Find ${\mathbf{\Delta S}}^{\mathbf{o}}$for the formation of ${\mathbf{CH}}_{\mathbf{3}}\mathbf{OH}$(l) from its elements.

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.

Hydrogen, Graphite, and oxygen combine to generate${\mathrm{CH}}_3\mathrm{OH}$. Write the reaction equation for the creation of$1$mole of${\mathrm{CH}}_3\mathrm{OH}$and balance it.

$\mathrm{C}\left(\mathrm{s},\mathrm{graphite}\right)+2{\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{CH}}_3\mathrm{OH}\left(\mathrm{l}\right)$

After that, make a list of each compound ${\mathrm{\Delta S}}^{\mathrm{o}}$ values:

$\begin{array}{c}\mathrm{S}°\left(\mathrm{C}\left(\mathrm{s},\mathrm{graphite}\right)\right)=5.686\mathrm{J}/\mathrm{mol}\mathrm{K}\\ \mathrm{S}°\left({\mathrm{H}}_2\left(\mathrm{g}\right)\right)=130.6\mathrm{J}/\mathrm{mol}\mathrm{K}\\ \mathrm{S}°\left({\mathrm{O}}_2\left(\mathrm{g}\right)\right)=205.0\mathrm{J}/\mathrm{mol}\mathrm{K}\\ \mathrm{S}°\left({\mathrm{CH}}_3\mathrm{OH}\left(\mathrm{l}\right)\right)=127\mathrm{J}/\mathrm{mol}\mathrm{K}\end{array}$

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

role="math" localid="1663362093876" $\begin{array}{c}{\mathrm{\Delta S}}^{\circ }=\sum {\mathrm{n}}_{\mathrm{p}}{\mathrm{S}}^{\circ }\left(\mathrm{product}\right)-\sum {\mathrm{n}}_{\mathrm{r}}{\mathrm{S}}^{\circ }\left(\mathrm{reactant}\right)......................\left(1\right)\\ =\left[{\mathrm{nS}}^{\circ }\left({\mathrm{CH}}_3\mathrm{OH}\left(\mathrm{g}\right)\right)\right]-\left[{\mathrm{nS}}^{\circ }\left(\mathrm{C}\right(\mathrm{s},\mathrm{graphite}\left)\right)+{\mathrm{nS}}^{\circ }\left({\mathrm{H}}_2\left(\mathrm{g}\right)\right)+\mathrm{nS}\left({\mathrm{O}}_2\right)\right]..............\left(2\right)\\ =\left(\left[1\right(127\left)\right]-\left[1(5.686)+2(130.6)+\frac{1}{2}(205.0)\right]\right)\mathrm{J}/\mathrm{K}........................\left(3\right)\\ {\mathrm{\Delta S}}^{\circ }=-242.39\mathrm{J}/\mathrm{K}...............................................................\left(4\right)\end{array}$

Therefore, the value is: ${\mathrm{\Delta S}}^{\circ }=-242.39\mathrm{J}/\mathrm{K}$.