Problem 45 - Chemistry: The Central Science in SI Units, Global Edition

Question

At \(120^{\circ} \mathrm{C}, K_{c}=0.090\) 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 \(0.100 \mathrm{M}\) and \(0.075 \mathrm{M}\), respectively. What is the partial pressure of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

Step-by-Step Solution

Answer

The partial pressure of Cl2 in the equilibrium mixture is 0.0434 atm.

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation is already given: \(\mathrm{SO}_{2}\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\)

Step 2: Write the Expression for the Equilibrium Constant, Kc

For the reaction, the expression for Kc is: \(K_c = \frac{[\mathrm{SO}_{2}][\mathrm{Cl}_{2}]}{[\mathrm{SO}_{2}\mathrm{Cl}_{2}]}\) We are given that \(K_c = 0.090\) at 120°C.

Step 3: Find the Change in Concentrations during the Reaction

Let the change in the concentration of reactants and products during the reaction be x. Since the balanced equation has a 1:1:1 stoichiometric ratio, we can find the change in concentration for each species. At equilibrium: \([SO_2Cl_2] = 0.100 - x\) \([SO_2] = 0.075 + x\) \([Cl_2] = x\)

Step 4: Substitute the Concentrations back into the Kc Expression

Substitute the concentrations into the Kc expression: \(0.090 = \frac{(0.075 + x)(x)}{(0.100 - x)}\)

Step 5: Solve for x

Solve the above equation for x: \(0.090 = \frac{0.075x + x^2}{0.100 - x}\) Multiply both sides by \((0.100 - x)\): \(0.009 - 0.090x = 0.075x + x^2\) Rearrange the equation: \(x^2 + 0.165x - 0.009 = 0\) Now, solve the quadratic equation for x. We can use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) In this case, \(a=1\), \(b=0.165\), and \(c=-0.009\). Upon solving for x, we find that there are two possible values: \(x = -0.996\) and \(x = 0.00906\). Since the negative value of x doesn't make sense in the context of concentrations, we choose the positive value for x: \(x = 0.00906 \, \mathrm{M}\) This means that the equilibrium concentration of Cl2 is 0.00906 M.

Step 6: Calculate the Partial Pressure of Cl2

To calculate the partial pressure of Cl2, first, we can assume that the total pressure (P) is 1 atm since it is not given. Then, we can use the mole fraction of Cl2 in the gas mixture to find its partial pressure. Mole fraction of Cl2, \(X_{Cl2} = \frac{[Cl_2]}{([SO_2] + [SO_2Cl_2] + [Cl_2])} = \frac{0.00906}{(0.075 + 0.100 - 0.00906 + 0.00906)}\) Calculate \(X_{Cl2}\): \(X_{Cl_2} = 0.0434\) Now, calculate the partial pressure of Cl2 using the total pressure and the mole fraction: \(P_{Cl_2} = X_{Cl_2} \times P\) \(P_{Cl_2} = 0.0434 \times 1 \, \mathrm{atm}\) \(P_{Cl_2} = 0.0434 \, \mathrm{atm}\) So, the partial pressure of Cl2 in the equilibrium mixture is 0.0434 atm.

Key Concepts

Equilibrium Constant (Kc)

The equilibrium constant, commonly denoted as \(K_c\), is a critical concept in chemical equilibrium, representing the ratio of product concentrations to reactant concentrations at equilibrium. It is calculated based on the concentration of gases or solutes in a reaction that has reached equilibrium. This ratio remains constant at a given temperature, simplifying the prediction of how the concentration of gases or solutes would be distributed when the reaction is at equilibrium.

For the reaction \(\mathrm{SO}_{2}\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\), the equilibrium constant expression is formulated as:
\[K_c = \frac{[\mathrm{SO}_{2}][\mathrm{Cl}_{2}]}{[\mathrm{SO}_{2}\mathrm{Cl}_{2}]}\]
In this exercise, \(K_c\) was provided as 0.090 at \(120^{\circ} \mathrm{C}\), allowing us to analyze how the concentrations of reactants and products change and settle into equilibrium. The equilibrium constant is fundamentally essential because knowing its value in a system helps chemists control reactions, predict product yields, and understand reaction dynamics.

Quadratic Equation in Chemistry

Quadratic equations often surface in chemistry, particularly when dealing with the equilibrium constant's calculations. When the reaction has a 1:1:1 stoichiometric ratio, as in our example, the expression for \(K_c\) results in a quadratic equation. To find the change in concentration of the species at equilibrium, solving this quadratic equation becomes necessary.

For our equation:
\[0.090 = \frac{(0.075 + x)(x)}{(0.100 - x)}\]
Rearranging this into standard quadratic form gives us:
\[x^2 + 0.165x - 0.009 = 0\]
The quadratic formula helps to find values for \(x\), which represent the change in concentration of \(\mathrm{Cl}_2\) through:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Upon deducing this, we select the physically meaningful solution (positive \(x\)) for concentration change, as concentrations cannot be negative. Thus, understanding quadratic equations facilitates the prediction and verification of chemical reaction dynamics.

Partial Pressure Calculation

Calculating partial pressures in a gas mixture is crucial for understanding gas equilibria. Partial pressure refers to the pressure exerted by a single gas type in a mixture of gases. The mole fraction, \(X\), is utilized to determine each gas's partial pressure in the mixture.

In our task, with the concentration of \(\mathrm{Cl}_2\) at equilibrium determined to be 0.00906 M, we calculate its mole fraction:

• Mole fraction of \(\mathrm{Cl}_2\), \(X_{Cl_2} = \frac{0.00906}{(0.075 + 0.100 - 0.00906 + 0.00906)}\)
• Which simplifies to \(X_{Cl_2} = 0.0434\)

The total pressure is assumed to be 1 atm for simplicity. The partial pressure of \(\mathrm{Cl}_2\) is then:
\[P_{Cl_2} = X_{Cl_2} \times P\]
Thus:
\[P_{Cl_2} = 0.0434 \times 1 = 0.0434\, \mathrm{atm}\]
This calculation helps in determining the behavior and interaction of gases in chemical reactions, crucial for practical applications in products and industrial processes.