Calculate The Equilibrium Concentrations Of Nitrogen Gas [N₂(g)] And Ammonia [NH₃(g)] For The Reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Given An Initial 1:3 Ratio Of Nitrogen Gas And Hydrogen Gas In A 1.0 Litre Vessel, An Equilibrium Constant Of 1.0 X 10⁻⁴, And An Equilibrium [H₂(g)] Of 0.12 Mol/L.

by ADMIN 293 views

Introduction: Delving into the Haber-Bosch Process

In the realm of chemical kinetics and equilibrium, the Haber-Bosch process, represented by the reversible reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), stands as a cornerstone of modern industrial chemistry. This reaction, crucial for the synthesis of ammonia (NH₃), a vital component in fertilizers, exemplifies the intricate interplay of reactants and products reaching a state of equilibrium. Understanding the factors that influence this equilibrium is paramount for optimizing the production of ammonia and other industrially significant compounds. This article will dissect a scenario involving this equilibrium, providing a step-by-step analysis of how to calculate equilibrium concentrations.

At the heart of this chemical dance lies the equilibrium constant, denoted as K, which quantifies the ratio of products to reactants at equilibrium. A small K value, as seen in our case (1.0 x 10⁻⁴), suggests that the equilibrium favors the reactants, meaning that at equilibrium, there will be a higher concentration of nitrogen (N₂) and hydrogen (H₂) gases compared to ammonia (NH₃). However, this doesn't preclude the formation of ammonia; it merely indicates the direction in which the equilibrium lies. To precisely determine the equilibrium concentrations of each species, we must employ the principles of chemical equilibrium and stoichiometry.

The initial conditions of the reaction are also pivotal in determining the final equilibrium state. In our scenario, we begin with a mixture of nitrogen gas and hydrogen gas in a 1:3 ratio, which mirrors the stoichiometric ratio in the balanced chemical equation. This strategic initial ratio can influence the rate at which equilibrium is attained and the ultimate yield of ammonia. Moreover, the reaction vessel's volume, in this case, 1.0 liter, directly impacts the concentrations of the gases, thereby affecting the equilibrium position. By meticulously analyzing these initial conditions in conjunction with the equilibrium constant, we can accurately predict the equilibrium concentrations of all species involved in the reaction.

Furthermore, the equilibrium value of [H₂(g)], provided as 0.12 mol/L, serves as a crucial anchor in our calculations. This piece of information allows us to reverse-engineer the equilibrium concentrations of the other species by utilizing the stoichiometry of the reaction and the equilibrium constant expression. Essentially, we are using the known equilibrium concentration of hydrogen gas to deduce the corresponding changes in the concentrations of nitrogen gas and ammonia as the system progresses towards equilibrium. This method underscores the interconnectedness of the concentrations of all species at equilibrium, highlighting the elegance and predictability of chemical systems.

Problem Statement: Unraveling the Equilibrium Concentrations

We are presented with a classic chemical equilibrium problem centered around the Haber-Bosch process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). The scenario involves a chemist introducing nitrogen gas (N₂) and hydrogen gas (H₂) into a 1.0-liter reaction vessel in a 1:3 molar ratio. The equilibrium constant (K) for this reaction is given as 1.0 x 10⁻⁴, indicating a preference for the reactants at equilibrium. A crucial piece of information is the equilibrium concentration of hydrogen gas, [H₂(g)], which is measured to be 0.12 mol/L. Our objective is to calculate the equilibrium concentrations of nitrogen gas [N₂(g)] and ammonia [NH₃(g)]. This task requires a thorough understanding of equilibrium principles, stoichiometry, and the application of the equilibrium constant expression.

The challenge lies in deciphering the changes in concentrations as the system moves from its initial state to the equilibrium state. We must carefully consider the stoichiometry of the reaction, which dictates the molar relationships between the reactants and products. For every mole of nitrogen gas that reacts, three moles of hydrogen gas are consumed, and two moles of ammonia are produced. This stoichiometric ratio is the cornerstone of our calculations, allowing us to relate the changes in concentrations of each species.

