91Ó°ÊÓ

Thermodynamics is the study of the connections between heat, work, temperature, and energy. Thermodynamic principles specify how energy develops within a system and whether it is capable of having a positive impact on its surroundings.

First, write the${\mathrm{CH}}_4$combustion reaction and balance the equation.

${\mathrm{CH}}_4\left(\mathrm{g}\right)+2{\mathrm{O}}_2\left(\mathrm{g}\right)→{\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

After that, make a list of the ${\mathrm{S}}^{\mathrm{o}}$values for each compound.

$\begin{array}{rcl}{\mathrm{S}}^{∘}\left({\mathrm{O}}_2\left(\mathrm{g}\right)\right)& =& 205.0\mathrm{J}/\mathrm{mol}\mathrm{K}\\ {\mathrm{S}}^{∘}\left({\mathrm{CH}}_4\left(\mathrm{g}\right)\right)& =& 186.1\mathrm{J}/\mathrm{mol}\mathrm{K}\\ {\mathrm{S}}^{∘}\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\right)& =& 69.94\mathrm{J}/\mathrm{mol}\mathrm{K}\\ {\mathrm{S}}^{∘}\left({\mathrm{CO}}_2\left(\mathrm{g}\right)\right)& =& 213.7\mathrm{J}/\mathrm{mol}\mathrm{K}\end{array}$

Solve for the ${\mathrm{Δ³\S }}^{\mathrm{o}}$now as:

$\begin{array}{rcl}{\mathrm{Δ³\S }}^{∘}& =& ∑{\mathrm{n}}_{\mathrm{p}}{\mathrm{S}}^{∘}\left(\mathrm{product}\right)-∑{\mathrm{n}}_{\mathrm{r}}{\mathrm{S}}^{∘}\left(\mathrm{reactant}\right)...........................\left(1\right)\\ & =& \left[{\mathrm{nS}}^{∘}\left({\mathrm{CO}}_2\left(\mathrm{g}\right)\right)+{\mathrm{nS}}^{∘}\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\right)\right]-\left[{\mathrm{nS}}^{∘}\left({\mathrm{CH}}_4\left(\mathrm{g}\right)\right)+{\mathrm{nS}}^{∘}\left({\mathrm{O}}_2\left(\mathrm{g}\right)\right)\right]....................\left(2\right)\\ & & =\left(1\right(213.7)+2(69.94\left)\right]-\left[1\right(186.1)+2(205.0\left)\right])\mathrm{J}/\mathrm{mol}×\mathrm{K}....................\left(3\right)\\ & =& -242.52\mathrm{J}/\mathrm{K}.....................\left(4\right)\end{array}$

The sign is not favorable. On the reactant side, all of the chemical species are gaseous compounds. Gaseous and liquid chemicals make up the chemical species found on the production side. Liquids, on the other hand, are more structured than gases. As a result, the reaction goes from being less structured to be more organized, with decreased entropy.

Therefore, the value is: ${\mathrm{Δ³\S }}^{∘}=-242.52\mathrm{J}/\mathrm{K}$.