Question

Find${\mathbf{\Delta S}}^{\mathbf{o}}$ for the formation of (g) from its elements.

Short answer

The ${\mathrm{\Delta S}}^{\mathrm{o}}$is obtained as:.

Step-by-step solution

Define Thermodynamics .

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.

Evaluating the ΔSo.

Phosphorus and Chlorine combine to generate ${\mathrm{PCl}}_5$. Write the reaction equation for the creation of $1$mole of ${\mathrm{PCl}}_5$and balance it.

$\mathrm{P}\left(\mathrm{s},\mathrm{white}\right)+\frac{5}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right)\to {\mathrm{PCl}}_5\left(\mathrm{l}\right)$

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

$\begin{array}{c}\mathrm{S}°\left(\mathrm{P}\left(\mathrm{s},\mathrm{white}\right)\right)=44.1\mathrm{J}/\mathrm{mol}\mathrm{K}\\ \mathrm{S}°\left({\mathrm{Cl}}_2\left(\mathrm{g}\right)\right)=223.0\mathrm{J}/\mathrm{mol}\mathrm{K}\\ \mathrm{S}°\left({\mathrm{PCl}}_5\left(\mathrm{g}\right)\right)=353\mathrm{J}/\mathrm{mol}\mathrm{K}\end{array}$

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

$\begin{array}{c}{\mathrm{\Delta S}}^{\circ }=\sum {\mathrm{n}}_{\mathrm{p}}{\mathrm{S}}^{\circ }\left(\mathrm{product}\right)-\sum {\mathrm{n}}_{\mathrm{r}}{\mathrm{S}}^{\circ }\left(\mathrm{reactant}\right)......................\left(1\right)\\ =\left[{\mathrm{nS}}^{\circ }\left({\mathrm{PCl}}_5\left(\mathrm{g}\right)\right)\right]-\left[{\mathrm{nS}}^{\circ }\left(\mathrm{P}\right(\mathrm{s},\mathrm{white}\left)\right)+{\mathrm{nS}}^{\circ }\left({\mathrm{Cl}}_2\right)\right]..............\left(2\right)\\ =\left(\left[1\right(353\left)\right]-\left[1(44.1)+\frac{5}{2}(223.0)\right]\right)\mathrm{J}/\mathrm{K}........................\left(3\right)\\ {\mathrm{\Delta S}}^{\circ }=-248.6\mathrm{J}/\mathrm{K}...............................................................\left(4\right)\end{array}$

Therefore, the value is: ${\mathrm{\Delta S}}^{\circ }=-248.6\mathrm{J}/\mathrm{K}$.