Question

On complete combustion of one mole of butane according to the equation: C$_4$H$_{10}$ + $\dfrac{13}{2}$ O$_2$ → 4CO$_2$ + 5H$_2$O, the standard enthalpy of the reaction is –2879 kJ/mol. If $\Delta H_f^\circ$ [CO$_2$] = –394, $\Delta H_f^\circ$ [H$_2$O] = –286, the standard enthalpy of formation of butane is

• A. –3006
• B. 3006
• C. 127
• D. –127

Hint:

Apply Hess's Law: Product – Reactant enthalpies, and carefully handle signs.

Answer 1

The Correct Option is D

Use Hess's law: $\Delta H^\circ_{rxn} = \sum \Delta H^\circ_f(\text{products}) - \sum \Delta H^\circ_f(\text{reactants})$
$\Rightarrow$ –2879 = [4(–394) + 5(–286)] – $\Delta H_f^\circ$(butane)
= [–1576 + (–1430)] = –3006
So, –2879 = –3006 – $\Delta H_f^\circ$(butane)
$\Rightarrow \Delta H_f^\circ$(butane) = –3006 + 2879 = –127 kJ/mol