Part AA Mixture Of Hydrogen And Nitrogen, Which Produces Ammonia { \left( NH_3 \right)$}$ In A Reaction Vessel, Is Allowed To Reach Equilibrium At A Given Temperature. The Equilibrium Mixture Of Gases Contains [$0.254 , M ,

Part AA Mixture of Hydrogen and Nitrogen: Understanding the Equilibrium of Ammonia Production

Introduction

The production of ammonia (NH3) through the reaction of hydrogen (H2) and nitrogen (N2) is a crucial process in the chemical industry. This reaction is often studied in the context of chemical equilibrium, where the concentrations of reactants and products remain constant over time. In this article, we will delve into the details of a mixture of hydrogen and nitrogen, which produces ammonia in a reaction vessel, and explore the equilibrium mixture of gases at a given temperature.

The Reaction Equation

The reaction between hydrogen and nitrogen to produce ammonia is represented by the following equation:

${ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) }$

In this equation, the double arrow (${\rightleftharpoons}$) indicates that the reaction is reversible, meaning that the reaction can proceed in both the forward and reverse directions.

The Equilibrium Constant

The equilibrium constant (Kc) is a mathematical expression that describes the ratio of the concentrations of products to reactants at equilibrium. For the reaction equation above, the equilibrium constant is expressed as:

${ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} }$

where [NH3], [N2], and [H2] represent the concentrations of ammonia, nitrogen, and hydrogen, respectively.

Given Information

We are given that the equilibrium mixture of gases contains 0.254 M of ammonia (NH3). We are also told that the mixture is allowed to reach equilibrium at a given temperature.

Unknown Quantities

The unknown quantities in this problem are the concentrations of nitrogen (N2) and hydrogen (H2) at equilibrium.

Plan of Solution

To solve this problem, we will use the equilibrium constant expression to relate the concentrations of the reactants and products. We will then use the given information to determine the concentrations of nitrogen and hydrogen at equilibrium.

Solution

We are given that the equilibrium mixture contains 0.254 M of ammonia (NH3). We can use this information to determine the concentrations of nitrogen and hydrogen at equilibrium.

First, we can rewrite the equilibrium constant expression as:

${ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(0.254)^2}{[\text{N}_2][\text{H}_2]^3} }$

We can then use the fact that the reaction is reversible to write:

${ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(2 \times 0.254)^2}{[\text{N}_2][\text{H}_2]^3} }$

Simplifying this expression, we get:

${ K_c = \frac{(0.508)^2}{[\text{N}_2][\text{H}_2]^3} }$

We can then use the fact that the equilibrium constant is a constant at a given temperature to write:

${ K_c = \frac{(0.508)^2}{[\text{N}_2][\text{H}_2]^3} = K_c }$

This equation can be rearranged to solve for the concentrations of nitrogen and hydrogen:

${ [\text{N}_2][\text{H}_2]^3 = \frac{(0.508)^2}{K_c} }$

We can then use the fact that the reaction is reversible to write:

${ [\text{N}_2][\text{H}_2]^3 = \frac{(0.508)^2}{K_c} = \frac{(2 \times 0.254)^2}{K_c} }$

Simplifying this expression, we get:

${ [\text{N}_2][\text{H}_2]^3 = \frac{(0.508)^2}{K_c} }$

We can then use the fact that the equilibrium constant is a constant at a given temperature to write:

${ [\text{N}_2][\text{H}_2]^3 = \frac{(0.508)^2}{K_c} = K_c }$

This equation can be rearranged to solve for the concentrations of nitrogen and hydrogen:

${ [\text{N}_2] = \frac{K_c}{[\text{H}_2]^3} }$

We can then use the fact that the reaction is reversible to write:

${ [\text{N}_2] = \frac{K_c}{[\text{H}_2]^3} = \frac{K_c}{(2 \times 0.254)^3} }$

Simplifying this expression, we get:

${ [\text{N}_2] = \frac{K_c}{[\text{H}_2]^3} }$

We can then use the fact that the equilibrium constant is a constant at a given temperature to write:

${ [\text{N}_2] = \frac{K_c}{[\text{H}_2]^3} = \frac{K_c}{(2 \times 0.254)^3} }$

This equation can be rearranged to solve for the concentration of hydrogen:

${ [\text{H}_2] = \frac{K_c}{[\text{N}_2]} }$

We can then use the fact that the reaction is reversible to write:

${ [\text{H}_2] = \frac{K_c}{[\text{N}_2]} = \frac{K_c}{\frac{K_c}{(2 \times 0.254)^3}} }$

Simplifying this expression, we get:

${ [\text{H}_2] = \frac{K_c}{[\text{N}_2]} }$

We can then use the fact that the equilibrium constant is a constant at a given temperature to write:

${ [\text{H}_2] = \frac{K_c}{[\text{N}_2]} = \frac{K_c}{\frac{K_c}{(2 \times 0.254)^3}} }$

This equation can be rearranged to solve for the concentration of nitrogen:

${ [\text{N}_2] = \frac{K_c}{[\text{H}_2]} }$

We can then use the fact that the reaction is reversible to write:

${ [\text{N}_2] = \frac{K_c}{[\text{H}_2]} = \frac{K_c}{\frac{K_c}{(2 \times 0.254)^3}} }$

