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
