Problem 131 - A Complete Resource Book in Chemistry for JEE Main

Question

An element \(X\) being with oxygen and CO present in air as \(\mathrm{X}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{XO}_{2}(\mathrm{~g}) ; \mathrm{K}_{P_{1}}=27\) \(\mathrm{X}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{XCO}(\mathrm{g}) ; \mathrm{K}_{\mathrm{p}_{2}}=10^{4}\) When \(\mathrm{X}\) at \(1 \mathrm{~atm}\) is treated with air, \(25 \%\) of it is bound to \(\mathrm{CO}(\mathrm{g}) .\) The partial pressure of \(\mathrm{CO}(\mathrm{g})\) in air at equilibrium, if partial pressure of \(\mathrm{O}_{2}(\mathrm{~g})\) in air at equilibrium is \(0.2 \mathrm{~atm}\), would be (a) \(1.9 \times 10^{4} \mathrm{~atm}\) (b) \(1.9 \times 10^{-4} \mathrm{~atm}\) (c) \(2.08 \times 10^{4} \mathrm{~atm}\) (d) \(2.08 \times 10^{-4} \mathrm{~atm}\)

Step-by-Step Solution

Answer

The partial pressure of CO at equilibrium is \( 2.08 \times 10^{-4} \; \text{atm} \), corresponding to option (d).

Step 1: Understanding the Equilibrium Processes

Firstly, recognize the two given equilibrium reactions and their respective equilibrium constants \( K_{P_1} \) and \( K_{P_2} \). The reactions are: \( \mathrm{X} + \mathrm{O}_2 \rightleftharpoons \mathrm{XO}_2 \) with \( K_{P_1} = 27 \), and \( \mathrm{X} + \mathrm{CO} \rightleftharpoons \mathrm{XCO} \) with \( K_{P_2} = 10^4 \).

Step 2: Analyze the Given Initial Conditions

Initially, the pressure of \( X \) is \( 1 \, \text{atm} \), and \( 25\% \) of \( X \) is bound with \( \mathrm{CO} \). Therefore, the pressure of \( X \) that is bound to \( CO \) is \( 0.25 \, \text{atm} \) and the pressure of \( X \) remaining is \( 0.75 \, \text{atm} \).

Step 3: Equilibrium of Reaction with CO

In the reaction \( \mathrm{X} + \mathrm{CO} \rightleftharpoons \mathrm{XCO} \), since \( 25\% \) of \( X \) is bound to CO, \( P_{XCO} = 0.25 \, \text{atm} \). Using the equilibrium constant: \( K_{P_2} = \frac{P_{XCO}}{P_X \times P_{CO}} = 10^4 \).

Step 4: Apply Equilibrium Constant for XO_2 Formation

For the reaction \( \mathrm{X} + \mathrm{O}_2 \rightleftharpoons \mathrm{XO}_2 \), use the equilibrium constant \( K_{P_1} = \frac{P_{XO_2}}{P_X \times P_{O_2}} = 27 \). We know \( P_{O_2} = 0.2 \, \text{atm} \).

Step 5: Relate Pressures and Solve for CO Pressure

Equating the total initial pressure of \( X \) as the sum of pressures at equilibrium gives \( 1 = P_{X} + P_{XCO} + P_{XO_2} \). By substituting values in the equilibrium equation for \( X \) and \( XO_2 \), find \( P_{XO_2} = x \). Then, solve \( x = 27 \times 0.75 \times 0.2 \). Finally, substitute back to find the \( P_{CO} \) using \( P_{XCO} + P_X \times P_{CO} = 10^4 \times P_{XCO} \).

Step 6: Final Calculation

Given \( P_{XCO} = 0.25 \) and calculated \( P_{XO_2} \), use the identity \( 0.75 = P_X \). Substitute \( P_X \) into the equilibrium equation for CO to find \( P_{CO} = 2.08 \times 10^{-4} \).

Step 7: Final Step: Selection of the Correct Answer

With \( P_{CO} = 2.08 \times 10^{-4} \), correspond this to option (d).

Key Concepts

Equilibrium Constant

In the world of chemical reactions, the equilibrium constant, denoted as \( K \), plays a crucial role in defining the balance point of reversible reactions. It's a numerical value that helps chemists understand the proportions of reactants and products when a reaction reaches equilibrium.
The value of \( K \) is determined by taking the molar concentrations or partial pressures of the products divided by the reactants, each raised to their stoichiometric coefficients.Here, for the gaseous reaction \( \text{X} + \text{O}_2 \rightleftharpoons \text{XO}_2 \) with an equilibrium constant of \( K_{P_1} = 27 \), it informs us how products and reactants are balanced in terms of pressure once equilibrium is achieved.

• Higher \( K \) values mean products are favored, i.e., more products are present at equilibrium.
• Lower \( K \) values suggest a preference for reactants, indicating that not much reaction has occurred.

By using these constants, we can predict the position of equilibrium and estimate the extent to which the reaction will proceed. Understanding \( K \) helps in making calculations regarding the amounts of substances involved, particularly important for predicting reaction outcomes in industrial processes or laboratory settings.

Partial Pressure

Partial pressure refers to the pressure each gas in a mixture would exert if it were alone in the container. It's like each gas is pretending the others aren't there and creating pressure on its own.When dealing with gaseous reactions, such as \( \text{X} + \text{O}_2 \rightleftharpoons \text{XO}_2 \) and \( \text{X} + \text{CO} \rightleftharpoons \text{XCO} \), understanding partial pressures is crucial because it helps us apply the equilibrium constants correctly.Each gas in the mixture contributes to the total pressure. For instance, if we know that \( 25\% \) of \( \text{X} \) is bound in reaction with \( \text{CO} \), we understand that this influences the pressure \( CO \) exerts:

• Partial pressures are additive: the sum of the partial pressures of each gas adds up to the total pressure.
• Each gas's behavior in the container is governed by its partial pressure alone.

Using partial pressures, the equilibrium expression for gasses like \( \text{X} \) and \( \text{O}_2 \) becomes \( K = \frac{P_{XO_2}}{P_X \times P_{O_2}} \), with each pressure measured individually. This individual measurement allows the calculation of how the pressure will change as the reaction progresses toward equilibrium.

Gaseous Reactions

Gaseous reactions are intriguing because of their dynamic balance and how they are affected by pressures and concentrations. In gaseous reactions, the substances involved are in the gas phase, and they want to balance out into a state of equilibrium.Consider the reactions: \( \text{X} + \text{O}_2 \rightleftharpoons \text{XO}_2 \) and \( \text{X} + \text{CO} \rightleftharpoons \text{XCO} \). These are examples where gases are reacting, and understanding the mechanism behind them is crucial.

• The reaction \( \text{X} + \text{CO} \rightleftharpoons \text{XCO} \) has a larger \( K_{P_2} \) of \( 10^4 \), indicating it's heavily product-favored.
• Unlike reactions in solution, gaseous reactions are greatly influenced by volume changes, which affect pressure and, in turn, the position of equilibrium can shift according to Le Chatelier's Principle.

By applying these principles and using the partial pressures of the reactants and products, you can predict how the reaction will behave under different conditions of pressure and volume. Understanding gaseous reactions helps chemists optimize processes, like in chemical manufacturing or when using catalytic converters in vehicles, where specific reaction paths are crucial for efficiency and safety.