Consider the Gibbs energies at 25 ∘C.

Substance 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=

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

idk

explanation:

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890