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Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Hf = –393.5 kJ/mol), and H2O(g) (Hf = –241.82) in the reaction:

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

What is the enthalpy of combustion, per mole, of butane?(A) -2,657.5 kJ/mol(B) -5315.0 kJ/mol(C) -509.7 kJ/mol(D) -254.8 kJ/mol

Respuesta :

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Answer:

[tex]\boxed{\text{(A) -2657.4 kJ/mol}}[/tex]

Explanation:

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

[tex]\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})[/tex]

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

ΔH°f/kJ·mol⁻¹:      -125.7            0            -393.5      -241.82

[tex]\begin{array}{rcl}\Delta_{\text{c}}H^{\circ} & = & [8(-393.5) + 10(-241.82)] - 2(-125.7)\\& = & [-3148.0 - 2418.2] +251.4\\& = & -5566.2 + 251.4\\& = & -5314.8\\\end{array}\\\\\text{This is the value for 2 mol of butane.}\\\text{The enthalpy of combustion is } \dfrac{-5314.8}{2} = \boxed{\textbf{-2657.4 kJ/mol}}[/tex]