Question

gibbs energy

Consider the following Gibbs energies at 25 degree C.(a) Calculate Delta G degree rxn for the dissolution of AgCl(s). (b) Calculate the solubility-product click to edit AgCl, Number Number kJ.mol^-1 K= (C) Calculate Delta G degree rxn for the dissolution of AgBr(s). (d) Calculate the solubility-product constant of AgBr. Number Number kJ.mol^-1 K= gibbs energy

Answer 1

The Gibbs free energy(4,6°) of the reaction can be calculated as follows: A,Gº = {n, 4Gº (product) – En, AG° (reactant) ...... (1) The relation between Gibb's free energy (AG°) and solubility constant (K) is given as follows: AG°=-RT In K (a) The dissolution of AgCl is given by the equation as follows: AgCl(s) → Ag+ (aq) +C1- (aq) The 4, Gº at 298 K can be calculated as follows: 4,6° = En, AG® (product) – En, AG° (reactant) = [1xAG* +1x AGⓇc-]-[1x AGⓇac] =[(1x 77.1)+(1x-131.2)]-[1x-109.8] kJ. mol-! = 55.7 kJ. mol-1 Hence, the value of 4, Gº for AgCl is 55.7 kJ. mol-?| (b) The value of gas constant, R = 8.314 J.Kº.mol- Temperature, T = 298 K Substitute the values in equation (2) as follows: AG°=-RT In Ky 55.7 kJ = (-8.314 J.K.mol-4)(298 K)* In Kop 55.7 kJ - mol? In Kop= sp (-8.314x10-kJ K-mol-')>(298 K) K.sp = exp(-22.48) =1.726x10-10 (1J=10-9 kJ) Hence, the solubility product constant of AgCl is 1.726 10-0

(C) The dissolution of AgBr is given by the equation as follows: AgBr(s) → Ag+ (aq)+Br(aq) The A,Gº at 298 K can be calculated as follows: 4,6° = n, AG° (product) – En, AG (reactant) =[1x AG 23. +1x AGør ]-[1x AG*Axbr] =[(1x 77.1)+(1x–104.0)]-[lx-96.9] kJ -mol = 70.0 kJ. mol-' Hence, the value of 4, Gº for AgBr is 70.0 kJ. mol-1 (d) The value of gas constant, R = 8.314 J.Kº.mol- Temperature, T = 298 K Substitute the values in equation (2) as follows: AG°=-RT In K 70.0 kJ =(-8.314 J.K'.mol)(298 K)* In Kg 70.0 kJ. mol- SP (-8.314 x 10-3 kJ-K-- mol-')>(298 K) Ksp = exp(-28.25) = 5.38x10-13 (15=10-kJ) Hence, the solubility product constant of AgBr is 5.38x10-13