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Consider the following Gibbs energies at 25 °C. Substance

Ag (aq) 77.1

Cl–(aq) -131.2

AgCl(s) -109.8

Br–(aq) -104.0

AgBr(s) -96.9

(a) Calculate ΔG°rxn for the dissolution of AgCl(s).

(b) Calculate the solubility-product constant of AgCl.

(c) Calculate ΔG°rxn for the dissolution of AgBr(s).

(d) Calculate the solubility-product constant of AgBr.

I tried this and got several answers that made not scene i have no idea were i went wrong please help.(please show work so i can see were i made a mistake please)

Update:

this did not quit work they asked for kj mol-1 i have no idea what that is

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(a) AgCl(s) ==> Ag+(aq) + Cl-(aq)

ΔG°rxn = sum (ΔG°f products) - sum (ΔG°f reactants) = ((1 mole Ag+(aq))(77.1 kJ/mole) + (1 mole Cl-(aq))(-131.2 kJ/mole)) - (1 mole AgCl(s)(-109.8 kJ/mole) = +55.7 kJ/mole

(b) The solubulity product constant is Ksp for the reaction in (a).

ΔG° = -RTlnK, or, K = e^(-ΔG°/RT) = e^(-55,700 J)/(8.31 J/mole K)(298 K) = e^-22.5 = 1.7 x 10^-10

(c) and (d) are done just like (a) and (b). You should get values of

ΔG°rxn = +70.0 kJ

Ksp AgBr = 5.3 x 10^-13

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