Q22.66CP - Chemistry: Molecular Nature Of Matter And Change

Question

The production of ${\text{S}}_8$ from the ${\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{ }\mathbf{\text{S}}\mathbf{(}\mathbf{ }\mathbf{\text{g}}\mathbf{)}$ for step 2, found in natural gas deposits occurs through the Claus process (Section 22.5):

(a) Use these two unbalanced steps to write an overall balanced equation for this process:

1. ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{S(g) + O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\to }{\mathbf{\text{S}}}_{\mathbf{\text{8}}}{\mathbf{\text{(g) + SO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g) + H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(g)}}$
   
2. ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{S(g) + SO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\to }{\mathbf{\text{S}}}_{\mathbf{\text{8}}}{\mathbf{\text{(g) + H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(g))}}$
   (b)  Write the overall reaction with ${\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}$ as the oxidizing agent instead of ${\mathbf{\text{O}}}_{\mathbf{\text{2}}}$. Use thermodynamic data to show whether ${\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}\left(g\right)$ can be used to oxidize  ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}\mathbf{\text{S(g).}}$
   (c) Why is oxidation by  preferred to oxidation by  ${\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}$?

Step-by-Step Solution

Answer

 

 (a) The overall balanced reaction should be: $\mathbf{8}{\text{H}}_2 \text{S}\left(\mathbf{g}\right)+\mathbf{2}{\mathbf{O}}_{\mathbf{2}}\left(\text{g}\right)\to {\text{S}}_{\mathbf{8}}\left(\text{g}\right)+\mathbf{8}{\text{H}}_{\mathbf{2}}\mathbf{O}\left(\text{g}\right)$.

(b) The overall reaction with ${\text{Cl}}_{\text{2}}$ as the oxidizing agent instead of $O_{\text{2}}$is: $8{\text{H}}_2 \text{S}( \text{g})+\mathbf{8}{\text{Cl}}_{\mathbf{2}}\left(\mathbf{g}\right)\to {\text{S}}_{\mathbf{8}}\left(\text{g}\right)+16\text{HCl}\left(\text{g}\right)$. It is also spontaneous at low temperatures because the reaction with ${\mathrm{Cl}}_2$is exothermic and the overall entropy of the process lowers. It is, for example, spontaneous at As a result, this reaction can likewise be used for the production.

(c) As a by-product of the reaction with $\text{C}{l}_2$ a large amount of $\mathrm{HCL}$ is produced. Because requires special treatment, it is preferable to use ${\mathrm{O}}_2$ which produces merely water

Step 1: Concept Introduction

The Claus method is used to handle gas streams with high concentrations of ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}\mathbf{\text{S}}$(over 50%). The units' chemistry includes hydrogen sulphide partial oxidation to sulphur dioxide and the catalytically stimulated coupling of ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}\mathbf{\text{S}}$and ${\mathbf{\text{SO}}}_{\mathbf{\text{2}}}$ to generate elemental sulphur.

Step 2: Combine two responses into a single one

(a)

We have to synchronizing every reaction at first, 

1. $16{\text{H}}_2 \text{S}\left(g\right)+16{\text{O}}_2\left(g\right)\to {\text{S}}_8\left(g\right)+8{\text{SO}}_2\left(g\right)+16{\text{H}}_2\text{O}\left(\text{g}\right)$
   
2. $16{\text{H}}_2 \text{S}\left(g\right)+8{\text{SO}}_2\left(g\right)\to 3 {\text{S}}_8\left(g\right)+16{\text{H}}_2\text{O}\left(g\right)$
   

Combining these two responses into a single reaction is now possible.

$$\begin{gathered} 16{\text{H}}_2 \text{S}( \text{g})+16{\text{O}}_2\left(g\right)+16{\text{H}}_2 \text{S}( \text{g})+8{\text{SO}}_2\left(g\right)\to \\ \to {\text{S}}_8( \text{g})+8{\text{SO}}_2( \text{g})+16{\text{H}}_2\text{O}\left(\text{g}\right)+3 {\text{S}}_8\left(g\right)+16{\text{H}}_2\text{O}\left(\text{g}\right) \end{gathered}$$

On both lines, cross out the terms that are repeated.

