Question

Methane, \(\mathrm{CH}_{4},\) reacts with \(\mathrm{I}_{2}\) according to the reaction \(\mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{I}(g)+\mathrm{HI}(g) .\) At \(630 \mathrm{K}, K_{p}\) 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 partial pressures are approximately: CH4: 104.688 torr I2: 7.548 torr CH3I: 0.412 torr HI: 0.412 torr

Step-by-step solution

List initial partial pressures and write the given Kp for the reaction

Initially, we have the partial pressures given as follows: Methane (CH4): 105.1 torr Iodine (I2): 7.96 torr Since there is no information about initial partial pressures of CH3I and HI, we can assume they are both 0. The given Kp at 630 K is 2.26 x 10^-4.

Define the change in partial pressures

To reach equilibrium, the partial pressures of CH4 and I2 will decrease, while the partial pressures of CH3I and HI will increase. We will use the variable x to represent the change in partial pressure of CH4 and I2 (since they have a 1:1 molar ratio). Change in partial pressure of CH4 = -x Change in partial pressure of I2 = -x Change in partial pressure of CH3I = +x Change in partial pressure of HI = +x

Write the Kp expression and substitute the equilibrium partial pressures

The Kp expression for the given reaction is: \(K_p = \frac{[CH_3I][HI]}{[CH_4][I_2]}\) Now, substitute the equilibrium partial pressures: \(2.26 \times 10^{-4} = \frac{(0 + x)(0 + x)}{(105.1 - x)(7.96 - x)}\)

Solve for x

Solving for x from the equation above can be a bit difficult due to the denominator having two terms with x. However, we can use the fact that Kp is very small, which implies that x will also be small compared to the initial pressures. This allows us to approximate the denominator to simplify the equation: \(2.26 \times 10^{-4} \approx \frac{x^2}{(105.1)(7.96)}\) Now, we can solve for x: \(x^2 \approx (2.26 \times 10^{-4})(105.1)(7.96)\) \(x \approx \sqrt{(2.26 \times 10^{-4})(105.1)(7.96)}\) \(x \approx 0.412\)

Find the equilibrium partial pressures

Using the calculated value of x, we can find the equilibrium partial pressures for each species: Partial pressure of CH4: P_CH4 = 105.1 - x = 105.1 - 0.412 ≈ 104.688 torr Partial pressure of I2: P_I2 = 7.96 - x = 7.96 - 0.412 ≈ 7.548 torr Partial pressure of CH3I: P_CH3I = 0 + x = 0.412 torr Partial pressure of HI: P_HI = 0 + x = 0.412 torr At equilibrium, the partial pressures are approximately: CH4: 104.688 torr I2: 7.548 torr CH3I: 0.412 torr HI: 0.412 torr

Key concepts

Equilibrium Constant

Understanding the equilibrium constant is pivotal when dealing with chemical reactions that can occur in both forward and reverse directions. This mathematical expression, typically denoted as 'K' for a given temperature, quantifies the relative amounts of products and reactants at the point of chemical equilibrium.

For gaseous reactions, we use the equilibrium constant in terms of partial pressures, denoted as 'Kp'. The formula for 'Kp' is based on the balanced chemical equation and the partial pressures of the gases involved. The expression for 'Kp' is formulated by taking the product of the partial pressures of the products, each raised to the power of its stoichiometric coefficient, and dividing by the same for the reactants. A key point is that solids and liquids don't appear in this expression; only gases and aqueous solutions are included.

Considering an equation with generic components A and B reacting to form C and D, the expression for Kp would look like this: \( K_p = \frac{{[C]^c [D]^d}}{{[A]^a [B]^b}} \) where A and B are reactants, C and D are products, and a, b, c, d represent their stoichiometric coefficients from the balanced equation. The equilibrium constant is fundamentally a ratio that tells us about the favorability of a reaction; a larger Kp signifies that the reaction favors the production of products, while a smaller Kp means that reactants are favored.

A critical consideration is that 'Kp' is constant only at a constant temperature. Any change in temperature leads to a new equilibrium constant. This reflects the innate link between the equilibrium position of a reaction and its thermal environment.

Partial Pressures

When we delve into the domain of gases, the concept of partial pressures becomes crucial. Every gas in a mixture exerts a pressure as if it were alone in the container, and this pressure is known as its partial pressure. It's instrumental to note that the total pressure of the gas mixture is the sum of all these individual pressures.

In the equation provided, methane and iodine react and are present as gases, so we consider their partial pressures. To calculate the changes in partial pressures as the reaction proceeds toward equilibrium, we assume changes to these pressures, represented by the variable 'x'. It's these changes that eventually bring the reaction to the point where the forward and reverse reaction rates are equal, which is chemical equilibrium.

An important piece of advice for students dealing with complex systems is not to overlook the initial conditions. In this problem, for example, we started with non-zero partial pressures for methane and iodine but zero for the products. These initial conditions set the stage for how 'x' will affect each species as the system reaches equilibrium. In practice, since 'Kp' is quite small, we presume that the change in pressure 'x' is small relative to the initial pressures, allowing for an approximation that simplifies the algebraic manipulation involved in solving for 'x'.

Le Chatelier's Principle

Le Chatelier's principle is a guiding light in the dynamic world of chemical reactions at equilibrium. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

This principle provides a qualitative understanding of how a system at equilibrium responds to changes in concentration, pressure, or temperature. For instance, increasing the pressure of a gaseous system will generally favor the reaction that produces fewer moles of gas, minimizing the change in pressure. Conversely, a decrease in pressure favors the side with more moles of gas.

Other factors such as the addition of a reactant or removal of a product will shift the equilibrium position to produce more products, as the system strives to rebalance. Temperature changes, however, have a distinct effect as they essentially change the value of the equilibrium constant - for endothermic reactions, an increase in temperature results in a higher 'K' value, favoring the products, while for exothermic reactions, the opposite is true.

Students should also note that Le Chatelier's principle assists in predicting the effect of a stress on a system at equilibrium, but it does not predict the rate at which the new equilibrium is reached or the extent of the change. Short and simple, it acts as a compass indicating the direction of the equilibrium shift but doesn't measure the distance traveled.