Chapter 18, Problem 99AP

The following reaction was described as the cause of sulfur deposits formed at volcanic sites:

${\text{2H}}_{\text{2}}\text{S}(g){\text{ + SO}}_{\text{2}}\text{(}g\text{) }\rightleftarrows \text{ 3S(}s{\text{) + 2H}}_{\text{2}}\text{O(}g\text{)}$

It may also be used to remove SO, from powerplant stack gases. (a) Identify the type of redox reaction it is. (b) Calculate the equilibrium constant $\left(K_{\text{p}}\right)$ at $\text{25}°\text{C,}$ and comment on whether this method is feasible for removing ${\text{SO}}_{\text{2+}}$ (c) Would this procedure become more effective or less effective at a higher temperature?

Interpretation Introduction

Interpretation:

The value of K_p, type of the given reaction, and whether the reaction is more or less effective at low temperature are to be determined, and the type of redox reaction is to be identified.

Concept introduction:

The disproportion reaction is a type of redox reaction. The compound of intermediate oxidation state converts into two different compounds, one is at a higher oxidation state and the other is at s lower oxidation state.

The relation between free energy change and standard free energy change is as follows:

$\Delta G=\Delta G^{\text{o}}+\text{R}T\ln Q$

Here, $\Delta G$ is free energy change, $\Delta G^o$ is standard free energy change, $Q$ is the reaction quotient, $\text{R}$ is the gas constant, and $T$ is the temperature.

At equilibrium the above equation is reduced to the expression: $\Delta G^{\text{o}}\text{=}-\text{R }T\text{lnK }$

The standard free energy change for the reaction is calculated using the following expression:

$\Delta G_{rxn}^o={\displaystyle \sum n}\Delta G_f^o\left(\text{product}\right)-{\displaystyle \sum m\Delta G_f^o}\left(\text{reactant}\right)$

Here, $\Delta G_{rxn}^o$ is the standard free energy change for the reaction, $\Delta G_f^o$ is the standard free energy of formation of the compound, $n$ is the stoichiometric coefficient of the product, and $m$ is the stoichiometric coefficient of the reactant.

The standard enthalpy change of the reaction $\Delta H_{rxn}^o$ is calculated using the expression as follows:

$\Delta H_{rxn}^o={\displaystyle \sum n}\Delta H_f^o\left(\text{product}\right)-{\displaystyle \sum m\Delta H_f^o}\left(\text{reactant}\right)$

Here, $\Delta H_{rxn}^o$ is the standard enthalpy change for the reaction, $\Delta H_f^o$ is the standard enthalpy change for the formation of the substance, $n$ is the stoichiometric coefficient of the product, and $m$ is the stoichiometric coefficient of the reactant.

The entropy changes of the system $\Delta S^o$ are calculated using the following expression:

$\Delta S^o=\Delta S_{rxn}^o$

The standard entropy change for this reaction is calculated using the expression as follows:

$\Delta S_{rxn}^o={\displaystyle \sum n}{S}^o\left(\text{product}\right)-{\displaystyle \sum m{S}^o}\left(\text{reactant}\right)$

Here, $\Delta S_{rxn}^o$ is the standard entropy change for the reaction, $\Delta S^o$ is the standard entropy change of the substance, $n$ is the stoichiometric coefficient of the product, and $m$ is the stoichiometric coefficient of the reactant.

Answer to Problem 99AP

Solution:

a)

${\text{2H}}_2\text{S}\left(g\right){\text{+ SO}}_2\left(g\right)\rightleftharpoons 3\text{S}\left(s\right)+2{\text{H}}_2\text{O}\left(g\right)$

This is the reverse of a disproportion redox reaction.

b)

The value of the equilibrium constant for the given reaction is $8.2\times {10}^{15}$. This method is feasible for removing ${\text{SO}}_2$.

c)

The reaction is less effective at high temperature.

Explanation of Solution

a) Type of redox reaction

The equation for the reaction of ${\text{H}}_2\text{S}\left(g\right)$ with ${\text{SO}}_2\left(g\right)$ is as follows:

${\text{2H}}_2\underset{-2}{\text{S}}\left(g\right)+\underset{+4}{\text{S}}{\text{O}}_2\left(g\right)\rightleftharpoons 3\underset{0}{\text{S}}\left(s\right)+2{\text{H}}_2\text{O}\left(g\right)$

Here, sulfur changes its oxidation state from $-2$ to $0$ and from $+4$ to $0$.