$32{\text{H}}_2 \text{S}\left(g\right)+16{\text{O}}_2\left(g\right)\to 4 {\text{S}}_8\left(g\right)+32{\text{H}}_2\text{O}\left(g\right)$

Therefore, after reducing the coefficients, the resulting balanced reaction would be:

$8{\text{H}}_2 \text{S}( \text{g})+2{\text{O}}_2( \text{g})\to {\text{S}}_8( \text{g})+8{\text{H}}_2\text{O}\left(\text{g}\right)$

Step 3: Write the overall reaction with Cl 2 as the oxidizing agent instead of O 2

(b)

Let’s synchronize every reaction at first, 

1. ${\text{H}}_2 \text{S}\left(g\right)+C{l}_2\left(g\right)\to {\text{S}}_8\left(g\right)+SC{l}_2\left(g\right)+\text{HCl}\left(g\right)$
   
2. ${\text{H}}_2 \text{S}\left(g\right)+SC{l}_2\left(g\right)\to S_8\left(g\right)+HCl\left(g\right)$
   

Combining these two responses into a single reaction is now possible.

$$\begin{gathered} H_{2}S\left(g\right)+C{l}_2\left(g\right)+H_{2}S\left(g\right)+SC{l}_2\left(g\right)\to \\ \to S_8\left(g\right)+SC{l}_2\left(g\right)+HCl\left(g\right)+S_8\left(g\right)+HCl\left(g\right) \end{gathered}$$

On both lines, cross out the terms that are repeated.

$2{\text{H}}_{2}S\left(g\right)+C{l}_2\left(g\right)\to 2S_8\left(g\right)+2HCl\left(g\right)$

By equating the number of  S molecules on both sides of the formula and rectifying the H number, the reaction can be balanced.

$16{\text{H}}_{2}S\left(g\right)+16{\text{Cl}}_2\left(g\right)\to 2S_8\left(g\right)+32\text{HCl}\left(g\right)$

Finally, the coefficients are minimized.

Therefore, the overall reaction with ${\text{Cl}}_{\text{2}}$ the oxidizing agent instead of ${\text{O}}_{\text{2}}$ is:

$8{\mathbf{H}}_2\mathbf{S}\left(\mathbf{g}\right)+\mathbf{8}{\mathbf{Cl}}_{\mathbf{2}}\left(\mathbf{g}\right)\to {\mathbf{S}}_{\mathbf{8}}\left(\text{g}\right)+\mathbf{16}\mathbf{HCl}\left(\mathbf{g}\right)$

To compare the thermodynamic data, first calculate the enthalpy and entropy of each conceivable reaction, then compare the Gibb's energy for both processes:

Computing the enthalpy and entropy of the process using Appendix B for the ${\text{O}}_2$ reaction.

$$\begin{gathered} \Delta H_{rxn}=\sum \Delta H_{products}-\sum \Delta H_{reac\tan ts} \\ \Delta H_{rxn}=\left[1 mol\times \Delta H\left(S_8\right)+8 mol\times \Delta H\left(H_{2}O\right)\right]- \\ -\left[8 mol\times \Delta H\left(H_{2}S\right)+2 mol\times \Delta H\left(O_2\right)\right] \end{gathered}$$

$$\begin{gathered} \Delta H_{rxn}=[1 mol\times 101.00 kJ/mol+8 mol\times (-241.826)kJ/mol] \\ -[8 mol\times (-20.20)kJ/mol+2 mol\times 0.00 kJ/mol] \\ \Delta H_{rxn}=-1672.008 kJ \end{gathered}$$

Therefore, the value of  $\Delta {\mathbf{H}}_{\text{rxn}}=-1672.008 \text{kJ}$

Then, find the value of $\Delta {\mathbf{S}}_{\mathbf{rxn}}$

$$\begin{gathered} \Delta S_{rxn}=\sum \Delta S_{\text{products }}-\sum \Delta S_{\text{reactants }} \\ \Delta S_{rxn}=\left[1 \text{mol}·\Delta S\left(S_8\right)+8 \text{mol}·\Delta S\left(H_2\text{O}\right)\right]- \\ -\left[8 \text{mol}·\Delta S\left(H_{2}S\right)+2 \text{mol}·\Delta S\left(O_2\right)\right] \\ \Delta S_{rxn}=\left[1 \text{mol}·430.211\frac{\text{J}}{\text{mol}·\text{K}}+8 \text{mol}·188.72\frac{\text{J}}{\text{mol}·\text{K}}\right]- \\ -\left[8 \text{mol}·205.60\frac{\text{J}}{\text{mol}·\text{K}}+2 \text{mol}·205.00\frac{\text{J}}{\text{mol}·\text{K}}\right] \\ \Delta {\text{S}}_{\text{rxn}}=-114.829\frac{\text{J}}{\text{K}}=-\mathbf{0}.\mathbf{114829}\frac{\text{kJ}}{\text{K}} \end{gathered}$$

Therefore, the value of $\Delta {\text{S}}_{\text{rxn}}=-\mathbf{0}.\mathbf{114829}\frac{\text{kJ}}{\text{K}}$

The Gibb's energy for is then,

$$\begin{gathered} \Delta G_{25}=-1672.008 \text{kJ}-(25+273)\text{K}·-0.114829\frac{\text{kJ}}{\text{K}} \\ \Delta G_{25}=-1637.789 \text{kJ} \end{gathered}$$

