Ask Question
25 September, 09:37

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

+5
Answers (1)
  1. 25 September, 09:58
    0
    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)
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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 ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers