Question

Question: From the data in Appendix D calculate $\mathrm{\Delta H}°$, $\mathrm{\Delta G}°$, and K, for the following reaction at 298 K:

$6{\mathrm{CH}}_4\left(\mathrm{g}\right)+\frac{9}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftarrows {\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{l}\right)+9{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Short answer

The values of $\mathrm{\Delta H}°=3070.36\mathrm{kJ}$, $\mathrm{\Delta G}°=2563.62\mathrm{kJ}$, and $\mathrm{K}=0.355$

Step-by-step solution

Calculation of ΔH°

According to the Appendix D the $\mathrm{\Delta H}{°}_f$ values are as follows,

role="math" localid="1660284854870" $$\begin{gathered} 6{\mathrm{CH}}_4\left(\mathrm{g}\right)+\frac{9}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftarrows {\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{l}\right)+9{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ \text{- 74.81049.03-285.83} \end{gathered}$$

From the above data we can calculate the vale of$\mathrm{\Delta H}°$,

\[\begin{gathered}

\Delta H^\circ = \Delta {H_f}^^\circ \left( {{C_6}{H_6}\left( l \right)} \right) + 9\Delta {H_f}^^\circ \left( {{H_2}O\left( l \right)} \right) - 6\Delta {H_f}^^\circ \left( {C{H_4}} \right) + \frac{9}{2}\Delta {H_f}^^\circ \left( {{O_2}} \right) \\

= 49.03 + 9 \times \left( { - 285.83} \right) - 6 \times \left( { - 74.81} \right) + \frac{9}{2} \times 0 \\

= 49.03 + 2572.47 + 448.86 + 0 \\

= 3070.36 kJ \\

\end{gathered} \]

Calculation of \[\Delta G^\circ \]

According to the Appendix D the\[\Delta {G_f}^^\circ \]values are as follows,

\[\begin{gathered}

6 C{H_4}\left( g \right) + \frac{9}{2} {O_2}\left( g \right) \rightleftarrows {C_6}{H_6}\left( l \right) + 9 {H_2}O\left( l \right) \hfill \\

- 50.75{\text{ 0 124}}{\text{.50 - 237}}{\text{.18}} \hfill \\

\end{gathered} \]

From the above data we can calculate the vale of\[\Delta G^\circ \],

\[\begin{gathered}

\Delta G^\circ = \Delta {G_f}^^\circ \left( {{C_6}{H_6}\left( l \right)} \right) + 9\Delta {G_f}^^\circ \left( {{H_2}O\left( l \right)} \right) - 6\Delta {G_f}^^\circ \left( {C{H_4}} \right) + \frac{9}{2}\Delta {G_f}^^\circ \left( {{O_2}} \right) \\

= 124.50 + 9 \times \left( { - 237.18} \right) - 6 \times \left( { - 50.75} \right) + \frac{9}{2} \times 0 \\

= 124.50 + 2134.62 + 304.5 + 0 \\

= 2563.62 kJ \\

\end{gathered} \]

Calculation of K

From the Gibb’s free energy formula we can calculate the vale of K as,

$$ $\begin{gathered}

ln K = \frac{{ - \Delta G^\circ }}{{RT}} \\

= \frac{{ - 2563.62 kJ}}{{\left( {8.315J{K^{ - 1}}mo{l^{ - 1}}} \right)\left( {298K} \right)}} \\

= \frac{{ - 2563.62 Jmo{l^{ - 1}}}}{{\left( {8.315J{K^{ - 1}}mo{l^{ - 1}}} \right)\left( {298K} \right)}} \\

= \frac{{ - 2563.62 }}{{2477.87}} \\

= - 1.034 \\

\Rightarrow K = {e^{ - 1.034}} = 0.355 \\

\end{gathered} $