The balanced chemical equation for the reaction is:
Cd(s) + Cu^2+(aq) -> Cd^2+(aq) + Cu(s)
The standard Gibbs free energy change, ∆G° for the reaction is given as 143 kJ.
The relationship between ∆G° and the equilibrium constant (K) for a reaction is given by:
∆G° = -RTlnK
where R is the gas constant (8.314 J/mol.K) and T is the temperature in Kelvin (25°C = 298 K).
Substituting the given values, we get:
143,000 J = -8.314 J/mol.K x 298 K x ln K
ln K = -143,000 J / (8.314 J/mol.K x 298 K) = -57.36
K = e^(-57.36) = 1.06 x 10^-25
Therefore, the equilibrium constant (K) at 25°C is 1.06 x 10^-25.
When cadmium metal is reduced Cu^2+ in solution Cd^2+ forms in addition to copper metal. If ∆G° =143kJ, Calculate K at 25°C
1 answer