Question

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.

Answer 1

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

Step-by-step explanation:

The balanced chemical reaction is,

`2C_4H_(10)(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)`

The expression for enthalpy change is,

`\Delta H=[n* H_f_(products)]-[n* H_f_(reactants)]`

Putting the values we get :

`\Delta H=[8* H_f_(CO_2)+10* H_f_(H_2O)]-[2* H_f_{C_4H_(10)+13* H_f_(O_2)}]`

`\Delta H=[(8* -393.5)+(10* -241.82)]-[(2* -125.7)+(13* 0)]`

`\Delta H=-5314.8kJ`

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat =
`(5314.8)/(2)* 1=2657.4kJ`

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