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Butane C4 H10 (g),(Delta.Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2 O(g) (Delta.Hf = –241.82) in the reaction:

2 upper C subscript 4 upper H subscript 10 (g) plus 13 upper O subscript 2 (g) right arrow 8 upper C upper O subscript 2 plus 10 upper H subscript 2 upper O (g).

What is the enthalpy of combustion, per mole, of butane?
Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..
–5,314.8 kJ/mol
–2,657.4 kJ/mol
2,657.4 kJ/mol
5,314.8 kJ/mol

Respuesta :

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Answer:

  • Second choice:

                       [tex]\Delta H_{rxn}=-2,657.4kJ/mol[/tex]

Explanation:

1. Combustion reaction

      [tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]

2. Enthalpy of reaction equation

     [tex]\Delta H_{rxn}=\sum\Delta H_f(products)-\sum\Delta H_f(reactants)[/tex]

3. Substitute and divide by the coefficient of butane that appears in the combustion reaction

      [tex]\Delta H_{rxn}=\dfrac{8(-393.5kJ/mol)+10(-241.82kJ/mol)-2(-125.7kJ/mol)}{2mol}[/tex]

      [tex]\Delta H_{rxn}=-2,657.4kJ/mol\leftarrow answer[/tex]

Answer:

B: –2,657.4 kJ/mol

Explanation:

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