Respuesta :
Answer:
(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
