Given the reactions, X(s)+12O2(g)⟶XO(s) Δ𝐻=−829.1 kJ XCO3(s)⟶XO(s)+CO2(g)Δ𝐻=+185.9 kJ what is Δ𝐻 for this reaction? X(s)+12O2(g)+CO2(g)⟶XCO3(s)

Answer 1

Problem using Hess' Law

X(s) + 12 O2(g) + CO2(g) ===> XCO3(s) ... TARGET EQUATION

Known equations:

(1) X(s) + 12 O2(g) ==> XO(s) ... ∆H = -829.1 kJ

(2) XCO3(s) ==> XO(s) + CO2(g) ... ∆H = +185.9 kJ

Procedure:

copy (1): X(s) + 12 O2(g) ==> XO(s) ... ∆H = -829.1 kJ

reverse (2): XO(s) + CO2(g) ==> XCO3(s) ... ∆H = -185.9 kJ (note to change sign of ∆H)

Add reactions together:

X(s) + 12 O2(g) + XO(s) + CO2(g) ==> XO(s) + XCO3(s)

After canceling and combining common items, we end up with the target equation...

X(s) + 12 O2(g) + CO2(g) ==> XCO3(s) ... TARGET EQUATION

∆H = -829.1 kJ + -185.9 kJ = -1015 kJ