##### J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 18.53

Write the equilibrium constant expression for each of the following reactions, and determine the numerical value of Keq at 25 oC.

(a) PbO2(s) + 4 H+(aq) + 2 Cl(aq)Pb2+(aq) + 2 H2O(l) + Cl2(g)

(b) 3 O2(g) + 2 Br(aq)2 BrO3(aq) (basic solution)

(a) PbO2(s) + 4 H+(aq) + 2 Cl(aq)Pb2+(aq) + 2 H2O(l) + Cl2(g)

The reaction is already balanced so the mass action expression for the equilibirum constant is easily found by inspection. Since a thermodynamic equilibrium constant is required, aqueous components are given as concentrations and gases are given as partial pressures (pure liquids and pure solids are ignored):

To find the value of this equilibrium constant, use the Table of Standard Reduction Potentials to find the cell potential and then use this value to find the equilibrium constant.

 PbO2(s) + 4 H+(aq) + 2 e–Pb2+(aq) + 2 H2O(l) Eored = 1.455 V 2 Cl–(aq) Cl2(g) + 2 e– Eoox = –1.358 V Eocell = 1.455 – 1.358 = 0.097 V Keq = exp(nFEocell/RT) n = 2 F = 96485 coul/mol R = 8.314 J/mol•K T = 298 K Keq = exp[(2)(96485)(0.097)/(8.314)(298)] = 1900
(b) 3 O2(g) + 2 Br(aq)2 BrO3(aq) (basic solution)

The reaction is already balanced so the mass action expression for the equilibirum constant is easily found by inspection. Since a thermodynamic equilibrium constant is required, aqueous components are given as concentrations and gases are given as partial pressures (pure liquids and pure solids are ignored):

To find the value of this equilibrium constant, use the Table of Standard Reduction Potentials to find the cell potential and then use this value to find the equilibrium constant.

 O2(g) + 2 H2O(l) + 4 e–4 OH–(aq) Eored = 0.401 V Br–(aq) + 6 OH–(aq)BrO3–(aq) + 3 H2O(l) + 6 e– Eoox = –0.584 V Eocell = 0.401 – 0.584 = –0.183 V Keq = exp(nFEocell/RT) n = 12 F = 96485 coul/mol R = 8.314 J/mol•K T = 298 K Keq = exp[(12)(96485)(–0.183)/(8.314)(298)] = 7×10–38