Give the δh value for the combustion of butane as shown in the reaction 2c4h10(g)+13o2(g)→8co2(g)+10h2o(g)+ 5315 kj.
express your answer using four significant figures. if the value is positive, do not include the + sign in your answer.

ΔH° = -3790 kJ

Explanation:

Let's consider the combustion of butane.

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

We can find the standard enthalpy of this combustion reaction (ΔH°) using the following expression.

ΔH° = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where,

n: mole

ΔH°f(): standard enthalpy of formation

p: product

r: reactant

ΔH° = 8 mol ×  ΔH°f(CO₂(g)) + 10 mol ×  ΔH°f(H₂O(g)) - 2 mol ×  ΔH°f(C₄H₁₀(g)) - 13 mol ×  ΔH°f(O₂(g))

ΔH° = 8 mol ×  (-393.5 kJ/mol) + 10 mol ×  (-241.8 kJ/mol) - 2 mol ×  (-888.0 kJ/mol) - 13 mol ×  0

ΔH° = -3790 kJ

Given ΔH° < 0, the reaction is exothermic.

2 C₄H₁₀(g) +13 O₂(g) → 8 CO₂(g) + 10 H₂O(g) + 3790 kJ

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