Answer:
The answer to your question is ΔH = -964.2 kJ
Explanation:
Data
Reaction 1 X + 12O₂ ⇒ XO ΔH = -607.3 kJ
Reaction 2 XCO₃ ⇒ XO + CO₂ ΔH = 356.9 kJ
ΔH for Reaction 3 X + 12O₂ + CO₂ ⇒ XCO₃
Process
1.- Write the Reaction 1 the same
2.- Invert the order of reaction 2, and change the sign of the Enthalpy
Reaction 1 X + 12O₂ ⇒ XO ΔH = -607.3 kJ
Reaction 2 XO + CO₂ ⇒ XCO₃ ΔH = - 356.9 kJ
3.- Sum up the reactions
X + 12O₂ + XO + CO₂ ⇒ XO + XCO₃
4.- Simplify
X + 12O₂ + CO₂ ⇒ XCO₃ ΔH = -964.2 kJ
