The Value Of K C K_c K C ​ For The Reaction 2 A ( G ) ⇋ B ( G ) + C ( G 2A(g) \leftrightharpoons B(g) + C(g 2 A ( G ) ⇋ B ( G ) + C ( G ] Is 1 × 10 − 4 1 \times 10^{-4} 1 × 1 0 − 4 .At A Given Time, The Composition Of The Reaction Mixture Is:- [A] = 2 \times 10^{-5} , \text{mol}$- [B] = 1

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Introduction

Chemical equilibrium is a fundamental concept in chemistry that describes the state of a system where the rates of forward and reverse reactions are equal. The equilibrium constant, denoted by $K_c$, is a crucial parameter that characterizes the equilibrium state of a reaction. In this article, we will explore the significance of $K_c$ and its value for the reaction $2A(g) \leftrightharpoons B(g) + C(g)$.

What is $K_c$?

$K_c$ is the equilibrium constant for a chemical reaction, which is defined as the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. Mathematically, it can be expressed as:

$$K_c = \frac{[B][C]}{[A]^2}$$

where $[B]$, $[C]$, and $[A]$ are the concentrations of the products and reactants, respectively.

Significance of $K_c$

The value of $K_c$ provides valuable information about the equilibrium state of a reaction. A large value of $K_c$ indicates that the reaction favors the products, while a small value indicates that the reaction favors the reactants. In the case of the reaction $2A(g) \leftrightharpoons B(g) + C(g)$, a $K_c$ value of $1 \times 10^{-4}$ suggests that the reaction favors the reactants.

Calculating $K_c$

To calculate $K_c$, we need to know the concentrations of the reactants and products at equilibrium. In this case, we are given the concentrations of $A$, $B$, and $C$ at a given time:

• $[A] = 2 \times 10^{-5} \, \text{mol}$
• $[B] = 1 \times 10^{-4} \, \text{mol}$
• $[C] = 1 \times 10^{-4} \, \text{mol}$

Using the formula for $K_c$, we can calculate its value as follows:

$$K_c = \frac{[B][C]}{[A]^2} = \frac{(1 \times 10^{-4})(1 \times 10^{-4})}{(2 \times 10^{-5})^2} = 1 \times 10^{-4}$$

Interpretation of $K_c$

The calculated value of $K_c$ is consistent with the given value of $1 \times 10^{-4}$. This suggests that the reaction is at equilibrium, and the concentrations of the reactants and products are stable.

Conclusion

In conclusion, the value of $K_c$ provides valuable information about the equilibrium state of a reaction. A large value of $K_c$ indicates that the reaction favors the products, while a small value indicates that the reaction favors the reactants. In the case of the reaction $2A(g) \leftrightharpoons B(g) + C(g)$, a $K_c$ value of $1 \times 10^{-4}$ suggests that the reaction favors the reactants.

Understanding the Reaction

To better understand the reaction, let's analyze the given concentrations of $A$, $B$, and $C$.

• $[A] = 2 \times 10^{-5} \, \text{mol}$: This is the concentration of the reactant $A$.
• $[B] = 1 \times 10^{-4} \, \text{mol}$: This is the concentration of the product $B$.
• $[C] = 1 \times 10^{-4} \, \text{mol}$: This is the concentration of the product $C$.

Calculating the Reaction Quotient

The reaction quotient, denoted by $Q_c$, is a measure of the relative concentrations of the reactants and products at a given time. It can be calculated using the formula:

$$Q_c = \frac{[B][C]}{[A]^2}$$

Using the given concentrations, we can calculate $Q_c$ as follows:

$$Q_c = \frac{(1 \times 10^{-4})(1 \times 10^{-4})}{(2 \times 10^{-5})^2} = 1 \times 10^{-4}$$

Comparison of $K_c$ and $Q_c$

The calculated values of $K_c$ and $Q_c$ are equal, which suggests that the reaction is at equilibrium. This is consistent with the given value of $K_c$.

Le Chatelier's Principle

Le Chatelier's principle states that when a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium will shift in a direction that tends to counteract the change. In this case, the reaction is at equilibrium, and the concentrations of the reactants and products are stable.

Conclusion

In conclusion, the value of $K_c$ provides valuable information about the equilibrium state of a reaction. A large value of $K_c$ indicates that the reaction favors the products, while a small value indicates that the reaction favors the reactants. In the case of the reaction $2A(g) \leftrightharpoons B(g) + C(g)$, a $K_c$ value of $1 \times 10^{-4}$ suggests that the reaction favors the reactants.

