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 \(\mathrm{K}\) .

Step by step solution

01

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

02

Calculate Kc

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

03

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: \(K_p = 0.71 × (8.314 \frac{\text{J}}{\text{mol K}} \times 303 \,\text{K})^1 = 1791\)

04

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.

05

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%.

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