Acetylene burns in air according to the following equation. C2H2(g) + 5/2 O2(g)

Question

Acetylene burns in air according to the following equation. C2H2(g) + 5/2 O2(g) 2 CO2(g) + H2O(g) H0rxn = -1255.8 kJ Given that H0f of CO2(g) = -393.5 kJ/mol and H0f of H2O(g) = -241.8 kJ/mol, what is H0f of C2H2(g)? ??????????????????????????? kJ/mol?

Explanation / Answer

C2H2(g) + 5/2 O2(g) ------> 2 CO2(g) + H2O(g) ; H0rxn = -1255.8 KJ --------1 C(s)+O2(g) ------> CO2(g) ; H0rxn= -393.5Kj --------2 H2+1/2 O2 -------------> H2O ; H0rxn = -241.8 Kj ---------3 enthalpy of formation of C2H2 is given by H0rxn of reaction: 2C(s) + H2(g) ------------> C2H2(g) ; H0rxn = ?? we can get this equation by reversing the 1st equation + 2 x (second equation) +3rd equation so H0rxn = 1255.8-(2*393.5) -241.8 =227 Kj/mol

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