Answer:
From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2(g)ΔH∘=−127.2kJ
ΔHf°(C2H2) = 227.4 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol and
ΔHf°(Ca(OH)2) = -985.2 kJ/mol
(Ans)
ΔHf° of CaC2 = -59.0 kJ/mol
Explanation:
CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2 (g) = −127.2kJ
ΔHrxn = −127.2kJ
ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - ΔHf°(CaC2)- 2ΔHf°(H2O);
ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn
Where
ΔHf°(C2H2) = 227.4 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol and
ΔHf°(Ca(OH)2) = -985.2 kJ/mol
ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol
ΔHf°(CaC2) = -59.0 kJ/mol
