Question

From the following reaction and data, find (a)${\mathit{S}}^{\mathbf{°}}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{S}\mathit{O}\mathit{C}{\mathit{l}}_{\mathbf{2}}$and (b) T at which the reaction becomes nonspontaneous:

$$\begin{gathered} {\mathbf{\text{SO}}}_{\mathbf{\text{3}}}{\mathbf{\text{(g) + SCl}}}_{\mathbf{\text{2}}}\mathbf{\text{(l}}\mathbf{\to }{\mathbf{\text{SOCl}}}_{\mathbf{\text{2}}}{\mathbf{\text{(l) + SO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g)∆G}}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}\mathbf{\text{=-75.2kJ}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{\text{SO}}}_{\mathbf{\text{3}}}{\mathbf{\text{(g)SCl}}}_{\mathbf{\text{2}}}{\mathbf{\text{(l)SOCl}}}_{\mathbf{\text{2}}}{\mathbf{\text{(l)SO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}} \\ \mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{°}}\left(kJ/mol\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{396}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{245}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{296}\mathbf{.}\mathbf{8} \\ {\mathit{S}}^{\mathbf{°}}\left(J/mol\times K\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{256}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{184}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{248}\mathbf{.}\mathbf{1}\mathbf{} \end{gathered}$$

Short answer

a) The entropy of $SOC{l}_2$production is $192.529J/mol\times K$.

b) Because$∆G_{rxn}^{°}>0$ , the reaction is non-spontaneous at temperatures above$1.355\times {10}^{6}K$ .

Step-by-step solution

Definition of Nonspontaneous reaction

A chemical reaction known as an endergonic reaction (also known as a nonspontaneous reaction) is one in which the standard change in free energy is positive and energy is absorbed. By linking endergonic processes to highly exergonic reactions, they can be pushed or pulled.

CalculatingS° of SOCl2

a) Because entropy is a state function, it can be stated in the following way:

$∆S_{rxn}^{°}=\sum _{}^{}∆S_{product}^{°}-\sum _{}^{}∆S_{reac\tan t}^{°}$

• In the case of the stated reaction,

$$\begin{gathered} ∆S_{rxn}^{°}=\left[1mol\times ∆S_{SOC{l}_2\left(I\right)}^{°}+1mol\times ∆S_{S{O}_2\left(g\right)}^{°}\right]- \\ -\left[1mol\times ∆S_{S{O}_3\left(g\right)}^{°}+1mol\times ∆S_{SC{l}_2\left(I\right)}^{°}\right] \end{gathered}$$

• As a result, the entropy of the$SOC{l}_2$ can be written as

$$\begin{gathered} ∆S_{rxn}^{°}=\left[1mol\times ∆S_{SOC{l}_2}^{°}+1mol\times ∆S_{S{O}_{2\left(g\right)}}^{°}\right]- \\ -\left[1mol\times ∆S_{S{O}_3\left(g\right)}^{°}+1mol\times ∆S_{SC{l}_2\left(I\right)}^{°}\right] \\ ∆S_{SOC{l}_2\left(I\right)}^{°}=∆S_{rxn}^{°}+\left[1mol\times ∆S_{S{O}_3\left(g\right)}^{°}+1mol\times ∆S_{SC{l}_2\left(I\right)}^{°}\right]-\left[1mol\times ∆S_{S{O}_2\left(g\right)}^{°}\right] \end{gathered}$$

• The reaction's total entropy is not given, but it can be calculated:

$$\begin{gathered} ∆G_{rxn}^{°}=∆H_{rxn}^{°}-T\times ∆S_{rxn}^{°} \\ ∆S_{rxn}^{°}=\frac{∆H_{rxn}^{°}-∆G_{rxn}^{°}}{T} \end{gathered}$$

Calculating reaction’s total enthalpy

• The reaction's total enthalpy can be calculated as follows:

• $$\begin{gathered} ∆H_{rxn}^{°}=\left[1mol\times ∆H_{SOC{l}_2}^{°}+1mol\times ∆H_{S{O}_2\left(g\right)}^{°}\right]- \\ -\left[1mol\times ∆H_{S{O}_3\left(g\right)}^{°}+1mol\times ∆H_{SC{l}_2\left(I\right)}^{°}\right] \\ ∆H_{rxn}^{°}=\left[1mol\times \left(-245.60kJ/mol\right)\_1mol\times \left(-296.80kJ/mol\right)\right]- \\ -\left[1mol\times \left(-396.00kJ/mol\right)+1mol\times \left(-50.00kJ/mol\right)\right] \\ ∆H_{rxn}^{°}=-96.40kJ \end{gathered}$$
• Now we can calculate the total reaction entropy:

$$\begin{gathered} ∆S_{rxn}^{°}=\frac{-96.40kJ-\left(-75.20kJ\right)}{298K} \\ ∆S_{rxn}^{°}=-0.07114J/K=-7.114\times {10}^{-2}J/K \end{gathered}$$

As a result, the standard entropy of a reaction is$-7.114\times {10}^{-2}J/K$ .

• Finally, the entropy of $SOC{l}_2$can be determined as follows:

$$\begin{gathered} ∆S_{SOC{l}_2\left(I\right)}^{°}=-7.114\times {10}^{-2}\frac{J}{mol\times K}+ \\ +\left[1mol\times 256.70\frac{J}{mol\times K}+1mol\times 184.00\frac{J}{mol\times K}\right]- \\ -\left[1mol\times 248.10\frac{J}{mol\times K}\right] \\ ∆S_{SOC{l}_2\left(I\right)}^{°}=192.529J/mol\times K \end{gathered}$$

$SOC{l}_2$has a formation entropy of$192.259J/mol\times K$.

Calculating T at which reaction become nonspontaneous

b) We must apply the following equation to determine the temperature at which the reaction becomes non-spontaneous:$∆G_{rxn}^{°}=∆H_{rxn}^{°}-T\times ∆S_{rxn}^{°}$

• When$∆G_{rxn}^{°}>0$, the reaction becomes non-spontaneous at T. To begin, determine the temperature at which$∆G_{rxn}^{°}=0$exists.

$$\begin{gathered} ∆G_{rxn}^{°}=∆H_{rxn}^{°}-T\times ∆S_{rxn}^{°} \\ ∆H_{rxn}^{°}-T\times ∆S_{rxn}^{°}=0 \\ T=\frac{∆H_{rxn}^{°}}{∆S_{rxn}^{°}} \end{gathered}$$

• Using the values determined in a) for normal reaction circumstances

$$\begin{gathered} T=\frac{-96.40\times {10}^{3}J}{-7.114\times {10}^{-2}J/K} \\ T=1.355\times {10}^{6}K \end{gathered}$$

• Because the reaction is at equilibrium at$1.355\times {10}^{6}K$, the forward and backward reactions have the same rate. The reaction is non-spontaneous above that temperature because $∆G_{rxn}^{°}>0$.