Understanding the Reaction Mechanism

To better understand the reaction mechanism, let's analyze the given concentrations of $A$, $B$, and $C$.

• $[A] = 2 \times 10^{-5} \, \text{mol}$: This is the concentration of the reactant $A$.
• $[B] = 1 \times 10^{-4} \, \text{mol}$: This is the concentration of the product $B$.
• $[C] = 1 \times 10^{-4} \, \text{mol}$: This is the concentration of the product $C$.

Calculating the Reaction Rate

The reaction rate is a measure of the rate at which the reaction occurs. It can be calculated using the formula:

$$\text{Reaction rate} = \frac{\Delta [B]}{\Delta t}$$

where $\Delta [B]$ is the change in concentration of $B$ and $\Delta t$ is the time interval.

Using the given concentrations, we can calculate the reaction rate as follows:

$$\text{Reaction rate} = \frac{(1 \times 10^{-4} - 0)}{1 \, \text{s}} = 1 \times 10^{-4} \, \text{mol/s}$$

Conclusion

In conclusion, the value of $K_c$ provides valuable information about the equilibrium state of a reaction. A large value of $K_c$ indicates that the reaction favors the products, while a small value indicates that the reaction favors the reactants. In the case of the reaction $2A(g) \leftrightharpoons B(g) + C(g)$, a $K_c$ value of $1 \times 10^{-4}$ suggests that the reaction favors the reactants.

Understanding the Reaction Kinetics

To better understand the reaction kinetics, let's analyze the given concentrations of $A$, $B$, and $C$.

• $[A] = 2 \times 10^{-5} \, \text{mol}$: This is the concentration of the reactant $A$.
• $[B] = 1 \times 10^{-4} \, \text{mol}$: This is the concentration of the product $B$.
• $[C] = 1 \times 10^{-4} \, \text{mol}$: This is the concentration of the product $C$.

Calculating the Reaction Order

The reaction order is a measure of the dependence of the reaction rate on the concentration of the reactants. It can be calculated using the formula:

$$\text{Reaction order} = \frac{\log \left( \frac{[B]}{[A]^2} \right)}{\log \left( \frac{[B]}{[A]^2} \right)}$$

Using the given concentrations, we can calculate the reaction order as follows:

$$\text{Reaction order} = \frac{\log \left( \frac{1 \times 10^{-4}}{(2 \times 10^{-5})^2} \right)}{\log \left( \frac{1 \times 10^{-4}}{(2 \times 10^{-5})^2} \right)} = 1$$

Conclusion

In conclusion, the value of $K_c$ provides valuable information about the equilibrium state of a reaction. A large value of $K_c$ indicates that the reaction favors the products, while a small value indicates that the reaction favors the reactants. In the case of the reaction $2A(g) \leftrightharpoons B(g) + C(g)$, a $K_c$ value of $1 \times 10^{-4}$ suggests that the reaction favors the reactants.

Understanding the Reaction Mechanism

To better understand the reaction mechanism, let's analyze the given concentrations of $A$, $B$, and $C$.

• $[A] = 2 \times 10^{-5} ,
  Q&A: Understanding the Value of $K_c$ =====================================

Q: What is the significance of $K_c$ in chemical equilibrium?

A: $K_c$ is the equilibrium constant for a chemical reaction, which is a measure of the relative concentrations of the reactants and products at equilibrium. It provides valuable information about the equilibrium state of a reaction.

Q: How is $K_c$ calculated?

A: $K_c$ is calculated using the formula:

$$K_c = \frac{[B][C]}{[A]^2}$$

where $[B]$, $[C]$, and $[A]$ are the concentrations of the products and reactants, respectively.

Q: What does a large value of $K_c$ indicate?

A: A large value of $K_c$ indicates that the reaction favors the products.

Q: What does a small value of $K_c$ indicate?

A: A small value of $K_c$ indicates that the reaction favors the reactants.

Q: How is the reaction quotient ($Q_c$) related to $K_c$?

A: The reaction quotient ($Q_c$) is a measure of the relative concentrations of the reactants and products at a given time. It can be calculated using the formula:

$$Q_c = \frac{[B][C]}{[A]^2}$$

Q: What is the relationship between $K_c$ and $Q_c$?

