chemistry question

Question

Given the reactions,

X(s) + 1/2 O2(g) ⟶ XO(s)  Δ𝐻= −710.1 kJ

XCO3(s) ⟶ XO(s) + CO2 (g) Δ𝐻=+470.3 kJ

what is Δ𝐻 for this reaction?

X(s) + 1/2 O2(g) + CO2(g) ⟶ XCO3(s)

Answer 1

Hess Law:

Target equation: X(s) + 1/2 O₂(g) + CO₂(g) ==> XCO₃(s)

copy eq. 1. X(s) + 1/2 O₂(g) ⟶ XO(s) Δ𝐻= −710.1 kJ

reverse eq. 2: XO(s) + CO₂(g) ==> XCO₃(s) ∆H = -470.3 kJ

Add them to get...

X(s) + 1/2 O₂(g) + XO(s) + CO₂(g) ==> XO(s) + XCO₃(s)

Combine and cancel items on left and right to end up with ...

X(s) + 1/2O₂(g) + CO₂(g) ==> XCO₃(s) THIS IS THE TARGET EQUATION

Add the ∆H values:

-710.1 kJ + (-470.3 kJ) = 1180.5 kJ = 1181 kJ (answer)