Question

A quantity of \(6.75 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) was placed in a \(2.00-\mathrm{L}\) flask. At \(648 \mathrm{~K},\) there is \(0.0345 \mathrm{~mol}\) of \(\mathrm{SO}_{2}\) present. Calculate \(K_{\mathrm{c}}\) for the reaction: $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftarrows \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$

Short answer

The equilibrium constant \( K_{c} \) is approximately 0.0384.

Step-by-step solution

Calculate Initial Moles of SO2Cl2

First, we need to calculate the initial number of moles of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \) using its molecular weight. The molar mass of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \) is \( 1 \times 32.07 + 2 \times 16.00 + 2 \times 35.45 = 134.97 \ g/mol \). Thus, the initial moles are:\[\text{moles of } \mathrm{SO}_{2}\mathrm{Cl}_{2} = \frac{6.75 \ g}{134.97 \ g/mol} = 0.0500 \ mol\]

Reaction Stoichiometry

The balanced chemical equation is:\[\mathrm{SO}_{2}\mathrm{Cl}_{2}(g) \rightleftarrows \mathrm{SO}_{2}(g) + \mathrm{Cl}_{2}(g)\]According to the reaction, each mole of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \) decomposes into one mole of \( \mathrm{SO}_{2} \) and one mole of \( \mathrm{Cl}_{2} \). Therefore, the change in moles for \(\mathrm{SO}_{2} \) is equal to the change in moles for \( \mathrm{Cl}_{2} \), and equal to the decomposed moles of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \).

Calculate Moles at Equilibrium

At equilibrium, 0.0345 mol of \( \mathrm{SO}_{2} \) is present. Given the stoichiometry, this means 0.0345 mol of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \) has decomposed. Therefore, the remaining moles of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \) are:\[0.0500 \ mol - 0.0345 \ mol = 0.0155 \ mol\]And the moles of \( \mathrm{Cl}_{2} \) are also 0.0345 mol.

Calculate Equilibrium Concentrations

To find the concentration, divide the moles of each species by the volume of the flask (2.00 L):\[[\mathrm{SO}_{2}] = \frac{0.0345 \ mol}{2.00 \ L} = 0.01725 \ M\]\[[\mathrm{Cl}_{2}] = \frac{0.0345 \ mol}{2.00 \ L} = 0.01725 \ M\]\[[\mathrm{SO}_{2}\mathrm{Cl}_{2}] = \frac{0.0155 \ mol}{2.00 \ L} = 0.00775 \ M\]

Calculate Equilibrium Constant Kc

The expression for the equilibrium constant \( K_{c} \) for this reaction is:\[K_{c} = \frac{[\mathrm{SO}_{2}][\mathrm{Cl}_{2}]}{[\mathrm{SO}_{2}\mathrm{Cl}_{2}]}\]Substituting the equilibrium concentrations:\[K_{c} = \frac{(0.01725)(0.01725)}{0.00775} = 0.0384\]

Final Computation of Kc

After computing the expression with the concentrations, the value of \( K_{c} \) is:\[K_{c} \approx 0.0384\]

Key concepts

Equilibrium Constant (Kc)

In chemical equilibrium, the equilibrium constant, denoted as \( K_c \), is an important value used to predict the position of the equilibrium. It is calculated from the concentrations of products and reactants at equilibrium. For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant expression is written as: \[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \] where \([C], [D], [A], \) and \([B]\) represent the molar concentrations of the chemical species. At a particular temperature, \( K_c \) serves as a constant that helps us understand the ratio of products to reactants. A larger \( K_c \) indicates that the reaction favors the formation of products, whereas a smaller \( K_c \) suggests the reactants are favored. In the given exercise, the calculation of \( K_c \) helped to determine the dynamic balance of the reaction involving \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \).

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationships between the amounts of reactants and products in a chemical reaction. For this process, it’s based on the balanced chemical equation, which indicates the proportions in which substances react. In our example, the reaction \( \mathrm{SO}_{2}\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g) + \mathrm{Cl}_{2}(g) \) shows that one mole of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \) decomposes to form one mole of \( \mathrm{SO}_{2} \) and one mole of \( \mathrm{Cl}_{2} \). This stoichiometric relationship is crucial because it allows us to predict the mole-to-mole conversion that occurs during the reaction. If you begin with 0.0500 moles of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \), you would expect the same total moles of \( \mathrm{SO}_{2} \) and \( \mathrm{Cl}_{2} \) to be formed, as observed at equilibrium with 0.0345 moles of each product.

Equilibrium Concentrations

Equilibrium concentrations are the concentrations of reactants and products in a chemical reaction when the reaction has reached equilibrium. At this point, the forward and reverse reaction rates are equal. To calculate these concentrations, the number of moles of each component at equilibrium is divided by the volume of the container. This gives the molarity, which is essential for calculating \( K_c \). In the given problem, knowing that 0.0345 mol of \( \mathrm{SO}_{2} \) was formed, we calculated the equilibrium concentrations as follows:

• For \( \mathrm{SO}_{2} \): \([\mathrm{SO}_{2}] = \frac{0.0345 \ mol}{2.00 \ L} = 0.01725 \ M \)
• For \( \mathrm{Cl}_{2} \): \([\mathrm{Cl}_{2}] = \frac{0.0345 \ mol}{2.00 \ L} = 0.01725 \ M \)
• For \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \): \([\mathrm{SO}_{2}\mathrm{Cl}_{2}] = \frac{0.0155 \ mol}{2.00 \ L} = 0.00775 \ M \)

These concentrations were then used to calculate the equilibrium constant.

Moles and Molarity

Understanding moles and molarity is fundamental to solving equilibrium problems. The concept of a "mole" is central in chemistry as it refers to the amount of substance containing the same number of entities as atoms in exactly 12 grams of carbon-12. The molarity, on the other hand, is a measure of the concentration of a solution and is defined as the number of moles of solute per liter of solution. When working on such exercises, you first need to convert the mass of reactants into moles using the molar mass (e.g., converting 6.75 g of \( \mathrm{SO}_{2}\mathrm{Cl}_{2} \) to moles). Then, using the volume of the solution (2.00 L in this case), you can determine the molarity of each substance. This approach helps in determining equilibrium concentrations, which is crucial for finding the \( K_c \). This clear understanding of moles and molarity facilitates solving advanced equilibrium problems efficiently.