Q22.76CP - Chemistry: Molecular Nature Of Matter And Change

Question

In the 1980 s, U.S. Fish and Wildlife Service researchers found high mortality among newbem coots and ducks in parts of California. U.S. Geological Survey scientists later found the cause to be a high selenium concentration agricultural drainage water. 

Balance these half-reactions in the selenium cycle:

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}{\mathbf{\text{Se}}}^{\mathbf{2}\mathbf{-}}\left(aq\right)\mathbf{\to }\mathbf{\text{Se}}\mathbf{\left(}\mathit{s}\mathbf{\right)} \\ \mathbf{\left(}\mathit{b}\mathbf{\right)}{\mathbf{\text{SeO}}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }{\mathbf{\text{SeO}}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)} \\ \mathbf{\left(}\mathit{c}\mathbf{\right)}{\mathbf{\text{SeO}}}_{\mathbf{4}}^{\mathbf{2}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }\mathbf{\text{Se}}\mathbf{\left(}\mathbf{\text{s}}\mathbf{\right)} \end{gathered}$$

Step-by-Step Solution

Answer

a) Increasing the pressure while lowering the temperature optimizes the production of  ${\text{SO}}_3$

b) The concentration of ${\text{SO}}_2$ rises, the reaction quotient drops.

c) The reaction rates are slower at lower temperatures, catalysis is utilized to accelerate the rate of the reaction.

Step 1: Concept introduction

Selenium is becoming more well recognised as a necessary component of life and medicine. Its biology is similar to that of sulphur, although it differs from it in terms of redox potentials and oxidation state stability. Selenium may replace the more common sulphur of cysteine, and as a result, it plays a key role in more than a dozen selenoproteins.

Step 2:Explanation of Reaction

a)  

Given reaction:

 $$\begin{gathered} {\text{SO}}_2\left(g\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\rightleftharpoons {\text{SO}}_3\left(g\right) \\ ∆H_{rxn}^{°}=-99.2 \text{kJ} \end{gathered}$$

The moles of gas are  $1.5 mol$on the reactant side. 

The moles of gas are $1mo$ lon the product side.

When the container's pressure is increased, the equilibrium position slips toward the product side.

This is an exothermic reaction since the value of $∆H_{rxn}^{°}$ is negative.

Because the reverse reaction is an endothermic reaction, lowering the temperature favours the product. 

As a result, increasing the pressure while lowering the temperature optimizes the production of${\text{SO}}_3$ .

Step 3: Concentration of reaction

(b)

A reactant is  ${\text{O}}_2$.When we add${\text{O}}_2$ ,the concentration of  ${\text{O}}_2$rises, as does the creation of${\text{SO}}_3$  when ${\text{SO}}_2$ reacts with${\text{SO}}_2$  .

As the result, the concentration of ${\text{SO}}_2$  rises, the reaction quotient drops.

Step 4: Forward reaction

(c) 

When the reactant concentration rises, so does the rate of the forward reaction, and the product concentration rises as well, until the ratio of product to reactant concentrations equals the equilibrium constant.

At equilibrium, the rate of forward reaction equals the rate of backward reaction, hence the equilibrium constant remains constant 

This is an exothermic process (forward reaction).

An endothermic reaction is the inverse reaction.

 Hence, the reaction rates are slower at lower temperatures, catalysis is utilized to accelerate the rate of the reaction.