Question

Hi, could someone please take a look at the following question?


Use the data given below to construct a Born-Haber cycle to determine the lattice energy of CaO.

DH°(kJ)
Ca(s) → Ca(g) 193
Ca(g) → Ca⁺(g) + e⁻ 590
Ca⁺(g) → Ca2⁺(g) + e⁻ 1010
2 O(g) → O2(g) -498
O(g) + e⁻ → O⁻(g) -141
O⁻(g) + e⁻ → O2⁻(g) 878
Ca(s) +  O2(g) → CaO(s) -635

A) -3414 kJ
B) +1397 kJ
C) -2667 kJ
D) +3028 kJ
E) -2144 kJ

I think that the lattice energy would be found by adding up the steps and then subtracting them from the standard enthalpy of formation of CaO which is -635 kJ/mol as is shown in the last step. However I think what's missing is the formation of oxygen atom from the oxygen molecule. The correct answer is A). Any help would be greatly appreciated.

Constantine

Answer 1

You have the O2 ==> 2O in the data BUT it is turned around wrong. So change the sign from -498 to +498 and divide by 2 to make +249 (that's 249 for EACH O). Then do what you suggest and you should obtain -3414

Answer 2

Thank you Dr. Bob.