Question

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

Answer 1

the answer for plato is –5318 kJ

Answer 2

The correct answer is -5318 KJ.

Step-by-step 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