Question

write the balanced chemical equation for when butane burns

I think it is 2c4h10+13o2 = 8co2 + 10h2o now how do I find the change in enthalpy

Answer 1

Yes you are correct. The balanced chemical equation for combustion of butane is

2C₄H₁₀(g)+ 13O₂(g)------> 8CO₂(g)+ 10H₂O

For standard enthalpy change according to Hess's law

ΔH°_reaction = Σ ΔH°_f (products) - Σ ΔH°_f (reactants)

Standard enthalpy of formation of components are given in below table:

ΔH°_comb = 8(-393.5) +10(-285.8) - {2(-125.5) +13(0)}

= -3148 - 2858 +251 = -5755 kJ per 2 moles of n-butane

Therfore, standard enthalpy of combustion of n-butane:

ΔH°_comb = -5755/2 = -2877.5 kJ/mol