Question

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O, Hf = –241.82 kJ/mol) according to the equation below. What is the enthalpy of combustion (per mole) of C4H10 (g)? Use .

Answer 1

-2657.5 kJ

Step-by-step explanation:

Answer 2

Answer: Enthalpy of combustion (per mole) of
`C_4H_(10) (g)` is -2657.5 kJ

Step-by-step explanation:

The chemical equation for the combustion of butane follows:

`2C_4H_(10)(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)`

The equation for the enthalpy change of the above reaction is:

`\Delta H^o_(rxn)=[(8* \Delta H^o_f_(CO_2(g)))+(10* \Delta H^o_f_(H_2O(g)))]-[(1* \Delta H^o_f_{C_4H_(10)(g)})+(4* \Delta H^o_f_(O_2(g)))]`

We are given:

`\Delta H^o_f_{(C_4H_(10)(g))}=-125.6kJ/mol\\\Delta H^o_f_((H_2O(g)))=-241.82kJ/mol\\\Delta H^o_f_((O_2(g)))=0kJ/mol\\\Delta H^o_f_((CO_2(g)))=-393.5kJ/mol\\\Delta H^o_(rxn)=?`

Putting values in above equation, we get:

`\Delta H^o_(rxn)=[(8* -393.5)+(10* -241.82)]-[(2* -125.6)+(4* 0)]\\\\\Delta H^o_(rxn)=-5315kJ`

Enthalpy of combustion (per mole) of
`C_4H_(10) (g)` is -2657.5 kJ