Thus, this reaction is the type of reverse disproportion reaction.

b) The equilibrium constant at $\text{25}°\text{C}$ and feasibility of the method for removing ${\text{SO}}_2\left(g\right)$

The equation for the reaction of ${\text{H}}_2\text{S}\left(g\right)$ with ${\text{SO}}_2\left(g\right)$ is as follows:

${\text{2H}}_2\text{S}\left(g\right){\text{+ SO}}_2\left(g\right)\rightleftharpoons 3\text{S}\left(s\right)+2{\text{H}}_2\text{O}\left(g\right)$

The standard free energy change for the reaction is calculated using the following expression:

$\Delta G_{rxn}^o={\displaystyle \sum n}\Delta G_f^o\left(\text{product}\right)-{\displaystyle \sum m\Delta G_f^o}\left(\text{reactant}\right)$

Here, $\Delta G_{rxn}^o$ is the standard free energy change for the reaction, $\Delta G_f^o$ is the standard free energy of formation of the compound, $n$ is the stoichiometric coefficient of the product, and $m$ is the stoichiometric coefficient of the reactant.

The Gibbs energy for the reaction is as follows:

$\Delta G_{rxn}^o=\left(3\times \Delta G_f^o\left[\text{S}\left(s\right)\right]+2\times \Delta G_f^o\left[{\text{H}}_2\text{O}\left(g\right)\right]\right)-\left(\Delta G_f^o\left[{\text{SO}}_2\left(g\right)\right]+2\times \Delta G_f^o\left[{\text{H}}_2\text{S}\left(g\right)\right]\right)$

The standard free energy of formation of any element in its stable allotropic form at $1\text{atm}$ and $\text{25}°\text{C}$ is zero.

From appendix $2$, the value of the standard free energies of formation of the compounds are as follows:

$\Delta G_f^o\left[\text{S}\left(s\right)\right]=0\text{kJ}/\text{mol}$

$\Delta G_f^o\left[{\text{SO}}_2\left(g\right)\right]=-300.4\text{kJ}/\text{mol}$

$\Delta G_f^o\left[{\text{H}}_2\text{S}\left(g\right)\right]=-33.0\text{kJ}/\text{mol}$

$\Delta G_f^o\left[{\text{H}}_2\text{O}\left(g\right)\right]=-228.6\text{kJ}/\text{mol}$

Substitute the values of the standard free energy of formation of the compound and $\Delta G_{rxn}^o$ in the above expression

$\begin{array}{l}\Delta G_{rxn}^o=\left(2\times \left[-228.6\text{kJ}/\text{mol}\right]+\left[0\right]\right)-\left(2\times \left[-33.0\text{kJ}/\text{mol}\right]+1\times \left[-300.4\text{kJ}/\text{mol}\right]\right)\\ =-90.8\text{kJ}/\text{mol}\end{array}$

Therefore, standard free energy change for the reaction is $-90.8\text{kJ}/\text{mol}$.

The equilibrium pressure constant is calculated using the expression as:

$\Delta G^o=-RT\ln \text{K}$

Here, $\text{R}$ is the constant, $T$ is the temperature, $\text{K}$ is the equilibrium constant, and $\Delta G^o$ is the Gibbs free energy change.

Substitute the values of $\text{R}$, $T$, and $\Delta G^o$ in the above expression

$\begin{array}{c}\ln \text{K}=\left(-\frac{-90.8\times {10}^3\text{J}/\text{mol}}{\left(8.314\text{J}/\text{K mol}\right)\left(\text{298 K}\right)}\right)\\ \ln \text{K}=36.64\\ \text{K}=e^{36.64}\\ \text{K}=8.2\times {10}^{15}\end{array}$

Therefore, the equilibrium constant for the given reaction is $8.2\times {10}^{15}$. The value of the equilibrium constant for this reaction is high. It is feasible to remove ${\text{SO}}_2$.

c) Effectiveness of procedure at a high temperature

The equation for the reaction of ${\text{H}}_2\text{S}\left(g\right)$ with ${\text{SO}}_2\left(g\right)$ is as follows:

${\text{2H}}_2\text{S}\left(g\right){\text{+ SO}}_2\left(g\right)\rightleftharpoons 3\text{S}\left(s\right)+2{\text{H}}_2\text{O}\left(g\right)$

The standard enthalpy change of the system, $\Delta H^o$, is calculated using the expression as follows:

$\Delta H^o=\Delta H_{rxn}^o$

The standard enthalpy change of the reaction $\Delta H_{rxn}^o$ is calculated using the expression as follows:

$\Delta H_{rxn}^o={\displaystyle \sum n}\Delta H_f^o\left(\text{product}\right)-{\displaystyle \sum m\Delta H_f^o}\left(\text{reactant}\right)$

Here, $\Delta H_{rxn}^o$ is the standard enthalpy change for the reaction, $\Delta H_f^o$ is the standard enthalpy change for the formation of the substance, $n$ is the stoichiometric coefficient of the product, and $m$ is the stoichiometric coefficient of the reactant.

