Question

For the reaction $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ the equilibrium constant \(K\) has the value \(8.1 \times 10^{-3}\) at a particular temperature. If the concentration of \(\mathrm{NO}_{2}(g)\) is found to be \(0.0021 \mathrm{M}\) in the equilibrium system, what is the concentration of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) under these conditions?

Short answer

The equilibrium concentration of \(\mathrm{N}_{2}\mathrm{O}_{4}(g)\) under these conditions is approximately \(0.000546 \mathrm{M}\).

Step-by-step solution

Write down the given information

We are given the following information: - The equilibrium reaction: \(\mathrm{N}_{2}\mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) - The equilibrium constant: \(K = 8.1 \times 10^{-3}\) - The concentration of \(\mathrm{NO}_{2}(g)\) at equilibrium: \([\mathrm{NO}_{2}]= 0.0021 \mathrm{M}\) Now we are ready to find the concentration of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) at equilibrium.

Write the expression for the equilibrium constant

The equilibrium constant \(K\) is defined as the ratio of the product of the equilibrium concentrations of the products (raised to their stoichiometric coefficients) to the product of the equilibrium concentrations of the reactants (raised to their stoichiometric coefficients). In this case, the expression for \(K\) for the given reaction can be written as: \[K = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{N}_{2} \mathrm{O}_{4}]}\] We will use this expression to find the unknown concentration.

Solve for the unknown concentration

Substitute the given values for \(K\) and \([\mathrm{NO}_{2}]\) into the expression for \(K\), then solve for \([\mathrm{N}_{2}\mathrm{O}_{4}]\): \[ 8.1 \times 10^{-3} = \frac{(0.0021 \mathrm{M})^2}{[\mathrm{N}_{2}\mathrm{O}_{4}]}\] Now, solve for \([\mathrm{N}_{2}\mathrm{O}_{4}]\). Rearrange the equation to isolate \([\mathrm{N}_{2}\mathrm{O}_{4}]\): \[[\mathrm{N}_{2}\mathrm{O}_{4}] = \frac{(0.0021 \mathrm{M})^2}{8.1 \times 10^{-3}}\] And calculate \([\mathrm{N}_{2}\mathrm{O}_{4}]\): \[[\mathrm{N}_{2}\mathrm{O}_{4}] = \frac{(0.0021 \mathrm{M})^2}{8.1 \times 10^{-3}} \approx 0.000546 \mathrm{M}\] Therefore, the equilibrium concentration of \(\mathrm{N}_{2}\mathrm{O}_{4}(g)\) under these conditions is approximately \(0.000546 \mathrm{M}\).

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted by the symbol \(K\), is a crucial value in the study of chemistry as it helps determine the extent of a reaction at equilibrium. It is calculated as the ratio of the concentration of products to the concentration of reactants, each raised to the power of their respective stoichiometric coefficients. For example, in the reaction \(\mathrm{N}_{2}\mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\), the expression of the equilibrium constant \(K\) is given by:\[K = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{N}_{2} \mathrm{O}_{4}]}\]This mathematical relationship provides valuable insights into how a system behaves under a set of conditions. A high equilibrium constant suggests that the reaction heavily favors products, whereas a low equilibrium constant, like \(8.1 \times 10^{-3}\) in this example, indicates that reactants are favored in the equilibrium mixture.
Knowing the equilibrium constant helps predict and calculate various concentrations within a reaction mixture, aiding in understanding and controlling chemical processes.

Reaction Equilibrium

Chemical reactions do not always go to completion; many exist in a state of balance called equilibrium. At equilibrium, the forward and reverse reaction rates are equal, leading to stable concentrations of both reactants and products. This doesn't mean the concentrations are equal; rather, they remain consistent over time.For the reaction \(\mathrm{N}_{2}\mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\), equilibrium is achieved when the rate of decomposition of \(\mathrm{N}_{2}\mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) is equal to the rate of formation of \(\mathrm{N}_{2}\mathrm{O}_{4}\) from \(\mathrm{NO}_{2}\). Thus, the concentrations stabilize even if they are very different from each other.Understanding this dynamic balance is critical, as it indicates how certain changes (like temperature or pressure shifts) can affect which direction the equilibrium might "shift", adjusting the relative concentrations accordingly. This is an essential concept in fields that depend on precise chemical reactions, such as pharmaceuticals and industrial chemistry.

Concentration Calculation

To find equilibrium concentrations in reactions, we utilize the equilibrium constant expression. Suppose you are given the equilibrium constant \(K\) and the concentration of one of the reaction species. You can rearrange the expression to solve for the others.In our problem with \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\), given \(K\) and \([\mathrm{NO}_{2}]=0.0021\ \mathrm{M}\), we can find \([\mathrm{N}_{2}\mathrm{O}_{4}]\) using:\[ \frac{(0.0021 \ \mathrm{M})^2}{[\mathrm{N}_{2}\mathrm{O}_{4}]} = 8.1 \times 10^{-3}\]Rearranging gives:\[\mathrm{N}_{2} \mathrm{O}_{4} = \frac{(0.0021 \ \mathrm{M})^2}{8.1 \times 10^{-3}} \approx 0.000546 \ \mathrm{M}\]This calculation demonstrates how equilibrium principles are applied to find missing concentrations, providing a clear view of the state of the system. Being proficient in this calculation process empowers you to handle a wide array of chemical equations and situations.