Question

(a) If, for the reaction in Equation (1),

NH4+(g) + Cl-(g) NH4Cl(aq)

we find that delta H= -159.3 kcal mol-1, find delta H for the reaction in Equation (2):

NH4Cl(s) NH4+(g) + Cl-(g).

The reactions given by Equations (1) and (2) are of sufficient importance that special names are attached to this type of reaction. The reaction of Equation (1) gives a hydration enthalpy, which is a measure of forces of attraction between dipolar water molecules and ions.

(b) If the hydration enthalpy of the ammonium ion NH4+ is -72.5 kcal mol^-1, what is the hydration enthalpy of Cl-?

The reaction of Equation (2) gives the lattice energy of a crystalline compound. The lattice energy is a measure of the attractive forces that maintain a compound in the solid state. The sign and magnitude of an enthalpy of dissolution are determined by the heat required to dissociate the solid into gaseous ionic species and the hydration enthalpy of the ions.

(c) What do the signs of the enthalpy changes for the reactions of Equations (1) and (2) indicate?

(d) In order for a solid compound to have a negative enthalpy of dissolution, what must be the relation between the hydration enthalpy and the lattice energy?

(e) The hydration enthalpy of K+ is -77.0 kcal mol^-1 and the lattice energy of KCl(s) is +167.6 kcal mol^-1. What is the enthalpy of dissolution for KCl(s)?

Answer 1

a) & c) i) Equation 1 is the formation of ammonium chloride and this is an exothermic reaction because heat is released, so H = -ve.

ii) Equation b is dissolution reaction or hydrolysis reaction and dissolution of ammonium chloride is an endothermic process because energy is absorbed from surrounding,so H_{(lattice energy)} = +ve.

b)    NH₄Cl_(s)    NH⁺_4(g) + Cl^-_(g) NH⁺_4(aq) + Cl^-_(aq)

H_(total) = - H_lattice ( NH₄Cl)   + H_hyd.(NH⁺₄) + H_hyd. (Cl^-)

             = - (159.3 kcal/mol) + (-72.5 kcal/mol) + H_hyd. (Cl^-)

d) Heat of dissolution = hydration enthalpy + lattice energy(enthalpy)

Hydration enthalpy is always exothermic(-ve).

lattice dissociation energy(enthalpy) is taken as +ve and lattice formation energy(enthalpy) is taken as -ve .Here lattice dissociation energy(enthalpy) is considered in equation 2.

Heat of dissolution will have -ve value if hydration energy value is more negative and lattice enthalpy is less positive.

e) Enthalpy of dissolution = hydration enthalpy + lattice energy(enthalpy)

    Enthalpy of dissolution = -77.0 kcal/mol + 167.6 kcal/mol

             Enthalpy of dissolution = 90.6 kcal/mol