Suppose A $500 \, \text{mL}$ Flask Is Filled With 1.5 Mol Of $N_2$ And 0.70 Mol Of $O_2$. The Following Reaction Becomes Possible:$N_2(g) + O_2(g) \rightleftharpoons 2 \, NO(g$\]The Equilibrium Constant $K$

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Introduction

Chemical equilibrium is a fundamental concept in chemistry that describes the balance between the forward and reverse reactions in a chemical reaction. The equilibrium constant, denoted by $K$, 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 discuss the concept of equilibrium constant and how it can be used to determine the equilibrium concentrations of the reactants and products in a chemical reaction.

The Equilibrium Constant Expression

The equilibrium constant expression is a mathematical expression that describes the relationship between the concentrations of the reactants and products at equilibrium. For the reaction $N_2(g) + O_2(g) \rightleftharpoons 2 \, NO(g)$, the equilibrium constant expression is given by:

$$K = \frac{[NO]^2}{[N_2][O_2]}$$

where $[NO]$, $[N_2]$, and $[O_2]$ are the concentrations of the products and reactants at equilibrium.

Calculating the Equilibrium Constant

To calculate the equilibrium constant, we need to know the concentrations of the reactants and products at equilibrium. In this case, we are given the initial concentrations of $N_2$ and $O_2$, which are 1.5 mol and 0.70 mol, respectively. We also know that the reaction is reversible, meaning that the equilibrium constant is a constant value that depends only on the temperature.

To calculate the equilibrium constant, we can use the following steps:

1. Write the equilibrium constant expression for the reaction.
2. Determine the initial concentrations of the reactants and products.
3. Use the equilibrium constant expression to calculate the equilibrium concentrations of the reactants and products.
4. Substitute the equilibrium concentrations into the equilibrium constant expression to calculate the equilibrium constant.

Calculating the Equilibrium Concentrations

To calculate the equilibrium concentrations of the reactants and products, we need to use the equilibrium constant expression. We can start by assuming that the equilibrium concentrations of the reactants and products are equal to the initial concentrations.

Let $x$ be the change in concentration of $N_2$ and $O_2$ at equilibrium. Then, the equilibrium concentrations of $N_2$ and $O_2$ are given by:

$$[N_2]_e = 1.5 - x$$

$$[O_2]_e = 0.70 - x$$

The equilibrium concentration of $NO$ is given by:

$$[NO]_e = 2x$$

Substituting the Equilibrium Concentrations into the Equilibrium Constant Expression

Now that we have the equilibrium concentrations of the reactants and products, we can substitute them into the equilibrium constant expression to calculate the equilibrium constant.

$$K = \frac{(2x)^2}{(1.5 - x)(0.70 - x)}$$

Simplifying the Equilibrium Constant Expression

To simplify the equilibrium constant expression, we can expand the numerator and denominator and then cancel out any common factors.

$$K = \frac{4x^2}{1.05 - 1.20x + x^2}$$

Calculating the Equilibrium Constant

To calculate the equilibrium constant, we need to know the value of $x$. We can find the value of $x$ by setting the numerator and denominator of the equilibrium constant expression equal to each other and solving for $x$.

$$4x^2 = (1.05 - 1.20x + x^2)(K)$$

Solving for $x$

To solve for $x$, we can use the quadratic formula.

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a$, $b$, and $c$ are the coefficients of the quadratic equation.

Substituting the Values of $a$, $b$, and $c$

We can substitute the values of $a$, $b$, and $c$ into the quadratic formula to solve for $x$.

$$a = 1$$

$$b = -1.20$$

$$c = 1.05 - K$$

Solving for $x$

Now that we have the values of $a$, $b$, and $c$, we can substitute them into the quadratic formula to solve for $x$.

$$x = \frac{1.20 \pm \sqrt{(-1.20)^2 - 4(1)(1.05 - K)}}{2(1)}$$

Simplifying the Expression for $x$

To simplify the expression for $x$, we can expand the square root and then cancel out any common factors.

$$x = \frac{1.20 \pm \sqrt{1.44 - 4.20 + 4K}}{2}$$

Simplifying the Expression for $x$

To simplify the expression for $x$, we can combine the constants in the square root.

$$x = \frac{1.20 \pm \sqrt{-2.76 + 4K}}{2}$$

Simplifying the Expression for $x$

To simplify the expression for $x$, we can rewrite the square root as a difference of squares.

$$x = \frac{1.20 \pm \sqrt{(2K - 1.38)^2}}{2}$$

Simplifying the Expression for $x$

To simplify the expression for $x$, we can take the square root of the difference of squares.

$$x = \frac{1.20 \pm (2K - 1.38)}{2}$$

Simplifying the Expression for $x$

To simplify the expression for $x$, we can combine the fractions.

$$x = 0.60 \pm (K - 0.69)$$

Finding the Value of $x$

To find the value of $x$, we need to know the value of $K$. We can find the value of $K$ by using the equilibrium constant expression.

$$K = \frac{[NO]^2}{[N_2][O_2]}$$

Finding the Value of $[NO]$

To find the value of $[NO]$, we need to know the value of $x$. We can find the value of $x$ by using the expression for $x$.

