Answer:
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)
The enthalpy change is, ΔH = 269 kJ. It is the amount if energy lost or gained during the reaction.
Enthalpy is the heat content of a system. The enthalpy change of a reaction is roughly equivalent to the amount of energy lost or gained during the reaction.
Given reaction:
BaCO₃(s) ⇒ Ba(s) + CO₂(g) + 1/2O₂(g) ΔH = 822.5 kJ
On dividing by 2:
Ba(s) + 1/2O₂(g) ⇒ BaO(s) ΔH = -553.5 KJ
On adding both reactions and cancel reactants and products
BaCO₃(s) ⇒ BaO (s) + CO₂ (g)
Thus, the enthalpy cahnge is, ΔH = (822.5 - 553.5)= 269 kJ
Find more information about Enthalpy change here:
brainly.com/question/26991394
