What is the value of ∆H° for this reaction?

Question

C4H4 (g) + 2H2 (g) -> C4H8 (g)

Enthalpies of combustion

C4H4(g) = -2341 Kj per mol

H2(g) = -286 Kj per mol

C4H8 (g) = -2755 kj per mol

Answer is -158 kj I don't know how to get it > my calculation gave me -128 kj

Please explain me in detail

Answer 1

all dHf's provide an energy relative to the same reference, elements in their most natural state.

since dH combustions provide energy differences all relative to the same combustion products,....

you can use enthalpies of combustion, the same as we used enthalpies of formations (dHf's),  & do Products - reactants.

C4H4 (g) + 2H2 (g)   ->    C4H8 (g)

-2341 Kj  & (2)-286 Kj  -->  -2755 kj

dH reaction = prod - reactants

dH rxn = -2755kJ - [(-2341) & (2)(-286)]

dH rxn = -2755kJ - (-2913)

dH rxn = -158kJ

your answer is -158kJ