Calculate the standard enthalpy change (ΔH⁰rxn ) for the reaction of TiCl4(g) and H2O(g) to form TiO2(s) and HCl(g) given the standard enthalpies of formation (ΔH⁰f ) shown in the table below. (Include the sign of the value in your answer.)
kJ
Compound |
ΔH⁰f (kJ/mol) |
TiCl4(g) |
−763.2 |
H2O(g) |
−241.8 |
TiO2(s) |
−944.0 |
HCl(g) |
−92.3 |
we have:
Hof(TiCl4(g)) = -763.2 KJ/mol
Hof(H2O(g)) = -241.8 KJ/mol
Hof(TiO2(s)) = -944.0 KJ/mol
Hof(HCl(g)) = -92.3 KJ/mol
we have the Balanced chemical equation as:
TiCl4(g) + 2 H2O(g) ---> TiO2(s) + 4 HCl(g)
deltaHo rxn = 1*Hof(TiO2(s)) + 4*Hof(HCl(g)) - 1*Hof( TiCl4(g)) - 2*Hof(H2O(g))
deltaHo rxn = 1*(-944.0) + 4*(-92.3) - 1*(-763.2) - 2*(-241.8)
deltaHo rxn = -66.4 KJ/mol
Answer: -66.4 KJ/mol
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