Solutions=
Let us calculate the lattice enthalpy of Magnesium chloride by the following steps given below:
1. Sublimation of Magnesium solid into gaseous state atom, sublimation enthalpy:
Mg(s)→Mg(g)
Δsub H° =148KJ/mol
2. The ionization of gaseous magnesium atom into an ion, ionization enthalpies:
Mg(g)→Mg1+(g)+1e-(g)
Δ1st lE H° = 738KJ/mol
Mg1+(g)→Mg2+(g)+1e−(g)
Δ2nd IEH° = 1451KJ/mol
3. The dissociation of chlorine molecule into gaseous atoms, bond dissociation enthalpy:
Cl2(g)→2Cl(g)
Δbond H° =2×122
= 244KJ/mol
4. The electrons gained by the chlorine atoms; electron gain enthalpy:
2Cl+2e−(g)→2Cl−
Δeg H° =2 ×−349
= -698KJ/mol
5. The enthalpy of formation of magnesium chloride is:
Mg2+(g)+2Cl−(g)→MgCl2
Δform H°= -641.8KJ/mol
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 1 images
