SbCl5 Decomposition Equilibrium Calculation A Step-by-Step Guide

by ADMIN 65 views

This article delves into the equilibrium of the reversible reaction involving antimony pentachloride (SbCl5) decomposing into antimony trichloride (SbCl3) and chlorine gas (Cl2). We'll break down the principles behind calculating equilibrium concentrations using the equilibrium constant Kp, and then apply these principles to solve a specific problem. This is a fundamental concept in chemistry, particularly in the study of chemical kinetics and thermodynamics. Mastering these concepts is essential for understanding and predicting the behavior of chemical reactions.

The Reaction Equilibrium

The reaction we're considering is:

SbCl5(g) ⇌ SbCl3(g) + Cl2(g)

This equation tells us that gaseous antimony pentachloride (SbCl5) can reversibly decompose into gaseous antimony trichloride (SbCl3) and chlorine gas (Cl2). The double arrow (⇌) indicates that the reaction can proceed in both the forward (decomposition of SbCl5) and reverse (formation of SbCl5 from SbCl3 and Cl2) directions. This dynamic equilibrium is established when the rates of the forward and reverse reactions are equal, and the net change in concentrations of reactants and products becomes zero. The equilibrium constant, Kp, is a numerical value that expresses the ratio of products to reactants at equilibrium, where each partial pressure is raised to the power of its stoichiometric coefficient in the balanced chemical equation.

Defining the Equilibrium Constant (Kp)

For the given reaction, the equilibrium constant in terms of partial pressures (Kp) is defined as:

Kp = (P(SbCl3) * P(Cl2)) / P(SbCl5)

Where:

  • P(SbCl3) is the partial pressure of SbCl3 at equilibrium.
  • P(Cl2) is the partial pressure of Cl2 at equilibrium.
  • P(SbCl5) is the partial pressure of SbCl5 at equilibrium.

The Kp value is temperature-dependent, meaning it changes with temperature. A larger Kp indicates that the equilibrium favors the products, while a smaller Kp suggests that the equilibrium favors the reactants. In our case, Kp = 1.48 at 448°C, indicating that at this temperature, the products are somewhat favored at equilibrium.

Relating Kp to Concentrations

While Kp is defined in terms of partial pressures, we are given the equilibrium concentrations in molarity (M). To relate these, we can use the ideal gas law:

PV = nRT

Where:

  • P is the pressure
  • V is the volume
  • n is the number of moles
  • R is the ideal gas constant (0.0821 L atm / (mol K))
  • T is the temperature in Kelvin

Rearranging the ideal gas law, we get:

P = (n/V)RT

Since n/V is the molar concentration (M), we can write:

P = MRT

This equation allows us to convert between partial pressure and molar concentration at a given temperature. This relationship is crucial for solving our problem, as we are given concentrations but need to use Kp, which is in terms of partial pressures.

Problem Setup and Solution

Problem Statement

Consider the following reaction:

SbCl5(g) ⇌ SbCl3(g) + Cl2(g) Kp = 1.48 at 448°C

If [SbCl3]eq = [Cl2]eq = 0.05 M, what is [SbCl5]eq?

Step-by-Step Solution

  1. Convert Temperature to Kelvin:

    First, we need to convert the temperature from Celsius to Kelvin:

    T(K) = T(°C) + 273.15

    T(K) = 448°C + 273.15 = 721.15 K

    Converting to Kelvin is essential because the ideal gas constant (R) uses Kelvin as the temperature unit. Using Celsius would lead to incorrect calculations. Always ensure consistent units in your calculations.

  2. Calculate Partial Pressures:

    Using the relationship P = MRT, we can calculate the partial pressures of SbCl3 and Cl2:

    P(SbCl3) = [SbCl3]eq * R * T = 0.05 M * 0.0821 L atm / (mol K) * 721.15 K ≈ 2.96 atm

    P(Cl2) = [Cl2]eq * R * T = 0.05 M * 0.0821 L atm / (mol K) * 721.15 K ≈ 2.96 atm

    We calculate the partial pressures of the known products first because we need these values to solve for the unknown partial pressure of the reactant, SbCl5. This step is crucial for bridging the gap between concentrations and the Kp expression.

