Question

At \(273^{\circ} \mathrm{K}\) and 1 atm, 1 liter of \(\mathrm{N}_{2} \mathrm{O}_{4}\) decomposes to \(\mathrm{NO}_{2}\) according to the equation $$ \mathrm{N}_{2} \mathrm{O}_{4} \rightleftarrows 2 \mathrm{NO}_{2} $$ To what extent has decomposition proceeded when the original volume is 25 percent less than the existing volume?

Short answer

The problem asks for the extent of decomposition of N₂O₄ into NO₂ when the volume decreases by 25%. The initial volume is 1L, so the final volume is 0.75L. We can set up the partial pressure expressions for N₂O₄ and NO₂ as \(P(N_2O_4) = \frac{V_0 - x}{V_0 + 0.75}\) and \(P(NO_2) = \frac{2x}{V_0 + 0.75}\), where x represents the extent of decomposition. However, we cannot solve for x without knowing the equilibrium constant K_p at 273K, which isn't provided in the given information. If the K_p value was known, we would be able to solve for x and find the extent of decomposition.

Step-by-step solution

Set up the initial volume and pressure table

In this problem, we are asked about the dissociation of N₂O₄ into 2 NO₂, at 273 K and 1 atm. We can set up the following initial volume and pressure table: Reactant/Products | Initial Volume/Pressure | Change | Equilibrium Volume/Pressure ----------------- | ---------------------- | ------ | -------------------------- N₂O₄ | V₀ | -x | V₀ - x NO₂ | 0 | +2x | 2x Here, V₀ represents the initial volume of N₂O₄ and x denotes the change in volume during the decomposition.

Calculate the decrease in volume

If the final volume is 25% less than the initial volume, we can determine the relationship between the volumes: Final Volume = Initial Volume - 0.25 × Initial Volume Since the initial volume is 1 L, the final volume would be: Final Volume = 1 L - 0.25 × 1 L = 0.75 L

Find the partial pressures at equilibrium

To find the partial pressures at equilibrium, we can use the expressions from Step 1: P(N₂O₄) = (V₀ - x) / (V₀ + 0.75) P(NO₂) = 2x / (V₀ + 0.75) (Note: The total volume change is 0.75 L as calculated in Step 2)

Calculate the K_p expression

The equilibrium constant, K_p, is related to the partial pressures of the reactants and products. For the given reaction, \[K_p = \frac{P(NO_2)^2}{P(N_2O_4)}\] Keep in mind that we're also given the conditions of T=273K and P=1 atm. In order to calculate K_p, we'd first have to know what K_p is at 273 K.

Solve for x

We don't have enough information to determine the value of K_p at 273 K, so the problem cannot be solved further. If given the K_p value or a means to determine K_p at 273 K, we would simply substitute the expressions from Step 3 into the K_p equation and solve for x, which would represent the extent of decomposition.

Key concepts

Le Chatelier's Principle

Imagine a seesaw on a playground, perfectly balanced with children of equal weight on each side. If a new child jumps on one end, the seesaw will tilt, and the children must adjust to restore balance. Similarly, Le Chatelier's Principle describes how chemical systems at equilibrium respond to disturbances. It asserts that if an external change is imposed on a system at equilibrium, the system will adjust itself to counteract that change and re-establish equilibrium.

When applying Le Chatelier's Principle to the decomposing of N₂O₄ into NO₂, if we were to increase pressure or decrease volume, say, by pushing the seesaw down on one side, N₂O₄, being on the pressured side, would react to form more NO₂ to lessen the stress. Conversely, if we reduce pressure or increase volume, the system would shift to produce more N₂O₄. Understanding this principle helps in predicting the behavior of the equilibrium position when external conditions like pressure or temperature change. This concept is crucial in optimizing industrial chemical processes, such as the Haber process for ammonia production.

Equilibrium Constant

In chemical warfare, think of the equilibrium constant, denoted as Kₚ, as the general who doesn't like surprises on the battlefield. It is a number that provides a ratio of the concentration or partial pressures of the products to those of the reactants at equilibrium condition for a reversible reaction. Importantly, Kₚ is temperature-dependent but does not change with variations in pressure or concentrations.

To put the equilibrium constant into context for the given exercise, we can consider Kₚ as an indicator of the position of equilibrium. If Kₚ were known, it would guide us to calculate 'x,' the extent of the decomposition of N₂O₄ into NO₂. A large Kₚ would hint at a highly product-favored reaction; whereas a small Kₚ would indicate a strong preference for reactant. Nonetheless, without knowing Kₚ at 273K, we're at a tactical standstill—like generals without intel on enemy positions.

Partial Pressure

Diving into the deep sea of gases, 'partial pressure' refers to the pressure a single gas in a mixture would exert if it alone occupied the entire volume. In a cocktail of gases, each gas contributes to the total pressure in proportion to its concentration—this is Dalton's Law of Partial Pressures. Think of it as a group project where each student's input contributes to the final grade.

Applying this to our exercise, the partial pressure of N₂O₄ and the formed NO₂ would tell us how much each gas is 'putting in the work' in this chemical project. Knowing their partial pressures at equilibrium is key to understanding the behavior of the system. We calculate these by taking the individual volume each gas would take at equilibrium compared to the total volume they share. Although the exercise does not provide us with enough data to ascertain the final equilibrium pressures, familiarizing with the concept of partial pressure is vital for students studying gas-phase reactions.