Given the reactions, X (s) + 1 2 O 2 (g) ⟶ XO (s) Δ H = - 607.3 kJ XCO 3 (s) ⟶ XO (s) + CO 2 (g) Δ H = + 356.9 kJ what is Δ H for this reaction? X (s) + 1 2 O 2 (g) + CO 2 (g) ⟶ XCO 3 (s)

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