Question

Al(OH)3(s) → Al3+(aq) + 3OH−(aq)

ΔfH∘/kJ⋅mol−1 −1284    −538.4 −230.02

S∘/J⋅mol−1⋅K−1 85.4 −325 -10.9

Part A Compute ΔrS (answer in J⋅mol−1⋅K−1)

Part B Compute ΔrG∘ at 298 K (answer in kJ⋅mol−1)

Part C Determine the equilibrium constant for the dissolution of Al(OH)3 at 298 K.

Part D Determine the Gibbs energy change for the dissolution of Al(OH)3 when the concentrations of Al3+ and OH− are both 2.60×10−7mol⋅L−1.(answer in kJ⋅mol−1)

Part E At what temperature will Al(OH)3(s) be at equilibrium with Al3+ and OH−, each at 2.60×10−7mol⋅L−1? (answer in K)

Answer 1

Al(OH)2 (A) = A13+ (aq) + 3 OH(aq) Aysº = Sproducts - Esreactants = (-325 + 3(-10.) - (85.4)] ank mol. K = -443 Jk' mol-! B) AxH = E AfH (producks) - E AFH°(reactants) =F538.4 +3(-230.02) - (-1284)] kJ 538.4 + mol = 55.54 kJ mol So, sh = AH - TAS° = 55.54 kJ mol 298 K x-443 Kimol 1000 J/k] → Ay6° -1187.6 kJ/mol Dr = -RT In K - 187.6 x 10 J Ink -- 8.314 J x298 K x komol mol -33 & Equilibrium constant, K = 11.33 X 10

[A13+][OH)3 Qc = = (2.60 x 10-7)(2.60 x107) = 4.57 X10-27 AG = Da + RT In Qc = . k 187.6 kJ mo I + 8314 J x 298 K x In 4.57X1027 Kimol - 1000 Jl kJ AG = 137.3 kJ/mol E) A+ Equilibrium AG = BH - TAS = 0 T= DH = 55.54 x103 J/mol 443 Ilk.moi Teq = [125.4 K