  3. Set up the Kp Expression:

    We know the Kp expression:

    Kp = (P(SbCl3) * P(Cl2)) / P(SbCl5)

    We are given Kp = 1.48, and we calculated P(SbCl3) and P(Cl2). Now we need to solve for P(SbCl5).

  4. Solve for P(SbCl5):

    Rearrange the Kp expression to solve for P(SbCl5):

    P(SbCl5) = (P(SbCl3) * P(Cl2)) / Kp

    Substitute the values:

    P(SbCl5) = (2.96 atm * 2.96 atm) / 1.48 ≈ 5.92 atm

    This step is the heart of the problem, where we apply the equilibrium constant to find the unknown partial pressure. The algebraic manipulation and substitution are key to arriving at the correct answer.

  5. Convert P(SbCl5) back to Concentration:

    Use the relationship P = MRT to convert the partial pressure of SbCl5 back to concentration:

    [SbCl5]eq = P(SbCl5) / (R * T) = 5.92 atm / (0.0821 L atm / (mol K) * 721.15 K) ≈ 0.10 M

    Converting back to concentration provides the final answer in the units requested by the problem. This step completes the cycle of conversions, ensuring we've answered the question fully.

The Answer

Therefore, [SbCl5]eq ≈ 0.10 M

The correct answer is D. 0.1 M.

Key Concepts and Considerations

Le Chatelier's Principle

This problem deals with chemical equilibrium, and it's important to understand Le Chatelier's Principle. This principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress. These conditions include changes in concentration, pressure, and temperature. For example, if we were to add more Cl2 to the system, the equilibrium would shift to the left, favoring the formation of SbCl5, to counteract the increase in Cl2 concentration. Understanding Le Chatelier's Principle provides a qualitative understanding of how equilibrium systems respond to disturbances.

Importance of Temperature

Kp is temperature-dependent. If the temperature were different, the Kp value would also be different, and the equilibrium concentrations would change. This highlights the crucial role of temperature in determining the position of equilibrium. For endothermic reactions, increasing the temperature favors the products, while for exothermic reactions, increasing the temperature favors the reactants. Temperature is a key factor in controlling chemical reactions.

The Significance of Equilibrium

Chemical equilibrium is a fundamental concept in chemistry with wide-ranging applications. It's crucial in industrial processes, where optimizing reaction conditions to maximize product yield is essential. It also plays a vital role in biological systems, where many biochemical reactions occur in equilibrium. Understanding equilibrium allows us to predict and control chemical reactions in various contexts.

Additional Practice Problems

To further solidify your understanding of equilibrium calculations, try solving similar problems with different Kp values, temperatures, or initial concentrations. You can also explore problems involving the equilibrium constant Kc, which is defined in terms of concentrations rather than partial pressures. Working through various problems will build your confidence and problem-solving skills in this area.

For example, consider these variations:

  1. What if the temperature was changed to 500°C? How would this affect the equilibrium if the reaction is endothermic?
  2. If the initial concentration of SbCl5 was 1.0 M, and the equilibrium concentrations of SbCl3 and Cl2 were both 0.1 M, what is the Kp at this temperature?
  3. How would adding an inert gas to the system at constant volume affect the equilibrium?

Conclusion

Calculating equilibrium concentrations is a fundamental skill in chemistry. By understanding the equilibrium constant Kp and its relationship to partial pressures and concentrations, we can solve a variety of problems related to chemical equilibrium. Remember to always pay attention to units, especially temperature, and to carefully apply the Kp expression. This detailed explanation should provide a clear understanding of the process and help you tackle similar problems with confidence. Mastering equilibrium calculations is essential for a strong foundation in chemistry.