Question

12. Given:   2Na(s) + O2(g) → Na2O2(s)       ΔG^o = -447 kJ

                 Na2O(s) + Cu(s) → CuO(s) + 2Na(s)   ΔG^o = +249 kJ

Calculate ΔG^o for: Na2O(s) + Cu(s) + O2(g) → Na2O2(s) + CuO(s)

Answer 1

2Na(s) + O2(g) → Na2O2(s) ΔGo = -447 kJ

Na2O(s) + Cu(s) → CuO(s) + 2Na(s)   ΔGo = +249 kJ

Na2O(s) + Cu(s) + O2(g) → Na2O2(s) + CuO(s) ??

we need... NA2O2 in the right side, so equation 1 remains the same; Na2O must be in the same side

so... simply add equation 1 and 2

2Na(s) + O2(g) + Na2O(s) + Cu(s) → CuO(s) + 2Na(s) + Na2O2(s)   ΔGo = +249 -447 kJ = -198kJ

cancel common terms

O2(g) + Na2O(s) + Cu(s) → CuO(s) +  Na2O2(s)   ΔGo = +249 -447 kJ = -198kJ

which is what we wanted, so  ΔGo =  -198kJ