Question

Methane, \(\mathrm{CH}_{4}\), reacts with \(\mathrm{I}_{2}\) according to the reaction \(\mathrm{CH}_{4}(g)+\mathrm{l}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{l}(g)+\mathrm{HI}(g)\). At \(630 \mathrm{~K}, K_{p}\) for this reaction is \(2.26 \times 10^{-4}\). A reaction was set up at \(630 \mathrm{~K}\) with initial partial pressures of methane of \(105.1\) torr and of \(7.96\) torr for \(\mathrm{I}_{2}\). Calculate the pressures, in torr, of all reactants and products at equilibrium.

Short answer

At equilibrium, the pressures for the reactants and products are as follows: CH4 has a pressure of 105.0 torr, I2 has a pressure of 7.86 torr, and both CH3I and HI have pressures of 0.10 torr.

Step-by-step solution

Write the balanced chemical equation

The given reaction is already balanced: \[CH_{4}(g) + I_{2}(g) \rightleftharpoons CH_{3}I(g) + HI(g)\]

Set up the ICE table

For the ICE table, we need to input the Initial pressures (in torr) and the change in pressures, represented by x: | | CH4 | I2 | CH3I | HI | |------------|------|------|------|-----| | Initial | 105.1| 7.96 | 0 | 0 | | Change | -x | -x | +x | +x | | Equilibrium| 105.1-x | 7.96-x | x | x |

Calculate the Kp expression

We can write the expression for the equilibrium constant Kp as follows. \[K_p = \frac{[CH_{3}I][HI]}{[CH_{4}][I_{2}]}\] Now, we can substitute the equilibrium pressures from the ICE table into the Kp expression: \[2.26 \times 10^{-4} = \frac{x^2}{(105.1-x)(7.96-x)}\]

Solve for x

To simplify the equation and solve for x, we multiply both sides by (105.1-x)(7.96-x): \[(2.26 \times 10^{-4})(105.1-x)(7.96-x) = x^2\] Now we have a quadratic equation, which can be solved using either factoring, completing the square, or the quadratic formula. Wolfram Alpha or other online solvers are suggested to find the roots. In this case, when solving for x, we'll get two possible solutions: \(x = 0.10\) and \(x = -6.74\). Since a negative pressure is not possible in this context, we discard this solution and select the positive root. Therefore, x = 0.10.

Determine the equilibrium pressures

Now, we can input the value of x into the equilibrium concentrations to get the final pressures at equilibrium: - CH4 equilibrium pressure: 105.1-x = 105.1-0.10 = 105.0 torr - I2 equilibrium pressure: 7.96-x = 7.96-0.10 = 7.86 torr - CH3I equilibrium pressure: x = 0.10 torr - HI equilibrium pressure: x = 0.10 torr Thus, at equilibrium, the pressures in torr are 105.0 for CH4, 7.86 for I2, and 0.10 for both CH3I and HI.

Key concepts

Le Chatelier's Principle

Le Chatelier's Principle is a guiding framework in chemistry that helps predict how a chemical equilibrium will shift when it's subjected to an external change. It states that if a dynamic equilibrium is disturbed by changing the conditions, such as pressure, temperature, or concentration, the system responds to counteract the imposed change and restores a new equilibrium.

This principle can be visualized with everyday examples. Think of it like a game of tug-of-war: when one side gains an extra player (akin to increasing concentration of a reactant), the balance tips to that side. To restore balance, the system 'shifts' the equilibrium to the other side, reducing the effect of the added player.

In the context of the exercise provided, if we were to increase the pressure of either reactant, Le Chatelier's Principle would suggest that the system will shift towards producing more products in order to reduce the increased pressure.

ICE Table

The ICE Table stands for Initial, Change, and Equilibrium. It is a systematic way to organize information about the concentrations or pressures of reactants and products over the course of a reaction headed towards equilibrium.

To break it down:

• Initial refers to the concentrations or partial pressures at the beginning of the reaction, before any reaction has occurred.
• Change represents the increase or decrease in concentrations or pressures of reactants and products as the reaction proceeds towards equilibrium.
• Equilibrium denotes the concentrations or pressures when the reaction has reached a state of balance -- when the rate of the forward reaction equals the rate of the reverse reaction.

For example, in solving this exercise, the ICE table helped in visualizing how the initial pressures change as the reaction proceeds, allowing us to determine the pressures at equilibrium. By indicating the change with a variable 'x', we can solve for 'x' and use it to calculate the equilibrium pressures.

Equilibrium Constant (Kp)

The equilibrium constant (\( K_p \) for gases) is a numerical value that characterizes the chemical equilibrium of a reaction. It is the ratio of the equilibrium partial pressures of the products to the equilibrium partial pressures of the reactants, each raised to the power of its coefficient from the balanced equation. For a reaction involving gaseous substances, such as the one in our exercise, we use partial pressures and the constant is denoted as \( K_p \).

The value of \( K_p \) is constant for a given reaction at a particular temperature. It provides a measure of how far the reaction will proceed to the right (towards products) under those specific conditions. A large \( K_p \) (>1) indicates that, at equilibrium, the reaction system consists mostly of products. Conversely, a small \( K_p \) (<1) implies that the reactants are favored at equilibrium.

In the exercise, we're given the \( K_p \) and we use it to find the equilibrium pressures. By arranging the ICE Table values into the \( K_p \) expression, we derive an equation that lets us calculate the changes in pressure and thus the final equilibrium state.

Partial Pressure

Partial pressure refers to the pressure that a single gas in a mixture of gases would exert if it alone occupied the entire volume. When dealing with chemical equilibria involving gases, we often discuss reactions in terms of the partial pressures of the reactants and products.

Particularly in the provided exercise, we work with the initial partial pressures of methane and iodine, and the changes in these pressures as the reaction approaches equilibrium. We assume that the reaction container does not change in volume or temperature, meaning that according to Dalton's Law of Partial Pressures, the total pressure in the system is the sum of the partial pressures of the individual gases involved. Therefore, by calculating the partial pressures of the products, we find the condition of the reaction at equilibrium.

Understanding the concept of partial pressures is crucial in calculating equilibrium positions for reactions involving gases, as evidenced by solving the exercise.