Question

Find Ksp for AgCl(s) where Ksp is the equilibrium consstant for AgCl(s)<=> Ag+ (aq) + Cl-(aq) dG respectively is -109.70, 77.11, -131.2 kj for mol. Answer: 1.8x10^-10 and what is the partial pressure of atomic chlorine that would be at equilibrium with Cl2(g) at 25*C and 1 atm? delta G =105.7 kJ mol-1. Answer: 3.0 x 10^-19.

Answer 1

Find Ksp for AgCl(s) where Ksp is the equilibrium consstant for AgCl(s)<=> Ag+ (aq) + Cl-(aq) dG respectively is -109.70, 77.11, -131.2 kj for mol. Answer: 1.8x10^-10 and what is the partial pressure of atomic chlorine that would be at equilibrium with Cl2(g) at 25*C and 1 atm? delta G =105.7 kJ mol-1. Answer: 3.0 x 10^-19.

AgCl(s)<=> Ag+ (aq) + Cl-(aq)

∆Go rxn=∑∆Go(products)-∑ ∆Go(reactants)

               =[77.11+(-131.2)]-[-109.70]

               =(-54.09)-(-109.70)

                =-54.09+109.70

               =55.61 KJ/mol

∆Go=-RT ln ksp

Where ksp=equilibrium constant

R=universal gas constant=8.314J/k mol

T=temperature=298K(standard T)

∆Go=-RT ln ksp

55.61 KJ/mol=-8.314 J/K mol * 298K *ln ksp

55.61 kj/mol=-2477.57 J/mo ln Ksp

55.61 kj/mol=-2.477 k J/mo ln Ksp

ln ksp=- 55.61/2.477=-22.450

ksp=exp(-22.450)=1.779*10^-10

ksp=1.779*10^-10(answer)