Problem 174 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction, \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})=2 \mathrm{NO}_{2}(\mathrm{~g})\), the degree of dissociation at equilibrium is \(0.2\) at 1 atm pressure. The reciprocal of equilibrium constant \(\left(\frac{1}{\mathrm{~K}_{\mathrm{P}}}\right)\) is \(\mathrm{atm}^{-1} .\)

Step-by-Step Solution

Answer

The reciprocal of the equilibrium constant \(\frac{1}{K_P}\) is \(6 \text{ atm}^{-1}\).

Step 1: Identify Known Values

For the given reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})=2 \mathrm{NO}_{2}(\mathrm{~g})\), the degree of dissociation \(\alpha\) is \(0.2\), and the total pressure \(P\) is \(1\, \mathrm{atm}\).

Step 2: Define Equilibrium Concentrations

Assume the initial amount of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is \(1\,\text{mole}\). At equilibrium, \(\mathrm{N}_{2} \mathrm{O}_{4}\) dissociates to a degree of \(\alpha = 0.2\), so \(\mathrm{N}_{2} \mathrm{O}_{4}\) becomes \(1 - \alpha = 0.8\). For every mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) that dissociates, 2 moles of \(\mathrm{NO}_{2}\) are formed. Therefore, moles of \(\mathrm{NO}_{2}\) are \(2 \times 0.2 = 0.4\).

Step 3: Calculate Total Moles at Equilibrium

The total moles at equilibrium is the sum of moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\), which is \(0.8 + 0.4 = 1.2\).

Step 4: Determine Partial Pressures

The partial pressure for each gas can be found using the mole fraction multiplied by the total pressure. For \(\mathrm{N}_{2} \mathrm{O}_{4}\), \(P_{\mathrm{N}_2\mathrm{O}_4} = \frac{0.8}{1.2} \times 1 = \frac{2}{3} \, \text{atm}\). For \(\mathrm{NO}_{2}\), \(P_{\mathrm{NO_2}} = \frac{0.4}{1.2} \times 1 = \frac{1}{3}\, \text{atm}\).

Step 5: Calculate Equilibrium Constant \(K_P\)

The equilibrium constant \(K_P\) for the reaction can be calculated using the equation \(K_P = \left(\frac{P_{\mathrm{NO}_2}^2}{P_{\mathrm{N}_2\mathrm{O}_4}}\right)\). Substitute the values: \(K_P = \left(\frac{(\frac{1}{3})^2}{\frac{2}{3}}\right) = \frac{1}{6}\).

Step 6: Compute Reciprocal of Equilibrium Constant

The reciprocal of the equilibrium constant is \(\frac{1}{K_P}\). Therefore, \(\frac{1}{K_P} = \frac{6}{1} = 6 \, \text{atm}^{-1}\).

Key Concepts

Degree of Dissociation

The degree of dissociation (\(\alpha\)) is a crucial concept in understanding how much a compound dissociates into its components in a reaction. It represents the fraction of the initial reactant molecules that have dissociated at equilibrium. This fraction can range between 0 and 1.
For instance, consider the reaction of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\equals2 \mathrm{NO}_{2}(\mathrm{~g})\). If the degree of dissociation is given as \(0.2\), this means 20% of \(\mathrm{N}_{2}\mathrm{O}_{4}\) has broken down into \(\mathrm{NO}_{2}\).
In practical terms, if you start with 1 mole of \(\mathrm{N}_{2}\mathrm{O}_{4}\), at equilibrium 0.2 moles will dissociate, leaving 0.8 moles of \(\mathrm{N}_{2}\mathrm{O}_{4}\) and forming 0.4 moles of \(\mathrm{NO}_{2}\). Understanding the degree of dissociation helps in predicting the amounts of the substances present at equilibrium, which is vital for calculating other chemical equilibrium parameters.

Equilibrium Constant (Kp)

The equilibrium constant, \(\mathrm{K_p}\), is pivotal in determining the position of equilibrium for a reaction involving gases. It quantifies the ratio of product partial pressures to reactant partial pressures, each raised to the power of their coefficients in the balanced chemical equation.
The formula is derived from the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})=2 \mathrm{NO}_{2}(\mathrm{~g})\) and is given by:\[ K_p = \left(\frac{P_{\mathrm{NO}_2}^2}{P_{\mathrm{N}_2\mathrm{O}_4}}\right) \]In the example, with partial pressures calculated as \(P_{\mathrm{NO}_2} = \frac{1}{3} \ ext{atm}\) and \(P_{\mathrm{N}_2\mathrm{O}_4} = \frac{2}{3} \ ext{atm}\), \(K_p\) becomes \(\frac{1}{6}\) atm.
This value expresses how far the reaction has proceeded. A larger \(K_p\) indicates a more product-favored equilibrium, while a smaller \(K_p\) suggests the reactants are favored. In this case, because \(K_p\) is less than 1, the reactants are more prevalent at equilibrium.

Partial Pressure Calculation

Calculating partial pressures in a gas mixture is essential for understanding chemical reactions involving gases. Partial pressure is the pressure each gas in a mixture would exert if it occupied the whole volume alone.
In a reaction involving gases like \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})\), we start by calculating the mole fraction of each component. The mole fraction is the ratio of the number of moles of a particular gas to the total number of moles of all gases present.

• For \(\mathrm{N}_{2}\mathrm{O}_{4}\): Mole fraction = \(\frac{0.8}{1.2}\)
• For \(\mathrm{NO}_{2}\): Mole fraction = \(\frac{0.4}{1.2}\)

Using the total pressure of \(1 \text{ atm}\), we determine the partial pressures:

• \(P_{\mathrm{N}_2\mathrm{O}_4} = \frac{2}{3} \ ext{atm}\)
• \(P_{\mathrm{NO}_2} = \frac{1}{3} \ ext{atm}\)

These calculations are vital in using the equilibrium constant expression and when predicting how changes in pressure, volume, or amount affect a chemical system.

Gas Phase Reactions

In gas phase reactions, understanding how reactants and products interact is essential for predicting reaction behavior. Gas phase reactions occur entirely in the gaseous state, where molecules move freely and collide to rearrange their atoms into new products.
Take the given reaction: \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\rightarrow2 \mathrm{NO}_{2}(\mathrm{~g})\). Such reactions are influenced by factors like temperature, pressure, and the nature of the gases involved.
In a closed system, if there is an increase in temperature, generally the equilibrium will shift to favor the endothermic direction. For this reaction, that would be toward more \(\mathrm{NO}_{2}\). Conversely, increasing pressure typically shifts the equilibrium toward the side with fewer gas molecules. In this case, it shifts toward \(\mathrm{N}_{2}\mathrm{O}_{4}\).
Gas phase reactions illustrate the principles of chemical equilibria and how various conditions affect them. They are a key focus in chemical thermodynamics due to their relevance across industrial, laboratory, and atmospheric processes.