Question

Given these reactions, X(s)+12O2(g)⟶XO(s)Δ=−954.5 kJ/molXCO3(s)⟶XO(s)+CO2(g)Δ=+378.1 kJ/mol X ( s ) + 1 2 O 2 ( g ) ⟶XO ( s ) Δ H=−954.5 kJ / mol XCO 3 ( s ) ⟶XO ( s ) + CO 2 ( g ) Δ H=+378.1 kJ / mol what is Δ Δ H for this reaction? X(s)+12O2(g)+CO2(g)⟶XCO3(s) X ( s ) + 1 2 O 2 ( g ) + CO 2 ( g ) ⟶ XCO 3 ( s )

Answer 1

ΔH = -1332,5 kJ/mol

Step-by-step explanation:

The general equation of the enthalpy of reaction is:

AB + CD → AC + BD

ΔH =( ΔH(AC) + ΔH(BD) ) - (ΔH(AB) + ΔH(CD) )

To simplify the equations, the states of aggregation can be omitted here.

So the first equation is:

X + 12 O2 → XO | ΔH = -954,5 kJ/mol

The equation of the enthalpy of reaction is thus:

-954,5 kJ/mol = ΔH(XO) - ( ΔH(X) + 12*ΔH(O2) )

Since the enthalpy of reaction of all pure elements is 0, the equation can be simplified:

-954,5 kJ/mol = ΔH(XO) - ( 0 + 12*0 )

ΔH(XO) = -954,5 kJ/mol

The next equation is:

XCO3 → XO + CO2 | ΔH = +378,1 kJ/mol

The equation of the enthalpy of reaction is thus:

+378,1 kJ/mol = ( ΔH(XO) + ΔH(CO2) ) - ΔH(XCO3)

+378,1 kJ/mol = -954,4 kJ/mol + ΔH(CO2) - ΔH(XCO3)

+1332,5 kJ/mol = ΔH(CO2) - ΔH(XCO3)

X + 12 O2 + CO2 → XCO3

The equation of the enthalpy of reaction is thus:

ΔH = ΔH(XCO3) - ΔH(CO2)

ΔH = -1332,5 kJ/mol