Question

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

Short answer

Derived expression: \(K_P = \frac{4x^2P}{(1-x)(1+x)}\). Increasing pressure shifts equilibrium towards \(\mathrm{N}_{2} \mathrm{O}_{4}\), aligning with Le Châtelier's principle.

Step-by-step solution

Write the equilibrium expressions for partial pressures

Start by noting that at equilibrium, \(x\) moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\) have dissociated.- Initial moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\): 1- Moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at equilibrium: \(1 - x\)- Moles of \(\mathrm{NO}_{2}\) at equilibrium: \(2x\)The partial pressures can be expressed using the mole fractions and total pressure \(P\) as follows:- \(P_{\mathrm{N}_{2} \mathrm{O}_{4}} = \frac{(1 - x)P}{1 + x}\)- \(P_{\mathrm{NO}_{2}} = \frac{2xP}{1 + x}\)

Derive the expression for equilibrium constant \(K_P\)

The equilibrium constant \(K_P\) for the reaction is given by the expression:\[K_P = \frac{(P_{\mathrm{NO}_{2}})^2}{P_{\mathrm{N}_{2} \mathrm{O}_{4}}}\]Substitute the expressions for the partial pressures:\[K_P = \frac{\left(\frac{2xP}{1 + x}\right)^2}{\frac{(1 - x)P}{1 + x}}\]Simplify the expression:\[K_P = \frac{4x^2P^2}{(1 + x)^2} \times \frac{(1 + x)}{(1 - x)P}\]Cancel out unnecessary terms and simplify further:\[K_P = \frac{4x^2P}{(1 - x)(1 + x)}\]

Analyze the effect of pressure change on equilibrium

The derived expression demonstrates that \(K_P\) is proportional to \(P\) in the numerator. According to Le Châtelier's principle, increasing the pressure will shift the equilibrium position towards the side with fewer moles of gas because the system will attempt to reduce the pressure.Since the reverse reaction (forming \(\mathrm{N}_{2} \mathrm{O}_{4}\)) results in fewer moles, increasing \(P\) shifts the equilibrium towards \(\mathrm{N}_{2} \mathrm{O}_{4}\), reducing \(x\). This prediction aligns with Le Châtelier's principle.

Key concepts

Le Châtelier's Principle

Le Châtelier's Principle is a fundamental concept in chemical equilibrium that helps us understand how a system at equilibrium reacts to external changes.
This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and restore balance.

• If the concentration of a reactant or product is changed, the equilibrium will shift in the direction that tends to reduce that change in concentration.
• If the pressure in a gaseous system is altered, the equilibrium will shift towards the side with fewer moles of gas to minimize the pressure change.
• If the temperature is adjusted, the equilibrium will favor the endothermic or exothermic side, depending on whether heat is added or removed.

Let's apply this to our reaction between \(_2O_4(g)\rightleftarrows 2NO_2(g)\).
When the pressure is increased, according to Le Châtelier's principle, the equilibrium will shift towards forming more \(_2O_4\) because it has fewer moles of gas, thus alleviating the pressure increase.
This helps maintain equilibrium in the dynamic system.

Equilibrium Constant (Kp)

The equilibrium constant, \(K_p\), is a crucial mathematical expression that quantifies the ratio of products to reactants at equilibrium in terms of partial pressures for a gaseous reaction.
For the reaction \(_2O_4(g)\rightleftarrows 2NO_2(g)\):

• The expression for \(K_p\) is formulated as \(K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}}\).
• It captures how the concentrations (or partial pressures) of reactants and products relate at equilibrium.

In the given exercise, we derived \(K_p\) based on the partial pressures of \(N_2O_4\) and \(NO_2\).
The expression was simplified to \[K_p = \frac{4x^2P}{(1 - x)(1 + x)}\], where \(x\) is the molar dissociation of \(N_2O_4\) and \(P\) is the total pressure.
This formula helps determine the effect pressure changes have on the equilibrium, illustrating the dependency of \(K_p\) on the system's conditions.

Partial Pressure

Partial pressure is an important concept when dealing with gaseous reactions and equilibrium.
It refers to the pressure that a particular gas in a mixture contributes to the total pressure exerted by the mixture.
In the context of our exercise:

• The partial pressure of \(N_2O_4\) at equilibrium was determined by \(P_{N_2O_4} = \frac{(1 - x)P}{1 + x}\).
• Similarly, for \(NO_2\), it was \(P_{NO_2} = \frac{2xP}{1 + x}\).

These formulas consider the total pressure of the system \((P)\) and the extent of dissociation \((x)\).
Understanding partial pressures allows us to apply the concept of \(K_p\) effectively, as they are integral to the equilibrium constant expression.
By analyzing how each gas’s pressure varies, it also aids in predicting the shifts in equilibrium positions based on changes to the system.