Question

Butane (C4H10(g), deltaHf =-125.6 kJ/mol reacts with oxygen to produce carbon dioxide (CO2 delta Hf =-393.5 kJ/mol)

Answer 1

The given question is incomplete. The complete question is:

Butane
`(C_4H_(10)(g)`, Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (
`CO_2` , Hf = –393.5 kJ/mol ) and water
`(H_2O`, Hf = –241.82 kJ/mol) according to the equation below. What is the enthalpy of combustion (per mole) of
`C_4H_(10) (g)`?

Answer: -2657.5 kJ

Step-by-step explanation:

The balanced chemical reaction is,

`C_4H_(10)(g)+(13)/(2)O_2(g)\rightarrow 4CO_2(g)+5H_2O(g)`

The expression for enthalpy change is,

`\Delta H=\sum [n* \Delta H_f(product)]-\sum [n* \Delta H_f(reactant)]`

`\Delta H=[(n_(CO_2)* \Delta H_(CO_2))+(n_(H_2O)* \Delta H_(H_2O))]-[(n_(O_2)* \Delta H_(O_2))+(n_{C_4H_(10)}* \Delta H_{C_4H_(10))]`

where,

n = number of moles

`\Delta H_(O_2)=0` (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

`\Delta H=[(4* -393.5)+(5* -241.82)]-[((13)/(2)* 0)+(1* -125.6)]`

`Delta H=-2657.5kJ`

Therefore, the enthalpy of combustion per mole of butane is -2657.5 kJ