Question

For the reaction \(2 \mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)+2 \mathrm{C}(g),\) a reaction vessel initially contains only \(\mathrm{A}\) at a pressure of \(P_{\mathrm{A}}=225 \mathrm{mm} \mathrm{Hg} .\) At equilibrium,\(P_{\mathrm{A}}=55 \mathrm{mmHg}\) . Calculate the value of \(K_{\mathrm{P}}\) . (Assume no changes in vol- ume or temperature.)

Short answer

For the reaction to reach equilibrium resulting in a pressure of 55 mmHg for A, Kp is calculated using the change in pressure for A and the stoichiometry of the reaction.

Step-by-step solution

Understanding the Reaction and Initial Conditions

We start by analyzing the balanced chemical equation and the initial state of the reaction. Initially, only A is present at a pressure of 225 mmHg. There is no B or C in the reaction vessel at the beginning.

Determining the Change in Pressure for A

Since the pressure of A at equilibrium is given to be 55 mmHg, we can calculate the change in pressure as the initial pressure minus the equilibrium pressure: \[ \Delta P_{\mathrm{A}} = P_{\mathrm{A_{initial}}} - P_{\mathrm{A_{equilibrium}}} \]

Calculating Equilibrium Partial Pressures of B and C

As per the stoichiometry of the reaction, for every 2 moles of A that react, 1 mole of B and 2 moles of C are formed. The change in pressure for B and C will be: \[ P_{\mathrm{B}} = P_{\mathrm{A}}/2 \] \[ P_{\mathrm{C}} = P_{\mathrm{B}} \times 2 \] We use the change in pressure of A to find the pressures for B and C at equilibrium.

Writing the Expression for Kp

The equilibrium constant for the reaction in terms of partial pressures is: \[ K_{\mathrm{P}} = \frac{P_{\mathrm{B}} \times P_{\mathrm{C}}^2}{P_{\mathrm{A}}^2} \]

Calculating Kp

We substitute the values for the pressures of A, B, and C at equilibrium into the expression for Kp and solve for Kp.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time. It's important to understand that this doesn't mean the reactants and products are in equal concentrations; rather, they are at concentrations that remain constant. For the given exercise, when the reaction of gas A to form gases B and C reaches equilibrium, the partial pressures of these gases do not change anymore despite the reaction still occurring in both directions.

This balance can be disturbed by changes in conditions such as pressure, volume, and temperature, also known as stressors, in accordance with Le Chatelier's principle. However, our exercise assumes constant volume and temperature, which simplifies the calculation as these potential stressors do not affect the equilibrium state.

Partial Pressures

In a mixture of gases, each gas exerts a pressure as if it occupied the volume alone. This individual pressure is known as partial pressure and is directly proportional to the mole fraction of the gas in the mixture. The contribution of each gas to the total pressure can be crucial in understanding equilibrium in gaseous reactions, as shown in the exercise. When calculating for partial pressures at equilibrium, it's essential to consider stoichiometry to determine how the pressure of one gas relates to the others. Notably, as partial pressures shift from their initial values due to the reaction, tracking these changes is key to finding the equilibrium constant, as demonstrated in the step-by-step solution.

Equilibrium Constant Expression

The equilibrium constant expression is a mathematical representation of the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients. For reactions involving gases, it's common to express the equilibrium constant in terms of partial pressures, denoted as \( K_{\mathrm{P}} \).

The step-by-step solution illustrates how to derive the expression for the equilibrium constant from the balanced chemical equation. In the expression \( K_{\mathrm{P}} = \frac{P_{\mathrm{B}} \times P_{\mathrm{C}}^2}{P_{\mathrm{A}}^2} \), \( P_{\mathrm{A}} \), \( P_{\mathrm{B}} \), and \( P_{\mathrm{C}} \) are the partial pressures of the gases at equilibrium. The expression is crucial as it quantitatively describes the position of equilibrium and is specific to the given reaction at a certain temperature.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves the use of the mole ratio derived from the balanced chemical equation to convert between amounts of different substances. Understanding stoichiometry is essential in solving problems related to chemical equilibrium, like the given exercise.

In the provided solution, stoichiometry is used to determine how the pressures of B and C change relative to the change in pressure of A. Since the reaction consumes 2 moles of A to produce 1 mole of B and 2 moles of C, the change in pressure for A directly relates to the changes in B and C, maintaining the stoichiometric ratio. This aspect is critical in accurately calculating the equilibrium constant for the system.