Problem 72 - Chemistry: The Central Science

Question

When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 2.00- \(\mathrm{L}\) flask at \(303 \mathrm{~K}\), \(56 \%\) 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 \(303 \mathrm{~K}\). (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 were transferred to a \(15.00\)-L vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes when \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(15.00\) - \(\mathrm{L}\) vessel at \(303 \mathrm{~K}\).

Step-by-Step Solution

Answer

In summary, the equilibrium constant Kc for this reaction is 0.71, and Kp is 1.77 at 303 K. According to Le Châtelier's principle, the percentage of SO₂Cl₂ decomposition will increase when the volume of the reaction mixture is increased. When 2.00 mol of SO₂Cl₂ is placed in a 15.00 L vessel at 303 K, the percentage of decomposition is 85.7%.

Step 1: Determine initial and equilibrium concentrations of reactants and products

We have 2.00 mol of SO₂Cl₂ in a 2.00 L flask at 303 K. Initially, there are no products. When the reaction reaches equilibrium, 56% of the SO₂Cl₂ decomposes to SO₂ and Cl₂. We need to determine the concentrations of all species at equilibrium. Initial concentrations: [SO₂Cl₂] = 2.00 mol / 2.00 L = 1.00 M [SO₂] = 0 M [Cl₂] = 0 M At equilibrium: [SO₂Cl₂] = (1 - 0.56) x 1.00 M = 0.44 M [SO₂] = [Cl₂] = 0.56 x 1.00 M = 0.56 M

Step 2: Calculate Kc

Now we have the equilibrium concentrations, we can calculate Kc for the reaction: Kc = [SO₂][Cl₂] / [SO₂Cl₂] Using the determined equilibrium concentrations: Kc = (0.56 M)(0.56 M) / (0.44 M) = 0.71

Step 3: Calculate Kp

To calculate Kp, we first need to find the relationship between Kc and Kp: Kp = Kc(RT)^(Δn) Where R is the gas constant (0.0821 L atm/mol K), T is the temperature (303 K), and Δn is the change in moles of gas (Δn=+1). Kp = 0.71 * (0.0821 L atm/mol K x 303 K)^(+1) = 1.77

Step 4: Le Châtelier's principle and volume change

According to Le Châtelier's principle, if the volume of the reaction mixture is increased, the reaction will shift in the direction that increases the number of moles of gas. In this case, it will shift towards the products (SO₂ and Cl₂) because there is an increase in the moles of gas in the forward direction (Δn=+1). So, the percentage of SO₂Cl₂ decomposition will increase.

Step 5: Calculate the percentage of decomposition in the 15.00-L vessel

Now we have Kc, we can set up a reaction table for the changed volume: Initial concentrations: [SO₂Cl₂] = 2.00 mol / 15.00 L = 0.133 M [SO₂] = 0 M [Cl₂] = 0 M Let x be the amount of SO₂Cl₂ that decomposes in the 15.00-L vessel: Equilibrium concentrations: [SO₂Cl₂] = 0.133 M - x M [SO₂] = x M [Cl₂] = x M Using Kc to solve for x: 0.71 = (x)(x) / (0.133 - x) Solve the quadratic equation for x: x = 0.114 M The percentage of SO₂Cl₂ that decomposes when 2.00 mol of SO₂Cl₂ is placed in a 15.00 L vessel at 303 K is: Percent = (0.114 M / 0.133 M) * 100% = 85.7%

Key Concepts

Le Châtelier's Principle

Le Châtelier's principle helps us predict how a change in conditions affects chemical equilibrium. It's like a chemical reaction's response to restore balance when disturbed. If we change the concentration, pressure, or temperature, the system will adjust to counteract that change.

For example, if we increase the volume of our reaction vessel, we reduce the pressure inside it. According to Le Châtelier's principle, our reaction will shift towards the side with more gas molecules to increase pressure. In the given reaction, decomposing \( \text{SO}_{2}\text{Cl}_{2}\) increases the number of gas molecules (from 1 to 2). Therefore, moving the reaction to a larger volume will increase the decomposition percentage.

Equilibrium Constant

The equilibrium constant, \( K_c \), quantifies the concentrations of reactants and products at equilibrium in a chemical reaction. For the reaction \( \text{SO}_{2}\text{Cl}_{2}(g) \rightleftharpoons \text{SO}_{2}(g) + \text{Cl}_{2}(g) \), \( K_c \) is calculated as follows: \[ K_c = \frac{[\text{SO}_2][\text{Cl}_2]}{[\text{SO}_2\text{Cl}_2]} \] Using the equilibrium concentrations, you can find \( K_c = \frac{(0.56)(0.56)}{0.44} = 0.71 \).

\( K_c \) remains constant at a given temperature, no matter the starting amounts of reactants and products. It allows us to predict concentrations at equilibrium for different initial conditions or vessel sizes.

Reaction Kinetics

Reaction kinetics involves studying the speed of a chemical reaction and the factors that affect it. In our equilibrium situation, kinetics can determine how quickly \( \text{SO}_2\text{Cl}_2 \) decomposes. However, kinetics is separate from equilibrium, as equilibrium refers to the constant concentration ratio at the end, whereas kinetics focuses on the path taken to reach that state.

Factors like temperature and concentration can speed up or slow down reactions, but they do not change the \( K_c \) of the reaction. A higher concentration of \( \text{SO}_2\text{Cl}_2 \) inside the flask might speed up the initial reaction rate but will still converge to the same equilibrium state, where \( K_c \) defines the concentrations.

Gas Laws

Gas laws describe the behavior of gases and play a crucial role in understanding reactions involving gases. They relate variables like volume, temperature, and pressure.

In our context, when the volume increases, the pressure decreases. According to the ideal gas law \( PV = nRT \), a decrease in pressure at constant temperature means an increase in volume. This impacts equilibrium as explained by Le Châtelier's principle, where the reaction shifts to produce more gas molecules at lower pressures.

The interconnection of these principles helps explain why changing the volume to 15.00 L increased the decomposition percentage to 85.7%. Larger volume allows for greater dissociation to achieve equilibrium, as more gas particles disperse throughout the increased space.