Answer to Question #140229 in Inorganic Chemistry for Christine

Question

3. Given the following thermochemical equations,
C2H2(g) + 5/2 O2(g) → 2CO2(g) + H2O(l) ∆H = -1299.5 KJ
C(s) + O2(g) → CO2(g) ∆H = -393.5 KJ
H2(g) + ½ O2(g) → H2O(l) ∆H = -285.8 KJ
Calculate ∆H for the decomposition of one mole of acetylene, C2H2(g), to its elements in their stable state at 25oC and 1 atm.

Answer 1

Standard formation of CO2 = - 393.5 kJ

Standard formation of H2O = - 285.8 kJ

Reaction of acetylene with oxygen will give

Hr = (2*H(CO2) + H(H2O)) - H(C2H2)

-1299.5 = (2*(-393.5) - 285.8)) - H(C2H2)

-1299.5 + 1072.8 = - H(C2H2)

H(C2H2) = 226.7 kJ - enthalpy of formation of C2H2

Then H for the decomposition of one mole of acetylene = - 226.7 kJ