The small value of the equilibrium constant (K = 1.0 x 10⁻⁴) provides an initial clue about the relative amounts of reactants and products at equilibrium. It suggests that the equilibrium position favors the reactants, meaning that only a small fraction of the nitrogen and hydrogen gases will convert into ammonia. This information can guide our approach to solving the problem, potentially allowing us to make simplifying assumptions to ease the calculations. However, it is crucial to validate these assumptions to ensure the accuracy of our results.

Moreover, the given equilibrium concentration of hydrogen gas ([H₂(g)] = 0.12 mol/L) provides a crucial foothold in our calculations. We can use this value, along with the stoichiometry of the reaction, to determine the changes in concentrations of nitrogen gas and ammonia. By working backward from the known equilibrium concentration of hydrogen, we can deduce the extent to which the reaction has proceeded towards product formation. This reverse engineering approach is a powerful tool in equilibrium calculations, allowing us to extract valuable information from limited data.

Methodology: A Step-by-Step Calculation

To accurately calculate the equilibrium concentrations of nitrogen gas [N₂(g)] and ammonia [NH₃(g)], we will employ a systematic approach that combines the principles of chemical equilibrium, stoichiometry, and the equilibrium constant expression. This method involves several key steps:

  1. Defining Initial Concentrations: First, let's denote the initial concentration of N₂(g) as 'x' mol/L. Given the 1:3 ratio of N₂(g) to H₂(g), the initial concentration of H₂(g) will be '3x' mol/L. The initial concentration of NH₃(g) is 0 mol/L as the reaction starts with only the reactants.

  2. Constructing the ICE Table: An ICE (Initial, Change, Equilibrium) table is instrumental in organizing the concentration changes. We know the equilibrium concentration of H₂(g) is 0.12 mol/L. This information allows us to determine the change in concentration of H₂(g), which in turn, helps us find the changes in concentrations of N₂(g) and NH₃(g) based on the stoichiometry of the reaction.

    N₂(g) 3H₂(g) 2NH₃(g)
    Initial (I) x 3x 0
    Change (C) -y -3y +2y
    Equilibrium (E) x-y 3x-3y 2y

    Here, 'y' represents the change in concentration of N₂(g) as the reaction reaches equilibrium.

  3. Determining the Change in Concentrations: From the ICE table and the given equilibrium concentration of H₂(g), we have 3x - 3y = 0.12 mol/L. This equation relates the initial concentration of H₂(g) to its equilibrium concentration and allows us to solve for 'y' in terms of 'x'.

  4. Expressing Equilibrium Constant: The equilibrium constant (K) expression for the reaction is given by:

    K = [NH₃(g)]² / ([N₂(g)] * [H₂(g)]³)

    Substituting the equilibrium concentrations from the ICE table, we get:

    1. 0 x 10⁻⁴ = (2y)² / ((x-y) * (0.12)³)

  5. Solving for Equilibrium Concentrations: We now have two equations: 3x - 3y = 0.12 and the equilibrium constant expression. We can solve these equations simultaneously to find the values of 'x' and 'y'. Solving for 'x' from the first equation gives x = 0.04 + y. Substituting this into the equilibrium constant expression yields a complex equation, which can be simplified by making an assumption. Since K is very small, we can assume that 'y' is much smaller than 'x', and thus, x - y ≈ x. This simplifies the equilibrium constant expression, allowing us to solve for 'y', and subsequently, 'x'.

  6. Calculating [N₂(g)] and [NH₃(g)]: Once we have the values of 'x' and 'y', we can calculate the equilibrium concentrations of N₂(g) and NH₃(g) using the expressions from the ICE table: [N₂(g)] = x - y and [NH₃(g)] = 2y.

This methodical approach ensures that we account for the stoichiometry of the reaction, the equilibrium constant, and the given information to accurately determine the equilibrium concentrations of all species involved.

Calculations: The Numerical Solution

Now, let's put our methodology into action and perform the calculations to determine the equilibrium concentrations of nitrogen gas [N₂(g)] and ammonia [NH₃(g)].

  1. Initial Concentrations: As defined earlier, let the initial concentration of N₂(g) be 'x' mol/L. The initial concentration of H₂(g) is '3x' mol/L, and the initial concentration of NH₃(g) is 0 mol/L.

