Problem 126 - A Complete Resource Book in Chemistry for JEE Main

Question

For the dissociation reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\) the degree of dissociation \((\alpha)\) in terms of \(K_{p}\) and total equilibrium pressure \(\mathrm{P}\) is: (a) \(\alpha=\sqrt{\frac{4 \mathrm{P}+\mathrm{K}_{\mathrm{p}}}{\mathrm{K}_{\mathrm{P}}}}\) (b) \(\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{p}}}{4 \mathrm{P}+\mathrm{K}_{\mathrm{p}}}}\) (c) \(\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{p}}}{4 \mathrm{P}}}\) (d) None of these

Step-by-Step Solution

Answer

The correct answer is (c).

Step 1: Write the expression for equilibrium

For the reaction \( \mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2} \), the equilibrium constant \( K_p \) is defined as:\[K_p = \frac{{(P_{\mathrm{NO}_2})^2}}{{P_{\mathrm{N}_2\mathrm{O}_4}}}\] where \( P_{\mathrm{NO}_2} \) and \( P_{\mathrm{N}_2\mathrm{O}_4} \) are the partial pressures at equilibrium.

Step 2: Express partial pressures in terms of α and total pressure P

Initially, let the pressure of \( \mathrm{N}_{2} \mathrm{O}_{4} \) be \( P \). Upon dissociation, let \( \alpha \) be the degree of dissociation. Hence,- The change in pressure due to dissociation: \( x = \alpha P/2 \).- \( P_{\mathrm{N}_2\mathrm{O}_4} = P - x = P - \alpha P/2 = P(1 - \alpha/2) \)- \( P_{\mathrm{NO}_2} = 2x = 2(\alpha P/2) = \alpha P \).

Step 3: Substitute the expressions into the equilibrium constant equation

Substitute the expressions from Step 2 into the equation for \( K_p \):\[K_p = \frac{(\alpha P)^2}{P(1 - \alpha/2)}\] Simplify:\[K_p = \frac{\alpha^2 P^2}{P(1 - \alpha/2)} = \frac{\alpha^2 P^2}{P - \alpha P/2}\] Thus,\[K_p = \frac{\alpha^2 P}{1 - \alpha/2}\]

Step 4: Solve for α in terms of Kp and P

Rearrange the equation from Step 3 to isolate \( \alpha \):\[\alpha^2 = \frac{K_p (1 - \alpha/2)}{P}\]Assume \( \alpha \) is small, so \( 1 - \alpha/2 \approx 1 \), which simplifies:\[\alpha^2 \approx \frac{K_p}{P}\] Therefore,\[\alpha = \sqrt{\frac{K_p}{4P}}\], which matches option (c).

Step 5: Conclusion

By comparing, we find that only option (c) matches our derived equation for \( \alpha \). So, option (c) is the correct answer.

Key Concepts

Degree of Dissociation

The degree of dissociation, denoted by \( \alpha \), is a measure of the extent to which a compound dissociates into its ions or simpler molecules. It is a dimensionless quantity and represents the fraction of the original compound that breaks down. In the context of chemical equilibrium, it is crucial to understand how changes in pressure and temperature affect \( \alpha \).

In our example involving the dissociation reaction \( \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) \), the initial pressure of \( \mathrm{N}_{2} \mathrm{O}_{4} \) starts at a given value \( P \). When it dissociates, it generates \( \alpha P \) for \( \mathrm{NO}_{2} \), illustrating how \( \alpha \) quantifies the conversion. Using the formula derived:
\( \alpha = \sqrt{\frac{K_p}{4P}} \), where \( K_p \) is the equilibrium constant and \( P \) is the total pressure at equilibrium, we can calculate how much of the compound dissociates under specific conditions.

Equilibrium Constant

The equilibrium constant, \( K_p \), is a fundamental concept in chemistry. It quantifies the ratio of the partial pressures of the products to the reactants at equilibrium. For a given chemical reaction, \( K_p \) provides insight into the position of the equilibrium and dictates how far the reaction proceeds.

In the case of the dissociation reaction \( \mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2} \), \( K_p \) is determined by the relationship:
\[K_p = \frac{{(P_{\mathrm{NO}_2})^2}}{{P_{\mathrm{N}_2\mathrm{O}_4}}}\]
This equation emphasizes that the equilibrium constant relies on the partial pressures of the dissociated molecules \( \mathrm{NO}_{2} \) and the remaining \( \mathrm{N}_{2} \mathrm{O}_{4} \). Larger \( K_p \) values suggest that the reaction favors product formation, while smaller values indicate that the reactants are favored.

Understanding \( K_p \) helps predict how changes in conditions, like pressure and temperature, might shift the equilibrium and affect the degree of dissociation.

Partial Pressure

Partial pressure refers to the pressure exerted by a gas in a mixture of gases. It's an essential concept for dealing with gaseous reactions like our dissociation example.

In the reaction \( \mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2} \), initially, the entire pressure \( P \) is due to \( \mathrm{N}_{2} \mathrm{O}_{4} \). As it dissociates, \( \mathrm{NO}_{2} \) is formed and contributes to the partial pressure. We express these pressures as:
- \( P_{\mathrm{N}_2\mathrm{O}_4} = P(1 - \alpha/2) \)
- \( P_{\mathrm{NO}_2} = \alpha P \)

These expressions highlight the distribution of total pressure across the different gases at equilibrium. The ability to calculate partial pressures helps in determining the equilibrium constant \( K_p \) and the degree of dissociation \( \alpha \), effectively linking the pressures with the chemical changes happening within the mixture.

Recognizing changes in partial pressures also allows chemists to assess how modifications in external conditions could influence equilibrium positions and reaction yields.