Determine the lattice energy for the formation of MgO(s) from Mg(s) and O_22(g) using the following equations:
Mg(s) \rightarrow→ Mg(g) , + 147 kJ/mol
Mg(g) \rightarrow→ Mg^{2+}2+(g) + 2e^-− , + 2189 kJ/mol
O_22(g) \rightarrow→ 2O(g) , + 498 kJ/mol
O(g) + 2e^-− \rightarrow→ O^{2-}2−(g) , +702 kJ/mol
Mg(s) + 1/2O_22(g) \rightarrow→ MgO(s) , -602 kJ/mol
Mg(s) → Mg(g) , H1=
+ 147 kJ/mol
Mg(g) → Mg^{2+}2+(g) + 2e^-− H2 = +2189
kJ/mol
O2(g) → 2O(g)
,
H3 = + 498
kJ/mol
O(g) + 2e^-− → O^2-
H4 = +702
kJ/mol
Mg ^2+ (g) + O2^- (g) ------------> MgO(s) H5 ( lattice
energy)=
Mg(s) + 1/2O2(g) → MgO(s) H= -602
kJ/mol
according to hess law
H =
H1 +
H2 +
H3 +
H4 +
H5
-602 = 147 +2189+498/2 +702 + H5
-602 = 147+2189+249 + 702+ H5
H5 =
-3889KJ/mole lattice energy of MgO
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