Calculate the second ionization energy for the Mg from the following data?
Thanks for showing your work.  We're given that 
$$\ce{\Delta H_{f} (MgO (s))~ =~ -601$.$8 ~kJ/mol}$$
So any balanced equation we write that has MgO as the product will have this same heat of formation.  An equation that produces MgO and allows us to solve the problem is
$$\ce{Mg^{+2}(g) + O^{-2}(g) -&gt; MgO (g)}$$
The heat of formation of $\ce{O^{-2}}$ is given by
$$\ce{\Delta H_{f} (O^{-2} (g))~=~1/2 the ~O2~ bond~ energy~ (only~ half~ because~ we~ only~ need~ 1~ oxygen~ atom) + the ~first ~ionization~ potential~ for~oxygen + the~second~ionization~ potential~ for~ oxygen}$$
or
$$\ce{\Delta H_{f} (O^{-2} (g))~=~1/2(498.7)~-~141+780~ = ~$888.4$~ kJ/mol}$$
Similarly the heat of formation of $\ce{Mg^{+2}}$ is given by
$$\ce{\Delta H_{f} (Mg^{+2} (g))~=~ \Delta H_{sublimation}Mg(s) + the ~first ~ionization~ potential~ for~Mg + the~second~ionization~ potential~ for~ Mg}$$
or
$$\ce{\Delta H_{f} (Mg^{+2} (g))~=~130 + $738.1$ + IP_2(Mg)}$$
Before solving for $\ce{IP_2(Mg)}$ there is one more adjustment to make.  Everything we've done so far has been for the gas phase except for our starting heat of formation for $\ce{MgO}$ which was for the solid, so let's adjust this heat of formation to its gas phase value.
$$\ce{\Delta H_{f} (MgO (g))~ =~ \Delta H_{f} (MgO (s)) + lattice~ energy~ MgO }$$
or
$$\ce{\Delta H_{f} (MgO (g))~ =~ $-601.8$+ 3800 ~=~$3198.2$~ kJ/mol}$$
Now, inserting all of these values into our equation we get,
$$\ce{130 + $738.1$ + IP_2(Mg) + $888.4$~=~$3198.2$}$$
or
$$\ce{IP_2(Mg)~=~$1441.7$~kJ/mol}$$