Question

Lighters burn butane as fuel according to the following balanced combustion equation: C4H10(g)+132O2(g)→4CO2(g)+5H2O(g),ΔH∘rxn=−2657.3kJ The standard enthalpies of formation are − 393.5 kJ/mol for CO2(g) and − 241.8 kJ/mol for H2O(g)

Answer 1

The enthalpy of combustion of butane for the formation of water vapor and carbon dioxide is -2613.5 kJ/mol.

Step-by-step explanation:

The enthalpy of combustion of butane, C4H10(g), for the formation of H2O(g) and CO2(g) can be calculated using the enthalpy of formation values of water vapor and carbon dioxide. The balanced combustion equation for butane is C4H10(g) + 13/2 O2(g) → 4 CO2(g) + 5 H2O(g).

Using the enthalpy of formation values (-393.5 kJ/mol for CO2 and -241.8 kJ/mol for H2O), we can calculate the enthalpy of combustion of butane as follows:

Enthalpy of combustion of butane = (4 x ∆H°f of CO2) + (5 x ∆H°f of H2O) - ∆H°rxn

= (4 x -393.5 kJ/mol) + (5 x -241.8 kJ/mol) - (-2657.3 kJ)

= -2613.5 kJ/mol

Answer 2

Standard enthalpy change of combustion of butane in the reaction C₄H₁₀(g) + 13/2O₂(g) → 4CO₂(g) + 5H₂O(g) is -2783 kJ/mol.

Step-by-step explanation:

Balanced Combustion Equation:

The balanced combustion equation for the burning of butane (C₄H₁₀) as fuel is as follows:

C₄H₁₀(g) + 13/2O₂(g) → 4CO₂(g) + 5H₂O(g)

Enthalpy Change:

The enthalpy change for the combustion of butane can be calculated using the given balanced equation and the standard enthalpies of formation for carbon dioxide (CO₂) and water (H₂O). The standard enthalpy of formation for CO₂ is -393.5 kJ/mol, and for H₂O, it is -241.8 kJ/mol.

Using the enthalpy change of formation values, we can calculate the enthalpy change for the combustion of butane:

ΔH∘rxn = ΣnΔH∘f(products) - ΣmΔH∘f(reactants)

Where:

ΔH∘rxn is the standard enthalpy change for the reaction

ΣnΔH∘f(products) is the sum of the standard enthalpies of formation of the products

ΣmΔH∘f(reactants) is the sum of the standard enthalpies of formation of the reactants

Substituting the values:

ΔH∘rxn = [4(-393.5 kJ/mol) + 5(-241.8 kJ/mol)] - [0 + 13/2(0)]

ΔH∘rxn = [-1574 kJ/mol + (-1209 kJ/mol)] - [0 + 0]

ΔH∘rxn = -2783 kJ/mol

Therefore, the standard enthalpy change for the combustion of butane is -2783 kJ/mol.