Question

\(\mathrm{SO}_{2}\) (sulfur dioxide) can be used as a preservative for meats, wine, and beer; as a bleaching agent for foods; in pulp and paper industry; and in tanning. Express the equilibrium constant, \(\mathrm{k}_{2}\), in terms of the equilibrium constant \(\mathrm{k}_{1}\), in the two gaseous equilibria involving \(\mathrm{SO}_{2}\). Namely \(\mathrm{SO}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightleftarrows \mathrm{SO}_{3}(\mathrm{~g}) \quad \mathrm{k}_{1}\) \(2 \mathrm{SO}_{3} \rightleftarrows 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \quad \mathrm{k}_{2}\)

Short answer

The relationship between the equilibrium constants \(k_1\) and \(k_2\) for the given gaseous equilibria involving \(\mathrm{SO}_2\) is: \[k_2 = \frac{1}{k_1^2}\]

Step-by-step solution

Write down the given reactions and equilibrium constants

We have the following reactions and their equilibrium constants: \[\mathrm{SO}_2(\mathrm{g}) + \frac{1}{2}\mathrm{O}_2(\mathrm{g}) \rightleftarrows \mathrm{SO}_3(\mathrm{g}) \qquad k_1\] \[\mathrm{2SO}_3(\mathrm{g}) \rightleftarrows 2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \qquad k_2\]

Adjust the first reaction to match the stoichiometry of the second reaction

To match the stoichiometry of the second reaction, we need an \(\mathrm{O}_2\) molecule on the reactant side, so we can multiply the first reaction by 2: \[2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftarrows 2\mathrm{SO}_3(\mathrm{g}) \]

Find the new equilibrium constant for the adjusted first reaction

When we multiply a balanced chemical equation, we raise its equilibrium constant to the power of the multiplication factor. In this case, we multiplied the first reaction by 2, so the new equilibrium constant is \(k_1^2\): \[2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftarrows 2\mathrm{SO}_3(\mathrm{g}) \quad k_1^2\]

Compare the adjusted first reaction and the second reaction

The adjusted first reaction is the exact inverse of the second reaction: \[2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftarrows 2\mathrm{SO}_3(\mathrm{g}) \qquad k_1^2\] \[2\mathrm{SO}_3(\mathrm{g}) \rightleftarrows 2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \qquad k_2\] This means the second reaction is the reverse reaction of the adjusted first reaction.

Find the relationship between \(k_1\) and \(k_2\) for the reverse reaction

The equilibrium constant of a reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. So, we have: \[k_2 = \frac{1}{k_1^2}\] This is the relationship between \(k_1\) and \(k_2\) that was asked for.

Key concepts

Sulfur Dioxide Equilibrium

Sulfur dioxide (\(\mathrm{SO}_2\)) is not just a simple molecule floating around; it plays a dynamic role in chemical equilibrium. In an equilibrium state, reactants and products coexist without any net change over time. Sulfur dioxide can react with oxygen to form sulfur trioxide (\(\mathrm{SO}_3\)), which is represented by the equation:

• \( \mathrm{SO}_2(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightleftarrows \mathrm{SO}_3(\mathrm{g}) \)

In this reversible reaction, sulfur dioxide on one side can convert into sulfur trioxide, while some sulfur trioxide can also go back to sulfur dioxide. The equilibrium constant (\(k\)), a crucial part of this system, indicates how favored the formation of products is compared to the reactants.

Understanding this balance is essential, especially because different conditions can shift the equilibrium, changing concentrations of the involved substances drastically.

Chemical Reactions

Chemical reactions like the ones involving sulfur dioxide are all about breaking and making bonds. In the given example, sulfur dioxide reacts with oxygen, showcasing a redox reaction where oxidation and reduction take place. Sulfur is oxidized as it gains oxygen to form sulfur trioxide. The interesting part here is that these reactions don't just go one way. They are reversible!

• The reactions have a forward direction (\(\mathrm{SO}_2 + \mathrm{O}_2 \to \mathrm{SO}_3\)) and a backward direction (\(\mathrm{SO}_3 \to \mathrm{SO}_2 + \mathrm{O}_2\)).
• This dynamic nature allows equilibria to be established, meaning the reaction reaches a steady state where the forward and reverse reaction rates are equal.

Such behavior is what makes chemical reactions both fascinating and useful in various industrial applications like those using sulfur dioxide in preservation or manufacturing processes.

Chemical Equilibrium Calculations

Calculating chemical equilibrium involves understanding the balance of concentrations. We use the equilibrium constant to express this balance numerically. In our system with sulfur dioxide, you must adjust for changes in stoichiometry.

Here’s a key point: If you change the equation by multiplying it, you must adjust the equilibrium constant by raising it to the power of the multiplication factor. For example, multiplying the reaction:

• \(2 \mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_3(\mathrm{g})\)
• results in adjusted equilibrium constant: \(k_1^2\)

The value of \(k_2\) in the reverse reaction becomes the reciprocal of this adjusted constant. Therefore, understanding both the mathematical and conceptual aspects of these calculations helps to predict and manipulate the outcomes of chemical processes accurately. Knowing the relationship between equilibrium constants helps in efficiently managing reactions in industrial settings.