Why do we write the activity of solids and pure liquids in equilibrium constants at standard conditions as #1#?
1 Answer
Because the standard state of solids and liquids is their molar density at room temperature and pressure, which is equal to their "concentrations" anyway.
When dividing by the standard state within the equilibrium constant, one gets approximately
Since solids and liquids are quite incompressible and their densities don't vary that much with temperature, "sufficiently close to
The standard equilibrium constant, or even the standard reaction quotient, is written to be unitless. Suppose we have this reaction:
#"AgCl"(s) rightleftharpoons "Ag"^(+)(aq) + "Cl"^(-)(aq)#
#K_(sp) = ((gamma_(Ag^(+))["Ag"^(+)]//c^@)(gamma_(Cl^(-))["Cl"^(-)]//c^@))/(gamma_(AgCl(s))["AgCl"(s)]//c_(AgCl(s))^@)# where
#c^@ = "1 M"# for aqueous solutions, and#c_(AgCl(s))^@ = barrho_(AgCl(s))# , the molar density of#"AgCl"(s)# .
We know that
The density of
#barrho = (5.56 cancel"g")/(cancel"1 mL") xx (1000 cancel"mL")/("1 L") xx ("1 mol AgCl")/(143.32 cancel"g AgCl")#
#=# #"38.79 mol/L"#
Now, to calculate its "concentration", we suppose we have
#cancel"1 L AgCl" xx (5.56 cancel"g")/(cancel"1 mL") xx (1000 cancel"mL")/(cancel"1 L") xx "1 mol AgCl"/(143.32 cancel"g AgCl")#
#=# #"38.79 mols"# contained in that#"1 L"#
Well, there you go, the concentration is equal to the molar density at the same temperature.
So, the activity of
#a_("AgCl"(s))^@ = gamma_("AgCl"(s)) cdot (["AgCl"(s)])/c^@ ~~ gamma_("AgCl"(s)) ~~ 1#
and
And so, the equilibrium constant could be written as:
#K_(sp) = ((gamma_(Ag^(+))["Ag"^(+)]//c^@)(gamma_(Cl^(-))["Cl"^(-)]//c^@))/(gamma_(AgCl(s))["AgCl"(s)]//c_(AgCl(s))^@)#
#~~ (gamma_(Ag^(+))["Ag"^(+)]//c^@)(gamma_(Cl^(-))["Cl"^(-)]//c^@)# where
#c^@ = "1 M"# , and#c_(AgCl(s))^@ = barrho_(AgCl(s)) = ["AgCl"(s)]# .
Particularly when the temperature is close to
At very nonstandard temperatures, eventually