Calcium carbide (CaC2) reacts with water to form acetylene (C2H2) and Ca (OH) 2.

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

ΔH∘f = kj?

The answer is - 60.57 = - 60.6 KJ.

CaC2 (s) + 2 H2O (l) - - - > Ca (OH) 2 (s) + C2H2 (g) H = - 127.2 KJ

Hf C2H2 = 226.77

Hf Ca (OH) 2 = - 986.2

Hf H2O = - 285.83

Now,

add them up. 226.77 - 986.2 + (2*285.83) = - 187.77

Add back the total enthalpy that is given in the question

-187.77+127.2 = - 60.57