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

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