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

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