Questions 3 & 4 Refer To The Following:At A Given Temperature, 0.300 Moles Of N O NO NO , 0.200 Moles Of C L 2 Cl_2 C L 2 ​ , And 0.500 Moles Of C L N O ClNO ClNO Were Placed In A 25.0-liter Container. The Following Equilibrium Is Established: $[2

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Equilibrium Constants and Molar Concentrations: A Comprehensive Analysis

In the realm of chemistry, understanding the concept of equilibrium constants and molar concentrations is crucial for predicting the behavior of chemical reactions. The given problem involves a specific scenario where three different substances, namely NONO, Cl2Cl_2, and ClNOClNO, are placed in a container and allowed to reach equilibrium. In this article, we will delve into the details of the problem and provide a step-by-step analysis to determine the equilibrium constant and molar concentrations of the substances involved.

The Given Equilibrium Reaction

The equilibrium reaction is given as:

2NO(g)+Cl2(g)⇌2ClNO(g){2NO(g) + Cl_2(g) \rightleftharpoons 2ClNO(g)}

This reaction involves the combination of nitrogen monoxide (NONO) and chlorine gas (Cl2Cl_2) to form dichloronitrosyl (ClNOClNO). The reaction is reversible, meaning that the products can also react to form the reactants.

Initial Molar Concentrations

The initial molar concentrations of the substances are given as:

  • NONO: 0.300 moles
  • Cl2Cl_2: 0.200 moles
  • ClNOClNO: 0.500 moles

The total volume of the container is 25.0 liters.

Calculating the Number of Moles

To calculate the number of moles of each substance, we can use the formula:

n=mM{n = \frac{m}{M}}

where nn is the number of moles, mm is the mass of the substance, and MM is the molar mass of the substance.

However, since we are given the initial molar concentrations, we can skip this step and proceed with the calculations.

Calculating the Molar Concentrations

The molar concentration of a substance is given by:

C=nV{C = \frac{n}{V}}

where CC is the molar concentration, nn is the number of moles, and VV is the volume of the container.

We can calculate the molar concentrations of each substance as follows:

  • NONO: CNO=0.300 mol25.0 L=0.012 MC_{NO} = \frac{0.300 \text{ mol}}{25.0 \text{ L}} = 0.012 \text{ M}
  • Cl2Cl_2: CCl2=0.200 mol25.0 L=0.008 MC_{Cl_2} = \frac{0.200 \text{ mol}}{25.0 \text{ L}} = 0.008 \text{ M}
  • ClNOClNO: CClNO=0.500 mol25.0 L=0.020 MC_{ClNO} = \frac{0.500 \text{ mol}}{25.0 \text{ L}} = 0.020 \text{ M}

Calculating the Equilibrium Constant

The equilibrium constant (KcK_c) is given by:

Kc=[ClNO]2[NO]2[Cl2]{K_c = \frac{[ClNO]^2}{[NO]^2[Cl_2]}}

where [ClNO][ClNO], [NO][NO], and [Cl2][Cl_2] are the molar concentrations of the substances at equilibrium.

Since the reaction is reversible, we can assume that the equilibrium constant is constant at a given temperature.

Determining the Equilibrium Molar Concentrations

To determine the equilibrium molar concentrations, we need to consider the stoichiometry of the reaction. The reaction involves a 2:1 ratio of NONO to ClNOClNO, and a 1:1 ratio of Cl2Cl_2 to ClNOClNO.

Let xx be the number of moles of NONO that react to form ClNOClNO. Then, the number of moles of ClNOClNO formed is 2x2x.

The number of moles of Cl2Cl_2 that react is also xx.

The equilibrium molar concentrations can be calculated as follows:

  • [NO]=0.300−x[NO] = 0.300 - x
  • [Cl2]=0.200−x[Cl_2] = 0.200 - x
  • [ClNO]=0.500+2x[ClNO] = 0.500 + 2x

Solving for the Equilibrium Constant

Substituting the equilibrium molar concentrations into the expression for the equilibrium constant, we get:

Kc=(0.500+2x)2(0.300−x)2(0.200−x){K_c = \frac{(0.500 + 2x)^2}{(0.300 - x)^2(0.200 - x)}}

Expanding and simplifying the expression, we get:

Kc=0.250+2x+4x20.060−0.600x+0.200x2{K_c = \frac{0.250 + 2x + 4x^2}{0.060 - 0.600x + 0.200x^2}}

Solving for the Equilibrium Constant (continued)

To solve for the equilibrium constant, we need to find the value of xx that satisfies the equation.

However, since the equation is quadratic, we can use the quadratic formula to find the value of xx.

The quadratic formula is given by:

x=−b±b2−4ac2a{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}

where aa, bb, and cc are the coefficients of the quadratic equation.

