What is the change in enthalpy for the combustion reaction of butane, c4h10?
2c4h10(g) + 13o2(g) → 10h2o(g) + 8co2(g)
given:
c4h10(g): δhf °= –125 kj
h2o(g): δhf °= –242 kj
co2(g): δhf °= –393.5 kj

The correct answer is -5318 KJ.

Explanation:

We calculate the reaction enthalphy (ΔHr) from formation enthalpies of reactants and products as follows:

ΔHr= ΔHfº (products) - ΔHfº (reactants)

ΔHr= (10 x ΔHfº(H₂O) + 8 x ΔHfº(CO₂)) - (2 x ΔHfº(C₂H₁₀) + 13 x ΔHfº(O₂))

As the formation enthalphy of an element in its standard state is zero⇒ ΔHfº(O₂)= 0

ΔHr= (10 x (-242 KJ) + 8 x (-393.5 KJ)) - (2 x (-125 KJ) + 13 x 0)

ΔHr= -2420 KJ - 3148 KJ + 250 KJ

ΔHr= 5318 KJ

50.23 i believe is correct but don't quote me on that.