Question

Determine the equilibrium constant, Kc, for the formation of the complex ion FeSCN2+ if 0.15 moles of Fe(NO3)3 and 0.25 moles of NaSCN are mixed in 1.0 L of water at 25°C and allowed to come to equilibrium. During the reaction, 0.05 moles of FeSCN2+ are formed.

Answer 1

To determine the equilibrium constant, Kc, we first need to write the balanced chemical equation for the formation of the complex ion FeSCN2+:Fe3+  aq  + SCN-  aq   FeSCN2+  aq Next, we need to determine the initial concentrations of the reactants and the product:[Fe3+] initial = 0.15 moles / 1.0 L = 0.15 M[SCN-] initial = 0.25 moles / 1.0 L = 0.25 M[FeSCN2+] initial = 0 moles / 1.0 L = 0 MNow, we can determine the change in concentrations during the reaction:[Fe3+] = -0.05 moles / 1.0 L = -0.05 M[SCN-] = -0.05 moles / 1.0 L = -0.05 M[FeSCN2+] = 0.05 moles / 1.0 L = 0.05 MNow, we can determine the equilibrium concentrations of the reactants and the product:[Fe3+] equilibrium = 0.15 M - 0.05 M = 0.10 M[SCN-] equilibrium = 0.25 M - 0.05 M = 0.20 M[FeSCN2+] equilibrium = 0 M + 0.05 M = 0.05 MFinally, we can calculate the equilibrium constant, Kc, using the equilibrium concentrations:Kc = [FeSCN2+] /  [Fe3+] * [SCN-] Kc =  0.05 M  /  0.10 M * 0.20 M Kc = 0.05 / 0.02Kc = 2.5The equilibrium constant, Kc, for the formation of the complex ion FeSCN2+ is 2.5.