Question

For the reaction \(\begin{aligned}&\mathrm{PQ}_{2} \rightleftharpoons \mathrm{PQ}+\mathrm{Q} \text { the degree of } \\\&' \mathrm{~g} & \mathrm{~g}\end{aligned} \mathrm{~g}\). dissociation \((\alpha)\) of \(\mathrm{PQ}_{2}\) can be related with \(\mathrm{P}\) or \(\mathrm{V}\) as: (a) \(\alpha \propto \frac{1}{\sqrt{\mathrm{P}}}\) (b) \(\alpha \propto \frac{1}{\mathrm{P}}\) (c) \(\alpha \propto \frac{1}{\sqrt{\mathrm{V}}}\) (d) \(\alpha \propto \mathrm{V}\)

Short answer

The correct option is (a) \(\alpha \propto \frac{1}{\sqrt{P}}\).

Step-by-step solution

Understanding the Reaction

For the reaction \(\mathrm{PQ}_{2} \rightleftharpoons \mathrm{PQ} + \mathrm{Q}\), we need to understand the dissociation. Initially, we start with one mole of \(\mathrm{PQ}_{2}\). When it dissociates, \(\alpha\) moles of \(\mathrm{PQ}_{2}\) dissociate to form \(\alpha\) moles of \(\mathrm{PQ}\) and \(\alpha\) moles of \(\mathrm{Q}\). Thus, the total number of moles at equilibrium is \(1 + \alpha\).

Writing the Equilibrium Constant

At equilibrium, the expression for the equilibrium constant \(K\) in terms of pressure is \(K = \frac{P_{\mathrm{PQ}} \cdot P_{\mathrm{Q}}}{P_{\mathrm{PQ}_{2}}}\). Given \(P_{\mathrm{PQ}_{2}} = (1-\alpha)P\), \(P_{\mathrm{PQ}} = \alpha P\), and \(P_{\mathrm{Q}} = \alpha P\), we substitute these into the expression to obtain: \(K = \frac{(\alpha P)^2}{(1-\alpha)P}\).

Simplifying the Equilibrium Constant

The expression simplifies to \(K = \frac{\alpha^2 P}{1-\alpha}\). For small values of \(\alpha\), where \(\alpha \ll 1\), \(1-\alpha \approx 1\). Thus the equation becomes \(K \approx \alpha^2 P\).

Determining the Relation Between α and P

Rearranging \(K \approx \alpha^2 P\), we get \(\alpha^2 = \frac{K}{P}\) or \(\alpha = \sqrt{\frac{K}{P}}\). Therefore, we find that \(\alpha \propto \frac{1}{\sqrt{P}}\).

Considering Volume-Related Options

We also consider volume \(V\), knowing \(P \cdot V = nRT\) from the ideal gas law. But in terms of direct relationships, this problem specifically gives us a relation between \(\alpha\) and \(P\), thus excluding direct relations with \(V\).

Conclusion

Based on the derivation, the degree of dissociation \(\alpha\) is proportional to \(\frac{1}{\sqrt{P}}\). Therefore, the correct choice is option (a).

Key concepts

Degree of Dissociation

The degree of dissociation, represented as \( \alpha \), is a measure of the fraction of a compound that dissociates into its components in a chemical reaction. It tells us how much of the initial compound has transformed into its products. For instance, in the reaction \( \text{PQ}_2 \rightleftharpoons \text{PQ} + \text{Q} \), \( \alpha \) signifies how much \( \text{PQ}_2 \) has split into \( \text{PQ} \) and \( \text{Q} \).
When we start with one mole of \( \text{PQ}_2 \), it dissociates into \( \alpha \) moles of \( \text{PQ} \) and \( \alpha \) moles of \( \text{Q} \). As the reaction reaches equilibrium, the total number of moles becomes \( 1 + \alpha \). The degree of dissociation is significant because it provides insight into the behavior of substances in reactions and helps understand equilibrium states.

Pressure Dependency

The extent to which a reaction proceeds is often influenced by pressure, especially in gaseous reactions. In the case of a dissociating gas such as \( \text{PQ}_2 \), pressure can affect the degree of dissociation \( \alpha \).
Using the equilibrium expression \( K = \frac{(\alpha P)^2}{(1-\alpha)P} \), we can see that as pressure \( P \) increases, \( \alpha \) tends to decrease if \( \alpha \) is small. Simplifying this gives us \( K \approx \alpha^2 P \), indicating that \( \alpha \) is related to \( \frac{1}{\sqrt{P}} \).
This means that the degree of dissociation \( \alpha \) decreases with increasing pressure, specifically following the relationship \( \alpha \propto \frac{1}{\sqrt{P}} \). It's interesting how this behavior showcases the inverse dependency, illustrating that higher pressures tend to favor the undissociated form of the compound.

Ideal Gas Law

The Ideal Gas Law, represented as \( PV = nRT \), is a fundamental equation in chemistry describing the behavior of ideal gases. It connects pressure \( P \), volume \( V \), and temperature \( T \) to the number of moles \( n \) and the ideal gas constant \( R \).
In applications related to dissociation, the Ideal Gas Law helps explain changes in volume and pressure. For instance, as the pressure changes in a system, it can directly be related to the number of moles and hence the degree of dissociation \( \alpha \).
Given that \( P = \frac{nRT}{V} \), we see how changes in temperature or volume would shift the pressure, which, as discussed, influences \( \alpha \). However, direct relationships between volume and \( \alpha \) are not straightforward without these considerations in a closed system with fixed temperatures.

Chemical Equilibrium

Chemical equilibrium occurs when a reversible reaction proceeds in such a way that the rates of the forward and reverse reactions are equal. At this point, the concentrations of reactants and products remain constant over time.
In the reaction \( \text{PQ}_2 \rightleftharpoons \text{PQ} + \text{Q} \), equilibrium is achieved when dissociation and recombination of \( \text{PQ}_2 \) balance out. The equilibrium constant \( K \), derived from the equilibrium concentrations or partial pressures of the reactants and products, is central to understanding chemical equilibrium.
The formula \( K = \frac{(\alpha P)^2}{(1-\alpha)P} \) tells us how equilibrium shifts with pressure changes and helps predict the concentrations of substances at equilibrium. Knowing \( \alpha \) through chemical equilibrium makes it easier to predict the reaction behavior under different conditions, helping in industrial, laboratory, and even natural processes.