`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`
`color{green}("Definition ")` The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.
`color{purple}(Na^(+) Cl^(-) (s) → Na^(+) (g) +Cl^(-) ; Delta_text(lattice) H^(⊖) = +788 kJ mol^(-1))`
`=>` Since it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a `color{red}("Born-Haber Cycle")` (Fig. 6.9).
`=>` Let us now calculate the lattice enthalpy of `color{purple}(Na^(+)` `Cl^(-) (s))` by following steps given below :
(i) `color{purple}(Na(s)→ Na(g))` , sublimation of sodium metal , `color{purple}(Delta_text(sub)H^(⊖) = 108.4 kJ mol^(-1))`
(ii) `color{purple}(Na(g) → Na^(+) (g) e^(-1) (g) )` , the ionization of sodium atoms, ionization enthalpy `color{purple}(Delta_t H^(⊖) = 496 kJ mol^(-1))`
(iii) `color{purple}(1/2 Cl_2 (g) → Cl (g)) `, the dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy.
`color{purple}(1/2 Delta_text(bond) H^(⊖) = 121 kJ mol^(-1))`
(iv) `color{purple}(Cl(g)+ e^(-1) (g) → C barl(g))` electron gained by chlorine atoms. The electron gain enthalpy, `color{purple}(Delta_(eg) H^(⊖)=
–348.6 kJ mol^(–1))` .
● Ionization enthalpy and electron gain enthalpy have been taken from thermodynamics.
● Earlier terms, ionization energy and electron affinity were in practice in place of the above terms.
(v) `color{purple}(Na^(+) (g) Cl^(-) (g) → Na^(+) Cl^(-) (s))`
● The sequence of steps is shown in Fig. 6.9, and is known as a Born-Haber cycle.
`text(Importance :)` The importance of the cycle is that, the sum of the enthalpy changes round a cycle is zero.
● Applying Hess’s law, we get,
`color{purple}(Delta_text(lattice) H^(⊖) = 411.2+108.4+121+496-348.6)`
`color{purple}(Delta_text(lattice) H^(⊖) = +788 kJ)`
for `color{purple}(NaCl (s) → Na^(+) + Cl^(-) (g))`
● Internal energy is smaller by `2RT` (because `color{purple}(Deltan_g =2)`) and is equal to `+ 783 kJ mol^(–1)`.
`=>` Now we use the value of lattice enthalpy to calculate enthalpy of solution from the expression :
`color{purple}(Delta_text(sol) H^(⊖) = Delta_text(lattice) H^(⊖) + Delta_text(hyd) H^(⊖))`
For one mole of `color{purple}(NaCl(s)),` lattice enthalpy `= + 788 kJ mol^(–1)` and `color{purple}(Delta_text(hyd)^(⊖)) = -784 kJ mol^(-1) ` ( from the literature)
`color{purple}(Delta_text(sol)H^(⊖) = +784 kJ mol^(-1) -784 kJ mol^(-1) = +4 kJ mol^(-1))`
● The dissolution of `color{purple}(NaCl(s))` is accompanied by very little heat change.
`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`
`color{green}("Definition ")` The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.
`color{purple}(Na^(+) Cl^(-) (s) → Na^(+) (g) +Cl^(-) ; Delta_text(lattice) H^(⊖) = +788 kJ mol^(-1))`
`=>` Since it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a `color{red}("Born-Haber Cycle")` (Fig. 6.9).
`=>` Let us now calculate the lattice enthalpy of `color{purple}(Na^(+)` `Cl^(-) (s))` by following steps given below :
(i) `color{purple}(Na(s)→ Na(g))` , sublimation of sodium metal , `color{purple}(Delta_text(sub)H^(⊖) = 108.4 kJ mol^(-1))`
(ii) `color{purple}(Na(g) → Na^(+) (g) e^(-1) (g) )` , the ionization of sodium atoms, ionization enthalpy `color{purple}(Delta_t H^(⊖) = 496 kJ mol^(-1))`
(iii) `color{purple}(1/2 Cl_2 (g) → Cl (g)) `, the dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy.
`color{purple}(1/2 Delta_text(bond) H^(⊖) = 121 kJ mol^(-1))`
(iv) `color{purple}(Cl(g)+ e^(-1) (g) → C barl(g))` electron gained by chlorine atoms. The electron gain enthalpy, `color{purple}(Delta_(eg) H^(⊖)=
–348.6 kJ mol^(–1))` .
● Ionization enthalpy and electron gain enthalpy have been taken from thermodynamics.
● Earlier terms, ionization energy and electron affinity were in practice in place of the above terms.
(v) `color{purple}(Na^(+) (g) Cl^(-) (g) → Na^(+) Cl^(-) (s))`
● The sequence of steps is shown in Fig. 6.9, and is known as a Born-Haber cycle.
`text(Importance :)` The importance of the cycle is that, the sum of the enthalpy changes round a cycle is zero.
● Applying Hess’s law, we get,
`color{purple}(Delta_text(lattice) H^(⊖) = 411.2+108.4+121+496-348.6)`
`color{purple}(Delta_text(lattice) H^(⊖) = +788 kJ)`
for `color{purple}(NaCl (s) → Na^(+) + Cl^(-) (g))`
● Internal energy is smaller by `2RT` (because `color{purple}(Deltan_g =2)`) and is equal to `+ 783 kJ mol^(–1)`.
`=>` Now we use the value of lattice enthalpy to calculate enthalpy of solution from the expression :
`color{purple}(Delta_text(sol) H^(⊖) = Delta_text(lattice) H^(⊖) + Delta_text(hyd) H^(⊖))`
For one mole of `color{purple}(NaCl(s)),` lattice enthalpy `= + 788 kJ mol^(–1)` and `color{purple}(Delta_text(hyd)^(⊖)) = -784 kJ mol^(-1) ` ( from the literature)
`color{purple}(Delta_text(sol)H^(⊖) = +784 kJ mol^(-1) -784 kJ mol^(-1) = +4 kJ mol^(-1))`
● The dissolution of `color{purple}(NaCl(s))` is accompanied by very little heat change.