At low temperatures, the reaction with $O_2$ is spontaneous because it is exothermic $(\Delta H<0)$ and the overall entropy of the process reduces. It is, for example, spontaneous at   $25°C$

Computing the enthalpy and entropy of the process using Appendix B for the reaction ${\text{Cl}}_2$.

$$\begin{gathered} \Delta H_{rxn}=\left[1 \text{mol}·\Delta H\left(S_8\right)+16 \text{mol}·\Delta H\left(HCl\right)\right]- \\ -\left[8 \text{mol}·\Delta H\left(H_{2}S\right)+8 \text{mol}·\Delta H\left(C{l}_2\right)\right] \end{gathered}$$

$$\begin{gathered} \Delta H_{rxn}=[1 \text{mol}·101.00 \text{kJ}/\text{mol}+16 \text{mol}·(-92.31)\text{kJ}/\text{mol}]- \\ -[8 \text{mol}·(-20.20)\text{kJ}/\text{mol}+8 \text{mol}·0.00 \text{kJ}/\text{mol}] \\ \Delta {\mathbf{H}}_{\mathbf{rxn}}=-1214.36 \text{kJ} \end{gathered}$$

Therefore, the value of  $\Delta {\mathbf{H}}_{\mathbf{rxn}}=-1214.36 \text{kJ}$

Find the value of $\Delta {\mathbf{S}}_{\mathbf{rxn}}$

$$\begin{gathered} \Delta S_{rxn}=\left[1 \text{mol}·\Delta S\left(S_8\right)+16 \text{mol}·\Delta S\left(HCl\right)\right]- \\ -\left[8 \text{mol}·\Delta S\left(H_{2}S\right)+8 \text{mol}·\Delta S\left(C{l}_2\right)\right] \\ \Delta S_{rxn}=\left[1 \text{mol}·430.211\frac{\text{J}}{\text{mol}·\text{K}}+16 \text{mol}·186.79\frac{\text{J}}{\text{mol}·\text{K}}\right]- \\ -\left[8 \text{mol}·205.60\frac{\text{J}}{\text{mol}·\text{K}}+8 \text{mol}·223.00\frac{\text{J}}{\text{mol}·\text{K}}\right] \\ \Delta {\mathbf{S}}_{\mathbf{rxn}}=-\mathbf{9}.\mathbf{949}\frac{\text{J}}{\text{K}}=-\mathbf{0}.\mathbf{009949}\frac{\text{kJ}}{\text{K}} \end{gathered}$$

Therefore, the value of  $\Delta {\mathbf{S}}_{\mathbf{rxn}}=-\mathbf{0}.\mathbf{009949}\frac{\text{kJ}}{\text{K}}$

The Gibb's energy for $25°C$ is then,

$$\begin{gathered} \Delta G_{25}=-1214.36 \text{kJ}-(25+273)\text{K}·-0.009949\frac{\text{kJ}}{\text{K}} \\ \Delta G_{25}=-1217.325 \text{kJ} \end{gathered}$$

At low temperatures, the reaction with $C{l}_2$ is spontaneous because it is exothermic $(\Delta H<0)$ and the overall entropy of the  process reduces. 

Step 6: Reason to prefer the oxidation by O 2

(c)

The ${\text{O}}_2$ reaction's equilibrium temperature is:

$$\begin{gathered} \Delta G=-1672.008 \text{kJ}-T \text{K}·-0.114829\frac{\text{kJ}}{\text{K}}=0 \\ T=\frac{-1672.008 \text{kJ}}{-0.114829\frac{\text{kJ}}{\text{K}}} \\ T=14560 \text{K} \end{gathered}$$

The response would be non-spontaneous over $14560 \text{K}$.

 The equilibrium temperature for the $C{l}_2$ process is:

$$\begin{gathered} \Delta G=-1214.36 \text{kJ}-T \text{K}·-0.009949\frac{\text{kJ}}{\text{K}} \\ T=\frac{-1214.36 \text{kJ}}{-0.009949\frac{\text{kJ}}{\text{K}}} \\ T=122058.5 \text{K} \end{gathered}$$

The response would be non-spontaneous over $122058.5 \text{K}$.

Because thermodynamics cannot provide a solution (because both reactions would occur spontaneously in industrial conditions), the results of the reactions should be evaluated.

The by-product of oxygen is water, whereas chlorine gas produces a large amount of hydrochloric acid $\mathbf{HCl}$

Because $\mathbf{HCl}$ is a hazardous chemical, working with it necessitates extra caution. Additionally, the manufacturing of large quantities of $\mathbf{HCl}$  necessitated specific storage and handling. 

Therefore, ${\text{O}}_2$ is the recommended oxidation method.