a) AgCl Ag+ + Cl-
Grxn0 = 77.1-131.2 + 109.8 = 55.7
b) G = G0 + RTln(K)
G = 0 at the equlribium
K = exp(-G0/RT)
K = 1.726 x 10-10
c) AgBr Ag+ + Br-
Grxn0= 77.1-104 +96.9 = 70
d) G = G0 + RTln(K)
G = 0 at the equlribium
K = exp(-G0/RT)
K = 5.384 x 10-13
Please show work Consider the following Gibbs energies at 25 degree C. calculate delta G degree...
Consider the following Gibbs energies at 25 degree C. Calculate Delta G degree _rxn for the dissolution of AgCl(s). Calculate the solubility-product constant of AgCl. K = Calculate Delta G degree_rxn for the dissolution of AgBr(s). Calculate the solubility-product constant of AgBr. K =
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
< Question 13 of 18 > Consider the Gibbs energies at 25 °C. Substance AG (kJ. mol-!) Ag+ (aq) 77.1 Cl(aq) -131.2 AgCl(s) -109.8 Br" (aq) -104.0 AgBr(s) -96.9 (a) Calculate AGran for the dissolution of AgCl(s). kJ. mol- (b) Calculate the solubility-product constant of AgCl. kJ. mol- (b) Calculate the solubility product constant of AgCl. K = Enter numeric value (c) Calculate AGtx for the dissolution of AgBr(s). kl. mol-? (d) Calculate the solubility-product constant of AgBr. K=
Consider the following Gibbs energies at 25 "C Substance Ag (aq) Cr(aq) AgCI(s) Br(aq) AgBr(s) 77.1 - 131.2 - 109.8 - 104.0 -96.9 (a) Calculate AG rn for the dissolution of AgCl(s). (b) Calculate the solubility-product constant of AgCl Number Number kJ mol (c) Calculate Δ3rxn for the dissolution of AgBr(s). (d) Calculate the solubility-product constant of AgBr Number Number kJ mol
Consider the Gibbs energies at 25 'C AGi (kJ mol) Substance Ag (aq) 77.1 CI (aq) -131.2 AgCls) -109.8 Br (aq) -104.0 -96.9 AgBr(s) (a) Calculate AGn for the dissolution of AgC1(s) kJ mol (b) Calculate the solubility-product constant of AgCl. K = (c) Calculate AGxn for the dissolution of AgBr(s). kJ mol (d) Calculate the solubility-product constant of AgBr. K =
Part c and d are incorrect Consider the following Gibbs energies at 25°C. AG (kJ mo (aq) cr(aq) AgCI(s) Br(aq) 77.1 -131.2 -109.8 -104.0 -96.9 (a) Calculate Δ. , for the dissolution of AgCl(s) (b) Calculate the solubility-product constant of AgCL Number Number 55.7 kJ mol K-11 1.745 × 10-10 (c) Calculate Δ.in for the dissolution of AgBr(s). (d) Calculate the solubility-product constant of AgBr. Number Number 82.9 kJ mol K2.996x 10 I O Previous Gve Up & View Solution...
Consider the Gibbs energies at 25 ∘C. SubstanceSubstance ΔG∘f (kJ⋅mol−1)ΔGf∘ (kJ·mol−1) Ag+(aq)Ag+(aq) 77.177.1 Cl−(aq)Cl−(aq) −131.2−131.2 AgCl(s)AgCl(s) −109.8−109.8 Br−(aq)Br−(aq) −104.0−104.0 AgBr(s)AgBr(s) −96.9−96.9 (a) Calculate ΔG∘rxn for the dissolution of AgCl(s)AgCl(s). kJ⋅mol−1 (b) Calculate the solubility-product constant of AgCl. K= (c) Calculate ΔG∘rxnΔGrxn∘ for the dissolution of AgBr(s)AgBr(s). kJ⋅mol−1kJ⋅mol−1 (d) Calculate the solubility-product constant of AgBr. K=K=
Use the data given here to calculate the values of delta G^degree _rxn at 25^degree C for the reaction described by the equation A + B double headed arrow C If delta D^degree _rxn and delta S^degree _rxn are both negative values, what drives the spontaneous reaction and in what direction at standard conditions? The spontaneous reaction is entropy-driven to the right. entropy-driven to the left. enthalpy-driven to the left. enthalpy-driven to the right.
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Use the standard reaction enthalpies given below to determine Delta H degree _rxn for the following reaction: 2 S(s) + 3 O_2(g) rightarrow 2 SO_3(g) Delta H degree _rxn = ? Given: SO_2(g) rightarrow S(s) + O_2(g) Delta H degree _rxn = +296.8 kJ 2 SO_2(g) + O_2(g) rightarrow 2 SO_3(g) Delta H degree rxn = -197.8 kJ Show all your work!