Question

The equilibrium constant for the reaction: \(\mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) is \(K_{c}=4.9 \times 10^{-2}\). the value of \(K\) for the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) will be (a) 416 (b) \(2.40 \times 10^{-3}\) (c) \(9.8 \times 10^{-2}\) (d) \(4.9 \times 10^{-2}\)

Short answer

The value of \( K \) for the target reaction is 416, so the answer is (a) 416.

Step-by-step solution

Understand the Given Reaction and Equilibrium Constant

The given reaction is:\[ \mathrm{SO}_3(\mathrm{g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g}) \]and its equilibrium constant is \( K_c = 4.9 \times 10^{-2} \). This tells us how the concentration of reactants and products are related at equilibrium under standard conditions.

Identify the Target Reaction and Its Relationship

The target reaction is:\[ 2 \mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{g}) \]Notice that this is the reverse of the given reaction with doubled coefficients.

Relate the Equilibrium Constants

For the reverse of a reaction, the equilibrium constant is the reciprocal of the original. Moreover, when the coefficients of a reaction are multiplied by a factor, the equilibrium constant for that reaction is raised to the power of the same factor. For the reaction \( \mathrm{SO}_2(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightleftharpoons \mathrm{SO}_3(\mathrm{g}) \), its equilibrium constant would be:\[ \frac{1}{K_c} = \frac{1}{4.9 \times 10^{-2}} \]Since \( 2\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{SO}_3(\mathrm{g}) \) is twice the original reverse reaction, the equilibrium constant should be squared:\[ K = \left( \frac{1}{4.9 \times 10^{-2}} \right)^2 \]

Calculate the Equilibrium Constant for the Target Reaction

First, calculate the reciprocal:\[ \frac{1}{4.9 \times 10^{-2}} = \frac{1}{0.049} \approx 20.41 \]Then square it to find the equilibrium constant for the target reaction:\[ K = (20.41)^2 \approx 416 \]

Choose the Correct Option

Since \( K \approx 416 \), the correct option is (a) 416.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a crucial concept in chemistry that describes a state where a reversible reaction reaches a point at which the rates of the forward and backward reactions are equal. At this stage, the concentrations of the reactants and products remain constant over time, though not necessarily equal. This steady state indicates that the system is in balance, but it doesn't mean that the reactions have stopped occurring; instead, they continue with the same frequency in both directions.

An understanding of chemical equilibrium is essential when studying reactions as it helps predict how a change in conditions, such as concentration, temperature, or pressure, can shift the equilibrium position. This is often referred to as Le Chatelier's principle, which states that a system will adjust to counteract any external changes imposed upon it.

• Key to achieving chemical equilibrium is the **equal rate** of the forward and reverse reactions.
• Only reversible reactions can achieve equilibrium.
• It is a **dynamic state** because reactions continue to happen.

Reversible Reactions

Reversible reactions are those chemical processes that can proceed in both the forward and reverse directions. Unlike irreversible reactions that proceed to completion, reversible reactions can attain a state of equilibrium where both forward and backward reactions occur at equal rates. For example, in the decomposition of sulfur trioxide ( \(\mathrm{SO}_3\mathrm{(g)} \rightleftharpoons \mathrm{SO}_2\mathrm{(g)} + \frac{1}{2} \mathrm{O}_2\mathrm{(g)} \)), both sulfur dioxide and oxygen can react to form sulfur trioxide, demonstrating the reversible nature.In reversible reactions, factors like temperature and pressure significantly impact the direction and extent of the reaction. By altering these conditions, it is possible to "push" the reaction in one direction, either favoring the formation of products or reactants, and thereby shift the equilibrium.**Important points about reversible reactions:**

• They can potentially lead to a state of equilibrium.
• Temperature and pressure affect their direction and extent.
• They are sensitive to changes in reaction conditions.

Equilibrium Expressions

The equilibrium expression is a mathematical representation that enables chemists to relate the concentrations of products and reactants at equilibrium. For a general reversible reaction, \[aA + bB \rightleftharpoons cC + dD\]the equilibrium constant expression is given by:\[K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]where brackets denote the concentration of each species, and the coefficients in the balanced equation become the exponents in the expression.The equilibrium constant, **\(K_c\)**, quantifies the ratio of product concentrations to reactant concentrations when the reaction reaches equilibrium under specific conditions. A higher \(K_c\) value suggests that products are favored, while a lower value indicates that reactants are more prevalent at equilibrium.In the context of different reactions, finding the equilibrium constant helps in understanding how the reaction will behave under changing conditions. The original exercise highlights how equilibrium constants \(K\) can be manipulated based on changes in reaction direction and coefficients. **Key takeaways about equilibrium expressions:**

• Concentrations in the expression come from the balanced chemical equation.
• \(K_c\) values help predict equilibrium behavior.
• They illustrate the relationship between products and reactants at equilibrium.