Question

11. Insoluble silver chloride dissolves in the presence of chloride ion. The equation for the overall reaction is AgCl(s) + Cl(aq) AgCl2 (aq) a) show that the overall reaction is the sum of two others: the ionization of AgCl(s) to give silver(I) and chloride ions, and the formation of AgCl2 (aq) from Ag (aq) and Cl(aq) ions. b) Calculate the equilibrium constant for the overall process from the equilibrium constants for the two steps. (refer back to Complex lon Formation Constants and Solubility Product Constants) c) Calculate the free energy change for each step and for the overall reaction. (AGI° (AgCl2 ) = -215.4 kJ/mol)

Answer 1

. Tonization 8 Ag at sy O Agci o = formation of Aglizcaq @ Mercopt acienda Agica + ci cand from Agt aq and cieq Agci caps equation 6 + equation ③ 0 Agar o = ASCOT + (aq) @ peopot fci cand a macı caq) po AgCl (s) + cícaq = Agca (aq) overall Reaction Agci o = Age + Ciced Ksp = (-8 x 10-10 Astept 2ci cand a agen caq kformation: 18x105 equation 0 + equation ③ o Agcios = 6) Sant Aci Cagl - ANCI (8) + cícaq = Asked + kit card Agch caq Agc5 (aq) when two equations are added, the new equilibrium constant is product Equilibrium constant of added Equations. - keq = ksp x Kformation Keq = 18x100 x 18 x105 [keq - 3.24 005) Overall Equation: AgCl (s) + cícaq = Agc aq Ikea - 3.24 x105) 1 (C) ohº = -RT in K R-8:314 J K'mol To 2981 Agci o = Agca) + Cīcaq Ksp = (-8 x 10-10 so = - RT enk = -2.303 RT log k se = -8.314 JK mol'x 39815 x 2.303 . loc = -8.314 Jk mol'x 398K 2303 X(-9.74) = 55601.89 J mol" SG - 55.6 kJ moin Ag (aq) + 200 (aq) ~ Agliz caq? krormation : 18x105 formation - RT en K = -2-303 RT log formation = -8.314 J["mo1'x 3985 2.303 log (1-8 x16M) = -8.314 Jk mo'x 398K * 2.303 x(5.26 ) - - 29985-8 Jmol." sa = -29.986 KJ moi" Agc (aq) AQCI (B) + cícaq = keq - 3.24 x105] AG' = - RT en K = -3.303 RT log Keg = -8-314 Jk'mol'x 29815 X 2303 = -8.314 Jk mol'x 3985 * 2.303 X(-4.49 ) - 25616.12 J mol" AG" = 25.616 KJ mol!