Question

What is the standard Gibbs free energy for the dissolution of AgI(s)? The Ksp = 8.3e-17.

Answer 1

Well, the standard change in Gibbs' free energy of reaction, `DeltaG_(rxn)^@` can be found at equilibrium since you have `K`.

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This is for the reaction

`"AgI"(s) rightleftharpoons "Ag"^(+)(aq) + "I"^(-)(aq)`

You have this equation from your notes:

`DeltaG = DeltaG^@ + RTlnQ`

At equilibrium, `DeltaG = 0` and `Q = K`. Therefore,

`DeltaG^@ = -RTlnK`

For dissolution of solids in water, `K -= K_(sp)`. As a result, using `R = "8.314 J/mol"cdot"K"` and the thermodynamic standard temperature,

`color(blue)(DeltaG_(rxn)^@) = -(8.314 cancel"J""/mol"cdotcancel"K")(298.15 cancel"K")ln(8.3 xx 10^(-17)) xx "1 kJ"/(1000 cancel"J")`

`=` `color(blue)"91.7 kJ/mol"`

The reaction significantly favors the reactants at equilibrium at room temperature. Does that agree with reality?