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 \(\mathrm{K}_{\mathrm{c}}=4.9 \times 10^{-2}\). The value of \(\mathrm{K}_{c}\) 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 \( K_c \) for the reaction is (a) 416.

Step-by-step solution

Understand the Given Reaction

The given reaction is \( \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \) with \( K_c = 4.9 \times 10^{-2} \). This means that at equilibrium, the concentrations are such that \( [\mathrm{SO}_2][\mathrm{O}_2]^{1/2} / [\mathrm{SO}_3] = 4.9 \times 10^{-2} \).

Write the Reaction for the Desired Equilibrium

We need to find \( K_c \) for the reaction \( 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) \).

Determine the Relationship between the Two Reactions

The desired reaction is the reverse of twice the initial reaction. To convert the initial reaction \( \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \) into \( 2 \mathrm{SO}_{2} \mathrm{~g} + \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) \), reverse the initial equation and multiply by 2.

Calculate the New Equilibrium Constant

Reversing a reaction inverts the equilibrium constant and multiplying all coefficients in a reaction by a factor \( n \) raises \( K_c \) to the power of \( n \). Thus, if we reverse and multiply the coefficients by 2, we get \( K_c' = (1/K_c)^2 = (1/(4.9 \times 10^{-2}))^2 \).

Simplify and Solve for \( K_c' \)

Calculate \( K_c' = (1/(4.9 \times 10^{-2}))^2 = (20.41)^2 \approx 416 \).

Choose the Correct Answer

The calculated \( K_c' = 416 \) matches option (a). Therefore, the answer is (a) 416.

Key concepts

Reversible Reactions

In a reversible reaction, the reactants are converted into products, but the products can also revert back to the original reactants. Such reactions are represented with a double arrow indicating the process can proceed in both directions: forward and backward. A good example of reversible reactions occurs when decomposing and reforming the molecule sulfur trioxide (\(\text{SO}_3\)) in a reaction with sulfur dioxide (\(\text{SO}_2\)) and oxygen (\(\text{O}_2\)).

These reactions will reach a state where the concentration of reactants and products remain constant over time. This state is known as chemical equilibrium. During this state, the rate at which the reactants are converted to products equals the rate at which products revert to reactants. However, it's important to note that the concentrations of reactants and products are not necessarily equal, but remain constant at equilibrium.

Reversible reactions allow chemists to understand processes such as synthesis and decomposition in various chemical and industrial applications, ensuring that systems can reach optimal performance under specified conditions.

Calculating Kc

The equilibrium constant, \( K_c \), is a dimensionless number that provides insight into the composition of a reaction at equilibrium. It essentially tells you how far a reaction will proceed in reaching equilibrium. This value is specific to temperature, so any change in temperature will lead to a new \( K_c \).

To calculate \( K_c \), consider the balanced chemical equation and use the formula:
\[K_c = \frac{[\text{products}]}{[\text{reactants}]}.\]Each concentration is raised to the power of its coefficient in the balanced equation. This reflects the extent of the reaction and the balance between reactants and products at equilibrium.

Let’s illustrate with an example: for the reaction \(2 \text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2 \text{SO}_3(\text{g})\), the \( K_c \) can be calculated using:
\[K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}.\]By knowing the initial \( K_c \) for a related reaction and understanding that altering the reaction (reversing or multiplying it) affects the \( K_c \), one can calculate a new \( K_c \). Multiplying coefficients by a factor means raising \( K_c \) to that power, and reversing the direction of the reaction means taking the reciprocal of \( K_c \). This key concept helps solve many equilibrium problems you might encounter.

Chemical Equilibrium Concepts

Chemical equilibrium is a pivotal concept in understanding how chemical reactions behave under closed conditions. At equilibrium, the forward and reverse reactions occur at the same rate, and the concentrations of reactants and products do not change with time.

Understanding equilibrium involves grasping a few key ideas:

• **Dynamic Process:** Even though it appears that everything has stopped when equilibrium is reached, the reactions at the molecular level continue to happen.
• **Equilibrium Position:** This term refers to the relative concentrations of reactants and products at equilibrium. The exact position depends on the inherent properties of the reaction described by \( K_c \).
• **Le Chatelier’s Principle:** When a system at equilibrium is disturbed, it will respond in a way to counteract the disturbance and restore a new equilibrium.

For example, adding more \( \text{SO}_2 \) to the equilibrium mixture of \(2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)\) will push the reaction towards forming more \( \text{SO}_3 \), in accordance with Le Chatelier’s Principle.

Grasping these concepts is crucial for understanding not just specific reactions but also broader chemical processes across various fields.