Question

The standard heat of combustion of propane is $-2220.1 \,kJ \,mol^{-1}.$ The standard heat of vaporisation of liquid water is $44.0\, kJ\, mol^{-1}$. What is $\Delta H^{\circ}$ of
$C_3H_8\left(g\right) + 5O_2\left(g\right)\to 3CO_2\left(g\right) + 4H_2O\left(g\right)$

• A. $-2220.1 \,kJ$
• B. $-2044.1\, kJ$
• C. $-2396.1\, kJ$
• D. $-2176.1 \,kJ$

Answer 1

Answer: (B)

$\Delta H^{\circ}_{\left(\text{vap}\right)} of H_{2}O =44.0 kJ$ mol $^{-1} $
$\therefore \Delta H^{\circ}_{\text{(vap)}}$ of $4\, mol$ $H_{2}O=176.0 \,kJ $
$\therefore \Delta H^{\circ}_{\text{Rxn}}=\Delta H^{\circ}_{\text{comb}}\left(C_{3}H_{8}\right)+\Delta H^{\circ}_{\text{vap}}\left(4H_{2}O\right)$
$=-2220.1+176.0$
$=-2044.1\, kJ$