Calculate the second ionization energy for the Mg from the following data?

Question

here is a problem :

What I did " as shown in the picture below "

I rewrote the information given in terms of reactants and products ( equations )// in Part 1

Then I used the Hess law to obtain the desired result //in part 2

in part 3 // I added and cancelled the terms that appear in both sides

but I left with $O_{(g)}$ !! and the enthalpy is 1192 which is inconsistent with the correct value 1442 !!

What's wrong?

Answer 1

You forgot that you only need 1/2 mole of $\ce{O2}$. Since the product is $\ce{MgO}$, the third equation in step one should be:

$$ \ce{1/2O2(g) -> O(g)} \space \space \rm{249.35 ~kJ/~mol} $$

Note that I divided the bond dissociation energy of $\ce{O2}$ by two as well.

This is a very common mistake in Born-Haber type problems - the way to catch it early is to write out the balanced reaction first, then adjust the coefficients of each step accordingly.

Answer 2

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 $$\ce{Mg^{+2}(g) + O^{-2}(g) -> MgO (g)}$$ 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 $$\ce{IP_2(Mg)~=~$1441.7$~kJ/mol}$$