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

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kobenhavn kobenhavn · Answer 1 · 2021-03-27T04:24:41+01:00

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

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

The expression for enthalpy change is,

[tex]\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}][/tex]

Putting the values we get :

[tex]\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}][/tex]

[tex]\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)][/tex]

[tex]\Delta H=-5314.8kJ[/tex]

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = [tex]\frac{5314.8}{2}\times 1=2657.4kJ[/tex]

Thus enthalpy of combustion per mole of butane is -2657.4 kJ