Q20.43 P - Chemistry: Molecular Nature Of Matter And Change

Question

Sulfur dioxide is released in the combustion of coal. Scrubbers use aqueous slurries of calcium hydroxide to remove the ${\mathbf{SO}}_{\mathbf{2}}$ from flue gases. Write a balanced equation for this reaction and calculate ${\mathbf{\Delta S}}^{\mathbf{o}}$ at 298 K [ ${\mathbf{S}}^{\mathbf{o}}$of  ${\mathbf{CaSO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}\mathbf{101}\mathbf{.}\mathbf{4}$J/mol K].

Step-by-Step Solution

Answer

The balanced equation is:  ${\mathrm{SO}}_2\left(\mathrm{g}\right)+\mathrm{Ca}(\mathrm{OH}{)}_2\left(\mathrm{s}\right)\to {\mathrm{CaSO}}_3\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$.

The  ${\mathrm{\Delta S}}^{\mathrm{o}}$ is obtained as:  $-160.15\mathrm{J}/\mathrm{K}$.

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: Writing the balanced equation and then evaluating the .

First, construct and balance the ${\mathrm{SO}}_2$ and $\mathrm{Ca}(\mathrm{OH}{)}_2$ reaction equation.

${\mathrm{SO}}_2\left(\mathrm{g}\right)+\mathrm{Ca}(\mathrm{OH}{)}_2\left(\mathrm{s}\right)\to {\mathrm{CaSO}}_3\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ 

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

$\begin{array}{c}\mathrm{S}°\left({\mathrm{SO}}_2\left(\mathrm{g}\right)\right)=248.1\mathrm{J}/\mathrm{mol} \mathrm{K}\\ \mathrm{S}°\left(\mathrm{Ca}(\mathrm{OH}{)}_2\left(\mathrm{s}\right)\right)=83.39\mathrm{J}/\mathrm{mol} \mathrm{K}\\ \mathrm{S}°\left({\mathrm{CaSO}}_3\left(\mathrm{s}\right)\right)=101.4\mathrm{J}/\mathrm{mol} \mathrm{K}\\ \mathrm{S}°\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\right)=69.94\mathrm{J}/\mathrm{mol} \mathrm{K}\end{array}$ 

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

$\begin{array}{c}{\mathrm{\Delta S}}^{\mathrm{o}}=\sum {\mathrm{n}}_{\mathrm{p}}{\mathrm{S}}^{\mathrm{o}}\left(\mathrm{product}\right)-\sum {\mathrm{n}}_{\mathrm{r}}{\mathrm{S}}^{\mathrm{o}}\left(\mathrm{reactant}\right)......................\left(1\right)\\ = \left[{\mathrm{nS}}^{\mathrm{o}}\left({\mathrm{CaSO}}_3\left(\mathrm{s}\right)\right)+{\mathrm{nS}}^{\mathrm{o}}({\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right))\right]-\left[{\mathrm{nS}}^{\mathrm{o}}(\mathrm{S}{\mathrm{O}}_2\left(\mathrm{g}\right))+\mathrm{n}{\mathrm{S}}^{\mathrm{o}}\left(\mathrm{Ca}(\mathrm{OH}{)}_2\left(\mathrm{s}\right)\right)\right]..............\left(2\right)\\ = \left(\left[1\right(101.4)+1(69.94\left)\right]-\left[1(248.1)+1(83.39)\right]\right)\mathrm{J}/\mathrm{K}........................\left(3\right)\\ {\mathrm{\Delta S}}^{\mathrm{o}}= -160.15\mathrm{J}/\mathrm{K}...............................................................\left(4\right)\end{array}$ 

Therefore, the value is ${\mathrm{\Delta S}}^{\mathrm{o}}= -160.15\mathrm{J}/\mathrm{K}$ .

The balanced equation is ${\mathrm{SO}}_2\left(\mathrm{g}\right)+\mathrm{Ca}(\mathrm{OH}{)}_2\left(\mathrm{s}\right)\to {\mathrm{CaSO}}_3\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$  .