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Problem 18.53

Write the equilibrium constant expression for each of the following reactions, and determine the numerical value of K_{eq} at 25 ^{o}C.

(a)PbO_{2}(s) + 4 H^{+}(aq) + 2 Cl^{–}(aq)Pb^{2+}(aq) + 2 H_{2}O(l) + Cl_{2}(g)(b)3 O_{2}(g) + 2 Br^{–}(aq)2 BrO_{3}^{–}(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.

PbO

_{2}(s) + 4 H^{+}(aq) + 2 e^{–}Pb^{2+}(aq) + 2 H_{2}O(l)E ^{o}_{red}= 1.455 V2 Cl

^{–}(aq) Cl_{2}(g) + 2 e^{–}E ^{o}_{ox}= –1.358 VE

^{o}_{cell}= 1.455 – 1.358 = 0.097 V

K

_{eq}= exp(nFE^{o}_{cell}/RT)n = 2

F = 96485 coul/mol

R = 8.314 J/mol•K

T = 298 K

K

_{eq}= exp[(2)(96485)(0.097)/(8.314)(298)] = 1900

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.

O

_{2}(g) + 2 H_{2}O(l) + 4 e^{–}4 OH^{–}(aq)E ^{o}_{red}= 0.401 VBr

^{–}(aq) + 6 OH^{–}(aq)BrO_{3}^{–}(aq) + 3 H_{2}O(l) + 6 e^{–}E ^{o}_{ox}= –0.584 VE

^{o}_{cell}= 0.401 – 0.584 = –0.183 V

K

_{eq}= exp(nFE^{o}_{cell}/RT)n = 12

F = 96485 coul/mol

R = 8.314 J/mol•K

T = 298 K

K

_{eq}= exp[(12)(96485)(–0.183)/(8.314)(298)] = 7×10^{–38}