$$x = 0.60 \pm (K - 0.69)$$

Finding the Value of $[NO]$

To find the value of $[NO]$, we can substitute the value of $x$ into the expression for $[NO]$.

$$[NO] = 2x$$

Finding the Value of $[NO]$

To find the value of $[NO]$, we can substitute the value of $x$ into the expression for $[NO]$.

$$[NO] = 2(0.60 \pm (K - 0.69))$$

Finding the Value of $[NO]$

To find the value of $[NO]$, we can simplify the expression for $[NO]$.

$$[NO] = 1.20 \pm 2(K - 0.69)$$

Finding the Value of $[NO]$

To find the value of $[NO]$, we can substitute the value of $K$ into the expression for $[NO]$.

$$[NO] = 1.20 \pm 2(0.69 - 0.69)$$

Finding the Value of $[NO]$

To find the value of $[NO]$, we can simplify the expression for $[NO]$.

$$[NO] = 1.20$$

Finding the Value of $K$

To find the value of $K$, we can substitute the value of $[NO]$ into the equilibrium constant expression.

$$K = \frac{[NO]^2}{[N_2][O_2]}$$

Finding the Value of $K$

To find the value of $K$, we can substitute the value of $[NO]$ into the equilibrium constant expression.

$$K = \frac{(1.20)^2}{(1.5 - 0.60)(0.70 - 0.60)}$$

Finding the Value of $K$

To find the value of $K$, we can simplify the expression for $K$.

$$K = \frac{1.44}{0.90 \times 0.10}$$

Finding the Value of $K$

To find the value of $K$, we can simplify the expression for $K$.

$$K = \frac{1.44}{0.09}$$

Finding the Value of $K$

To find the value of $K$, we can simplify the expression for $K$.

$$K = 16$$

Conclusion

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Q&A

Q: What is the equilibrium constant expression for the reaction $N_2(g) + O_2(g) \rightleftharpoons 2 \, NO(g)$?

A: The equilibrium constant expression for the reaction $N_2(g) + O_2(g) \rightleftharpoons 2 \, NO(g)$ is given by:

$$K = \frac{[NO]^2}{[N_2][O_2]}$$

Q: How do we calculate the equilibrium constant?

A: To calculate the equilibrium constant, we need to know the concentrations of the reactants and products at equilibrium. We can use the equilibrium constant expression to calculate the equilibrium concentrations of the reactants and products.

Q: What is the initial concentration of $N_2$ and $O_2$?

A: The initial concentration of $N_2$ is 1.5 mol and the initial concentration of $O_2$ is 0.70 mol.

Q: How do we calculate the equilibrium concentrations of the reactants and products?

A: We can use the equilibrium constant expression to calculate the equilibrium concentrations of the reactants and products. We can start by assuming that the equilibrium concentrations of the reactants and products are equal to the initial concentrations.

Q: What is the equilibrium concentration of $NO$?

A: The equilibrium concentration of $NO$ is given by:

$$[NO]_e = 2x$$

Q: How do we calculate the value of $x$?

A: We can use the quadratic formula to calculate the value of $x$.

Q: What is the value of $K$?

A: The value of $K$ is 16.

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

A: We can use the equilibrium constant expression to calculate the equilibrium concentrations of the reactants and products. We can substitute the equilibrium concentrations into the equilibrium constant expression to calculate the equilibrium constant.

Q: What is the significance of the equilibrium constant?

A: The equilibrium constant is a measure of the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. It is a fundamental concept in chemistry that describes the balance between the forward and reverse reactions in a chemical reaction.

Q: How do we determine the equilibrium concentrations of the reactants and products?

A: We can use the equilibrium constant expression to determine the equilibrium concentrations of the reactants and products. We can substitute the equilibrium concentrations into the equilibrium constant expression to calculate the equilibrium constant.

Q: What is the relationship between the equilibrium constant and the equilibrium concentrations of the reactants and products?

A: The equilibrium constant is a measure of the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. It is a fundamental concept in chemistry that describes the balance between the forward and reverse reactions in a chemical reaction.

Additional Questions and Answers

Q: What is the relationship between the equilibrium constant and the temperature?

A: The equilibrium constant is a function of the temperature. It is a fundamental concept in chemistry that describes the balance between the forward and reverse reactions in a chemical reaction.

Q: How do we calculate the equilibrium constant at different temperatures?

A: We can use the van't Hoff equation to calculate the equilibrium constant at different temperatures.

Q: What is the van't Hoff equation?

A: The van't Hoff equation is a mathematical expression that describes the relationship between the equilibrium constant and the temperature.

Q: How do we use the van't Hoff equation to calculate the equilibrium constant at different temperatures?

A: We can use the van't Hoff equation to calculate the equilibrium constant at different temperatures. We can substitute the values of the equilibrium constant and the temperature into the van't Hoff equation to calculate the equilibrium constant at different temperatures.

Conclusion

In this article, we have discussed the concept of equilibrium constant and how it can be used to determine the equilibrium concentrations of the reactants and products in a chemical reaction. We have also calculated the equilibrium constant for the reaction $N_2(g) + O_2(g) \rightleftharpoons 2 \, NO(g)$ and found the value of $K$ to be 16. We have also answered additional questions and provided more information on the relationship between the equilibrium constant and the temperature.