Question

Consider the hypothetical reaction at \(325^{\circ} \mathrm{C}\) $$ \mathrm{R}(g)+\mathrm{Q}(g) \rightleftharpoons 2 \mathrm{Z}(g) \quad K=2.71 $$ What are the equilibrium partial pressures of all the gases if all the gases (products and reactants) have an initial partial pressure of 0.228 atm?

Short answer

Answer: The equilibrium partial pressures of the gases are R = 0.1441 atm, Q = 0.1441 atm, and Z = 0.1678 atm.

Step-by-step solution

Write the expression for the equilibrium constant

The equilibrium constant (K) for the given reaction is the ratio of the product of the molar concentrations (or partial pressures) of the products to the product of the molar concentrations (or partial pressures) of the reactants at equilibrium. For this reaction, the expression for K is: $$ K = \frac{[\mathrm{Z}]^2}{[\mathrm{R}][\mathrm{Q}]} $$

Set up the equation for the changes in partial pressures

Let x be the amount each of R and Q will be consumed at equilibrium. Then, 2x will be the amount of Z produced. We can express the equilibrium partial pressures of the gases in terms of x: $$ [\mathrm{R}]_{eq} = [\mathrm{R}]_0 - x = 0.228 - x \\ [\mathrm{Q}]_{eq} = [\mathrm{Q}]_0 - x = 0.228 - x \\ [\mathrm{Z}]_{eq} = [\mathrm{Z}]_0 + 2x = 0 + 2x $$

Substitute the equilibrium partial pressures into the K expression

Now, we can substitute the equilibrium partial pressures into the expression for K: $$ 2.71 = \frac{(2x)^2}{(0.228 - x)(0.228 - x)} $$

Solve the quadratic equation

First, we can simplify the equation by multiplying both sides by \((0.228 - x)^2\) and then expanding: $$ 2.71(0.228 - x)^2 = (2x)^2 $$ Expand the brackets and then simplify the equation by combining like terms: $$ 0.1406 - 1.2352x + x^2 = 4x^2 $$ Now, move all the terms to one side to set the equation to zero: $$ 3x^2 + 1.2352x - 0.1406 = 0 $$ Now, we solve for x using the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Substitute the coefficients a = 3, b = 1.2352, and c = -0.1406 into the formula and solve for x: $$ x_1 = 0.0839, \quad x_2 = -0.5623 $$ We reject the negative value for x, as it doesn't have a physical meaning in this context.

Calculate the equilibrium partial pressures

Using the calculated value for x (x = 0.0839 atm), find the equilibrium partial pressures for R, Q, and Z: $$ [\mathrm{R}]_{eq} = 0.228 - x = 0.228 - 0.0839 = 0.1441 \ \mathrm{atm} \\ [\mathrm{Q}]_{eq} = 0.228 - x = 0.228 - 0.0839 = 0.1441 \ \mathrm{atm} \\ [\mathrm{Z}]_{eq}= 2x = 2\times 0.0839 = 0.1678 \ \mathrm{atm} $$ Therefore, the equilibrium partial pressures of the gases are: R = 0.1441 atm; Q = 0.1441 atm; Z = 0.1678 atm.

Key concepts

Equilibrium Constant

Understanding the equilibrium constant (K) is crucial for unraveling the dynamic balance between reactants and products in a chemical reaction that has reached equilibrium. By definition, the equilibrium constant is a numerical value that represents the ratio of the concentration of products to reactants, each raised to the power of their respective coefficients in the balanced equation, when the reaction has reached a stable balance.

In our example, the reaction \( \text{R}(g) + \text{Q}(g) \rightleftharpoons 2 \text{Z}(g) \) with an equilibrium constant \( K=2.71 \) suggests that at \(325^\circ \text{C}\), the ratio of the square of the molar concentration of Z to the product of the molar concentrations of R and Q, will equal 2.71. The equilibrium constant not only predicts the direction of the reaction but also gives insights into the extent of the reaction - the larger the value of K, the more the reaction favors the formation of products.

It's important to mention that K is constant only at a given temperature; any change in temperature can result in a different K value, signifying a new equilibrium position.

Partial Pressures

In the context of gases, partial pressures play the role often taken by concentrations in liquids. The partial pressure of a gas is the pressure that it would exert if it alone occupied the entire volume of the mixture at the same temperature. For a reaction involving gases, the equilibrium constant can also be expressed in terms of partial pressures. This is especially useful since it's often easier to measure the pressure of a gas rather than its concentration.

When dealing with equilibrium in gaseous systems, it is crucial to consider how changes in pressure can influence the system. The exercise demonstrates this by asking for the equilibrium partial pressures of all gases, assuming they each start with an initial pressure of 0.228 atm. Here, the concept of changes in partial pressures due to the progression of the reaction is applied to find equilibrium conditions. It's noteworthy that the process of reaching equilibrium does not depend on the initial amounts of reactants or products but on the change in pressure (or concentration) from the initial state.

Equilibrium Expressions

The equilibrium expression is the formula that allows us to calculate the equilibrium constant (K) for a given reaction. For reactions in gas phase, it involves the equilibrium partial pressures of the reactants and products. In our example, the equilibrium expression is derived from the balanced chemical equation and presented as \( K = \frac{[\text{Z}]^2}{[\text{R}][\text{Q}]} \).

Writing the equilibrium expression requires the understanding that the concentrations (or partial pressures) of the reactants and products are raised to their stoichiometric coefficients from the chemical equation. This is reflected in how \( [\text{Z}] \) is squared in the expression because two moles of Z are produced for every mole of R and Q that reacts.

As students venture into equilibrium calculations, they must become proficient in constructing such expressions and realizing how the equilibrium constant physically represents the state of the reaction mixture when the rates of the forward and reverse reactions are equal. This allows them to not only solve numerical problems but also to predict how a system in equilibrium will respond to various disturbances such as changes in concentration, pressure, or temperature.