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) Repeat these calculations for \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a \(15.00-\mathrm{L}\) vessel at \(303 \mathrm{~K}\).

Step-by-Step Solution

Answer

For given $2.00\,\text{mol}$ of $\text{SO}_{2}\text{Cl}_{2}$ in a $2.00\,\text{L}$ flask, at equilibrium: $[\text{SO}_{2}] = [\text{Cl}_{2}] = (0.56 \times 2.00)\frac{\text{mol}}{2.00\,\text{L}}$ $[\text{SO}_{2}\text{Cl}_{2}] = (2.00 - 0.56 \times 2.00)\frac{\text{mol}}{2.00\,\text{L}}$ Kc can be found using the expression: $K_c = \frac{[\text{SO}_{2}][\text{Cl}_{2}]}{[\text{SO}_{2}\text{Cl}_{2}]}\approx 1.42$ Kp can be found using the expression: $K_p = K_c(RT)^n \approx 3.32 \,\text{atm}$ For the given $2.00\,\text{mol}$ of $\text{SO}_{2}\text{Cl}_{2}$ in a $15.00\,\text{L}$ vessel, at equilibrium: $[\text{SO}_{2}] = [\text{Cl}_{2}] = (0.56 \times 2.00)\frac{\text{mol}}{15.00\,\text{L}}$ $[\text{SO}_{2}\text{Cl}_{2}] = (2.00 - 0.56 \times 2.00)\frac{\text{mol}}{15.00\,\text{L}}$ Kc can be found using the expression: $K_c = \frac{[\text{SO}_{2}][\text{Cl}_{2}]}{[\text{SO}_{2}\text{Cl}_{2}]}\approx 0.602$ Kp can be found using the expression: $K_p = K_c(RT)^n \approx 1.40 \,\text{atm}$

Step 1: Calculate Initial Concentrations and Change in Concentration

The initial concentration of SO₂Cl₂ can be calculated using the formula: \[c = \frac{n}{V}\] where c is concentration, n is number of moles, and V is volume. For the reaction, let x be the concentration of SO₂Cl₂ that will decompose. At equilibrium, we will have: \[ SO₂Cl₂ : 2.00\frac{mol}{2.00L} - x \] \[ SO₂ : x \] \[ Cl₂ : x \] As given, 56% of SO₂Cl₂ decomposes: x = \( 0.56 \times 2.00\frac{mol}{2.00L} \)

Step 2: Calculate Equilibrium Concentrations

Now that we have x, we will find the concentration of each species at equilibrium: \[ SO₂Cl₂ : 2.00\frac{mol}{2.00L} - x = (2.00 - 0.56 \times (2.00))\frac{mol}{2.00L}\] \[ SO₂ : x = (0.56 \times 2.00)\frac{mol}{2.00L}\] \[ Cl₂ : x = (0.56 \times 2.00)\frac{mol}{2.00L}\]

Step 3: Calculate Kc

Now, we can use the concentrations to find the Kc: \[ K_c = \frac{[SO₂][Cl₂]}{[SO₂Cl₂]} = \frac{(0.56 \times 2.00)(0.56 \times 2.00)}{(2.00 - 0.56 \times 2.00)}\] (b) Calculate the Kp

Step 4: Relate Kc to Kp

First, we need to find the relationship between Kc and Kp using the expression: \[ K_p = K_c(RT)^n \] where R is the gas constant, T is the temperature, and n is the change in the number of moles of gaseous products and reactants. Here, n = (1 + 1) - 1 = 1 R = 0.0821 \(L \cdot atm \cdot K^{-1} \cdot mol^{-1}\) T = 303 K Now, we can calculate Kp using the value of Kc we calculated. (c) Kc and Kp for 2.00 mol of SO₂Cl₂ in a 15.00-L vessel:

Step 5: Repeat the Process for Different Volume

Follow the same process for Step 1 to Step 4, replacing the V in step 1 with 15L. Remember to update the calculation of change in concentration (x) and equilibrium concentrations before calculating the Kc, and Kp for the new volume.

