Q20.40 P - Chemistry: Molecular Nature Of Matter And Change

Question

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

Step-by-Step Solution

Answer

The  ${\mathrm{\Delta S}}^{\mathrm{o}}$ is obtained as:  $82.63\mathrm{J}/\mathrm{K}$.

Step 1: 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.

Step 2: Evaluating the ΔS o .

Hydrogen and Iodine combine to generate HI. Write the reaction equation for the creation of   mole of HI and balance it.

 $\frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{I}}_2\left(\mathrm{s}\right)\to \mathrm{HI}\left(\mathrm{g}\right)$

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

 $\begin{array}{c}\mathrm{S}°\left({\mathrm{H}}_2\left(\mathrm{g}\right)\right)=130.6\mathrm{J}/\mathrm{mol} \mathrm{K}\\ \mathrm{S}°\left({\mathrm{I}}_2\left(\mathrm{g}\right)\right)=260.58\mathrm{J}/\mathrm{mol} \mathrm{K}\\ \mathrm{S}°\left(\mathrm{HI}\left(\mathrm{g}\right)\right)=206.33\mathrm{J}/\mathrm{mol} \mathrm{K}\end{array}$

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

  $\begin{array}{c}\mathrm{\Delta S}°=\sum {\mathrm{n}}_{\mathrm{p}}\mathrm{S}°\left(\mathrm{product}\right)-\sum {\mathrm{n}}_{\mathrm{r}}\mathrm{S}°\left(\mathrm{reactant}\right)...........................\left(1\right)\\ =\left[\mathrm{nS}°\left(\mathrm{HI}\left(\mathrm{s}\right)\right)\right]-\left[\mathrm{nS}°({\mathrm{H}}_2\left(\mathrm{s}\right))+\mathrm{nS}°\left({\mathrm{I}}_2\left(\mathrm{g}\right)\right)\right].............................\left(2\right)\\ =\left(\left[1\right(206.33\left)\right]-\left[\frac{1}{2}(130.6)+\frac{1}{2}(116.14)\right]\right)\mathrm{J}/\mathrm{K}...................\left(3\right)\\ \mathrm{\Delta S}°=82.63\mathrm{J}/\mathrm{K}....................................................................\left(4\right)\end{array}$

Therefore, the value is:  $\mathrm{\Delta S}°=82.63\mathrm{J}/\mathrm{K}$.