Question #dbcb5

The first thing to note here is that the problem provides you with the enthalpy change of hydration, #DeltaH_"hyd"#, Not with the enthalpy change of formation, #DeltaH_f#, for the potassium cation and chloride anion.

The enthalpy change of hydration represents the change in enthalpy when one mole of gaseous ions dissolves in water to produce a solution of infinite dilution.

The enthalpy change of hydration is always negative because it represents the energy released when the dissolved ions form bonds with water molecules.

In your case, you will have

#color(purple)("K"_text((g])^(+) -> "K"_text((aq])^(+)" " " "DeltaH_text(hyd K) = - "322 kJ mol"^(-1))#

#color(purple)("Cl"_text((g])^(-) -> "Cl"_text((aq])^(-)" " " "DeltaH_text(hyd Cl) = -"364 kJ mol"^(-1))#

Now, this part is really important. The problem provides you with the lattice enthalpy of potassium chloride, #"KCl"#

#DeltaH_"latt" = -"701 kJ mol"^(-1)#

When lattice enthalpy carries a negative sign, it represents the energy released when one mole of a solid compound is formed from its constituent gaseous particles.

Simply put, when one mole of gaseous potassium cations combines with one mole of gaseous chloride anions to form one mole of potassium chloride, #"701 kJ"# of heat is being released

#"K"_text((g])^(+) + "Cl"_text((g])^(-) -> "KCl"_text((s])" " " "DeltaH_text(latt) = -"701 kJ mol"^(-1)#

In order to be able to determine the enthalpy change of solution, you will need to use the lattice dissociation enthalpy, which represents the heat needed to break up a lattice into its constituent gaseous particles.

Since it represents energy required, the lattice dissociation enthalpy will carry a positive sign.

In your case, you will have

#color(blue)("KCl"_text((s]) -> "K"_text((g])^(+) + "Cl"_text((g])^(-)" " " "DeltaH_text(dis latt) = +"701 kJ mol"^(-1))#

Now, the standard enthalpy change of solution, #DeltaH_"sol"^@#, represents the enthalpy change when one mole of a solid ionic compound dissolves in water to form a solution of infinite dilution under standard conditions.

#"KCl"_text((s]) -> "K"_text((aq])^(+) + "Cl"_text((aq])^(-)" " " "DeltaH_text(sol)^@#

In order to find #DeltaH_"sol"#, you need to use Hess' Law.

More specifically, you will go from solid to gaseous ions, then from gaseous ions to aqueous ions, since this is equivalent to going directly from solid to aqueous ions

You will thus have something like this

#color(blue)("KCl"_text((s]) ->color(white)(a) color(red)(cancel(color(black)("K"_text((g])^(+)))) + color(red)(cancel(color(black)("Cl"_text((g])^(-))))" " " "DeltaH_text(dis latt) = +"701 kJ mol"^(-1))#
#color(purple)(color(red)(cancel(color(black)("K"_text((g])^(+)))) color(white)(a)->color(white)(a) "K"_text((aq])^(+)" " " " " " " " " "DeltaH_text(hyd K) = - "322 kJ mol"^(-1))#
#color(purple)(color(red)(cancel(color(black)("Cl"_text((g])^(-)))) color(white)(a)-> color(white)(a)"Cl"_text((aq])^(-)" " " " " " " " " "DeltaH_text(hyd Cl) = -"364 kJ mol"^(-1))#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#

#color(blue)("KCl"_text((s])) -> color(purple)("K"_text((aq])^(+)) + color(purple)("Cl"_text((aq])^(-))" " " " " "DeltaH_text(sol)^@#

You will thus have

#DeltaH_text(sol)^@ = color(blue)(DeltaH_text(dis latt)) + color(purple)(DeltaH_text(hyd K)) + color(purple)(DeltaH_text(hyd Cl))#

#DeltaH_"sol"^@ = +"701 kJ mol"^(-1) + (-"322 kJ mol"^(-1) - "364 kJ mol"^(-1))#

#DeltaH_"sol"^@ = color(green)(+ "15 kJ mol"^(-1))#

Now, the problem wants you to deduce the sign of the entropy change for the dissolution of potassium chloride, #DeltaS^@#.

The trick here is to realize that since the enthalpy change of solution is positive, and since the solid dissolves readily at #"298 K"#, the entropy change must be positive.

In order for the dissolution of the salt to be spontaneous, i.e. for the salt to readily dissolve at that temperature, you need to have a negative change in Gibbs free energy, #DeltaG^@#.

#color(blue)(DeltaG^@ = DeltaH^@ - T * DeltaS^@)#

A positive #DeltaH_"sol"^@# and a spontaneous dissolution implies that this reaction is entropy driven, meaning that the change in entropy overpowers the positive change in enthalpy of solution.

Therefore, you can say that the entropy change will be positive, #color(green)(DeltaS^@ > 0)#.

You can prove this logic by calculating #DeltaS^@#

#color(blue)(DeltaS^@ = (DeltaH^@)/T)#

For #"T = 298 K"#, you will have

#DeltaS^@ = (+"15 kJ mol"^(-1))/"298 K" = color(green)(+"50.3 J mol"^(-1)"K"^(-1))#

No surprise here, the entropy change is positive.

Now, what does that positive sign tell you?

The dissolution of potassium chloride increases the disorder of the system. The ions went from being stuck in place in the lattice structure of the solid to being able to move around in solution.

In fact, this increase in disorder is large enough to overpower the positive enthalpy change of solution.

Mind you, this happens despite the fact that hydrated ions actually decrease the entropy of the solution by causing an arrangement of the surrounding water molecules.

The overall change in entropy is positive, which is why the dissolution is spontaneous at the given temperature.