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Calculation of lattice enthalpy of MgBr_2 from the given data |
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Answer» SOLUTION :The enthalpy of formation of MgBr2 according to the reaction `Mg (s)+ Br_2(l )to MgBr_2(2) ,Delta_f H^@=- 524 KJ// mol` ` Delta H^@ ""_1`for ` Mg (s)toMg (g)=+148mol^(-1)` `Delta H^@ ""_2`for `Mg(g)to Mg^(2-)(g)+ 2e^(-)+ 2187KJ mol^(-1)` ` DeltaH^@ ""_3`for `Br_2(1)toBr_2(g)= 31KJmol ^(-1)` ` DeltaH^@ ""_4` for `Br_2(g)to 2 Br(g)= 193KJ mol^(-1)` ` Delta H^(@) ""_5` for ` Br(g)+e^( - )(g)Br^(- )=- 331KJ mol ^(-1)` `DeltaH^@""_6` for`Mg^(2+ )(g)+ 2 Br ^-(g)toMgBr_2 (s) =? ` `Delta_fH^@ =DeltaH^@ ""_1+DeltaH^@""_2+ DeltaH^@""_3 + DeltaH^@ ""_4 +DeltaH^@ ""_5+DeltaH^@ ""_6 ` `-524KJ mol^(-1)= ( +148+ 2187 +31+ 193 - 2 (331) +DeltaH^@""_6) KJmol ^(-1)` `=- 2421KJ mol ^(-1)=Delta^@ ""_6` hencelattive enthalpyof `MgBr_2 = DeltaH^@ ""_6= 24 21KJ mol ^(-1)` |
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