Question

Methane, \(\mathrm{CH}_{4}\), reacts with \(I_{2}\) according to the reaction \(\mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{I}(g)+\mathrm{HI}(g)\). At \(600 \mathrm{~K}, K_{p}\) for this reaction is \(1.95 \times 10^{-4}\). A reaction was set up at 600 \(\mathrm{K}\) with initial partial pressures of methane of \(13.3 \mathrm{kPa}\) and of \(6.67 \mathrm{kPa}\) for \(\mathrm{I}_{2}\). Calculate the pressures, in \(\mathrm{kPa}\), of all reactants and products at equilibrium.

Short answer

The equilibrium pressures of all reactants and products are approximately: \[\mathrm{CH}_4: 13.169\,\mathrm{kPa}\] \[\mathrm{I}_2: 6.539\,\mathrm{kPa}\] \[\mathrm{CH}_3\mathrm{I}: 0.131\,\mathrm{kPa}\] \[\mathrm{HI}: 0.131\,\mathrm{kPa}\]

Step-by-step solution

1. Write down the balanced equation and the equilibrium constant expression.

The balanced equation is: \[\mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3}\mathrm{I}(g)+\mathrm{HI}(g)\] The equilibrium constant expression for this reaction is: \[K_p = \frac{[\mathrm{CH}_3\mathrm{I}][\mathrm{HI}]}{[\mathrm{CH}_4][\mathrm{I}_2]}\] Where \([\mathrm{CH}_4]\), \([\mathrm{I}_2]\), \([\mathrm{CH}_3\mathrm{I}]\) and \([\mathrm{HI}]\) represent the equilibrium pressures of each species.

2. Set up a table with initial, change, and equilibrium pressures.

We will set up a table with the initial pressures, the changes in pressures due to the reaction and the equilibrium pressures. ``` CH4 I2 CH3I HI Initial 13.3 6.67 0 0 Change -x -x +x +x Equilibrium 13.3-x 6.67-x x x ``` Where x is the amount of both reactants consumed at equilibrium.

3. Plug in the equilibrium concentrations into the equilibrium constant expression.

Now we substitute the equilibrium pressures into the equilibrium constant expression: \[1.95 \times 10^{-4} = \frac{x^2}{(13.3 - x)(6.67 - x)}\]

4. Solve for x

It is not a simple quadratic equation, but it can be solved using a numerical method or approximation. Additionally, since \(K_p\) is much smaller than 1, we can make an assumption to simplify the calculation. We can assume that the change in pressure x is relatively small compared to the initial pressures of methane and iodine. So we can rewrite the expression as: \[1.95 \times 10^{-4} = \frac{x^2}{(13.3)(6.67)}\] Now solve for x: \[x^2 = 1.95 \times 10^{-4} \times (13.3)(6.67)\] \[x^2 = 1.715 \times 10^{-2}\] \[x = \sqrt{1.715 \times 10^{-2}}\] \[x = 0.131\]

5. Find the equilibrium pressures for all species.

As we assumed that the change in pressure x is relatively small compared to the initial pressures of methane and iodine, the equilibrium pressures of all species are: \[\mathrm{CH}_4: 13.3 - x = 13.3 - 0.131 \approx 13.169\,\mathrm{kPa}\] \[\mathrm{I}_2: 6.67 - x = 6.67 - 0.131 \approx 6.539\,\mathrm{kPa}\] \[\mathrm{CH}_3\mathrm{I}: x \approx 0.131\,\mathrm{kPa}\] \[\mathrm{HI}: x \approx 0.131\,\mathrm{kPa}\] So, the equilibrium pressures of all reactants and products are approximately: \[\mathrm{CH}_4: 13.169\,\mathrm{kPa}\] \[\mathrm{I}_2: 6.539\,\mathrm{kPa}\] \[\mathrm{CH}_3\mathrm{I}: 0.131\,\mathrm{kPa}\] \[\mathrm{HI}: 0.131\,\mathrm{kPa}\]

Key concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant (\(K_p\) for reactions involving gases) is a numerical value that expresses the ratio of the concentration of products to reactants at equilibrium. This value remains constant for a given reaction at a specific temperature. The equilibrium constant helps to understand whether a reaction favors the formation of products or reactants when equilibrium is achieved.

• If the value of \(K_p\) is greater than 1, the reaction tends to produce more products.
• If \(K_p\) is less than 1, as in our example with \(K_p = 1.95 \times 10^{-4}\), the reaction favors the reactants.

In our exercise, \(K_p\) for the reaction involving methane and iodine is provided. This small value indicates that, at 600 K, the reaction favors the presence of the reactants (\(\text{CH}_4\) and \(\text{I}_2\)) over the products (\(\text{CH}_3\text{I}\) and \(\text{HI}\)).Understanding the equilibrium constant is crucial for predicting the extent of a reaction and aligning expectations with real-world results.

Partial Pressures

Partial pressures refer to the pressure exerted by individual gas components in a mixture of gases. In a chemical reaction involving gases, each gas contributes to the total pressure based on its mole fraction in the mixture. Partial pressures are essential when analyzing reactions under equilibrium, as they allow us to calculate the equilibrium constant for reactions involving gaseous substances.

To calculate partial pressures at equilibrium conditions:

• Identify initial partial pressures of each reactant and product.
• Account for changes in these pressures as the reaction proceeds to equilibrium.
• Use the changes to find the equilibrium partial pressures, considering individual gas consumption and formation rates.

In the given exercise, the initial partial pressures of methane and iodine are specified, allowing us to calculate changes as the reaction approaches equilibrium. This method aids in determining the proportion of each reactant and product at equilibrium, thus providing insight into the reaction dynamics.

Reaction Quotient

The Reaction Quotient (\(Q_p\) in terms of partial pressures) is a ratio similar to the equilibrium constant, but it's used for any state of the reaction, not just at equilibrium. Calculating the reaction quotient helps to determine the direction a reaction will proceed to reach equilibrium.

To calculate \(Q_p\), use the expression:\[ Q_p = \frac{[\text{Products}]}{[\text{Reactants}]} \]Before the reaction reaches equilibrium:

• If \(Q_p < K_p\), the reaction will proceed forward to produce more products until equilibrium is reached.
• If \(Q_p > K_p\), the reaction will shift to produce more reactants until equilibrium is achieved.
• If \(Q_p = K_p\), the system is already at equilibrium.

Understanding \(Q_p\) and its comparison with \(K_p\) in this exercise provides practical insights into the state of the reaction throughout its progress, indicating whether adjustments are needed to achieve equilibrium.