Calculate the lattice enthalpy of `MgBr_(2)` from the given date:
`{:(Mg(s)+Br_(2)(l)rarrMgBr_(2)(s),,Delta_(f)H^(@)=-524 kJ mol^(-1)),(Mg(s)rarrMg(g),,Delta_(1)H^(@)=+148 kJ mol^(-1)),(Mg(g)rarrMg^(2+)(g)+2e^(-),,Delta_(2)H^(@)=+2187 kJ mol^(-1)),(Br_(2)(l)rarrBr_(2)(g),,Delta_(3)H^(@)=+31 kJ mol^(-1)),(Br_(2)(g)rarr2Br(g),,Delta_(4)H^(@)=+193 kJ mol^(-1)),(2Br(g)+2e^(-)rarr2Br(g),,Delta_(5)H^(@)=-662 kJ mol^(-1)):}`
Strategy : The thermochemical equation corresponding to lattice enthalpy of `MgBr_(2)` is
`{:(Mg^(2+)(g)+2Br^(-)(g)rarrMgr_(2)(s),,,Delta_("Lattice")H^(@)=?):}`
Add the last five thermochemical equations to the thermochemical equation corresponding to lattice enthalpy to get the thermochemical equation for the formation of `MgBr_(2) (s)` from its constituent element. Finally, calculate `Delta_(Lattice) H^(@)`, using the concept of Hess's law.

Explanation

Born-Haber cycle for the calculation of lattice enthalpy of `Mg Br _(2)` is shown below:

`Delta _(r) H ^(Θ) = Delta H _(1) ^(Θ) + Delta _(2) ^(Θ) + Delta H _(3)^(Θ) + Delta H _(4) ^(Θ) + Detla H_(5) ^(o+) + Delta H _(6) ^(Θ)- 524 = 148 + 31+193+ 2187+2(-131)+ Delta H _(6) ^(Θ) Delta H _(0) ^(Θ) =-2821 kJ mol ^(-1)`
Hence lattice enthalpy `=- Delta H _(0) ^(Θ) =+2821 kJ mol ^(-1)`
`i.e., Mg B r_(2) (s) + underset(("lattice enthalpy of " MgBr _(2)))(2821kJ) to Mg ^(2+) (g) + 2Br ^(-) (g)`