Question

Below are some thermochemical data for the dissolution equilibrium of solid aluminum hydroxide in water:

	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 ΔrH^∘ (Express your answer in kilojoules per mole as an integer.)

Part B Compute ΔrS^∘ (Express your answer in joules per mole per kelvin to one decimal place.)

Part C Compute ΔrG^∘ at 298 K (Express your answer in kilojoules per mole as an integer.)

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

Part F Determine the Gibbs energy change for the dissolution of Al(OH)₃ when the concentrations of Al³⁺ and OH⁻ are both 1.20×10⁻⁷mol⋅L⁻¹ (Express your answer in kilojoules per mole as an integer.)

Part H At what temperature will Al(OH)₃(s) be at equilibrium with Al³⁺ and OH⁻, each at 1.20×10⁻⁷mol⋅L⁻¹?

Part I Explain why the entropy of dissolution for this reaction, solid Al(OH)₃(s) dissolving to form Al³⁺ and 3OH⁻, is so negative. Match the words in the left column to the appropriate blanks in the sentences on the right.

lose, holds on to, low, is released from, gain, high

The Al ³⁺ ion has a quite __ charge so that it __ water molecules, which causes the water molecules to __ translational and rotational degrees of freedom so that the entropy associated with Al ³⁺ in the aqueous phase is very negative.

Answer 1

Here basic concept of thermodynamics is applied. Enthaply and entropy are calculated by difference of enthalpies of product and reactant and in this stochiometric cosfficient is also taken into account.

Gibbs energy is calculated by applying its definition G= H-TS . Tempurature is given 298k . Enthalpy and entropy are used from part A and Part B

Equilibrium constant is calculated by applying the Vant hoff equation . Equilbrium constant comes out to be very very small . It means equilibrium will be almost in reverse direction and the dissolution of Al (OH)3 does not takes place.