10. Equilibrium constant calculations (see Krasuskopf & Bird, Chapt. 1 or Christ

Question

10. Equilibrium constant calculations (see Krasuskopf & Bird, Chapt. 1 or Christian, Chapt. 6) Cadmium (Cd) is a very toxic contaminant at low concentrations that may be found in natural waters as a result of mining or industrial activities. Its concentration in solution (its “solubility”) as the ion Cd2+ could be controlled in some natural systems by equilibrium with its carbonate mineral form. That is, it dissolves according to the reaction: CdCO3(s) = Cd2+ + CO3 2- K1 = 10-12.1 at 25°C, 1 atm where CdCO3(s) is the solid mineral form and K1 is the equilibrium constant for the dissolution reaction. (a) Write an expression in logarithmic form stating the Law of Mass Action for this reaction. (b) Calculate the standard free energy of the reaction (G°r) at 25°C, 1 atm (note units) from the equilibrium constant.

Explanation / Answer

a)

the given reaction is

CdC03 (s) ----> Cd+2 + C032-

now

the law of mass action is given by

K= [Cd+2] [C032-]

applying log on both sides

we get

ln K = ln [Cd+2] [C032-]

lnK = ln[Cd+2] + ln [C032-]

b) we know that

the standard free energy is given by the reaction

dGo = -RT lnKeq

given

T = 25 c

T = 25 + 273

T = 298 K

now

Keq = 10-12

so

we get

dGo = -8,314 x 298 x ln 10-12

dGo = 68.46 x 1000 J

dGo = 68.46 kJ

so

the standard free energy is 68.46 kJ

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