Respuesta :
The given question is incomplete. The complete question is:
Butane [tex](C_4H_{10}(g)[/tex], Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide ([tex]CO_2[/tex] , Hf = –393.5 kJ/mol ) and water [tex](H_2O[/tex], Hf = –241.82 kJ/mol) according to the equation below. What is the enthalpy of combustion (per mole) of [tex]C_4H_{10} (g)[/tex]?
Answer: -2657.5 kJ
Explanation:
The balanced chemical reaction is,
[tex]C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(g)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
[tex]\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_4H_{10}}\times \Delta H_{C_4H_{10})][/tex]
where,
n = number of moles
[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
[tex]\Delta H=[(4\times -393.5)+(5\times -241.82)]-[(\frac{13}{2}\times 0)+(1\times -125.6)][/tex]
[tex]Delta H=-2657.5kJ[/tex]
Therefore, the enthalpy of combustion per mole of butane is -2657.5 kJ