  2. ICE Table: The ICE table is as follows:

    N₂(g) 3H₂(g) 2NH₃(g)
    Initial (I) x 3x 0
    Change (C) -y -3y +2y
    Equilibrium (E) x-y 3x-3y 2y

  3. Change in Concentrations: We are given that the equilibrium concentration of H₂(g) is 0.12 mol/L. Therefore, from the ICE table, we have:

    3x - 3y = 0.12

    Dividing by 3, we get:

    x - y = 0.04

    So, x = 0.04 + y

  4. Equilibrium Constant Expression: The equilibrium constant (K) expression is:

    K = [NH₃(g)]² / ([N₂(g)] * [H₂(g)]³)

    Substituting the equilibrium concentrations from the ICE table:

    1. 0 x 10⁻⁴ = (2y)² / ((x-y) * (0.12)³)

  5. Solving for Equilibrium Concentrations: Substitute x = 0.04 + y into the equilibrium expression:

    1. 0 x 10⁻⁴ = (4y²) / (0.04 * (0.12)³)

    Since K is small, we assume y << 0.04, simplifying the expression:

    1. 0 x 10⁻⁴ ≈ (4y²) / (0.04 * (0.12)³)

    Solving for y²:

    y² ≈ (1.0 x 10⁻⁴ * 0.04 * (0.12)³) / 4

    y² ≈ 1.728 x 10⁻⁹

    Taking the square root:

    y ≈ 4.16 x 10⁻⁵ mol/L

    Now, we can calculate x:

    x = 0.04 + y ≈ 0.04 + 4.16 x 10⁻⁵ ≈ 0.04 mol/L

  6. Calculating [N₂(g)] and [NH₃(g)]: Finally, we calculate the equilibrium concentrations:

    [N₂(g)] = x - y ≈ 0.04 - 4.16 x 10⁻⁵ ≈ 0.03996 mol/L

    [NH₃(g)] = 2y ≈ 2 * 4.16 x 10⁻⁵ ≈ 8.32 x 10⁻⁵ mol/L

Therefore, the equilibrium concentration of nitrogen gas [N₂(g)] is approximately 0.03996 mol/L, and the equilibrium concentration of ammonia [NH₃(g)] is approximately 8.32 x 10⁻⁵ mol/L.

Conclusion: Equilibrium Achieved

In this comprehensive analysis, we have successfully navigated the intricacies of a chemical equilibrium problem involving the Haber-Bosch process. By meticulously applying the principles of chemical equilibrium, stoichiometry, and the equilibrium constant expression, we have calculated the equilibrium concentrations of nitrogen gas [N₂(g)] and ammonia [NH₃(g)]. Our calculations revealed that at equilibrium, [N₂(g)] is approximately 0.03996 mol/L, and [NH₃(g)] is approximately 8.32 x 10⁻⁵ mol/L.

Our step-by-step approach, beginning with the definition of initial concentrations and culminating in the calculation of equilibrium concentrations, highlights the power of systematic problem-solving in chemistry. The use of the ICE table proved invaluable in organizing the changes in concentrations, while the equilibrium constant expression provided the mathematical framework for relating the concentrations of reactants and products at equilibrium. The small value of the equilibrium constant (K = 1.0 x 10⁻⁴) guided our simplification, allowing us to make a reasonable assumption that significantly eased the calculations.

The results of our calculations align with the initial indication provided by the small equilibrium constant value. The equilibrium concentrations show that the system favors the reactants, with a significantly higher concentration of nitrogen gas compared to ammonia. This observation underscores the importance of the equilibrium constant as a predictor of the relative amounts of reactants and products at equilibrium. It is crucial to recognize that the equilibrium position is a dynamic state, where the forward and reverse reactions occur at equal rates, maintaining a constant ratio of reactants and products.

This exercise also demonstrates the practical applications of chemical equilibrium principles in industrial processes. The Haber-Bosch process, which we have analyzed, is a cornerstone of ammonia production, a vital component in fertilizers. Understanding and manipulating the equilibrium conditions of this reaction is essential for optimizing ammonia synthesis and meeting the global demand for fertilizers. By mastering the concepts and techniques presented in this analysis, students and professionals alike can gain a deeper appreciation for the elegance and utility of chemical equilibrium in various scientific and industrial contexts.