Consider an ionic compound, MX, composed of generic metal M and generic, gaseous halogen X.

2 answers:

8 0

As the given values are

<span>ΔHf° = –423 kJ/mol
</span>ΔHsub = 119 kJ/mol
IE = 469 kJ/mol
ΔHEA = –301 kJ/mol
BE = 161 kJ/mol

The lattice energy of the compound can be determined by applying formula

U = <span>ΔHf° - </span>ΔHsub - BE - IE - ΔHEA
by putting values
U = -423 - 119 - 161 - 469 - (-301)
solving further
U = -871 kJ/mol

so the lattice energy is 871 kJ/mol
hope it helps

dybincka [34]3 years ago

7 0

The lattice energy of MX is $\boxed{ - {\text{871 kJ/mol}}}$ .

Further explanation:

Lattice energy:

It is defined as the energy released when an ionic compound is formed from its constituent ions or when the compound gets dissociated into its gaseous ions. It is denoted as $\Delta {H_{{\text{lattice}}}}$ .

Born Haber cycle:

The reaction energies are evaluated by this cycle. It is used to calculate the lattice energy that is difficult to measure directly. The various steps involved in the Born Haber cycle of an ionic compound, MX are as follows:

1. Sublimation of solid metal (M) to the gaseous state. The reaction occurs as follows:

${\text{M}}\left( s \right)\xrightarrow{{\Delta {H_{{\text{sublimation}}}}}}{\text{M}}\left( g \right)$

2. Bond breakage of diatomic halogen molecule $\left( {{{\text{X}}_{\text{2}}}} \right)$ . The reaction occurs as follows:

$\frac{1}{2}{{\text{X}}_2}\left( g \right)\xrightarrow{{{\text{BE}}}}{\text{X}}\left( g \right)$

3. Ionization of metal into positive ions. The reaction occurs as follows:

${\text{M}}\left( {\text{g}} \right)\xrightarrow{{{\text{IE}}}}{{\text{M}}^{\text{ + }}}\left( {\text{g}} \right)$

4. Electron affinity of halogen atom to form negative ions. The reaction occurs as follows:

${\text{X}}\left( g \right)\xrightarrow{{{\text{EA}}}}{{\text{X}}^ - }\left( g \right)$

5. Formation of an ionic compound, MX. The reaction occurs as follows:

${{\text{M}}^{\text{ + }}}\left( {\text{g}} \right) + {{\text{X}}^ - }\left( {\text{g}} \right)\xrightarrow{{\Delta {H^0}_{{\text{formation}}}}}{\text{MX}}\left( s \right)$

The formula to calculate the enthalpy of formation of MX is as follows:

$\Delta H_{{\text{formation}}}^0 = \Delta {H_{{\text{sublimation}}}} + {\text{IE}} + {\text{BE}} + {\text{EA}} + \Delta {H_{{\text{lattice}}}}$ ......(1)

Rearrange equation (1) to calculate $\Delta {H_{{\text{lattice}}}}$ .

$\Delta {H_{{\text{lattice}}}} = \Delta H_{{\text{formation}}}^0 - \Delta {H_{{\text{sublimation}}}} - {\text{IE}} - {\text{BE}} - {\text{EA}}$ ......(2)

The value of $\Delta H_{{\text{formation}}}^0$ is -423 kJ/mol.

The value of $\Delta {H_{{\text{sublimation}}}}$ is 119 kJ/mol.

The value of IE is 469 kJ/mol.

The value of BE is 161 kJ/mol.

The value of EA is -301 kJ/mol.

Substitute these values in equation (2).

$\begin{aligned}\Delta {H_{{\text{lattice}}}}&= \left( { - 42{\text{3 kJ/mol}}} \right) - \left( {119{\text{ kJ/mol}}} \right) - \left( {469{\text{ kJ/mol}}} \right) - \left( {{\text{161 kJ/mol}}} \right) - \left( {-301{\text{kJ/mol}}}\right)\\&=871{\text{kJ/mol}}\\\end{aligned}$