In this case, the quadratic equation is:

4x2+6.2x−0.25=0{4x^2 + 6.2x - 0.25 = 0}

Solving for xx, we get:

x=−6.2±6.22−4(4)(−0.25)2(4){x = \frac{-6.2 \pm \sqrt{6.2^2 - 4(4)(-0.25)}}{2(4)}}

x=−6.2±38.44+48{x = \frac{-6.2 \pm \sqrt{38.44 + 4}}{8}}

x=−6.2±42.448{x = \frac{-6.2 \pm \sqrt{42.44}}{8}}

x=−6.2±6.58{x = \frac{-6.2 \pm 6.5}{8}}

Since xx must be a positive value, we take the positive root:

x=−6.2+6.58{x = \frac{-6.2 + 6.5}{8}}

x=0.38{x = \frac{0.3}{8}}

x=0.0375{x = 0.0375}

Calculating the Equilibrium Molar Concentrations (continued)

Now that we have found the value of xx, we can calculate the equilibrium molar concentrations as follows:

  • [NO]=0.300−0.0375=0.2625 M[NO] = 0.300 - 0.0375 = 0.2625 \text{ M}
  • [Cl2]=0.200−0.0375=0.1625 M[Cl_2] = 0.200 - 0.0375 = 0.1625 \text{ M}
  • [ClNO]=0.500+2(0.0375)=0.575 M[ClNO] = 0.500 + 2(0.0375) = 0.575 \text{ M}

Calculating the Equilibrium Constant (continued)

Now that we have found the equilibrium molar concentrations, we can calculate the equilibrium constant as follows:

Kc=(0.575)2(0.2625)2(0.1625){K_c = \frac{(0.575)^2}{(0.2625)^2(0.1625)}}

Kc=0.32906250.0137{K_c = \frac{0.3290625}{0.0137}}

Kc=24.0{K_c = 24.0}

In this article, we have analyzed the given problem and provided a step-by-step solution to determine the equilibrium constant and molar concentrations of the substances involved. We have used the quadratic formula to solve for the equilibrium constant and have found that the equilibrium constant is 24.0. We have also calculated the equilibrium molar concentrations of the substances involved. The results of this analysis can be used to predict the behavior of the chemical reaction and to understand the concept of equilibrium constants and molar concentrations.
Equilibrium Constants and Molar Concentrations: A Comprehensive Analysis - Q&A

In the previous article, we analyzed the given problem and provided a step-by-step solution to determine the equilibrium constant and molar concentrations of the substances involved. In this article, we will provide a Q&A section to address any questions or concerns that readers may have.

Q: What is the equilibrium constant?

A: The equilibrium constant is a value that describes the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium.

Q: How is the equilibrium constant calculated?

A: The equilibrium constant is calculated using the formula:

Kc=[ClNO]2[NO]2[Cl2]{K_c = \frac{[ClNO]^2}{[NO]^2[Cl_2]}}

where [ClNO][ClNO], [NO][NO], and [Cl2][Cl_2] are the molar concentrations of the substances at equilibrium.

Q: What is the significance of the equilibrium constant?

A: The equilibrium constant is a measure of the extent to which a reaction occurs. A large equilibrium constant indicates that the reaction favors the products, while a small equilibrium constant indicates that the reaction favors the reactants.

Q: How do you determine the equilibrium molar concentrations?

A: To determine the equilibrium molar concentrations, you need to consider the stoichiometry of the reaction. The reaction involves a 2:1 ratio of NONO to ClNOClNO, and a 1:1 ratio of Cl2Cl_2 to ClNOClNO.

Q: What is the relationship between the equilibrium constant and the equilibrium molar concentrations?

A: The equilibrium constant is related to the equilibrium molar concentrations through the formula:

Kc=[ClNO]2[NO]2[Cl2]{K_c = \frac{[ClNO]^2}{[NO]^2[Cl_2]}}

This formula shows that the equilibrium constant is a function of the equilibrium molar concentrations.

Q: How do you solve for the equilibrium constant?

A: To solve for the equilibrium constant, you need to find the value of xx that satisfies the equation:

4x2+6.2x−0.25=0{4x^2 + 6.2x - 0.25 = 0}

This equation can be solved using the quadratic formula.

Q: What is the significance of the quadratic formula?

A: The quadratic formula is a mathematical tool that allows you to solve quadratic equations. In this case, the quadratic formula is used to solve for the value of xx that satisfies the equation.

Q: How do you calculate the equilibrium molar concentrations?

A: To calculate the equilibrium molar concentrations, you need to substitute the value of xx into the expressions for the equilibrium molar concentrations.

Q: What is the relationship between the equilibrium molar concentrations and the equilibrium constant?

A: The equilibrium molar concentrations and the equilibrium constant are related through the formula:

Kc=[ClNO]2[NO]2[Cl2]{K_c = \frac{[ClNO]^2}{[NO]^2[Cl_2]}}

This formula shows that the equilibrium constant is a function of the equilibrium molar concentrations.

Q: How do you use the equilibrium constant to predict the behavior of a chemical reaction?

A: The equilibrium constant can be used to predict the behavior of a chemical reaction by determining the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium.

In this article, we have provided a Q&A section to address any questions or concerns that readers may have. We have discussed the significance of the equilibrium constant, the relationship between the equilibrium constant and the equilibrium molar concentrations, and how to use the equilibrium constant to predict the behavior of a chemical reaction.