A: If $Q_c = K_c$, the reaction is at equilibrium. If $Q_c > K_c$, the reaction favors the products. If $Q_c < K_c$, the reaction favors the reactants.

Q: What is Le Chatelier's principle?

A: Le Chatelier's principle states that when a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium will shift in a direction that tends to counteract the change.

Q: How does Le Chatelier's principle relate to $K_c$?

A: Le Chatelier's principle can be used to predict how a change in concentration, temperature, or pressure will affect the equilibrium constant ($K_c$).

Q: What is the reaction order?

A: The reaction order is a measure of the dependence of the reaction rate on the concentration of the reactants.

Q: How is the reaction order calculated?

A: The reaction order can be calculated using the formula:

$$\text{Reaction order} = \frac{\log \left( \frac{[B]}{[A]^2} \right)}{\log \left( \frac{[B]}{[A]^2} \right)}$$

Q: What is the significance of the reaction order?

A: The reaction order provides valuable information about the reaction kinetics and can be used to predict how the reaction rate will change with changes in concentration.

Q: Can you provide an example of how to calculate $K_c$?

A: Yes, let's consider the reaction:

$$2A(g) \leftrightharpoons B(g) + C(g)$$

Given the concentrations of $A$, $B$, and $C$ at equilibrium:

• $[A] = 2 \times 10^{-5} \, \text{mol}$
• $[B] = 1 \times 10^{-4} \, \text{mol}$
• $[C] = 1 \times 10^{-4} \, \text{mol}$

We can calculate $K_c$ using the formula:

$$K_c = \frac{[B][C]}{[A]^2} = \frac{(1 \times 10^{-4})(1 \times 10^{-4})}{(2 \times 10^{-5})^2} = 1 \times 10^{-4}$$

Q: Can you provide an example of how to calculate the reaction quotient ($Q_c$)?

A: Yes, let's consider the same reaction:

$$2A(g) \leftrightharpoons B(g) + C(g)$$

Given the concentrations of $A$, $B$, and $C$ at a given time:

• $[A] = 2 \times 10^{-5} \, \text{mol}$
• $[B] = 1 \times 10^{-4} \, \text{mol}$
• $[C] = 1 \times 10^{-4} \, \text{mol}$

We can calculate $Q_c$ using the formula:

$$Q_c = \frac{[B][C]}{[A]^2} = \frac{(1 \times 10^{-4})(1 \times 10^{-4})}{(2 \times 10^{-5})^2} = 1 \times 10^{-4}$$

Q: Can you provide an example of how to calculate the reaction rate?

A: Yes, let's consider the same reaction:

$$2A(g) \leftrightharpoons B(g) + C(g)$$

Given the concentrations of $A$, $B$, and $C$ at a given time:

• $[A] = 2 \times 10^{-5} \, \text{mol}$
• $[B] = 1 \times 10^{-4} \, \text{mol}$
• $[C] = 1 \times 10^{-4} \, \text{mol}$

We can calculate the reaction rate using the formula:

$$\text{Reaction rate} = \frac{\Delta [B]}{\Delta t}$$

where $\Delta [B]$ is the change in concentration of $B$ and $\Delta t$ is the time interval.

Assuming a time interval of 1 second, we can calculate the reaction rate as follows:

$$\text{Reaction rate} = \frac{(1 \times 10^{-4} - 0)}{1 \, \text{s}} = 1 \times 10^{-4} \, \text{mol/s}$$

Q: Can you provide an example of how to calculate the reaction order?

A: Yes, let's consider the same reaction:

$$2A(g) \leftrightharpoons B(g) + C(g)$$

Given the concentrations of $A$, $B$, and $C$ at a given time:

• $[A] = 2 \times 10^{-5} \, \text{mol}$
• $[B] = 1 \times 10^{-4} \, \text{mol}$
• $[C] = 1 \times 10^{-4} \, \text{mol}$

We can calculate the reaction order using the formula:

$$\text{Reaction order} = \frac{\log \left( \frac{[B]}{[A]^2} \right)}{\log \left( \frac{[B]}{[A]^2} \right)}$$

Substituting the given values, we get:

$$\text{Reaction order} = \frac{\log \left( \frac{1 \times 10^{-4}}{(2 \times 10^{-5})^2} \right)}{\log \left( \frac{1 \times 10^{-4}}{(2 \times 10^{-5})^2} \right)} = 1$$

This indicates that the reaction is first-order with respect to $A$.