Problem 105 - Chemistry

Question

In a given experiment, 5.2 moles of pure NOCl were placed in an otherwise empty \(2.0-\mathrm{L}\) container. Equilibrium was established by the following reaction: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K=1.6 \times 10^{-5}$$ a. Using numerical values for the concentrations in the Initial row and expressions containing the variable \(x\) in both the Change and Equilibrium rows, complete the following table summarizing what happens as this reaction reaches equilibrium. Let \(x=\) the concentration of \(\mathrm{Cl}_{2}\) that is present at equilibrium. b. Calculate the equilibrium concentrations for all species.

Step-by-Step Solution

Answer

The equilibrium concentrations for all species are: NOCl: \(2.5996 \thinspace M\) NO: \(0.0004 \thinspace M\) Cl₂: \(0.0002 \thinspace M\)

Step 1: Create an ICE table

To create an ICE table, the first step is to calculate the initial concentrations of each participant involved in the chemical reaction. For now, only NOCl has a given amount (5.2 moles), while the others are simply zero. Calculate NOCl's initial concentration: Initial Concentration of NOCl = Moles of NOCl / Volume of Container \(C_0(NOCl) = \frac {5.2 \thinspace mol }{2.0 \thinspace L}\) \(C_0(NOCl) = 2.6 \thinspace M\) Step 2: Complete the Initial Row of ICE Table

Step 2: Fill out the ICE table for the initial concentrations

Now that we have the initial concentration of NOCl, we can fill out the initial row in the ICE table as follows: | | NOCl | 2 NO | Cl₂ | |---|------|------|------| | I | 2.6 | 0 | 0 | | C | | | | | E | | | | Step 3: Formulate the Change and Equilibrium Rows

Step 3: Write the expressions for Change and Equilibrium rows

We know that, for every 2 moles of NOCl that react, we produce 2 moles of NO and 1 mole of Cl₂. So, when x moles of Cl₂ are formed, x moles of NO are formed and 2x moles of NOCl are consumed. Let's now write the expressions for the Change and Equilibrium rows: | | NOCl | 2 NO | Cl₂ | |---|---------|---------|-------| | I | 2.6 | 0 | 0 | | C | -2x | 2x | x | | E | 2.6-2x | 2x | x | Step 4: Write the Equilibrium Expression

Step 4: Set up the equilibrium expression

Now that we have expressions for the equilibrium concentrations of all species, we will write the equilibrium expression using the equilibrium constant, K: \[K = \frac{[NO]^2 \times [Cl_2]}{[NOCl]^2}\] Plug in the equilibrium concentrations: \[1.6 \times 10^{-5} = \frac{(2x)^2 \times x}{(2.6-2x)^2}\] Step 5: Solve for x

Step 5: Solve the equation to find x

By solving for x, we can find the concentration of Cl₂ at equilibrium, and also the equilibrium concentrations for the other species. Multiplying both sides by (2.6-2x)^2: \[(2.6-2x)^2= (2x)^2 \times x \times (1.6 \times 10^{-5})\] This is a quadratic equation in x, which we can solve to get: \(x \approx 0.0002 \thinspace M\) Step 6: Calculate Equilibrium Concentrations

Step 6: Find the equilibrium concentrations

Using x, which is approximately 0.0002 M, we can now find the equilibrium concentrations of all species: \[NOCl = 2.6 - 2x \approx 2.6 - (2 \times 0.0002) \approx 2.5996 \thinspace M\] \[NO = 2x \approx 2 \times 0.0002 \approx 0.0004 \thinspace M\] \[Cl_2 = x \approx 0.0002 \thinspace M\] Thus, the equilibrium concentrations for all species are: NOCl: 2.5996 M NO: 0.0004 M Cl₂: 0.0002 M

Key Concepts

ICE Table

The ICE table is a valuable tool in chemistry used to simplify equilibrium calculations. Its name comes from Initial, Change, and Equilibrium—the three stages it represents. The goal of the ICE table is to clearly lay out how the concentrations of reactants and products evolve during a reaction until equilibrium is achieved.

In an ICE table:

• Initial - Displays the initial concentrations or pressures of the chemical species involved. These values are either given or can be calculated if the moles and volume are known.
• Change - Represents the change in concentrations as the reaction progresses towards equilibrium. This is often represented using a variable, such as 'x', to denote changes.
• Equilibrium - Shows the concentrations after the system has reached equilibrium. This is calculated by applying the changes to the initial concentrations.

The ICE table for the exercise begins with the initial concentration of NOCl, calculated using the formula \( C_0(NOCl) = \frac{5.2 \thinspace mol}{2.0 \thinspace L} = 2.6 \thinspace M \). Then, as the reaction progresses, changes occur based on stoichiometry, represented by \(-2x\), \(2x\), and \(x\) for each species.

Equilibrium Concentration

Equilibrium concentrations are the concentrations of reactants and products when a reaction has reached a state where the rates of the forward and reverse reactions are equal. In this scenario, the molar concentrations remain constant over time, although the reactions continue to occur.

In the context of the problem, after setting up the ICE table, we found expressions for equilibrium concentrations:

• For NOCl: \(2.6 - 2x\)
• For NO: \(2x\)
• For Cl₂: \(x\)

These expressions rise from stoichiometry rules that relate the consumption and formation of moles. Specifically, because 2 moles of NOCl generates 2 moles of NO and 1 mole of Cl₂, this informs the "Change" row, thereby helping us calculate equilibrium concentration terms.

Quadratic Equation

Solving for equilibrium concentration often leads to forming and solving a quadratic equation. This is an algebraic process where you solve for the unknown using the quadratic formula or factoring, depending on the equation structure.

The equilibrium constant expression derived from the ICE table was:\[ K = \frac{[NO]^2 \times [Cl_2]}{[NOCl]^2} \] Plugging in the equilibrium expressions, we get:\[ 1.6 \times 10^{-5} = \frac{(2x)^2 \times x}{(2.6-2x)^2} \]Multiplying both sides by \((2.6-2x)^2\) results in a quadratic equation in terms of \(x\):\[ (2.6 - 2x)^2 = (2x)^2 \times x \times (1.6 \times 10^{-5}) \] This kind of an equation, once simplified, can be approached using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) or possibly simplified further for easier solving.

Equilibrium Constant

The equilibrium constant \( K \) provides insight into the position of equilibrium of a chemical reaction. A small \( K \) value, like in this exercise, means reactants are favored at equilibrium, whereas a large \( K \) shows products are favored.

The equilibrium constant expression for the given reaction is:\[ K = \frac{[NO]^2 \times [Cl_2]}{[NOCl]^2} \]Inserting the equilibrium expressions into this formula allows us to solve for equilibrium concentrations, verifying our solution's consistency with the system's original \( K \) value. In this problem, the reported system constant \( K = 1.6 \times 10^{-5} \) indicates a stronger favoring of the reactants, aligning with the small \( x \) values obtained for products' concentrations at equilibrium. This demonstrates the typical chemical equilibrium where reactions can progress forward or backward but stabilize at a certain concentration ratio.