Question

Calculate the lattice energy of magnesium sulfide from the data given below. Mg(s) → Mg(g) ΔH° = 148 kJ/mol Mg(g) → Mg2+(g) + 2e– ΔH° = 2186 kJ/mol S8(s) → 8S(g) ΔH° = 2232 kJ/mol S(g) + 2e– → S2–(g) ΔH° = 450 kJ/mol 8Mg(s) + S8(s) → 8MgS(s) ΔH° = –2744 kJ/mol MgS(s)→Mg2+(g) + S2–(g) ΔH°lattice = ?

Answer 1

The lattice energy of magnesium sulfide is -3,406 kJ/mol.

Step-by-step explanation:

`Mg(s)\rightarrow Mg(g) ,\Delta H^o_(1) = 148 kJ/mol`..[1]

`Mg(g) \rightarrow Mg^(2+)(g) + 2e^- ,\Delta H^o_(2)  = 2,186 kJ/mol`..[2]

`S_8(s) \rightarrow 8S(g) ,\Delta H^o_(3) = 2,232 kJ/mol`..[3]

`S(g) + 2e^- \rightarrow S^(2-)(g),\Delta H^o_(4)  = 450 kJ/mol`..[4]

`8Mg(s) + S_8(s) \rightarrow 8MgS(s),\Delta H^o_(5) = -2,744 kJ/mol`..[5]

`MgS(s)\rightarrow Mg^(2+)(g) + S^(2-)(g),\Delta H^o_(lattice)=?`..[6]

By using Hess's law:

`[6]=(1)/(8)* [5]-[1]-[2]-(1)/(8)* [3]-[4]`

`\Delta H^o_(lattice) =`

`=(1)/(8)* \Delta H^o_(5)-\Delta H^o_(1)-\Delta H^o_(2)-(1)/(8)* \Delta H^o_(3)- \Delta H^o_(4)`

`\Delta H^o_(lattice)=(1)/(8)* (-2,744 kJ/mol)-148 kJ/mol-2,186 kJ/mol-(1)/(8)* 2,232 kJ/mol-450 kJ/mol`

`\Delta H^o_(lattice)=-3406 kJ/mol`

The lattice energy of magnesium sulfide is -3,406 kJ/mol.