Question

Consider the following Gibbs energies at 25 "C Substance Ag (aq) Cr(aq) AgCI(s) Br(aq) AgBr(s) 77.1 - 131.2 - 109.8 - 104.0 -96.9 (a) Calculate AG rn for the dissolution of AgCl(s). (b) Calculate the solubility-product constant of AgCl Number Number kJ mol (c) Calculate Δ3rxn for the dissolution of AgBr(s). (d) Calculate the solubility-product constant of AgBr Number Number kJ mol

Answer 1

(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 above

ΔG° = -RTlnK or K = e^^(-ΔG°/RT)

K = e^^(-ΔG°/RT)  

=e^^{(-55,700 J)/(8.31 J/mole K)(298 K)}

= e^^-22.5

= 1.7 x 10^-10

^(C)

ΔG°rxn = sum (ΔG°f products) - sum (ΔG°f reactants)

= ((1 mole Ag+_(aq))(77.1 kJ/mole) + (1 mole Br-_(aq))(-104.0 kJ/mole)) - (1 mole AgBr_(s)(-96.9 kJ/mole)

ΔG°rxn = +70.0 kJ

(d)

K = e^^(-ΔG°/RT)  

=e^^{(-70000 J)/(8.31 J/mole K)(298 K)}

= 5.3 x 10^-13