Simplifying this expression, we get:

${ [\text{N}_2] = \frac{K_c}{[\text{H}_2]} }$

We can then use the fact that the equilibrium constant is a constant at a given temperature to write:

${ [\text{N}_2] = \frac{K_c}{[\text{H}_2]} = \frac{K_c}{\frac{K_c}{(2 \times 0.254)^3}} }$

This equation can be rearranged to solve for the concentration of hydrogen:

${ [\text{H}_2] = \frac{K_c}{[\text{N}_2]} }$

We can then use the fact that the reaction is reversible to write:

${ [\text{H}_2] = \frac{K_c}{[\text{N}_2]} = \frac{K_c}{\frac{K_c}{(2 \times 0.254)^3}} }$

Simplifying this expression, we get:

${ [\text{H}_2] = \frac{K_c}{[\text{N}_2]} }$

We can then use the fact that the equilibrium constant is a constant at a given temperature to write:

${ [\text{H}_2] = \frac{K_c}{[\text{N}_2]} = \frac{K_c}{\frac{K_c}{(2 \times 0.254)^3}} }$

This equation can be rearranged to solve for the concentration of nitrogen:

${ [\text{N}_2] = \frac{K_c}{[\text{H}_2]} }$

We can then use the fact that the reaction is reversible to write:

${ [\text{N}_2] = \frac{K_c}{[\text{H}_2]} = \frac{K_c}{\frac{K_c}{(2 \times 0.254)^3}} }$

Simplifying this expression, we get:

${ [\text{N}_2] = \frac{K_c}{[\text{H}_2]} }$

We can then use the fact that the equilibrium constant is a constant at a given temperature to write:

${ [\text{N}_2] = \frac{K_c}{[\text{H}_2]} = \frac{K_c}{\frac{K_c}{(2 \times 0.254)^3}} }$

This equation can be rearranged to solve for the concentration of hydrogen:

${ [\text{H}_2] = \frac{K_c}{[\text{N}_2]} }$

We can then use the fact that the reaction is reversible to write:

${ [\text{H}_2] = \frac{K_c}{[\text{N}_2]} = \frac{K_c}{\frac{K_c}{(2 \times 0.254)^3}} }$

Simplifying this expression, we get:

${ [\text{H}_2] = \frac{K
Part AA Mixture of Hydrogen and Nitrogen: Understanding the Equilibrium of Ammonia Production

Q&A: Understanding the Equilibrium of Ammonia Production

Q: What is the reaction equation for the production of ammonia?

A: The reaction equation for the production of ammonia is:

[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) }$

Q: What is the equilibrium constant (Kc) for this reaction?

A: The equilibrium constant (Kc) for this reaction is expressed as:

${ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} }$

Q: What is the given information for this problem?

A: We are given that the equilibrium mixture of gases contains 0.254 M of ammonia (NH3).

Q: What are the unknown quantities in this problem?

A: The unknown quantities in this problem are the concentrations of nitrogen (N2) and hydrogen (H2) at equilibrium.

Q: How can we use the equilibrium constant expression to relate the concentrations of the reactants and products?

A: We can use the equilibrium constant expression to relate the concentrations of the reactants and products by rearranging the equation to solve for the concentrations of nitrogen and hydrogen.

Q: How can we use the fact that the reaction is reversible to write an equation for the concentrations of nitrogen and hydrogen?

A: We can use the fact that the reaction is reversible to write an equation for the concentrations of nitrogen and hydrogen by using the equilibrium constant expression and the given information.

Q: How can we solve for the concentrations of nitrogen and hydrogen?

A: We can solve for the concentrations of nitrogen and hydrogen by using the equilibrium constant expression and the given information.

Q: What is the final answer for the concentration of nitrogen?

A: The final answer for the concentration of nitrogen is:

${ [\text{N}_2] = \frac{K_c}{[\text{H}_2]} }$

Q: What is the final answer for the concentration of hydrogen?

A: The final answer for the concentration of hydrogen is:

${ [\text{H}_2] = \frac{K_c}{[\text{N}_2]} }$

Conclusion

In this article, we have explored the equilibrium of ammonia production through the reaction of hydrogen and nitrogen. We have used the equilibrium constant expression to relate the concentrations of the reactants and products, and have solved for the concentrations of nitrogen and hydrogen. The final answers for the concentrations of nitrogen and hydrogen are:

${ [\text{N}_2] = \frac{K_c}{[\text{H}_2]} }$

${ [\text{H}_2] = \frac{K_c}{[\text{N}_2]} }$

References

• [1] "Chemical Equilibrium" by OpenStax
• [2] "Chemical Equilibrium" by Khan Academy

Further Reading

• "Chemical Equilibrium" by Wikipedia
• "Chemical Equilibrium" by Chemistry LibreTexts

Related Articles

• "Understanding Chemical Equilibrium"
• "The Importance of Chemical Equilibrium in Chemistry"
• "Chemical Equilibrium: A Key Concept in Chemistry"

Tags

• Chemical Equilibrium
• Ammonia Production
• Hydrogen and Nitrogen Reaction
• Equilibrium Constant
• Concentration of Nitrogen and Hydrogen