Question

urgent

2) Thermochemical data are given below (at T = 25.0 °C) and may be useful in doing the following problem. Substance ΔΗ, (kJ/mol Sº (J/mol K) AG (kJ/mol) Mg?"(aq) Mg(OH)2(5) OH(aq) 466.9 924.7 230.0 138.1 63.1 10.8 -454.8 833.9 - 1572 Using only the data above, find AG rxn, AS man, and K for the reaction Mg(OH)2(8) = Mg?" (aq) + 2 OH(aq) [18 points)

Answer 1

The reaction at T = 25°C = 298 K:

Mg(OH)2(s) <==> Mg²⁺(aq) + 2OH^-(aq)

Using Hess's law:

• ∆H°rxn = (∆H°_f)_Mg+2 + 2×(∆H°_f)_OH- - (∆H°_f)_MgOH

∆H°rxn = -466.9 kJ/mol +2×(-230.0 kJ/mol) -(-924.7 kJ/mol)

∆H°rxn = (-466.9 - 460.0 + 924.7) kJ/mol

∆H°rxn = -2.2 kJ/mol

• ∆S°rxn = (S°_f)_Mg+2 + 2×(S°_f​​​​​​​)_OH- - (S°_f)_MgOH

∆S°rxn = -138.1 J/molK + 2×(-10.8 J/molK) -(63.1 J/molK)

∆S°rxn = (-138.1 -21.6 - 63.1) J/mol K

∆S°rxn = -222.8 J/mol K

Use Gibb's equation:

• ∆G°rxn = ∆H°rxn -T∆S°rxn

∆G°rxn = -2.2 kJ/mol - 298 K×(-222.8 J/mol K)

∆G°rxn = -2.2 kJ/mol + 663944 J/mol

∆G°rxn = -2.2 kJ/mol + 66.4 kJ/mol

∆G°rxn = 64.2 kJ/mol

• ∆G°rxn = -RTlnK

64.2 kJ/mol = -(8.314 J/mol K)(298 K)(lnK)

64200 J/mol = -2477.572 J/mol×lnK

lnK = -25.912

K = e^-25.912

K = 5.58×10^-12

•( ∆G°rxn = 64.2 kJ/mol, ∆S°rxn = -222.8 J/mol K, K = 5.58×10^-12 )