I got the delta gs right now i need the ks Consider the foll

I got the delta g\'s right now i need the k\'s

Consider the following Gibbs energies at 25 Degree C. Calculate Delta G degree rxn for the dissolution of AgCI(S). Calculate the solubility -product constant of AgCI . Calculate Delta G Degree rxn for the dissolution of AgBr(s). Calculate the solubility -product constant of AgBr.

Solution

I am telling you for the first reaction and same thing can be done for the 2nd reaction.

First we calculate the Gibbs free energy (?G) of solution. This is ?G of products - ?G of reactants or:

77.1 kJ/mol - (-131.2) kJ/mol -[ -109.8] kJ/mol = 318.1 kJ/mol = 318100 =

for 2nd reaction 77.1 kJ/mol - (-104) kJ/mol -[ -96.9] kJ/mol = 278 kJ/mol

Now, we know that the solubility product constant is related to the Gibbs free energy by the expression

Ksp = e^[-?G/(RT)]

Substituting in the given values we obtain

Ksp = e^[(3.18*10^5 J*mol^-1)/(8.314 J*mol^-1*K^-1 * 298 K)]

With the Ksp we can now determine the molar solubility of AgCl. To do this, we remember that

Ksp = [Ag+][Cl-]

For every mole of Agcl that dissociates, we get 1 mole of Ag+ and 1 mole of Cl- in aqueous solution. Thus, if we let the variable \'s\' represent the amount of AgCl that dissolves, the expression becomes

Ksp = s^2

Finally, solving for s we get

s = ?(Ksp)