Problem 80 - Chemistry The Central Science

Question

A \(0.831-\mathrm{g}\) sample of \(\mathrm{SO}_{3}\) is placed in a 1.00 - \(\mathrm{L}\) container and heated to \(1100 \mathrm{~K}\). The \(\mathrm{SO}_{3}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) : $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the total pressure in the container is \(1.300 \mathrm{~atm}\). Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at \(1100 \mathrm{~K}\).

Step-by-Step Solution

Answer

The equilibrium constants for the given reaction at 1100 K are \(K_c = 17.64\) and \(K_p = 15.60 \: atm^{-1}\).

Step 1: Determine the initial moles of SO₃

First, convert the given mass (0.831 g) of SO₃ into moles using the formula: moles = (mass)/(molar mass) The molar mass of SO₃ = S (32.06 g/mol) + 3 × O (16.00 g/mol) = 80.06 g/mol. Thus, moles of SO₃ = (0.831 g) / (80.06 g/mol) = 0.0104 mol.

Step 2: Find initial concentrations

We are given that the container has a volume of 1.00 L. We can find the initial concentrations of the reactants and products as follows: [SO₃]₀ = moles of SO₃ / volume = 0.0104 mol / 1.00 L = 0.0104 M Initially, there are no products formed, so [SO₂]₀ = 0 and [O₂]₀ = 0.

Step 3: Set up ICE (Initial, Change, Equilibrium) table

Set up an ICE table to keep track of the changes in concentrations as the reaction proceeds towards equilibrium. ``` [SO₃] [SO₂] [O₂] Initial 0.0104 M 0 0 Change -2x +2x +x Equilibrium 0.0104-2x 2x x ```

Step 4: Use total pressure to find equilibrium concentrations

We are given that the total pressure at equilibrium is 1.300 atm. We can use the relationship between partial pressures and concentrations (PV=nRT) to find the partial pressures of the gases. Let P_SO₃, P_SO₂, and P_O₂ be the partial pressures of SO₃, SO₂, and O₂ at equilibrium, respectively. Then: P_SO₃ = x_SO₃ * R * T / V = (0.0104 - 2x) * (0.0821 L atm / K mol) * 1100 K / 1.00 L P_SO₂ = x_SO₂ * R * T / V = (2x) * (0.0821 L atm / K mol) * 1100 K / 1.00 L P_O₂ = x_O₂ * R * T / V = (x) * (0.0821 L atm / K mol) * 1100 K / 1.00 L Total pressure at equilibrium = P_SO₃ + P_SO₂ + P_O₂ = 1.300 atm (0.0104 - 2x) * 90.71 + 2x * 90.71 + x * 90.71 = 1.300

Step 5: Solve for x

Solve the above equation to find the value of x. After solving, x = 0.0042

Step 6: Calculate equilibrium concentrations

Substitute the value of x into the equilibrium concentrations from the ICE table. [SO₃] = 0.0104 - 2x = 0.0104 - 2 * 0.0042 = 0.0020 M [SO₂] = 2x = 2 * 0.0042 = 0.0084 M [O₂] = x = 0.0042 M

Step 7: Calculate Kc

Calculate Kc using the equilibrium concentrations. Kc = ([SO₂]^2 * [O₂]) / ([SO₃]^2) = (0.0084^2 * 0.0042) / (0.0020^2) = 17.64

Step 8: Calculate Kₚ

Since Kc and Kₚ are related through the equation Kₚ = Kc(RT)^(Δn), calculate Kₚ using Kc and the given temperature. Δn = (moles of gaseous products - moles of gaseous reactants) = (2 + 1) - (2) = 1 Kₚ = Kc(RT)^Δn = 17.64 * (0.0821 L atm / K mol * 1100 K)^1 = 15.60 atm^{-1} The equilibrium constants for the given reaction at 1100 K are Kc = 17.64 and Kₚ = 15.60 atm^{-1}.

Key Concepts

Equilibrium Constant (Kc)

The equilibrium constant, denoted as \(K_c\), is a vital concept in chemical equilibrium that measures the ratio of the concentrations of products to reactants at equilibrium. In the provided exercise, we have a balanced chemical equation:\[2 \text{SO}_3(g) \rightleftharpoons 2 \text{SO}_2(g) + \text{O}_2(g)\]To calculate \(K_c\), we use the equilibrium concentrations determined from an ICE (Initial, Change, Equilibrium) table. The expression for \(K_c\) is derived from the general reaction quotient, considering the stoichiometry of the reaction:

• \([\text{SO}_2]^2 [\text{O}_2]\) as the concentration of products.
• \([\text{SO}_3]^2\) as the concentration of reactants.

This yields:\[K_c = \frac{[\text{SO}_2]^2 \cdot [\text{O}_2]}{[\text{SO}_3]^2}\]Substituting the values from the equilibrium concentrations gives:\[K_c = \frac{(0.0084)^2 \cdot 0.0042}{(0.0020)^2} = 17.64\]This calculation shows that the forward reaction (decomposition of \(\text{SO}_3\)) is favorable under the given conditions, as indicated by a high \(K_c\) value.

Equilibrium Constant (Kp)

The equilibrium constant for gases can also be expressed in terms of partial pressures, indicated as \(K_p\). \(K_p\) relates to the partial pressures of gases involved in the reaction at equilibrium. The relationship between \(K_c\) and \(K_p\) is given by the equation:\[K_p = K_c(RT)^\Delta n\]Where:

• \(R\) is the universal gas constant, \(0.0821\, \text{L atm }/\text{K mol}\).
• \(T\) is the temperature in Kelvin.
• \(\Delta n\) is the change in moles of gases (products minus reactants).

For the reaction \(2 \text{SO}_3(g) \rightleftharpoons 2 \text{SO}_2(g) + \text{O}_2(g)\), \(\Delta n = 1\) since there are a total of 3 moles of product gases and 2 moles of reactant gases.Substituting values calculated earlier:\[K_p = 17.64 \times (0.0821 \times 1100)^1 = 15.60\, \text{atm}^{-1}\]This demonstrates that \(K_p\), like \(K_c\), suggests a tendency towards product formation at equilibrium under the conditions of this reaction.

Le Chatelier's Principle

Le Chatelier's Principle is an essential guideline for predicting the effect of changes in conditions on a chemical equilibrium. It states that if a dynamic equilibrium is disturbed by an external change in conditions, the position of equilibrium shifts to counteract the change and restore a new equilibrium.For the decomposition reaction of \(\text{SO}_3\), consider changes such as:

• **Pressure**: An increase in pressure would shift the equilibrium towards the side with fewer gas moles, favoring the formation of \(\text{SO}_3\). Conversely, a decrease in pressure would shift it towards \(\text{SO}_2\) and \(\text{O}_2\).
• **Temperature**: Since the decomposition of \(\text{SO}_3\) is endothermic, increasing the temperature favors the forward reaction (more \(\text{SO}_2\) and \(\text{O}_2\)). Reducing temperature would favor the reverse reaction.
• **Concentration**: Adding more \(\text{SO}_3\) would shift the equilibrium towards products (\(\text{SO}_2\) and \(\text{O}_2\)), while removing \(\text{SO}_3\) would shift it toward reactants.

Le Chatelier's Principle aids in anticipating the behavior of chemical systems under stress, without calculating exact concentrations or pressures.