Problem 23 - General Chemistry: Principles and Modern Applications

Question

Equilibrium is established at \(1000 \mathrm{K},\) where \(K_{\mathrm{c}}=281\) for the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) .\) The equilibrium amount of \(\mathrm{O}_{2}(\mathrm{g})\) in a \(0.185 \mathrm{L}\) flask is 0.00247 mol. What is the ratio of \(\left[\mathrm{SO}_{2}\right]\) to \(\left[\mathrm{SO}_{3}\right]\) in this equilibrium mixture?

Step-by-Step Solution

Answer

The ratio of the concentrations of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) in the equilibrium mixture is \(\frac{1}{\sqrt{281 * 0.0134}}\). Simply put, if you want the numerical value for the ratio, just evaluate this expression.

Step 1: Calculate the concentration of \(\mathrm{O}_{2}\)

Given that the number of moles of \(\mathrm{O}_{2}\) is 0.00247 mol and the volume is 0.185 L, the concentration of \(\mathrm{O}_{2}\) can be calculated using the formula \(c = n/V\), where \(n\) is the number of moles and \(V\) is the volume. Thus, \([O_2] = \frac{0.00247}{0.185} = 0.0134 \, \text{M}\).

Step 2: Substitute the values into the \(K_c\) expression

The \(K_c\) for the reaction is given as 281. The general form of the equilibrium expression for the reaction is \(K_c = \frac{[\mathrm{SO}_{3}]^2}{[\mathrm{SO}_{2}]^2[\mathrm{O}_{2}]}\). Substitute \(K_c = 281\) and \([O_2] = 0.0134\) into the equation. We get \(281 = \frac{[\mathrm{SO}_{3}]^2}{[\mathrm{SO}_{2}]^2 * 0.0134}\). The target of this exercise is to find the ratio \(\frac{[\mathrm{SO}_{2}]}{[\mathrm{SO}_{3}]}\), so it's better to rewrite this expression as \(\frac{[\mathrm{SO}_{3}]}{[\mathrm{SO}_{2}]} = \sqrt{\frac{281 * 0.0134}{[\mathrm{SO}_{2}]^2}\). From this expression, it is evident that the ratio requested in the exercise is the reciprocal of the square root of \(\frac{281 * 0.0134}{[\mathrm{SO}_{2}]^2}\).

Step 3: Simplify the expression for the ratio

Take the reciprocal of the square root from the previous step, which gives \(\frac{[\mathrm{SO}_{2}]}{[\mathrm{SO}_{3}]} = \frac{1}{\sqrt{281 * 0.0134}}\). This will be the final expression for the ratio. Algebraically, this cannot be further simplified without additional information.

Key Concepts

Equilibrium Constant (Kc)

The concept of the equilibrium constant, represented as \(K_c\), is central to understanding chemical equilibrium in a closed system. For a given reaction at a specific temperature, the \(K_c\) value provides insight into the ratio of concentrations of products to reactants when the system is at equilibrium.

In our example, the reaction \(2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)\) has a known \(K_c\) of 281 at 1000 K. The equation for calculating \(K_c\) is given by the formula:

• \(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}\)

This expression mathematically relates the concentrations of sulfur trioxide (\(\text{SO}_3\)), sulfur dioxide (\(\text{SO}_2\)), and oxygen (\(\text{O}_2\)) at equilibrium.

The value of \(K_c\) provides an indication of the extent of a reaction. A large \(K_c\) means the equilibrium favors the formation of products, while a small \(K_c\) indicates that the reactants are favored. Here, a \(K_c\) of 281 indicates a significant formation of \(\text{SO}_3\) from \(\text{SO}_2\) and \(\text{O}_2\).

Concentration Calculations

Concentration plays a key role in chemical reactions, as it affects the rate and equilibrium position of the reaction. Concentration is typically measured in molarity (M), which denotes moles of solute per liter of solution.

To find the concentration of a substance, use the formula:

• \([X] = \frac{n}{V}\)

where \(n\) is the number of moles and \(V\) is the volume of the solution in liters. In the provided exercise, we calculated the concentration of \(\text{O}_2\) in a 0.185 L container as follows:

Given 0.00247 mol of \(\text{O}_2\), the concentration becomes:

• \([\text{O}_2] = \frac{0.00247}{0.185} = 0.0134 \text{ M}\)

This calculation allows us to substitute the concentration value into the equilibrium constant expression and solve for the desired ratio.

Understanding how to compute concentrations enables you to determine the state of equilibrium in a chemical reaction and analyze how changes in concentration can shift the equilibrium.

Reaction Quotient

The Reaction Quotient, denoted as \(Q\), is a measure used to determine the progress of a reaction at any point before reaching equilibrium. It utilizes the same expression as the equilibrium constant (\(K_c\)) but can be evaluated with non-equilibrium concentrations.

For the reaction \(2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)\), the reaction quotient \(Q\) is expressed as:

• \(Q = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}\)

Comparing \(Q\) with \(K_c\) helps predict the direction in which the reaction will shift to achieve equilibrium:

• If \(Q < K_c\), the reaction will proceed forward, increasing the concentration of products.
• If \(Q > K_c\), the reaction will shift backward, increasing the concentration of reactants.
• If \(Q = K_c\), the system is already at equilibrium.

While in our specific scenario, \(Q\) is not directly calculated, understanding \(Q\) provides valuable insight into the dynamics of chemical reactions and their tendency to adjust concentration toward equilibrium.