g Leaming Consider the followling Gibbs energles at 25 \"C Substance AGr (kJ mol
ID: 304263 • Letter: G
Question
g Leaming Consider the followling Gibbs energles at 25 "C Substance AGr (kJ mol Ag (aq) Cl(aq AgCI(s) Br (aq AgBr(s 77.1 -131.2 -109.8 104.0 -96.9 (a) Calculate AGnn for the dissolution of AgCl(s). (b) Calculate the solubility-product constant of Agcl Number Number -55.7 kJ mol" K5.79x 10 (c) Calculate AGrn for the dissolution of AgBr(s).(d) Calculate the solubility-product constant of AgBr. 3 Number Number K 1.85x 1012 70 There is additional feedback available nol on the iew this feedback b ottom divider bar. Click on the divider ar again to hide the additional feedback Close Incorrect.Explanation / Answer
deltaG(rxn) can be calculated by subtracting the sum of deltaG(f) of reactants from that of products,
i.e. deltaG(rxn) = sum[deltaG(f) products] - sum[deltaG(f)reactants]
deltaG is related with K(here Ksp) though the formula deltaGo = -RTlnK, K = e^(-deltaGo/RT)
R = 8.314 J/mol.K = 8.413 x 10^-3 kJ/mol
T = Temperature = 25 oC = 298.15 K
(a) Dissolution of AgCl(s) ---> Ag+(aq) + Cl-(aq)
deltaG(rxn) = [(77.1) +(-131.2)] -[(-109.8)] = +55.7 kJ/mol
(b) Solubility product Ksp of AgCl,
K = e^[-deltaGo/RT] = e^[(-55.7)/(8.413 x 10^-3 x 298.15)] = e^(-22.47) = 1.7428 x 10^-10.
(c) Dissolution of AgBr(s) ---> Ag+(aq) + Br-(aq)
deltaG(rxn) = [(77.1) +(-104.0)] -[(-96.9)] = +70.0 kJ/mol
(d) Solubility product Ksp of AgBr,
K = e^[-deltaGo/RT] = e^[(-70.0)/(8.413 x 10^-3 x 298.15)] = e^(-28.24) = 5.44 x 10^-13.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.