Question

3. Chemical reactions and equilibrium constants

(a) An alternative way of writing the reaction for the dissolution of CdCO3(s) is to form the species HCO3- rather than CO32-.
Write a balanced chemical reaction for CdCO3(s) dissolution with HCO3- rather than CO32- as a product species.
Calculate an equilibrium constant and standard free energy of reaction for your new reaction. You will need to look up some equilibrium constants.

(b) How would you decide which reaction to use to describe the equilibrium of CdCO3(s) with water? From the standpoint of chemical equilibrium, does it matter? Explain.

Answer 1

CdCO3(s) + H+(aq)HCO3- + Cd2+

HCO3- + H+ H2O(aq) + CO2(g)

overall reaction CdCO3(s) + 2H+CO2(g) + Cd2+ +

The required equation is CdCO3(s) + H+(aq)HCO3- + Cd2+

    compound                               Gfo (kj/mol)  

CdCO3(s)    -674.2

H+(aq) 0.00

Cd2+ (aq)    -77.73

HCO3- (aq)    -586.85

Gorxn=Gofproduct-        Gofreactant = (-586.85-    -77.73) - ( -674.2 +0)= -664.58 +674.2= 9.62 KJ/mol

All Gof values are measured at 25oC

Also we know Gorxn=-RT ln Keq=-8.314J/K mol* 10^-3 KJ/J 298K ln Keq

                        9.62 =2477.57* 10^-3 ln keq

                         9.62=-2.477 ln keq

                           ln keq=-3.883

                              keq=0.02058=2.058*10^-2

b) the product of the reaction of CdCO3(s) with water is CO2(g) and Cd(OH)2 in this case. It is a sparingly soluble salt and so does not dissociate completely.The product decides the reaction to be used in equilibrium condition. The product species present during the reaction are at equilibrium with the reactant species. So the correct species at equilibrium conditions needs to be decided.