Chapter 15.6, Problem 15.15WE

Hydrogen sulfide (H₂S) is a contaminant commonly found m natural gas. It is removed by reaction with oxygen to produce elemental sulfur.

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

For each of the following scenarios, determine whether the equilibrium will shift to the right, shift to the left, or neither: (a) addition of O₂(g), (b) removal of H₂S(g), (c) removal of H₂O(g), and (d) addition of S(s).

Interpretation Introduction

Interpretation:

The equilibrium directions should be identified for the given Hydrogen Sulfide (H₂S) equilibrium reaction with different conditions.

Concept Introduction:

Concept of equilibrium process: Any system at equilibrium subject to change in concentration, temperature, volume or pressure then the system readjusts itself to partly counteract the effect of the applied change and new equilibrium is formed.  This equilibrium can be explained in a simple way, a system equilibrium is distributed the system will adjust itself in such a way that the effect of the change will be reduced.

Le Chatelier's Principle: A change in one of the variables that describe a system at equilibrium produces a shift in the position of the equilibrium that counteracts the effect of this change. In other words the closed system is an increase in pressure; the equilibrium will shift towards the sides of the reaction with some moles of gas.  The decrease in pressure the equilibrium will shift towards the side of the reaction with high moles of gas.

Forward Reaction: This type of reaction has involved irreversible, if obtained product cannot be converted back in to respective reactants under the same conditions.

Backward Reaction: This type of reaction process involved a reversible, if the products can be converted into a back to reactants.

Homogeneous equilibrium: A homogeneous equilibrium involved has an everything present in the same phase and same conditions, for example reactions where everything is a gas, or everything is present in the same solution.

Answer to Problem 15.15WE

The equilibrium directions for given the statements of addition, removal and temperature reactions are given below.

$\begin{array}{l}\text{a)}\text{.}2{\text{H}}_{\text{2}}\text{S(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2S(s)}+2H_{2}O(g)\text{----------Equilibrium shift to right}\\ [Additionofoxygen{O}_2]\\ \text{b)}\text{.}2{\text{H}}_{\text{2}}\text{S(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2S(s)}+2H_{2}O(g)\text{----------Equilibrium shift to left}\\ [Removalof{\text{H}}_{\text{2}}\text{S}]\\ \text{c)}\text{.}2{\text{H}}_{\text{2}}\text{S(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2S(s)}+2H_{2}O(g)----\text{Equilibrium shift to right}\\ [Removalof{\text{H}}_{\text{2}}\text{O}]\\ \text{d)}\text{.}2{\text{H}}_{\text{2}}\text{S(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2S(s)}+2H_{2}O(g)\text{----------Equilibrium unaffected}\\ [Additionof{\text{S}}_2]\end{array}$

Explanation of Solution

To Identify: Given the equilibrium reactions (a and b) addition, removal of (O₂ and H₂S) should be analyzed.

Given both reactions (a and b) equilibrium directions must be analyzed.

Before identification of equilibrium directions we writing the reaction quotient (Qc) expression fallowed.

$\begin{array}{l}2{\text{H}}_{\text{2}}\text{S(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2S(s)}+2H_{2}O(g)\\ \text{Qc=}\frac{{[H_{2}O]}^2}{{[H_{2}O]}^2[O_2]}\end{array}$

Here (S) will not appear in the expression, because sulfur is solid.

$\begin{array}{l}\text{a)}\text{.}2{\text{H}}_{\text{2}}\text{S(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2S(s)}+2H_{2}O(g)\text{----------Equilibrium shift to right}\\ [Additionofoxygen{O}_2]\\ \text{b)}\text{.}2{\text{H}}_{\text{2}}\text{S(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2S(s)}+2H_{2}O(g)\text{----------Equilibrium shift to left}\\ [Removalof{\text{H}}_{\text{2}}\text{S}]\end{array}$

Let us consider the reaction process, Le Chatelier's Principle to determine which of the following process will cause the equilibrium shift to the right or left.

The given equilibrium reaction ( ${\text{H}}_{\text{2}}\text{S}$) gas phase molecule converted into sulfur ( $\text{S}$) and water ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ here system is no longer at equilibrium, so evaluating the magnitude of the $\left(\text{Qc}\right)$ allowed to predict the direction of the resulting equilibrium shifted, the $\left(\text{Qc}\right)$ for this system as showed above.

Analyzing for reaction (a): The given reaction if we addition of sufficient amount (O₂) this process will undergoes for right side, process will be re-established in such a way that Qc is again to the equilibrium constant, highest mole amount of product (S) will form. Hence the system of equilibrium shifted to the right side to established concentration equilibrium.

Analyzing for reaction (b): The given reaction if we addition of sufficient amount hydrogen sulfide this process will undergoes for left side, process will be re-established in such a way that Qc is again to the equilibrium constant, lowest amount of product (S) will form. Hence the system of equilibrium shifted to the left side to established concentration equilibrium.

Explanation:

To Identify: Given the equilibrium reactions (c and d) addition, removal of ( ${\text{S}}_{\text{2}}{\text{and H}}_{\text{2}}\text{O}$) should be analyzed.

Given both reactions (c and d) equilibrium directions must be analyzed.

$\begin{array}{l}\text{c)}\text{.}2{\text{H}}_{\text{2}}\text{S(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2S(s)}+2H_{2}O(g)----\text{Equilibrium shift to right}\\ [Removalof{\text{H}}_{\text{2}}\text{O}]\\ \text{d)}\text{.}2{\text{H}}_{\text{2}}\text{S(g)}\text{+}{\text{O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2S(s)}+2H_{2}O(g)\text{----------Equilibrium unaffected}\\ [Additionof{\text{S}}_2]\end{array}$

Here also we will identify given the reaction process, Le Chatelier's Principle to determine which of the following process will cause the equilibrium shift to the right or left.

Reaction (c): Above the removal process of ( ${\text{H}}_{\text{2}}\text{O}$) given reactant or addition of product that appears in the expression for (Qc) will shifted to the right side.

Reaction (b): The given reaction if we addition of sufficient amount ( ${\text{S}}_{\text{2}}$) this equilibrium process will not affect any changes (or) migrated to either left or right side.

Conclusion

The equilibrium directions were identified given the hydrogen sulfide (H₂S) addition and removal reaction with respective conditions.