Question

Given that $C + O _{2} \longrightarrow CO _{2}: \Delta H ^{\circ}=- xkJ$
$2 CO + O _{2} \longrightarrow 2 CO _{2}: \Delta H ^{\circ}=-y kJ$
the enthalpy of formation of carbon monoxide will be

• A. $\frac{2x-y}{2}$
• B. $\frac{y-2x}{2}$
• C. 2x - y
• D. y - 2x

Answer 1

Answer: (B)

Given $C + O _{2} \rightarrow CO _{2}, \Delta H ^{\circ}=- xkJ$...(i)
$2 CO _{2} \rightarrow 2 CO + O _{2} \Delta H ^{\circ}=+ y kJ \ldots$ (ii)
or $CO _{2} \rightarrow CO +1 / 2 O _{2}, \Delta H ^{\circ}=+ y / 2 kJ \ldots$..(iii)
By adding eq. (i) and (iii), we get
$C + O _{2}+ CO _{2} \longrightarrow CO _{2}+ CO +\frac{1}{2} O _{2} .$
$C +\frac{1}{2} O _{2} \longrightarrow CO$,
$\Delta H ^{\circ}= y / 2- x =\frac{ y -2 x}{-2} kJ$