Question

At \(\mathrm{Kp}\) for the following reaction is 1 atm \(\mathrm{X}(\mathrm{g}) \rightleftharpoons \mathrm{Y}(\mathrm{g})+\mathrm{Z}(\mathrm{g})\) At equilibrium, \(50 \%\) of \(\mathrm{X}(\mathrm{g})\) is dissociated. The total pressure of the equilibrium system is ' \(\mathrm{P}\) ' atm. what is the partial pressure (in atm) of \(X(g)\) at equilibrium: (a) 1 (b) 4 (c) 2 (d) \(0.5\)

Short answer

The partial pressure of \( X(g) \) at equilibrium is 1 atm.

Step-by-step solution

Understand the Reaction Setup

The reaction provided is \( \mathrm{X}(\mathrm{g}) \rightleftharpoons \mathrm{Y}(\mathrm{g}) + \mathrm{Z}(\mathrm{g}) \). At equilibrium, 50% of \( \mathrm{X} \) is dissociated, which means that half of the initial moles of \( \mathrm{X} \) have reacted to form \( \mathrm{Y} \) and \( \mathrm{Z} \). The total equilibrium pressure of the system is \( P \) atm.

Set Up the Initial and Equilibrium Conditions

Let the initial number of moles of \( X \) be \( n \). Initially, the moles of \( Y \) and \( Z \) are zero. At equilibrium, since 50% of \( X \) is dissociated, the number of moles of \( X \) will be \( 0.5n \). The moles of \( Y \) and \( Z \) formed will be equal to \( 0.5n \) each, due to the stoichiometry of the reaction.

Express Total Pressure in Terms of Moles

The total pressure at equilibrium is given as \( P \) atm. Total moles at equilibrium are \( 0.5n + 0.5n + 0.5n = 1.5n \). Therefore, \( P = 1.5n \).

Calculate Partial Pressures of Each Gas

Partial pressure is calculated as the mole fraction of each gas multiplied by the total pressure. For \( X \), the mole fraction is \( \frac{0.5n}{1.5n} = \frac{1}{3} \) and thus \( P_X = \frac{1}{3} P \). For \( Y \) and \( Z \), the mole fraction is the same, and their partial pressures will also be \( \frac{1}{3} P \) each.

Use Equilibrium Constant to Solve for \( P \)

The equilibrium constant \( K_p \) given is 1 atm. Using the equation \( K_p = \frac{P_Y \cdot P_Z}{P_X} = \frac{\left(\frac{1}{3}P\right)\left(\frac{1}{3}P\right)}{\frac{1}{3}P} = \frac{1}{3}P = 1 \). Solve for \( P \) to find \( P = 3 \) atm.

Determine \( P_X \)

Now, using \( P = 3 \) atm and the expression for \( P_X \) as \( \frac{1}{3} P \), we find \( P_X = \frac{1}{3} \times 3 = 1 \) atm.

Key concepts

Partial Pressure

Partial pressure is a concept in chemistry used to describe the pressure that each gas in a mixture would exert if it were alone in the container. In a gaseous reaction, understanding partial pressures helps us determine how gases interact at equilibrium. The total pressure of a gaseous system is a sum of all the partial pressures of individual gases present.

• Partial pressure (\( P_X \)) of component X is calculated as its mole fraction multiplied by the total pressure of the system.
• The mole fraction is the ratio of the number of moles of the gas to the total moles in the system.
• This concept is particularly important in equilibrium calculations, like solving for the partial pressure of unfavored reactants or products.

In the provided example, partial pressure allows us to express how much of each component is present at equilibrium, thus helping confirm that the equilibrium constant is accurate.

Equilibrium Constant

The equilibrium constant, represented as \( K_p \) for gaseous reactions, indicates the ratio of product pressures to reactant pressures at equilibrium. It offers insights into the extent to which a reaction progresses and helps predict the direction of the reaction.

• \( K_p \) is defined for reactions involving gases and uses partial pressures in its expression.
• A \( K_p \) value equal to 1, as in our scenario, indicates that at equilibrium, neither reactants nor products are favored.
• This constant can guide us in calculating unknown pressures or concentrations, as demonstrated by rearranging the formula to find other variables.

Understanding \( K_p \) is crucial for correctly predicting how gas reactions behave under various conditions, ultimately aiding in calculating necessary quantities like total and partial pressures.

Stoichiometry

Stoichiometry involves the quantitative relationships of reactants and products in a chemical reaction. It's foundational for understanding how much of each chemical is involved and produced during a reaction and is essential for calculating equilibriums.

• Stoichiometry is used to determine the amounts of reactants needed and the quantity of products formed.
• In equilibrium reactions, this concept helps calculate how much of each gas will be present at equilibrium, using the initial conditions and known percentage dissociation.
• It also assists in expressing the total pressure by accounting for the mole proportions of gases formed or consumed.

For the given reaction, stoichiometry allows us to predict the number of moles remaining or formed, which is key in finding partial pressures and verifying equilibrium conditions.

Gaseous Reactions

Gaseous reactions involve chemical processes where reactants and products are primarily in the gaseous state. These reactions can significantly change pressure and volume conditions, directly affecting chemical equilibrium.

• In gases, the relationship between pressure, volume, and temperature is governed by ideal gas laws, which also impact how equilibrium states are reached.
• The reaction between gases often leads to changes in total pressure, which need to be measured to understand equilibrium conditions.
• These reactions require considering partial pressures and total atmospheric pressure to analyze how close a system is to equilibrium.

In the exercise, understanding the nature of gaseous reactions helps us comprehend changes in pressure due to dissociation and how partial pressures of gases contribute to calculating the equilibrium constant, \( K_p \).