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

Explanation:

The balanced chemical reaction is,

$C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

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

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

$\Delta H=[(4\times -393.5)+(5\times -241.82)]-[(\frac{13}{2}\times 0)+(1\times -125.6)]$

$Delta H=-2657.5kJ$

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