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Ammonia, a very important industrial chemical, is produced by the direct combination of the elements under carefully controlled conditions. $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ Suppose, in an experiment, that the reaction mixture is analyzed after equilibrium is reached and it is found, at a particular temperature, that \(\left[\mathrm{NH}_{3}(g)\right]\) \(=0.34 M,\left[\mathrm{H}_{2}(g)\right]=2.1 \times 10^{-3} \mathrm{M},\) and \(\left[\mathrm{N}_{2}(g)\right]=4.9\) \(\times 10^{-4} M .\) Calculate the value of \(K\) at this temperature.

Short Answer

Expert verified
The value of the equilibrium constant (K) for the production of ammonia at the given temperature is approximately \(2.57 \times 10^{11}\).

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the production of ammonia is: \( \mathrm{N}_{2}(g) + 3\mathrm{H}_{2}(g) \rightleftharpoons 2\mathrm{NH}_{3}(g) \)
02

Identify the concentrations at equilibrium

At equilibrium, we are given the following concentrations: \( [\mathrm{NH_{3}}] = 0.34 \, M \) \( [\mathrm{H_{2}}] = 2.1 \times 10^{-3} \, M \) \( [\mathrm{N_{2}}] = 4.9 \times 10^{-4} \, M \)
03

Write the expression for the equilibrium constant, K

For the balanced chemical equation, the equilibrium constant (K) is given by: \( K = \dfrac{[NH_{3}]^2}{[N_{2}][H_{2}]^3} \)
04

Substitute the equilibrium concentrations into the expression for K

After substituting the equilibrium concentrations into the K expression, we get: \( K = \dfrac{(0.34)^2}{(4.9 \times 10^{-4})(2.1 \times 10^{-3})^3} \)
05

Calculate the value of K

Simplify the expression and calculate the value of K: \( K = \dfrac{0.1156}{(4.9 \times 10^{-4})(9.261 \times 10^{-9})} = \dfrac{0.1156}{4.49 \times 10^{-12}} \) \( K \approx 2.57 \times 10^{11} \) The value of the equilibrium constant (K) for the production of ammonia at the given temperature is approximately \(2.57 \times 10^{11}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant Calculation
The concept of chemical equilibrium is a cornerstone in understanding chemical reactions, particularly, how to calculate the equilibrium constant, symbolized as \( K \). This value quantifies the ratio of the concentrations of products to reactants at equilibrium and is a direct indicator of the position of equilibrium.

To calculate \( K \), you first need a balanced chemical equation. From there, you write the equilibrium expression where the concentrations of products are divided by the concentrations of reactants, each raised to the power of their respective stoichiometric coefficients.

For the production of ammonia (\( NH_3 \)), the balanced equation is \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \), and thus the equilibrium constant expression is \( K = \frac{[NH_3]^2}{[N_2][H_2]^3} \). You plug in the equilibrium concentrations provided and solve for \( K \). The calculated equilibrium constant is essential as it reflects the extent to which a reaction will proceed under given conditions, and in this case, it suggests a strong bias toward the formation of ammonia.
Ammonia Synthesis
Ammonia synthesis is one of the most significant industrial chemical processes, known as the Haber process. It produces ammonia by combining nitrogen and hydrogen gases under high pressure and temperature in the presence of a catalyst.

The balanced chemical equation for ammonia synthesis is fundamental to understanding the reaction: \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \). Efficient ammonia synthesis requires careful control of reaction conditions to maximize yield while considering energy and cost efficiency.

Understanding the principles of reaction equilibrium and le Chatelier's principle is essential for optimizing ammonia production. This process is crucial for fertilizers and has a global impact on agricultural productivity and sustainability.
Le Chatelier's Principle
Le Chatelier's principle is a guiding concept in chemistry that describes how a system at equilibrium responds to a change in conditions. If a change is imposed on a system at equilibrium, the system will adjust to counteract that change and restore a new equilibrium.

This principle applies to ammonia synthesis when, for instance, pressure, temperature, or concentrations are altered. An increase in pressure or a decrease in temperature will shift the equilibrium towards the products, favoring the formation of ammonia. Similarly, removing ammonia from the system, increasing the concentration of reactants, or using a catalyst will also drive the reaction towards product formation.

Le Chatelier's principle helps industrial chemists to manipulate conditions to increase the yield of ammonia, optimizing the process both economically and practically.
Equilibrium Concentrations
Equilibrium concentrations are the amounts of reactants and products present when a chemical reaction reaches a state of balance. At this point, the rate of the forward reaction equals the rate of the reverse reaction, and no further change in concentrations occurs.

The values of these concentrations are used in the calculation of the equilibrium constant (\( K \)) and are indicators of the reaction's extent under specific conditions. These values aren't static and can change in response to alterations in conditions such as temperature and pressure, following le Chatelier's principle.

In the exercise provided, finding the equilibrium concentrations allowed for the calculation of \( K \), demonstrating that even from seemingly straightforward observations, complex and essential quantitative information about the reaction can be deduced. This knowledge is paramount for tailoring conditions to achieve desired chemical outcomes.

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Most popular questions from this chapter

The three common silver halides (AgCl, AgBr, and AgI) are all sparingly soluble salts. Given the values for \(K_{\mathrm{sp}}\) for these salts below, calculate the concentration of silver ion, in \(\mathrm{mol} / \mathrm{L},\) in a saturated solution of each salt. $$ \text {Silver Halide } \quad \quad\quad\quad \text {\(K_{\mathrm{sp}}\)} $$ $$ \text {\(\mathrm{AgCl}\) } \quad \quad \text {\(1.8 \times 10^{-10}\)} $$ $$ \text {\(\mathrm{AgBr}\) } \quad \quad \text {\(5.0 \times 10^{-13}\)} $$ $$ \text {\(\mathrm{Agl}\) } \quad \quad \text {\(8.3 \times 10^{-17}\)} $$

Write the equilibrium expression for each of the following reactions. a. \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) b. \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) c. \(\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)\)

At high temperatures, elemental bromine, \(\mathrm{Br}_{2}\), dissociates into individual bromine atoms. $$\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)$$ Suppose that in an experiment at \(2000^{\circ} \mathrm{C},\) it is found that \(\left[\mathrm{Br}_{2}\right]=0.97 \mathrm{M}\) and \([\mathrm{Br}]=0.034 \mathrm{M}\) at equilibrium. Calculate the value of \(K\).

Suppose the reaction system $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ has already reached equilibrium. Predict the effect of each of the following changes on the position of the equilibrium. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected. a. Additional oxygen is injected into the system. b. \(\mathrm{NO}_{2}\) is removed from the reaction vessel. c. 1.0 mol of helium is injected into the system.

Write the equilibrium expression for each of the following reactions. a. \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) b. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) c. \(2 \mathrm{NO}_{2}(g)+\mathrm{O}_{3}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{O}_{2}(g)\)

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