gibbs energy Consider the following Gibbs energies at 25 degree C.(a) Calculate Delta G degree rxn…
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Question “gibbs energy Consider the following Gibbs energies at 25 degree C.(a) Calculate Delta G degree rxn…”
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
The Gibbs free energy(4,6deg) of the reaction can be calculated as follows: A,Go = {n, 4Go (product) – En, AGdeg (reactant) …… (1) The relation between Gibb’s free energy (AGdeg) and solubility constant (K) is given as follows: AGdeg=-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: 4,6deg = En, AG(r) (product) – En, AGdeg (reactant) = [1xAG* +1x AGRc-]-[1x AGRac] =[(1x 77.1)+(1x-131.2)]-[1x-109.8] kJ. mol-! = 55.7 kJ. mol-1 The value of 4, Go For AgCl is 55.7 KJ. (b) The value of gas constant, R = 8.314 J.Ko.mol- Temperature, T = 298 K Substitute the values in equation (2) as follows: AGdeg=-RT In Ky 55.7 kJ = (-8.314 J.K.mol-4)(298 K)* In Kop 55.7 kJ – mol? Kop= sp -8.314×10 kJ K.mol 298 K) K.sp = exp (-22.48), =1.726×10-10 (1J=10-9 KJ) AgCl’s solubility product constant is 1.726 10-0
(C) AgBr’s dissolution can be described as: AgBr (s)-Ag+ (aq),+Br (aq). The A,Go at 298K can be calculated as: 4,6deg = N, AGdeg Product – En AG (reactant). =[1x AG 23. +1x AGor ]-[1x AG*Axbr] =[(1x 77.1)+(1x-104.0)]-[lx-96.9] kJ -mol = 70.0 kJ. mol-‘ The value of 4, Go For AgBr is 70.0 KJ. mol-1 (d). The value of the gas constant, R= 8.314 J.Ko.mol. Temperature, T= 298 K. Substitute these values into equation (2): AGdeg=RT In K 70.0 J =(-8.314 J.K.mol)(298 K).* In Kg 70.0 J. mol- SP (-8.314×10-3 kJK– Mol 298K) Ksp = exp(0) = 5.38×10-13 (15=10–kJ)
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