Problem 74 - Chemistry: The Central Science in SI Units, Global Edition

Question

When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 5.00 -Lflaskat \(310 \mathrm{~K}\), \(40 \%\) 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 \(1.00-\mathrm{L}\) vessel at \(310 \mathrm{~K}\).

Step-by-Step Solution

Answer

In summary, for the given reaction, the equilibrium constants are Kc = 0.107 and Kp = 274.699 at 310 K. According to Le Châtelier's principle, the percentage of SO2Cl2 decomposition would decrease when transferred to a smaller vessel. The actual percentage of decomposition in a 1.00 L vessel at 310 K is around 14.45%.

Step 1: 1. Write the balanced reaction equation

The balanced reaction equation is already given in the prompt: \(\mathrm{SO}_{2}\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) \)

Step 2: 2. Calculate initial and equilibrium concentrations

First, let us find the initial concentration of SO2Cl2 in the 5.0 L flask: Initial concentration of SO2Cl2 = \(\frac{2.00\, \text{mol}}{5.00\, \text{L}} = 0.40\, \text{M}\) Then, we find the change in concentrations at equilibrium by utilizing the percentage of decomposition provided. Since \(40 \%\) of the SO2Cl2 decomposes, the change in concentration would be 0.4 times the initial concentration. Concentration change of SO2Cl2 = \(0.40 \times 0.40 = 0.16\, \text{M}\) Therefore, at equilibrium: [SO2Cl2] = \(0.40\, \text{M} - 0.16\, \text{M} = 0.24\, \text{M}\) [SO2] = [Cl2] = \(0.16\, \text{M}\)

Step 3: 3. Calculate the equilibrium constant Kc and Kp

Now, calculate Kc with the equilibrium concentrations: \(K_c = \frac{\text{[SO2]}\times\text{[Cl2]}}{\text{[SO2Cl2]}} = \frac{0.16\, \text{M} \times 0.16\, \text{M}}{0.24\, \text{M}} = 0.107\) To find conversion factor to calculate Kp, we need the given temperature, which is 310 K. \(K_p = K_c(RT)^{\Delta n}\) where R is the ideal gas constant (\(8.314\, \frac{\text{J}}{\text{mol K}}\)). In this case, the reaction involves 1 mole of reactant and 2 moles of products, so ∆n = 2 - 1 = 1. Now calculate Kp: \(K_p = 0.107 (8.314\, \text{J/mol K} \times 310\, \text{K})^1 = 274.699\) Thus, Kc=0.107 and Kp=274.699.

Step 4: 4. Applying Le Châtelier's Principle

Le Châtelier's Principle states that if a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the reactants or products, the system will adjust to counteract the change and restore equilibrium. In this case, we are decreasing the volume of the flask, causing an increase in pressure. Since the forward reaction produces more moles of gas (2 moles) than the reverse reaction (1 mole), the system will tend to move in the direction of the reverse reaction (consuming the products and producing more reactants) to counteract the increased pressure. Therefore, the percentage of SO2Cl2 decomposition would decrease.

Step 5: 5. Calculating percentage decomposition in a 1.00 L vessel

Initial concentration of SO2Cl2 in a 1 L flask: \(\frac{2.00\, \text{mol}}{1.00\, \text{L}} = 2.00\, \text{M}\) Let x represent the change in concentration of SO2Cl2 due to decomposition: Equilibrium concentrations: [SO2Cl2] = \(2.00 \, \text{M} - x \, \text{M}\) [SO2] = [Cl2] = \(x\, \text{M}\) Now, we will use the calculated Kc value (0.107) to determine x: \(0.107 = \frac{x^2}{(2.00 - x)}\) Solving for x (using a quadratic equation solver), we get two possible values: ~0.803 M (rounded) or ~0.289 M. Since 0.803 M decomposition is not feasible and x should be lower than 1, we choose x = 0.289 M. The percentage of decomposition in a 1.00 L vessel can be calculated with the formula: \(\text{Percentage of decomposition} = \frac{\text{Decomposed SO2Cl2 concentration}}{\text{Initial SO2Cl2 concentration}} \times 100\%\) Inserting the respective values, we get: \(\text{Percentage of decomposition} = \frac{0.289\, \text{M}}{2.00\, \text{M}} \times 100\% = 14.45\% \) (rounded to 2 decimals)

Key Concepts

Le Châtelier's Principle

When a system at equilibrium is disturbed, it adjusts to lessen the disturbance and restore a new equilibrium. This idea, known as **Le Châtelier's Principle**, is a fundamental aspect of chemical equilibrium. It's like a balance trying to stabilize after being pushed.
In the reaction given, \[\mathrm{SO}_{2}\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\]changing the volume of the container affects the pressure inside. In this case, reducing the volume increases the pressure. The system responds by favoring the side of the equation with fewer moles of gas to decrease the pressure.
In this reaction, the reactants side has one mole, while the products side has 2 moles. So, increasing pressure shifts the balance toward the reactants, reducing the decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). This results in a decreased percentage of decomposition, as confirmed in the solution.

Equilibrium Constant (Kc and Kp)

Equilibrium constants are crucial for predicting how a reaction behaves at equilibrium. **Kc** is used when dealing with concentrations in a solution, while **Kp** applies to gases and is calculated using pressures.
In the reaction provided, \[\mathrm{SO}_{2}\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\]the equilibrium constant \(K_c\) is calculated by dividing the product of the concentrations of the products by the concentration of the reactants at equilibrium:

• \(K_c = \frac{[\mathrm{SO}_2][\mathrm{Cl}_2]}{[\mathrm{SO}_2\mathrm{Cl}_2]}\)

Substituting the numbers gives \(K_c = 0.107\). To convert \(K_c\) to \(K_p\), we use the formula:

• \(K_p = K_c(RT)^{\Delta n}\)

where \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas. Here \(\Delta n = 1\), leading to \(K_p = 274.699\). This calculation helps understand how changes in pressure or volume can affect equilibrium.

Reaction Stoichiometry

Stoichiometry involves using the ratio of reactants to products — provided by the balanced chemical equation — to calculate the amounts needed or produced in a chemical reaction. In our problem, knowing the stoichiometry of the reaction \(\mathrm{SO}_{2}\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\) allowed us to find the concentrations at equilibrium.
Initially, we calculate the concentration of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\) as \(0.40\,M\) from 2.00 mol in a 5.00 L flask. As \(40\%\) decomposes, it changes by \(0.16\,M\), leaving \(0.24\,M\) of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\) and forming \(0.16\,M\) each of \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\).
Reaction stoichiometry ensures that the changes in concentration corresponding to the decomposition and formation of new products is consistent with the coefficients in the balanced equation. This understanding allows us to write expressions needed to solve for equilibrium constants and predict changes in reaction conditions.