The Reaction of Chlorine and Water

Oct 13, 2025 by pynkocean.com 77 views

Introduction

In chemistry, the equilibrium constant ( $K_{\text{eq}}$ ) is a crucial concept that helps us understand the balance between reactants and products in a chemical reaction. It is a measure of the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. In this article, we will explore the concept of $K_{\text{eq}}$ and its application to a specific reaction: the reaction between chlorine and water.

The Reaction of Chlorine and Water

The reaction between chlorine and water is a complex process that involves the formation of hydrogen chloride (HCl) and oxygen gas (O2). The balanced chemical equation for this reaction is:

2 Cl2(g) + 2 H2O(g) ⇌ 4 HCl(g) + O2(g)

This reaction is an example of a heterogeneous equilibrium, where the reactants and products are in different phases (gas and liquid).

The Equilibrium Constant ( $K_{\text{eq}}$ )

The equilibrium constant ( $K_{\text{eq}}$ ) is a dimensionless quantity that is calculated using the concentrations of the reactants and products at equilibrium. It is defined as:

K_{\text{eq}} = \frac{[\text{HCl}]^4 [\text{O}_2]}{[\text{Cl}_2]^2 [\text{H}_2\text{O}]^2}

where [HCl], [O2], [Cl2], and [H2O] are the concentrations of hydrogen chloride, oxygen, chlorine, and water, respectively.

Calculating the Value of $K_{\text{eq}}$

Given that the value of $K_{\text{eq}}$ is $7.52 \cdot 10^{-2}$ at $480.0^{\circ} C$ , we can use this information to calculate the concentrations of the reactants and products at equilibrium.

Step 1: Write the equilibrium expression

The equilibrium expression is:

K_{\text{eq}} = \frac{[\text{HCl}]^4 [\text{O}_2]}{[\text{Cl}_2]^2 [\text{H}_2\text{O}]^2}

Step 2: Rearrange the equilibrium expression

To solve for the concentrations of the reactants and products, we need to rearrange the equilibrium expression:

[\text{HCl}]^4 [\text{O}2] = K{\text{eq}} [\text{Cl}_2]^2 [\text{H}_2\text{O}]^2

Step 3: Take the fourth root of both sides

To isolate the concentration of HCl, we take the fourth root of both sides:

[\text{HCl}] = \sqrt[4]{K_{\text{eq}} [\text{Cl}_2]^2 [\text{H}_2\text{O}]^2}

Step 4: Take the square root of both sides

To isolate the concentration of O2, we take the square root of both sides:

[\text{O}2] = \sqrt{K{\text{eq}} [\text{Cl}_2]^2 [\text{H}_2\text{O}]^2}

Step 5: Substitute the given values

We are given that the value of $K_{\text{eq}}$ is $7.52 \cdot 10^{-2}$ at $480.0^{\circ} C$ . We can substitute this value into the equations:

[\text{HCl}] = \sqrt[4]{7.52 \cdot 10^{-2} [\text{Cl}_2]^2 [\text{H}_2\text{O}]^2}

[\text{O}_2] = \sqrt{7.52 \cdot 10^{-2} [\text{Cl}_2]^2 [\text{H}_2\text{O}]^2}

Step 6: Simplify the equations

We can simplify the equations by canceling out the common factors:

[\text{HCl}] = \sqrt[4]{7.52 \cdot 10^{-2} [\text{Cl}_2]^2}

[\text{O}_2] = \sqrt{7.52 \cdot 10^{-2} [\text{Cl}_2]^2}

Step 7: Solve for the concentrations

We can solve for the concentrations of HCl and O2 by substituting the given values:

[\text{HCl}] = \sqrt[4]{7.52 \cdot 10^{-2} (1)^2} = \sqrt[4]{7.52 \cdot 10^{-2}} = 0.27

[\text{O}_2] = \sqrt{7.52 \cdot 10^{-2} (1)^2} = \sqrt{7.52 \cdot 10^{-2}} = 0.27

Conclusion

In this article, we have explored the concept of the equilibrium constant ( $K_{\text{eq}}$ ) and its application to a specific reaction: the reaction between chlorine and water. We have calculated the value of $K_{\text{eq}}$ using the given concentrations of the reactants and products at equilibrium. The results show that the concentration of HCl is 0.27 and the concentration of O2 is 0.27.

References

Atkins, P. W., & de Paula, J. (2010). Physical chemistry (9th ed.). Oxford University Press.
Chang, R. (2010). Chemistry (10th ed.). McGraw-Hill.
Levine, I. N. (2012). Physical chemistry (6th ed.). McGraw-Hill.

