Chemistry - Thermochemistry Question with Solution | TestHub

The combustion reaction can be written as follows,

${\mathrm{C}}_3{\mathrm{H}}_{8_{\left(\mathrm{g}\right)}}+5{\mathrm{O}}_2\stackrel{}{\to }3{\mathrm{CO}}_2+4{\mathrm{H}}_2\mathrm{O}; ∆\mathrm{H}=-2220.0 \mathrm{kJ} {\mathrm{mol}}^{-1}$

${\mathrm{C}}_{\mathrm{graphite}}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\stackrel{}{\to }{{\mathrm{CO}}_2}_{\left(\mathrm{g}\right)} ; ∆\mathrm{H}=-393.5 \mathrm{kJ} {\mathrm{mol}}^{-1}$

${{\mathrm{H}}_2}_{\left(\mathrm{g}\right)}+\frac{1}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\stackrel{}{\to }{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)} ; ∆\mathrm{H}=-285.8 \mathrm{kJ} {\mathrm{mol}}^{-1}$

Now, the formation reaction of propane is

$$\begin{gathered} 3{\mathrm{C}}_{\left(\mathrm{graphite}\right)}+4{\mathrm{H}}_{2\left(\mathrm{g}\right)}\to {\mathrm{C}}_3{\mathrm{H}}_{8\left( \mathrm{g}\right)} \\ ∆{\mathrm{H}}_{\mathrm{f}}=3∆{\mathrm{H}}_{\mathrm{C}}^{°}\left(\mathrm{graphite}\right)+4∆{\mathrm{H}}_{\mathrm{C}}^{°}\left({\mathrm{H}}_{2\left(\mathrm{g}\right)}\right)-∆{\mathrm{H}}_{\mathrm{C}}^{°}\left(\mathrm{propane}\right) \\ ∆{\mathrm{H}}_{\mathrm{f}}=3\times \left(-393.5\right)+4\times \left(-285.8\right)-\left(-2220\right)=-103.7 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$