ΔHrxn reactions

Question

Given these reactions,

X(s) +1/2 O2(g) --> XO(s)

ΔH=-539.5 kJ

XCO3(s) --> XO(s)+CO2(g)

ΔH=+266.3 kJ

what is ΔHrxn for this reaction?

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

Answer 1

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

this is your goal equation.

X(s) +1/2 O2(g) --> XO(s) ΔH=-539.5 kJ

this equation is ok

XCO3(s) --> XO(s)+CO2(g)

this equation must be flipped and the H sign changed:

XO(s)+CO2(g) --> XCO3(s) ΔH=-266.3 kJ

Sum equations to derive goal equation. Sum H's = -805.8kJ

Answer 2

X(s) +1/2 O2(g) --> XO(s) ΔH = -539.5 kJ

XCO3(s) --> XO(s)+CO2(g) ΔH = +266.3 kJ

what is ΔHrxn for this reaction? X(s)+1/2 O2(g)+CO2(g) --> XCO3(s)

Equation #2 has XCO3 as a reactant, but in the net thermochemical equation, we want XCO3 as a product. Therefore, reverse equation #2 and the ΔH value for this reaction will then = -266.3 kJ

XO(s) + CO2(g) --> XCO3(s) ΔH = -266.3 kJ

Leave Equation #1 "as is" since X and 1/2 O2 are on the reactant side.

X(s) +1/2 O2(g) --> XO(s) ΔH = -539.5 kJ

XO(s) + CO2(g) --> XCO3(s) ΔH = -266.3 kJ

Add both equations,

X(s) +1/2 O2(g) + XO(s) + CO2(g) --> XO(s) + XCO3(s) ΔH = -539.5 kJ + (-266.3 kJ)

XO (s) will cancel from both sides of the equation giving the net equation

X(s) +1/2 O2(g) + CO2(g) --> XCO3(s) ΔH = -539.5 kJ + (- 266.3 kJ) = -805.8 kJ