Consider the gibbs energies at 25 ∘ c. substance î” g ∘ f ( kj â‹… mol − 1 ) ag + ( aq ) 77.1 cl − ( aq ) − 131.2 agcl ( s ) − 109.8 br − ( aq ) − 104.0 agbr ( s ) − 96.9 (a) calculate î” g ∘ rxn for the dissolution

(a)+55.7 KJ/mol

(b)1.74 x 10-10

(c)+70 KJ/mol

(d)5.44 x 10-13

Explanation:

(a)∆G reaction = n ∆Gproducts - m ∆Greactants

Where, = sigma = sum of,

∆ = delta = change in,

n and m = stoichiometric coefficients of the products and reactants from the balanced equation respectively.

For AgCl, it ionizes in the form:

AgCl(s) ↔ Ag+(aq) + Cl-(aq)

∆G reaction = [ {1 x ∆Gf(Ag+(aq)) } + { 1 x ∆Gf(Cl-(aq) } ] – [ { 1 x ∆Gf(AgCl(s)) } ]

∆G reaction = [ { 1 x 77.1 } + { 1 x (-131.2) } ] – [ 1 x (-109.8) ]

∆G reaction = [ 77.1 – 131.2 ] – [ -109.8 ]

∆G reaction = - 54.1 + 109.8

∆G reaction = + 55.7 KJ/mol

(b)Ksp = [Ag+] [Cl-]

At equilibrium,

∆G = -RT InK

Ksp = e (-∆G/RT)

Where, ∆G = 55.7 KJ/mol = 55.7 x 1000 J/mol = 55,700 J/mol,

R = 8.314 J/mol.K,

T = 25oC = 25 + 273.15 = 298.15 K

Ksp = e [ - 55,700 J/mol / 8.314 J/mol.K x 298.15 K ]

Ksp = e [ - 22.470 ]

Ksp = 1.74 x 10-10

(c) For AgBr, ionizing in the form:

AgBr(s) ↔ Ag+(aq) + Br-(aq)

∆G reaction = [ {1 x ∆Gf(Ag+(aq)) } + { 1 x ∆GfBr-(aq) } ] – [ { 1 x ∆Gf(AgBr(s)) } ]

∆G reaction = [ { 1 x 77.1 } + { 1 x (-104) } ] – [ 1 x (-96.9) ]

∆G reaction = [ 77.1 – 104 ] – [ -96.9 ]

∆G reaction = - 26.9 + 96.9

∆G reaction = + 70 KJ/mol

(d)Ksp = [Ag+] [Br-]

At equilibrium,

∆G = -RT InK

Ksp = e (-∆G/RT)

Where, ∆G = 70 KJ/mol = 70 x 1000 J/mol = 70,000 J/mol,

R = 8.314 J/mol.K,

T = 25oC = 25 + 273.15 = 298.15 K

Ksp = e [ - 70,000 J/mol / 8.314 J/mol.K x 298.15 K ]

Ksp = e [ - 28.2393 ]

Ksp = 5.44 x 10-13

answer: mole of the gas

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