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

1 Answer
Truong-Son N.
Apr 25, 2018

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

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?

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