The enthalpy change for the given reaction is as follows:

$\Delta H_{rxn}^o=\left(3\times \Delta H_f^o\left[\text{S}\left(s\right)\right]+2\times \Delta H_f^o\left[{\text{H}}_2\text{O}\left(g\right)\right]\right)-\left(2\times \Delta H_f^o\left[{\text{H}}_2\text{S}\left(g\right)\right]+\Delta H_f^o\left[{\text{SO}}_2\left(g\right)\right]\right)$

From appendix 2, the standard enthalpy changes for the formation of the substances are as follows:

$\Delta H_f^o\left[{\text{SO}}_2\left(g\right)\right]=\text{}-296.4\text{kJ}/\text{mol}$

$\Delta H_f^o\left[\text{S}\left(s\right)\right]=0\text{kJ}/\text{mol}$

$\Delta H_f^o\left[{\text{H}}_2\text{O}\left(g\right)\right]=-241.8\text{kJ}/\text{mol}$

$\Delta H_f^o\left[{\text{H}}_2\text{S}\left(g\right)\right]=20.15\text{kJ}/\text{mol}$

Substitute the standard enthalpy change of the formation value of the substances in the above expression

$\begin{array}{l}\Delta H_{rxn}^o=\left[2\times \left(-241.8\text{kJ}/\text{mol}\right)+\left(0\right)\right]-\left[2\times \left(-20.15\text{kJ}/\text{mol}\right)+1\times \left(0\right)\right]\\ =-146.9\text{kJ}/\text{mol}\\ \Delta H^o=-146.9\text{kJ}/\text{mol}\end{array}$

The entropy changes of the system $\Delta S^o$ are calculated using the expression as:

$\Delta S^o=\Delta S_{rxn}^o$

The standard entropy change for this reaction is calculated using the expression as follows:

$\Delta S_{rxn}^o={\displaystyle \sum n}{S}^o\left(\text{product}\right)-{\displaystyle \sum m{S}^o}\left(\text{reactant}\right)$

Here, $\Delta S_{rxn}^o$ is the standard entropy change for the reaction, $\Delta S^o$ is the standard entropy change of the substance, $n$ is the stoichiometric coefficient of the product, and $m$ is the stoichiometric coefficient of the reactant.

The entropy change for the reaction is as follows:

$\Delta S_{rxn}^o=\left(3\times S^0\left[\text{S}\left(s\right)\right]+2\times S^0\left[{\text{H}}_2\text{O}\left(g\right)\right]\right)-\left(2\times S^o\left[{\text{H}}_2\text{S}\left(g\right)\right]+S^o\left[{\text{SO}}_2\left(g\right)\right]\right)$

From appendix $2$, the standard entropy values of the substances are as follows:

$S^o\left[{\text{SO}}_2\left(g\right)\right]=\text{}205.64\text{J}/\text{K}\cdot \text{mol}$

$S^o\left[{\text{H}}_2\text{S}\left(g\right)\right]=248.5\text{J}/\text{K}\cdot \text{mol}$

$S^o\left[{\text{H}}_2\text{O}\left(g\right)\right]=31.88\text{kJ}/\text{mol}$

$S^o\left[\text{S}\left(s\right)\right]=0\text{J}/\text{K}\cdot \text{mol}$

Substitute the standard entropy values of the substances in the above expression

$\begin{array}{l}\Delta S_{rxn}^o=\left[2\times \left(188.7\text{J}/\text{K}\cdot \text{mol}\right)+3\times \left(31.88\text{J}/\text{K}\cdot \text{mol}\right)\right]-\left[2\times \left(205.64\text{J}/\text{K}\cdot \text{mol}\right)+\left(248.5\text{J}/\text{K}\cdot \text{mol}\right)\right]\\ =-186.7\text{J}/\text{K}\cdot \text{mol}\\ \Delta S^o=-186.7\text{J}/\text{K}\cdot \text{mol}\end{array}$

The standard Gibbs free energy change for the given reaction at temperature $300°\text{C}$ is calculated using the following expression:

$\Delta G^o=\Delta H^o-T\times \Delta S^o$

Here, $\Delta H^o$ is the standard enthalpy changes, $T$ is the temperature, $\Delta S^o$ is the standard entropy change, and $\Delta G^o$ is the standard Gibbs free energy change.

Substitutes the value of $\Delta H^o$, $T$, and $\Delta S^o$ in the above expression

$\begin{array}{l}\Delta G^o=\left(-90.8\text{J}/\text{mol}\right)-\left(298\text{K}\right)\times \left(-186.7\text{J}/\text{mol}\right)\\ =55,545.8\text{J}/\text{mol}\end{array}$

Therefore, standard Gibbs free energy change for the given reaction is $55,545.8\text{J}/\text{mol}$

The change in entropy for this reaction is negative. The change in Gibbs free energy will become positive at high temperature. The reaction will be less effective at high temperature.