Question

At a particular temperature, a 3.0-L flask contains \(2.4 \mathrm{~mol} \mathrm{Cl}_{2}\), \(1.0 \mathrm{~mol} \mathrm{NOCl}\), and \(4.5 \times 10^{-3} \mathrm{~mol}\) NO. Calculate \(K\) at this temperature for the following reaction: $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$

Short answer

We have set up the equilibrium expression \(K = \frac{[NO]^2 [Cl_{2}]}{[NOCl]^2}\) and calculated the initial concentrations of the species involved in the reaction: \([Cl_{2}]_{initial} = 0.8 \, M\), \([NOCl]_{initial} = 0.333 \, M\), and \([NO]_{initial} = 1.5 \times 10^{-3} \, M\). However, we cannot find the value of K in this exercise because there is not enough information provided to solve for the change in concentration (x). Additional information, such as the equilibrium concentrations of one or more species, is required to complete the problem.

Step-by-step solution

1. Write the balanced chemical equation and the equilibrium expression (K)

First, we need to write down the balanced chemical equation for the reaction: \[ 2NOCl(g) \rightleftharpoons 2NO(g) + Cl_{2}(g) \] From the balanced chemical equation, we can write the equilibrium expression (K): \[K = \frac{[NO]^2 [Cl_{2}]}{[NOCl]^2}\]

2. Convert moles to concentrations

To convert moles to concentrations, we use the formula: \[ Concentration = \frac{Moles}{Volume} \] Given the volume of the flask is 3.0 L, we can calculate the initial concentrations of the species: \[ [Cl_{2}]_{initial} = \frac{2.4 \, mol}{3.0 \, L} = 0.8 \, M \] \[ [NOCl]_{initial} = \frac{1.0 \, mol}{3.0 \, L} = 0.333 \, M \] \[ [NO]_{initial} = \frac{4.5 \times 10^{-3} \, mol}{3.0 \, L} = 1.5 \times 10^{-3} \, M \]

3. Calculate the equilibrium concentrations of the species

Let x be the change in the concentration of NO (it increases by x M). Then, the change in concentration of NOCl is -2x (it decreases by 2x). At equilibrium, we have: \[ [NOCl]_{eq} = 0.333 - 2x \] \[ [NO]_{eq} = 1.5 \times 10^{-3} + x \] \[ [Cl_{2}]_{eq} = 0.8 - x \]

4. Substitute the equilibrium concentrations into the equilibrium expression (K)

Substitute the expressions for the equilibrium concentrations into the equilibrium expression and solve for x: \[K = \frac{[(1.5 \times 10^{-3} + x)^2 (0.8 - x)]}{(0.333 - 2x)^2}\] However, we need more information to solve for x. The exercise does not provide enough information to calculate K from this expression.

5. Conclusion

In this exercise, we have set up the equilibrium expression (K) and calculated the initial concentrations of the species involved in the reaction. Unfortunately, we cannot find the value of K because there is not enough information provided in the exercise to solve for x. Additional information, such as the equilibrium concentrations of one or more species, is required to complete the problem.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals that of the reverse reaction.
In such a state, the concentrations of the reactants and products remain constant, not because the reactions have stopped, but because they are occurring at equal rates.

• This balance does not necessarily mean that the amounts of reactants and products are equal.
• Equilibrium can shift with changes in conditions like temperature, pressure, and concentration.

Understanding chemical equilibrium is crucial as it helps predict how changes in conditions affect a reaction's position, depicted by Le Chatelier's Principle. Thus, it's essential to consider the forward and reverse reactions while calculating equilibrium.

Equilibrium Expression

The equilibrium expression is a mathematical representation of the concentrations of products and reactants at equilibrium.
It is crucial when calculating the equilibrium constant, often denoted as 'K'.

• The expression is derived from the balanced chemical equation.
• The concentrations of the products are placed in the numerator, and those of the reactants go in the denominator, each raised to the power of their coefficients in the balanced equation.

For the given reaction: \[ 2 ext{NOCl}(g) ightleftharpoons 2 ext{NO}(g) + ext{Cl}_{2}(g) \] the equilibrium expression is: \[ K = \frac{[\text{NO}]^2 [\text{Cl}_{2}]}{[\text{NOCl}]^2} \] This expression allows us to determine how the concentrations of various species at equilibrium relate to one another.

Concentration Calculation

Calculating concentrations allows us to determine the amount of a substance in a particular volume.
This is fundamental when dealing with chemical equilibria, as equilibrium positions are expressed in terms of concentrations.

• The formula to calculate the concentration is: \[ ext{Concentration} = \frac{\text{Moles}}{\text{Volume}} \]
• For accurate equilibrium calculations, such conversion is essential.

Given the substances and a flask of 3.0 L, we derived the initial concentrations:

• \([\text{Cl}_{2}] = 0.8 \, M\)
• \([\text{NOCl}] = 0.333 \, M\)
• \([\text{NO}] = 1.5 \times 10^{-3} \, M\)

These values become the basis on which changes at equilibrium are considered. Calculating initial concentrations is a critical step in analyzing the system's behavior as it progresses to equilibrium.

Balanced Chemical Equation

A balanced chemical equation is vital for understanding any reaction chemistry.
It ensures the law of conservation of mass is satisfied, meaning the number of atoms for each element is the same on both sides of the equation.

• Balancing an equation requires adjusting the coefficients before the chemical formulas.
• It does not involve altering the subscripts within the chemical formulas, as doing so would change the substances themselves.

In the given problem, the equation is: \[ 2\text{NOCl}(g) \rightleftharpoons 2\text{NO}(g) + \text{Cl}_{2}(g) \]Balancing is critical when writing the equilibrium expression. From the covered material, we clearly see how the balanced equation guides us in forming the expression: \[ K = \frac{[\text{NO}]^2 [\text{Cl}_{2}]}{[\text{NOCl}]^2} \]Understanding how to balance equations is key to mastering equilibrium calculations.