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The degree of dissociation is a vital concept in chemical equilibrium, especially when dealing with gases. It represents the fraction of a substance that has dissociated, or split, into its product forms in a chemical reaction. This can be calculated as the ratio of the moles of a dissociated component to the initial moles of the substance. For example, in the dissociation of dinitrogen tetroxide, \( \mathrm{N}_2\mathrm{O}_4 \rightarrow 2 \mathrm{NO}_2 \), the degree of dissociation (\( \alpha \)) can indicate how much \( \mathrm{N}_2\mathrm{O}_4 \) has converted into \( \mathrm{NO}_2 \) at a given condition.

Consider an initial amount of one mole of \( \mathrm{N}_2\mathrm{O}_4 \). If \( 25\% \) is dissociated, this implies \( \alpha = 0.25 \). Consequently, at the onset, 0.75 moles remain as \( \mathrm{N}_2\mathrm{O}_4 \), and 0.5 moles turn into \( \mathrm{NO}_2 \).

This concept is essential when calculating changes in a system when external conditions such as pressure are altered, as it helps predict the system's response under new conditions.

Le Chatelier's Principle is a fundamental rule in understanding how equilibria respond to changes in conditions. It states that if a dynamic equilibrium is disturbed by changing the temperature, pressure, or concentration, the system will shift its equilibrium position to counteract the effect of the disturbance and restore a new equilibrium.

For instance, consider the equilibrium of \( \mathrm{N}_2\mathrm{O}_4 \leftrightarrow 2 \mathrm{NO}_2 \). According to Le Chatelier's Principle, reducing the pressure shifts the equilibrium towards the side with a greater number of moles of gas. Since the right side of the equation (\( 2 \mathrm{NO}_2 \)) has more moles, the equilibrium shifts right as pressure decreases.

This shift is why, when the pressure is lowered to 0.1 atm from 1 atm, the degree of dissociation increases. The system adapts by favoring the forward reaction (more dissociation), resulting in a higher percentage of \( \mathrm{N}_2\mathrm{O}_4 \) dissociating, as seen in the exercise.

The equilibrium constant, \( K_P \), is an essential concept that quantifies the ratio of the concentrations of products to reactants at equilibrium for a given reaction and is specifically valuable in gas equilibria. It remains constant at a fixed temperature, regardless of initial concentrations or pressure, providing a snapshot of the equilibrium state.

For the dissociation reaction \( \mathrm{N}_2\mathrm{O}_4 \rightarrow 2 \mathrm{NO}_2 \), the equilibrium constant \( K_P \) is calculated using partial pressures. At 1 atm, \( K_P \) is \( \frac{(2 \alpha)^2}{1 - \alpha} \), resulting in a value of \( \frac{1}{3} \) when \( \alpha = 0.25 \).

When investigating the effect of pressure changes (such as reducing pressure to 0.1 atm), it’s important to maintain the same \( K_P \). This constant provides a relationship to solve for the new degree of dissociation. As seen from the calculations in the exercise, \( K_P \) is conserved, guiding the determination of the new equilibrium position and ultimately leading to the higher dissociation percentage.