Problem 95 - Chemistry An Atoms First Approach

Question

In a given experiment, 5.2 moles of pure NOCl was placed in an otherwise empty \(2.0-\mathrm{L}\) container. Equilibrium was established by the following reaction: $$2 \operatorname{Nocl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K=1.6 \times 10^{-5}$$, 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 s present at equilibrium. b. Calculate the equilibrium concentrations for all species.

Step-by-Step Solution

Answer

The equilibrium concentrations of all species in the given reaction are NOCl: \(2.54\,M\), NO: \(0.06\,M\), and Cl₂: \(3.0 \times 10^{-2}\,M\).

Step 1: Convert the initial moles of NOCl to concentration and construct the ICE table.

First, let's convert the initial moles of NOCl into concentration by dividing it by the volume of the container : Initial concentration of NOCl = \(\frac{5.2 \, moles}{2.0 \, L}\) = \(2.6 \, M\) Now, we will construct an ICE table using the given information: | Species | Initial | Change | Equilibrium | |:-----------:|:-------:|:-------:|:-----------:| | NOCl | 2.6 M | -2x M | 2.6 - 2x M | | NO | 0 M | 2x M | 2x M | | Cl₂ | 0 M | x M | x M |

Step 2: Write the equilibrium expression.

Next, we will write the expression for the equilibrium constant, K, using the concentrations of species from the ICE table: \(K = \frac{[NO]^2[Cl_{2}]}{[NOCl]^2}\) Plug in the numerical value of K and the concentrations from the ICE table: \(1.6 \times 10^{-5} = \frac{(2x)^2(x)}{(2.6-2x)^2}\)

Step 3: Solve the equation for x.

Now, we need to solve this equation for x. This will give us the concentration of Cl₂ at equilibrium: First, simplify the equation: \(1.6 \times 10^{-5} = \frac{4x^3}{(2.6-2x)^2}\) To make it easier to solve, we can cross-multiply: \(1.6 \times 10^{-5}(2.6-2x)^2 = 4x^3\) This equation is a cubic equation in x. Solving this equation can be quite complex, so we can make an approximation by assuming that the value of x will be much smaller than 2.6, thus: \(1.6 \times 10^{-5} \approx \frac{4x^3}{(2.6)^2}\) Solve for x: \(x^3 = \frac{(1.6\times 10^{-5})(2.6)^2}{4}\) \(x^3 = \frac{1.08704 \times 10^{-4}}{4}\) \(x^3 = 2.7176 \times 10^{-5}\) Thus, \(x = \sqrt[3]{2.7176 \times 10^{-5}}\) \(x \approx 3.0 \times 10^{-2}\, M\)

Step 4: Calculate the equilibrium concentrations of all species.

Now, we will use the value of x to calculate the equilibrium concentrations of all species in the reaction: Equilibrium concentration of NOCl: \[2.6 - 2x\, M = 2.6 - 2(3.0 \times 10^{-2})\, M= 2.6 - 0.06\, M = 2.54\, M\] Equilibrium concentration of NO: \[2x\, M = 2(3.0 \times 10^{-2})\, M = 0.06\, M \] Equilibrium concentration of Cl₂: \[x\, M = 3.0 \times 10^{-2}\, M\]

Step 5: Final Results:

The equilibrium concentrations of all species are as follows: NOCl: 2.54 M NO: 0.06 M Cl₂: 3.0 x 10⁻² M

Key Concepts

ICE table

The ICE table is a useful tool for organizing information while solving equilibrium problems. ICE stands for Initial, Change, and Equilibrium, which are the stages used to determine the concentrations of reactants and products in a chemical reaction. In our example, the initial concentration of NOCl is calculated by dividing the initial moles by the volume of the container. This gives:

• Initial concentration of NOCl: \(2.6 \, M\)
• Initial concentration of NO and Cl₂: \(0 \, M\)

To represent the changes as the system reaches equilibrium, we introduce a variable \( x \). Here, \( x \) stands for the concentration of \( \text{Cl}_2 \) produced. The changes for each species in terms of \( x \) are as follows:

• NOCl decreases by \(-2x\)
• NO increases by \(2x\)
• Cl₂ increases by \(x\)

Finally, we calculate the equilibrium concentrations by applying these changes to the initial concentrations, which can be seen in the respective expressions in the ICE table. This structure breaks the problem into manageable pieces, allowing a systematic approach to solving equilibrium problems.

Equilibrium constant

The equilibrium constant, \( K \), is a value that expresses the ratio of concentrations of products to reactants at equilibrium for a given chemical reaction at a specific temperature. For a balanced chemical equation\[ 2 \text{NOCl}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Cl}_2(g) \]the equilibrium constant expression would be:\[ K = \frac{[\text{NO}]^2 [\text{Cl}_2]}{[\text{NOCl}]^2} \]This equation helps us understand how far a reaction proceeds to form products before reaching equilibrium. Large \( K \) values indicate that products are favored, whereas small \( K \) values, like our example with \(1.6 \times 10^{-5}\), indicate that reactants are favored.In our problem, we use the concentrations from the ICE table in the equilibrium constant expression to set up an equation involving \( x \), which represents the concentration of \( \text{Cl}_2 \). This equation allows us to solve for the unknown \( x \), ultimately helping us determine the system's equilibrium state.

Reaction stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is crucial for balancing chemical equations and calculating the amount of each substance involved. In our example reaction, \(2 \text{NOCl} \) decomposes into \(2 \text{NO} \) and \(\text{Cl}_2 \). The stoichiometry tells us:

• For every 2 moles of NOCl that react, 2 moles of NO and 1 mole of Cl₂ are produced.
• This relationship is reflected in the coefficients of the balanced equation.

These stoichiometric relationships help us determine the changes in concentration for each species involved during the reaction. When integrating into the ICE table, it allows for the accurate calculation of the equilibrium concentrations.Understanding stoichiometry ensures that the equations for changes in concentration are consistent with the balanced chemical reaction, and it is essential when writing the equilibrium expression and solving for \( x \). This understanding simplifies complex equilibrium problems, like ours, by reducing the problem to basic arithmetic ratios.

Cubic equation approximation

Often, solving equilibrium equations leads to high-degree polynomial equations, such as cubic equations, which can be labor-intensive without approximation. In this exercise, we see:\[ 1.6 \times 10^{-5} = \frac{4x^3}{(2.6-2x)^2} \]Due to the small equilibrium constant and large initial concentration of NOCl, we assume that \( x \) is small compared to 2.6 M. With this assumption, the equilibrium equation simplifies significantly:

• \((2.6 - 2x)^2 \approx 2.6^2\)

Simplifying the equation makes solving for \( x \) manageable. We calculate:\( x^3 = \frac{1.08704 \times 10^{-4}}{4} \)and find the cubic root, which gives \( x \approx 3.0 \times 10^{-2} \, M \).These approximations acknowledge that changes in very small quantities have negligible impact in certain contexts, especially when dealing with equilibrium constants much smaller in magnitude compared to initial concentrations. This approach makes solving cubic equations more practical in real-world scenarios.