Question

Consider the reaction \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\). If the initial partial pressure of \(\mathrm{NO}_{2}(\mathrm{~g})\) is \(3.0\) bar, and \(x\) is the equilibrium concentration of \(\mathrm{O}_{2}(\mathrm{~g})\), what is the correct equilibrium relation? (a) \(K=x^{3}\); (b) \(K=2 x^{2} /(3.0-2 x)^{2}\); (c) \(K=4 x^{3} /(3.0-2 x)^{2}\); (d) \(K=x^{2} /(3.0-x)\) (e) \(K=2 x /(3.0-x)^{2}\).

Short answer

The correct equilibrium relation is (c) \(K=\frac{4x^{3}}{(3.0-2x)^{2}}\).

Step-by-step solution

Write down the balanced chemical equation

Firstly, start by writing down the balanced chemical equation for the reaction: \[2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g})\].

Derive the expression for the equilibrium constant

The equilibrium constant expression (K) for a reaction is given by the ratio of the products raised to the power of their stoichiometric coefficients to the reactants raised to the power of their stoichiometric coefficients. This expression should only include gaseous and aqueous species involved in the reaction.For this particular reaction, the expression is given by: \[ K = \frac{[\mathrm{NO}]^{2} \times [\mathrm{O}_{2}]}{[\mathrm{NO}_{2}]^{2}} \]

Set up the initial concentrations and changes

Since initially, there is no NO and O2 present, their initial concentrations are zero, while NO2 has an initial concentration of 3.0 bar. At equilibrium, NO2 will have lost 2x partial pressure while NO will have gained 2x and O2 will have gained x bars of pressure.The change in concentrations can be expressed as follows:\[\begin{align*}\text{NO2} &: 3.0 - 2x \text{NO} &: 2x \text{O2} &: x\end{align*}\]

Substitute the equilibrium concentrations

Now, we can substitute the equilibrium concentrations into the equilibrium expression:\[ K = \frac{(2x)^{2} \times x}{(3.0 - 2x)^{2}} \]This simplifies to:\[ K = \frac{4x^{3}}{(3.0 - 2x)^{2}} \]

Identify the correct equilibrium relation

By examining the choices given, we can conclude that the correct equilibrium relation that fits our derived equation is:\[ K=\frac{4x^{3}}{(3.0-2x)^{2}} \]Therefore, the correct answer is (c).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that occurs when the rates of the forward and reverse reactions are equal, leading to a constant concentration of reactants and products. It is important to note that being at equilibrium does not imply that the reactants and products are at equal concentrations, but rather that their concentrations no longer change with time.

In the given exercise involving nitrogen dioxide (NO2) and nitrogen monoxide (NO), we see a dynamic process where NO2 is converted to NO and oxygen (O2), and vice versa. At equilibrium, the amount of NO2 being transformed into NO and O2 is equal to the amount of NO and O2 recombining to form NO2. This balance is described quantitatively by the equilibrium constant (K), which expresses the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients.

Understanding equilibrium helps students predict the direction in which a reaction will shift when conditions are changed, by applying Le Chatelier's principle. This principle states that a change in conditions (such as concentration, temperature, or pressure) will cause an equilibrium to shift in a direction that counteracts the change, eventually establishing a new equilibrium.

Partial Pressure

Partial pressure is a term used in chemistry to describe the pressure that each gas in a mixture contributes to the total pressure. It's particularly relevant when discussing gases at equilibrium. According to Dalton's Law of Partial Pressures, the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases.

In the exercise, we're asked to consider the initial partial pressure of NO2 and the equilibrium concentration of O2. The partial pressure of each gas after the reaction can be determined by stoichiometry, taking into account the volume and moles of gas involved. In the scenario provided, as the reaction proceeds towards equilibrium, NO2 is consumed while NO and O2 are produced, and their respective partial pressures change accordingly.

Properly understanding partial pressures is crucial for students to accurately calculate and predict the behavior of gases in various chemical reactions, which includes determining the equilibrium state in the reaction under study. This comprehension aids in visualizing how gases behave individually and as part of a system in closed containers.

Stoichiometry

Stoichiometry is the area of chemistry that involves calculating the quantities of reactants and products involved in chemical reactions. It is based on the laws of conservation of mass and the conservation of atoms, which state that in a chemical reaction, atoms are neither created nor destroyed, merely rearranged.

Stoichiometry uses the coefficients in the balanced chemical equation to relate the amount of one substance to the amount of another. In our example, the balanced chemical equation shows a stoichiometric ratio of 2:2:1 between NO2, NO, and O2 respectively. This ratio is vital for understanding the changes in concentration or partial pressure of each species as the reaction approaches equilibrium.

By employing stoichiometry, we can calculate the amount of reactants needed to produce desired product quantities or determine the amount of products formed from given reactants. In understanding chemical equilibrium, stoichiometry also allows us to express changes in concentration or pressure of the substances involved. For instance, if 2 moles of NO2 decrease, 2 moles of NO and 1 mole of O2 are produced, which is essential for setting up the equilibrium expression, as seen in the step-by-step solution.