Problem 46 - Chemistry: The Central Science

Question

At \(100{ }^{\circ} \mathrm{C}, K_{c}=0.078\) for the reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ In an equilibrium mixture of the three gases, the concentrations of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and \(\mathrm{SO}_{2}\) are \(0.108 \mathrm{M}\) and \(0.052 \mathrm{M}\), respectively. What is the partial pressure of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

Step-by-Step Solution

Answer

In the given equilibrium reaction, we find that the equilibrium concentration of Cl2 is 0.042 M (rounded to 3 decimal places). However, without the given volume, the exact partial pressure of Cl2 cannot be determined.

Step 1: Write the equilibrium expression for the reaction

For the given reaction, the equilibrium constant Kc can be written as: \[K_c = \frac{[\mathrm{SO}_2] [\mathrm{Cl}_2]}{[\mathrm{SO}_{2}\mathrm{Cl}_2]}\]

Step 2: Define variables for change and equilibrium concentrations

Let x be the moles of Cl2 formed at equilibrium. Since the reaction consumes 1 mole of SO2Cl2 and produces 1 mole of SO2 and 1 mole of Cl2, the increase in the concentrations of SO2 and Cl2 is also x. We can now write the equilibrium concentrations as: \[[\mathrm{SO}_2] = 0.052 + x\] \[[\mathrm{Cl}_2] = x\] \[[\mathrm{SO}_{2}\mathrm{Cl}_2] = 0.108 - x\]

Step 3: Substitute the expressions in Kc equation

Now, substitute these expressions in the Kc equation from Step 1: \[0.078 = \frac{(0.052+x)(x)}{(0.108-x)}\]

Step 4: Solve for x, the equilibrium concentration of Cl2

Solving the equation in Step 3 for x, we can determine the equilibrium concentration of Cl2: \[0.078(0.108-x) = (0.052+x)(x)\] \[x^2 + 0.052x - 0.006084 = 0\] This is a quadratic equation with coefficients a=1, b=0.052, and c=-0.006084. To determine x, use the quadratic formula: \[x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\] On calculating, \[x = 0.042\, \mathrm{(rounded\, to\, 3\, decimal\, places)}\] Now we have the equilibrium concentration of Cl2.

Step 5: Find the partial pressure of Cl2

Now, we can find the partial pressure of Cl2 using the ideal gas law formula, provided we have the total volume. \(P_{Cl_2}=\frac{(n_{Cl_2})*R*T}{V}\) Using the provided equilibrium concentration of Cl2, and assuming R and T are in appropriate units, properly converting the concentration values to partial pressure will yield the desired result. However, without the given volume, the exact partial pressure of Cl2 cannot be determined.

Key Concepts

Equilibrium Constant

Understanding the equilibrium constant, denoted as \(K_c\), is essential in predicting how a chemical system behaves at equilibrium. It provides a ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients in a balanced chemical equation. For instance, in the reaction \(\mathrm{SO}_{2} \mathrm{Cl}_{2} \rightleftharpoons \mathrm{SO}_2 + \mathrm{Cl}_2\), \(K_c\) would be expressed as: \[K_c = \frac{[\mathrm{SO}_2][\mathrm{Cl}_2]}{[\mathrm{SO}_2\mathrm{Cl}_2]}\] This equation tells us how product concentrations influence the equilibrium mixture.

• A \(K_c\) greater than 1 implies that, at equilibrium, the products are favored.
• A \(K_c\) less than 1 indicates that the reactants are favored at equilibrium.

The magnitude of \(K_c\) gives insight into the position of equilibrium but does not affect the speed at which equilibrium is reached.

Partial Pressure

In gas mixtures, partial pressures are crucial for understanding individual gas behavior. Each gas within a mixture contributes to the total pressure, based on its amount and the container's temperature. The partial pressure of a gas is directly proportional to its mole fraction, an expression of its composition in the mixture. For our reaction at equilibrium, \(\mathrm{Cl}_2\)'s partial pressure can be calculated using its concentration and the ideal gas law: \[P_{Cl_2} = \frac{n_{Cl_2}RT}{V}\] Here, \(n_{Cl_2}\) stands for moles of \(\mathrm{Cl}_2\), \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(V\) is the volume. Once you know \(Cl_2\)'s concentration and either the temperature or volume, you can find its partial pressure.

• Partial pressure concepts are frequently utilized in equilibrium constants expressed in terms of pressure (\(K_p\)) rather than concentration.
• It's important to remember that total pressure within a mixture is the sum of all partial pressures.

Quadratic Equation

Quadratic equations can pop up in chemical equilibrium scenarios when we try to solve for unknown concentrations at equilibrium. In this exercise, we ended up with a quadratic equation to solve for \(x\), the equilibrium concentration of \(\mathrm{Cl}_2\). The equation appeared as: \[x^2 + 0.052x - 0.006084 = 0\] A quadratic equation is generally written as \(ax^2 + bx + c = 0\). The quadratic formula used to find \(x\) is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Here's how this formula breaks down:

• \(-b\): Changes the sign of \(b\).
• \(\sqrt{b^2 - 4ac}\): Determines the nature of solutions (real or complex). Positive values offer two real solutions, zero a single solution, and negative values indicate no real solutions.
• \(2a\): Denominator that divides the entire expression. Adjusts the solution based on the leading coefficient.

This formula is invaluable in solving equations like the one in the reaction study guide. Successfully solving such equations allows us to predict concentrations and, in turn, the behavior of reactions at equilibrium.