Question

Give the ΔH value for the combustion of butane as shown in the reaction 2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)+5315 kJ.

Answer 1

Answer:

ΔH = - 5315 kJ.

Explanation:

The given chemical reaction is as follows -

2C₄H₁₀ (g) + 13 O₂ (g)  →  8 CO₂ (g) + 10 H₂O (g) + 5315 kJ

In the above equation , the amount of energy i.e. 5315 kJ is released , i.e. it is in the product side , hence , the reaction is an example of an exothermic reaction .

Hence ,  

The value of the change in enthalphy , i.e. , the enthalpy of product minus the enthalpy of the product .

Therefore ,  

The value of the change in enthalphy = - ve .

Hence ,  

ΔH = - 5315 kJ.

Answer 2

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