Problem 92 - A Complete Resource Book in Chemistry for JEE Main

Question

At 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 \(X(g)\) is dissociated. The total pressure of the equilibrium system is 'P' atm. what is the partial pressure (in \(\mathrm{atm}\) ) of \(\mathrm{X}(\mathrm{g})\) at equilibrium? (a) 1 (b) 4 (c) 2 (d) \(0.5\)

Step-by-Step Solution

Answer

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

Step 1: Identify Initial Conditions

Let's assume the initial pressure of \(X\) is \(P_0\) atm. At equilibrium, 50% of \(X\) is dissociated, meaning the partial pressure of dissociated \(X\) is \(\frac{P_0}{2}\) atm, leaving \(\frac{P_0}{2}\) atm of \(X\) remaining.

Step 2: Determine Partial Pressures at Equilibrium

For each dissociated \(X\), one \(Y\) and one \(Z\) is produced. Thus, the partial pressures of \(Y\) and \(Z\) both become \(\frac{P_0}{2}\) atm. The equilibrium pressures are: \(P_X = \frac{P_0}{2}\), \(P_Y = \frac{P_0}{2}\), \(P_Z = \frac{P_0}{2}\).

Step 3: Write the Expression for Kp

Using the equilibrium expression for the reaction, \[ K_p = \frac{P_Y \cdot P_Z}{P_X} = \frac{\left(\frac{P_0}{2}\right) \cdot \left(\frac{P_0}{2}\right)}{\frac{P_0}{2}} = \frac{\left(\frac{P_0^2}{4}\right)}{\frac{P_0}{2}} = \frac{P_0}{2}. \] Given \(K_p = 1\) atm, we have \(\frac{P_0}{2} = 1\).

Step 4: Solve for Initial Pressure P0

From \(\frac{P_0}{2} = 1\), solving for \(P_0\), we get \(P_0 = 2\) atm. Since initially, 50% of \(X\) was dissociated, at equilibrium, the remaining partial pressure of \(X\) is \(\frac{P_0}{2} = 1\) atm.

Key Concepts

Partial Pressure

Partial pressure is an important concept in chemical equilibrium, especially when dealing with gases. It describes the pressure contributed by a single gas in a mixture of gases. The total pressure of the gas mixture is the sum of the partial pressures of all individual gases present.
For a gas X with an initial pressure of P_0 atm, if 50% dissociates, the partial pressure of the remaining X is \( \frac{P_0}{2} \) atm. Each product species in the dissociation will also contribute its own partial pressure, adding up to the total equilibrium pressure. In simple terms, consider partial pressure as how much pressure each gas component contributes to the overall pressure in a container.

• It is calculated for each gas using the formula: \( P_{i} = \frac{n_{i}RT}{V} \), where \( n_{i} \) is the number of moles of the gas, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( V \) is the volume.
• In the provided exercise, the emphasis is on how these partial pressures add up to contribute to the equilibrium state, specifically utilizing the concentration changes during dissociation.

Equilibrium Constant (Kp)

The equilibrium constant for partial pressures, K_p, is a dimensionless number that provides insights into the balance between reactants and products in a gaseous reaction at equilibrium. It is specifically defined for reactions involving gases, with its expression incorporating the partial pressures of the gases involved.
In the reaction: \( \mathrm{X}(\mathrm{g}) \rightleftharpoons \mathrm{Y}(\mathrm{g})+\mathrm{Z}(\mathrm{g}) \), K_p tells us the relationship between the equilibrium pressures of products and reactants. It is calculated using the formula:

• \[ K_p = \frac{P_Y \cdot P_Z}{P_X} \]
• This formula signifies the pressure of products raised to their stoichiometric coefficients, divided by the pressure of reactants raised to their coefficients.

The value of K_p helps predict the direction of the reaction under given conditions. A large K_p value suggests the formation of more products, while a small K_p indicates more reactants at equilibrium.

In this exercise, given that K_p equals 1 atm, it indicates a balanced situation, where neither side is strongly favored. The calculated pressures at equilibrium ensure that this balance is maintained.

Dissociation Reaction

Dissociation reactions are chemical reactions where a compound breaks apart into two or more components. In gaseous reactions, this often means a single substance decomposes into smaller gaseous products.
For example, in the exercise, \( \mathrm{X}(\mathrm{g}) \rightleftharpoons \mathrm{Y}(\mathrm{g})+\mathrm{Z}(\mathrm{g}) \), \( \mathrm{X} \) dissociates into \( \mathrm{Y} \) and \( \mathrm{Z} \).

• The extent of dissociation can be expressed as a percentage. Here, 50% of X is dissociated, meaning half of the original amount has been converted to products.
• The dissociation process alters the concentration of reactants and products, directly affecting the partial pressures and subsequently the equilibrium condition.

This reaction follows the principle of conservation of mass. The number of moles of products formed corresponds to the number of moles of reactant that has dissociated. Recognizing these changes is essential to solving equilibrium problems, and understanding dissociation helps predict and calculate changes in pressure and concentration over time.