Problem 74 - Chemistry The Central Science In Si Units

Question

When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 5.00-L flask at \(310 \mathrm{~K}\), \(40 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ (a) Calculate \(K_{c}\) for this reaction at this temperature. (b) Calculate \(K_{P}\) for this reaction at \(310 \mathrm{~K}\). (c) According to Le Châtelier's principle, would the percent of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes increase, decrease or stay the same if the mixture was transferred to a 1.00-L vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes when 2.00 mol of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 1.00-L vessel at \(310 \mathrm{~K}\).

Step-by-Step Solution

Answer

(a) \( K_c = 0.107 \), (b) \( K_p = 2.73 \), (c) Decrease, (d) \( 21.6\% \) decomposes.

Step 1: Determine Initial Concentrations

The initial concentration of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \) in the 5.00-L flask is calculated by \( \frac{2.00 \text{ mol}}{5.00 \text{ L}} = 0.400 \text{ M} \). Initially, the concentrations of \( \mathrm{SO}_{2} \) and \( \mathrm{Cl}_{2} \) are both \( 0 \text{ M} \).

Step 2: Calculate Changes in Concentrations

Since 40% of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \) decomposes, the change in concentration for \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \) is \( 0.400 \times 0.40 = 0.160 \text{ M} \). At equilibrium, \( \mathrm{SO}_{2}\mathrm{Cl}_{2} = 0.400 - 0.160 = 0.240 \text{ M} \), \( \mathrm{SO}_{2} = \mathrm{Cl}_{2} = 0.160 \text{ M} \).

Step 3: Calculate Kc

The equilibrium constant \( K_c \) is given by the equation \( K_c = \frac{[\mathrm{SO}_{2}][\mathrm{Cl}_{2}]}{[\mathrm{SO}_{2}\mathrm{Cl}_{2}]} \). Substituting the equilibrium concentrations: \( K_c = \frac{0.160 \times 0.160}{0.240} = 0.107 \).

Step 4: Calculate Kp using the Ideal Gas Law

For the reaction, \( \Delta n = 1 + 1 - 1 = 1 \). The relationship between \( K_c \) and \( K_p \) is given by: \( K_p = K_c (RT)^{\Delta n} \). Here, \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \) and \( T = 310 \text{ K} \). Calculating gives \( K_p = 0.107 \times (0.0821 \times 310)^{1} = 2.73 \).

Step 5: Predict the Effect of Decreasing Volume

According to Le Châtelier's principle, decreasing the volume increases the pressure, favoring the side with fewer moles of gas. Since the reaction results in more moles of gas (2 moles on the product side vs. 1 mole on the reactant side), the system will shift to the left, decreasing the decomposition of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \).

Step 6: Calculate Decomposition in a 1.00-L Vessel

Initially, \( 2.00 \text{ mol} \) of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \) in a 1.00-L vessel results in a concentration of \( 2.00 \text{ M} \). Suppose \( x \text{ M} \) decomposes, then at equilibrium, \([\mathrm{SO}_{2}\mathrm{Cl}_{2}] = 2.00 - x\), \([\mathrm{SO}_{2}] = [\mathrm{Cl}_{2}] = x\). Solving \( K_c = 0.107 = \frac{x^2}{2.00-x} \) yields \( x \approx 0.432 \). So, \( 100 \times \frac{0.432}{2.00} \approx 21.6 \% \) decomposes.

Key Concepts

Le Châtelier's Principle

Le Châtelier's Principle is a fundamental concept used to predict the behavior of a chemical system when it experiences changes such as pressure, temperature, or concentration. It helps us understand the "push and pull" occurring in a reaction to maintain equilibrium. In simple terms, Le Châtelier's Principle states that if you apply a change to a system at equilibrium, the system will adjust itself to counteract that change and re-establish equilibrium.

For instance, in our reaction where \[\text{SO}_2\text{Cl}_2 (g) \rightleftharpoons \text{SO}_2 (g) + \text{Cl}_2 (g),\]when we decrease the volume of the container, the pressure increases. A reaction at equilibrium will try to reduce this pressure. Since the products side has more moles of gas (2 moles total) compared to the reactants (1 mole), the system shifts toward the reactants. This means less \(\text{SO}_2\text{Cl}_2\) decomposes when the volume is reduced, as observed in the exercise. So, Le Châtelier's Principle predicts a shift left when volume decreases.

Equilibrium Constant (Kc)

The Equilibrium Constant \( K_c \) is a value that expresses the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. It provides insight into the extent of a reaction and whether the reaction favored the reactants or products at a given temperature.

In our exercise, the decomposition of \(\text{SO}_2\text{Cl}_2,\)produces a specific equilibrium constant calculated through:\[K_c = \frac{[\text{SO}_2][\text{Cl}_2]}{[\text{SO}_2\text{Cl}_2]}\]Using the given concentrations at equilibrium: \(\text{SO}_2 = 0.160 \text{ M}, \text{Cl}_2 = 0.160 \text{ M},\) and \(\text{SO}_2\text{Cl}_2 = 0.240 \text{ M},\)we find \( K_c = 0.107.\)

This relatively small \( K_c \) value reveals that, under these conditions, the reaction does not proceed extensively to the right, indicating that less product is formed in comparison to the amount of reactants.

Reaction Quotients and Kp

The reaction quotient \( Q \) is similar to the equilibrium constant but applies to non-equilibrium conditions. It helps to predict the direction in which a reaction will proceed to reach equilibrium. However, in this exercise, we focus more on \( K_p,\)another form of the equilibrium constant applicable to gases.

Given our gaseous reaction, the relationship connecting \( K_c \) and \( K_p \) is given by:\[K_p = K_c (RT)^{\Delta n},\]where \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas.For our reaction, \(\Delta n \) equals 1 (2 moles products - 1 mole reactant), allowing us to compute \(K_p = 2.73.\)This \( K_p \) value indicates that at 310 K and equilibrium, the pressure relationship among the gases is defined by this ratio.

Gas Law Calculations

Gas Law Calculations connect our understanding of chemical reactions involving gases with fundamental principles from physics, namely the Ideal Gas Law:\[PV = nRT.\]These calculations are vital when relating pressure, volume, temperature, and moles of a gas.

In this exercise, while calculating \( K_p \) from \( K_c \), we used the Ideal Gas Law principles to adjust for changes in moles (\( \Delta n \)) and temperature through the expression:\[K_p = K_c (RT)^{\Delta n}.\]
This showcases how gas reactions vary under different conditions, such as pressure or volume changes, providing essential insights into predicting their behavior. Moreover, understanding these relationships allows chemists to determine required conditions for both industrial processes and laboratory experiments.