Question

Consider the reaction in a closed container: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ Initially, 1 mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is present. At equilibrium, \(\alpha\) mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) has dissociated to form \(\mathrm{NO}_{2}\). (a) Derive an expression for \(K_{P}\) in terms of \(\alpha\) and \(P,\) the total pressure. (b) How does the expression in (a) help you predict the shift in equilibrium caused by an increase in \(P ?\) Does your prediction agree with Le Châtelier's principle?

Short answer

The derived expression for \(K_{P}\) is \(K_{P} = P{4\alpha^{2} \over (1-\alpha)(1+\alpha)}\). As per this expression, an increase in total pressure will shift the equilibrium towards the right (towards \(NO_{2}\)), which is in agreement with Le Châtelier's principle.

Step-by-step solution

Initial conditions

Initially, there is 1 mole of \(N_{2}O_{4}\) gas and none of the \(NO_{2}\) gas in the system. Since the reaction hasn't started, no \(N_{2}O_{4}\) has dissociated.

Conditions at equilibrium

At equilibrium, \(\alpha\) mole of \(N_{2}O_{4}\) has dissociated into \(NO_{2}\). Since each mole of \(N_{2}O_{4}\) produces 2 moles of \(NO_{2}\), there will be \(2\alpha\) moles of \(NO_{2}\) gas formed and \(1-\alpha\) mole of \(N_{2}O_{4}\) remaining.

Expressing partial pressures

The partial pressure of a gas in a mixture is given by the mole fraction of the gas multiplied by the total pressure of the system. So, we can express the partial pressures of \(N_{2}O_{4}\) and \(NO_{2}\) as follows: \[ P(N_{2}O_{4}) = P{1-\alpha \over 1 + \alpha} \] \[ P(NO_{2}) = P{2\alpha \over 1 + \alpha} \] Here \(P\) is the total pressure of the system.

Deriving the expression for \(K_{P}\)

The equilibrium constant \(K_P\) is given by the ratio of the products of the partial pressures of the products to the reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation. For this reaction: \[ K_{P} = {(P(NO_{2}))^2 \over P(N_{2}O_{4})} \] Substituting the expressions for partial pressures from step 3 into this equation, we get \[ K_{P} = P{(2\alpha)^{2} \over (1 - \alpha)}{1 \over (1 + \alpha)} \] which simplifies to \(K_{P} = P{4\alpha^{2} \over (1-\alpha)(1+\alpha)} \)

Predicting the shift in equilibrium with change in total pressure

From the derived expression, it can be seen that \(K_{P}\) is directly proportional to the total pressure \(P\). Hence, increasing \(P\) will increase \(K_{P}\), and to maintain the equilibrium (because \(K_{P}\) is a constant at a given temperature), the reaction will shift in the direction where the total gas moles (and hence pressure) are increased. In this case, it is towards the right (towards \(NO_{2}\)).

Comparison with Le Châtelier's principle

Le Châtelier's principle states that if a system at equilibrium is subjected to a change, the system will adjust itself in such a way as to counteract that change. In this case, an increase in pressure would shift the equilibrium towards the side with more moles of gas, which is consistent with our prediction from the derived expression. This confirms that the derived expression and Le Châtelier's principle are in agreement.

Key concepts

Equilibrium Constant (Kp)

The equilibrium constant, known as \(K_p\), is essential in understanding chemical reactions at equilibrium, especially in systems involving gases. For reactions in a gaseous state, \(K_p\) relates the partial pressures of the gaseous products and reactants at equilibrium. This allows chemists and students alike to grasp how different components are present at equilibrium.

The expression for \(K_p\) is derived from the balanced chemical equation. For the given reaction \({\text{N}_2 \text{O}_4 (g) \rightleftharpoons 2\text{NO}_2 (g)}\). The equilibrium constant in terms of pressure can be expressed as:

• \(K_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}}\)

By substituting the partial pressures expressed as fractions of the total pressure \(P\), you can calculate \(K_p\) as \({P\frac{4\alpha^2}{(1-\alpha)(1+\alpha)}}\), where \(\alpha\) represents the mole fraction of dissociated \(\text{N}_2\text{O}_4\). Understanding this allows students to predict how changes in pressure affect the position of equilibrium by looking at \(K_p\).

Le Châtelier's Principle

Le Châtelier's principle is an essential concept that helps explain how a system at equilibrium responds to external changes. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the imposed change and establish a new equilibrium.

In this context, when pressure is increased in the setup involving \(\text{N}_2\text{O}_4}\) and \(\text{NO}_2\), the system will shift the equilibrium towards the direction that accommodates more gas molecules to alleviate the pressure increase. This is because the side with more moles of gas helps reduce the impact of the pressure change. In the given equation, increasing pressure shifts the equilibrium to the right, favoring the formation of \(\text{NO}_2\), which aligns perfectly with what we predicted using the expression of \(K_p\). Understanding Le Châtelier's principle enables students to comprehend the behavior of chemical systems undergoing stress, enhancing their ability to predict reaction adjustments.

Partial Pressure

Partial pressure is a fundamental concept in gases. It refers to the pressure each gas exerts in a mixture. In chemical equilibrium, partial pressures help determine the concentrations of reactants and products. When a reaction takes place in a closed container, like in our example, calculating the partial pressures of each gas is crucial for finding the equilibrium position.

For each gas in the reaction \({\text{N}_2 \text{O}_4\rightleftharpoons 2\text{NO}_2}\), the partial pressure can be articulated as:

• \(P_{\text{N}_2\text{O}_4} = P\frac{1-\alpha}{1+\alpha}\)
• \(P_{\text{NO}_2} = P\frac{2\alpha}{1+\alpha}\)

These expressions are derived considering the mole fraction and the overall pressure \(P\) of the system. Understanding partial pressure is crucial because these values are integrally used in the \(K_p\) expressions, which further elucidates how changes in reaction conditions can shift the equilibrium.