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \(K_c\) for concentrations or \(K_p\) for partial pressures, is a vital concept in chemical equilibrium. It represents the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their coefficients from the balanced chemical equation. These constants are crucial for predicting the direction of a reaction and understanding how systems respond to changes in conditions.

For the decomposition of \(\text{SO}_2\text{Cl}_2\), the equilibrium constant focuses on this decomposition process:

• Products: \(\text{SO}_2\) and \(\text{Cl}_2\)
• Reactant: \(\text{SO}_2\text{Cl}_2\)

The expression for \(K_c\) in this context is:\[ K_c = \frac{[\text{SO}_2][\text{Cl}_2]}{[\text{SO}_2\text{Cl}_2]} \]
Understanding \(K_c\) allows chemists to determine how a change in conditions, like concentration or temperature, could shift the equilibrium, following Le Chatelier's Principle.

Kc and Kp Calculations

Calculating \(K_c\) and \(K_p\) involves specific steps that depend on the reaction conditions and context. The first step is calculating the concentrations of the reactants and products at equilibrium. In the given reaction, you start with 2.00 moles of \(\text{SO}_2\text{Cl}_2\) in a 2.00 L flask. The decomposition of 56% must be noted for accurate calculations. So, the concentrations at equilibrium for all species are computed, taking into account the changes in each component’s concentration as the system reaches equilibrium.

Once the equilibrium concentrations of \(\text{SO}_2\), \(\text{Cl}_2\), and \(\text{SO}_2\text{Cl}_2\) are known, \(K_c\) is calculated with their appropriate values plugged into the equilibrium expression provided earlier.
The conversion to \(K_p\) from \(K_c\) involves the equation:\[ K_p = K_c(RT)^\Delta n \]where \(\Delta n\) is the difference in moles of gas between products and reactants. This formula helps bridge the concentration-based \(K_c\) to pressure-based \(K_p\), employing the gas constant \(R = 0.0821\) L atm K\(^{-1}\) mol\(^{-1}\) and temperature \(T\) in Kelvin.

Decomposition Reaction

In chemistry, a decomposition reaction involves breaking down a compound into simpler substances. It is a reversible process where the initial reactant dissociates into its component products. The equation for the decomposition of \(\text{SO}_2\text{Cl}_2\) can be expressed as:\[ \text{SO}_2\text{Cl}_2(g) \rightleftharpoons \text{SO}_2(g) + \text{Cl}_2(g) \]
Decomposition requires energy input, which can be in the form of heat, light, or electricity. Here, the decomposition of sulfuryl chloride into sulfur dioxide and chlorine is significant as it contributes to establishing a chemical equilibrium.
From a molecular perspective, these reactions can also involve complexities such as:

• Recognizing when equilibrium is reached
• Understanding reversibility where products can recombine to form original reactants

Such insights into decomposition reactions help predict the conditions under which a reactant may dissociate, guiding design and implementation in labs and industry.

Reaction Stoichiometry

Reaction stoichiometry involves using a balanced chemical equation to determine the relationships between reactants and products in a chemical reaction. It provides the quantitative relationship between the substances consumed and produced. When calculating quantities in reactions, stoichiometry is your guiding principle.
In decomposition reactions, stoichiometry helps us understand the ratios involved. For the reaction:

• The balanced equation:\[ \text{SO}_2\text{Cl}_2(g) \rightleftharpoons \text{SO}_2(g) + \text{Cl}_2(g) \]
• The initial amount of \(\text{SO}_2\text{Cl}_2\) directly influences the amounts of \(\text{SO}_2\) and \(\text{Cl}_2\) formed.

By applying stoichiometric relations, we can accurately calculate how much of each material is present at equilibrium after a certain percentage decomposes.
Understanding reaction stoichiometry can aid in predicting the outcome of reactions under different conditions, ensuring precise calculations in demonstrating how to achieve specific product yields and manage reactant use efficiently.