Q: What is the equilibrium constant ( $K_{\text{eq}}$ )?

A: The equilibrium constant ( $K_{\text{eq}}$ ) is a dimensionless quantity that is calculated using the concentrations of the reactants and products at equilibrium. It is a measure of the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium.

Q: How is the equilibrium constant ( $K_{\text{eq}}$ ) calculated?

A: The equilibrium constant ( $K_{\text{eq}}$ ) is calculated using the concentrations of the reactants and products at equilibrium. The formula for calculating $K_{\text{eq}}$ is:

K_{\text{eq}} = \frac{[\text{HCl}]^4 [\text{O}_2]}{[\text{Cl}_2]^2 [\text{H}_2\text{O}]^2}

where [HCl], [O2], [Cl2], and [H2O] are the concentrations of hydrogen chloride, oxygen, chlorine, and water, respectively.

Q: What is the significance of the equilibrium constant ( $K_{\text{eq}}$ )?

A: The equilibrium constant ( $K_{\text{eq}}$ ) is a crucial concept in chemistry that helps us understand the balance between reactants and products in a chemical reaction. It is a measure of the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium.

Q: How is the equilibrium constant ( $K_{\text{eq}}$ ) related to the concentrations of the reactants and products?

A: The equilibrium constant ( $K_{\text{eq}}$ ) is related to the concentrations of the reactants and products through the formula:

K_{\text{eq}} = \frac{[\text{HCl}]^4 [\text{O}_2]}{[\text{Cl}_2]^2 [\text{H}_2\text{O}]^2}

This formula shows that the equilibrium constant ( $K_{\text{eq}}$ ) is a function of the concentrations of the reactants and products.

Q: Can the equilibrium constant ( $K_{\text{eq}}$ ) be used to predict the direction of a chemical reaction?

A: Yes, the equilibrium constant ( $K_{\text{eq}}$ ) can be used to predict the direction of a chemical reaction. If the value of $K_{\text{eq}}$ is greater than 1, the reaction will proceed in the forward direction. If the value of $K_{\text{eq}}$ is less than 1, the reaction will proceed in the reverse direction.

Q: How is the equilibrium constant ( $K_{\text{eq}}$ ) affected by temperature?

A: The equilibrium constant ( $K_{\text{eq}}$ ) is affected by temperature. As the temperature increases, the value of $K_{\text{eq}}$ also increases. This is because the rate of reaction increases with temperature, which means that the equilibrium constant ( $K_{\text{eq}}$ ) also increases.

Q: Can the equilibrium constant ( $K_{\text{eq}}$ ) be used to calculate the concentrations of the reactants and products?

A: Yes, the equilibrium constant ( $K_{\text{eq}}$ ) can be used to calculate the concentrations of the reactants and products. By rearranging the formula for $K_{\text{eq}}$ , we can solve for the concentrations of the reactants and products.

Q: What is the relationship between the equilibrium constant ( $K_{\text{eq}}$ ) and the reaction quotient ( $Q$ )?

A: The equilibrium constant ( $K_{\text{eq}}$ ) and the reaction quotient ( $Q$ ) are related through the formula:

Q = K_{\text{eq}} \cdot \frac{[\text{HCl}]^4 [\text{O}_2]}{[\text{Cl}_2]^2 [\text{H}_2\text{O}]^2}

This formula shows that the reaction quotient ( $Q$ ) is a function of the equilibrium constant ( $K_{\text{eq}}$ ) and the concentrations of the reactants and products.

Q: Can the equilibrium constant ( $K_{\text{eq}}$ ) be used to predict the stability of a chemical reaction?

A: Yes, the equilibrium constant ( $K_{\text{eq}}$ ) can be used to predict the stability of a chemical reaction. If the value of $K_{\text{eq}}$ is greater than 1, the reaction is stable. If the value of $K_{\text{eq}}$ is less than 1, the reaction is unstable.

Conclusion

In this Q&A article, we have explored the concept of the equilibrium constant ( $K_{\text{eq}}$ ) and its application to a specific reaction: the reaction between chlorine and water. We have answered questions about the significance of the equilibrium constant ( $K_{\text{eq}}$ ), its relationship to the concentrations of the reactants and products, and its use in predicting the direction and stability of a chemical reaction.

References

Atkins, P. W., & de Paula, J. (2010). Physical chemistry (9th ed.). Oxford University Press.
Chang, R. (2010). Chemistry (10th ed.). McGraw-Hill.
Levine, I. N. (2012). Physical chemistry (6th ed.). McGraw-Hill.