Find ΔH for: BaCO3 (s) → BaO (s)+ CO2 (g) given 2 Ba (s) + O2 (g) → 2 BaO (s) ΔH = – 1107.0 kJ Ba (s) + CO2 (g) + ½ O2 (g) → BaCO3 (s) ΔH = – 822.5 kJ

The answer to your question is ΔH = 269 kJ

Explanation:

Change the sense of the third reaction

           BaCO₃(s)  ⇒  Ba(s)  + CO₂(g)  + 1/2O₂(g)       ΔH = 822.5 kJ

Divide the second reaction by 2

           Ba(s)   +  1/2O₂(g)  ⇒  BaO(s)                         ΔH = -553.5 KJ

Add both reactions and cancel reactants and products

            BaCO₃(s) ⇒  BaO (s)  +  CO₂ (g)                  ΔH = 269  (822.5 - 553.5)

it's all four options.