7 kJ⋅mol−1
b) Then;
ΔG∘rxn = -RTlnKsp
Ksp = e^-(ΔG∘rxn/RT)
Ksp = e^-(55.7 × 10^3J⋅mol−1 /8.314 × 298 K)
Ksp = 1.7 × 10^-10
c) AgBr(s) ⇄ Ag+(aq) + Br−(aq)
ΔG∘rxn = 77.1 kJ·mol−1 + (−104.0 kJ·mol−1 ) - (−96.9 kJ·mol−1)
ΔG∘rxn = 70 kJ·mol−1
d) Then
ΔG∘rxn = -RTlnKsp
Ksp = e^-(ΔG∘rxn/RT)
Ksp =e^-(70 × 10^3J·mol−1 /8.314 × 298 K)
Ksp = 5.4 × 10^-13
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