Question

When 2.00 \(\mathrm{mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 2.00 -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 Chatelier'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 \(15.00-\mathrm{L}\) . 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 \(15.00-\mathrm{L}\) vessel at 303 \(\mathrm{K}\) .

Short answer

In summary, at 303 K, the equilibrium constant Kc for the decomposition reaction of SO₂Cl₂ is 0.71, and Kp is 1791. When the mixture is transferred to a 15.00-L vessel, the percentage of SO₂Cl₂ decomposition increases to approximately 64.7%, according to Le Chatelier's principle.

Step-by-step solution

Determine initial, change, and equilibrium concentrations

Initially, 2.00 moles of SO₂Cl₂ are placed in a 2.00-L flask. To determine the concentration, divide the moles of SO₂Cl₂ by the volume of the flask: Initial concentration of SO₂Cl₂ = \(\frac{2.00 \,\text{moles}}{2.00\, \text{L}}\) = 1.00 M Since the reaction decomposes 56% of the SO₂Cl₂, we can find the change in concentrations: Change in SO₂Cl₂ = -0.56 × Initial concentration of SO₂Cl₂ = -0.56 × 1.00 M = -0.56 M According to the stoichiometry of the reaction, the change in SO₂ and Cl₂ concentrations will be equal to the change in SO₂Cl₂: Change in SO₂ = Change in Cl₂ = +0.56 M Now, we can determine the equilibrium concentrations: Equilibrium [SO₂Cl₂] = Initial [SO₂Cl₂] + Change in [SO₂Cl₂] = 1.00 M - 0.56 M = 0.44 M Equilibrium [SO₂] = Initial [SO₂] + Change in [SO₂] = 0 M + 0.56 M = 0.56 M Equilibrium [Cl₂] = Initial [Cl₂] + Change in [Cl₂] = 0 M + 0.56 M = 0.56 M

Calculate Kc

Using the equilibrium concentrations, we can calculate the equilibrium constant Kc: \(K_c = \frac{[\text{SO}_2][\text{Cl}_2]}{[\text{SO}_{2}\text{Cl}_{2}]}\) \(K_c = \frac{(0.56)(0.56)}{0.44} = 0.71\)

Convert Kc into Kp

Using the ideal gas law, we can convert Kc into Kp: \(K_p = K_c \times (RT)^{\Delta n}\) The change in the number of moles (\(\Delta n\)) for this reaction is: \(\Delta n = \text{moles of products} - \text{moles of reactants} = (1+1) - 1 = 1\) Now we can plug in the values to find Kp: \(K_p = 0.71 × (8.314 \frac{\text{J}}{\text{mol K}} \times 303 \,\text{K})^1 = 1791\)

Le Chatelier's Principle

According to Le Chatelier's principle, when the volume of the vessel increases to 15.00 L, the equilibrium will shift to counteract the change. In this case, since the number of moles of products is greater than reactants, the equilibrium will shift towards the products, decomposing more SO₂Cl₂.

Determine new percentage of decomposition

To find the new percentage of decomposition, we will use the same expression for Kc, this time using a generic value x to represent the equilibrium concentrations: Initial concentration of SO₂Cl₂ = \(\frac{2.00 \,\text{moles}}{15.00 \,\text{L}}\) = 0.133 M Equilibrium concentrations in new vessel: [SO₂Cl₂] = 0.133 - x [SO₂] = x [Cl₂] = x \(K_c = \frac{x^2}{0.133 - x}\) Using the calculated value of Kc: \(0.71 = \frac{x^2}{0.133 - x}\) Solving for x: \(x = 0.086\,\text{M}\) Since the 2.00 moles of SO₂Cl₂ is placed in a 15.00 L vessel, the initial concentration has changed. To find the new percentage of decomposition, we can use the following formula: \(\text{Percentage of decomposition} = \frac{0.086}{0.133} \times 100\,\% = 64.7\,\%\) Thus, when 2.00 mol of SO₂Cl₂ is placed in a 15.00-L vessel at 303 K, the percentage of SO₂Cl₂ decomposed is approximately 64.7%.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_c\) or \(K_p\), is a key concept in chemical equilibrium. It tells us how favored the products are compared to the reactants at equilibrium. For a generic reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \(K_c\) is expressed in terms of the molar concentrations:

• \(K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\)

For reactions involving gases, we can also express the equilibrium constant in terms of partial pressures, known as \(K_p\).
The relationship between \(K_c\) and \(K_p\) is given by the expression:

• \(K_p = K_c (RT)^{\Delta n}\)

where \(R\) is the gas constant (8.314 J/mol K), \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gases during the reaction.
Understanding \(K_c\) and \(K_p\) helps in predicting the extent of a reaction under different conditions, making them essential tools in chemical equilibrium analysis.

Le Chatelier's Principle

Le Chatelier's principle is a fundamental rule that predicts how a system at equilibrium will respond to disturbances. When an external stress, such as a change in pressure, temperature, or concentration, is applied to a chemical system at equilibrium, the system adjusts to restore equilibrium.
This principle can be applied to determine how equilibrium shifts:

• If the concentration of a reactant or product is changed, the equilibrium will shift to counter that change.
• Increasing temperature for an endothermic reaction shifts equilibrium towards products, while for an exothermic reaction, it shifts towards reactants.
• If volume increases, the equilibrium will shift towards the side with more gas molecules.

In our exercise, increasing the volume led the system to shift towards producing more products, resulting in greater decomposition of \(\text{SO}_2\text{Cl}_2\). This adjustment aligns with the prediction made by Le Chatelier's principle given the reaction and its conditions.

Ideal Gas Law

The ideal gas law is a powerful equation that relates the pressure, volume, temperature, and number of moles of an ideal gas:

• \(PV = nRT\)

where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.
This equation is extremely helpful in converting between \(K_c\) and \(K_p\), as it provides a relationship between concentration and pressure. It is essential to remember that this law assumes ideal behavior, which is a good approximation in many scenarios, but not always perfect under high pressure or low temperature.
In the exercise, the ideal gas law was fundamental in the calculation where \(K_c\) was converted to \(K_p\), demonstrating how closely intertwined these gas laws and equilibrium concepts can be.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationships between the reactants and products in a chemical reaction. It is essential for determining how moles of one substance relate to moles of another, governed by the balanced chemical equation.

• In our exercise, the reaction \(\text{SO}_2\text{Cl}_2 \rightleftharpoons \text{SO}_2 + \text{Cl}_2\) shows a 1:1:1 stoichiometric ratio.

Using stoichiometry, we deduce changes in concentrations during the reaction: for every mole of \(\text{SO}_2\text{Cl}_2\) that decomposes, one mole each of \(\text{SO}_2\) and \(\text{Cl}_2\) is produced.
Stoichiometry is crucial in calculating the equilibrium concentrations required for determining \(K_c\). By applying these stoichiometric principles, students can predict and quantify reaction behaviors, which is fundamental for understanding and solving equilibria problems.