The solubility product constant at 25°C for AgI(s) in water has the value 8.3 × 10–17. Calculate ∆Grxn at 25°C for the process AgI(s) <--> Ag+(aq) + I– (aq) where [Ag+] = 9.1 × 10–9 and [I–] = 9.1 × 10–9. –91.7 kJ/mol +91.7 kJ/mol 0.0 kJ/mol –4.4 kJ/mol +4.4 kJ/mol

+91.7 KJmol-1

Explanation:

Recall that ∆G= -RTlnK

Since ∆G in this case is ∆Grxn and K is the Ksp

Note that the Ksp is the solubility product (as shown by the reaction equation)

∆Grxn is the change in free energy for the reaction, in this case the ionization of the silver iodide into silver and iodide ions.

R= 8.314JK-1 and T =25°C +273 = 298 K (the centigrade temperature must be appropriately converted to its corresponding absolute absolute before proceeding with the calculation)

Hence we can substitute values accordingly;

∆Grxn = -(8.314 × 298 × ln 8.3×10^-17)

∆Grxn = +91.7 KJmol-1

b.final

the answer is a and have a great day