Question

At a particular temperature, \(K=4.0 \times 10^{-7}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ In an experiment, \(1.0 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{4}\) is placed in a 10.0-L vessel. Calculate the concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) when this reaction reaches equilibrium.

Short answer

At equilibrium, the concentration of N₂O₄ is approximately 0.0998 M, and the concentration of NO₂ is approximately 4.0 × 10⁻⁴ M.

Step-by-step solution

Find the initial concentrations

First, we need to convert the given amount of N₂O₄ (in moles) to concentration (in moles per liter). We are given 1.0 mol of N₂O₄ and a 10.0 L vessel. To find the initial concentration, divide the amount by the volume: Initial concentration of N₂O₄ = \(\frac{1.0 ~mol}{10.0 ~L}\) = 0.1 M. There is no NO₂ initially present, so its initial concentration is 0 M.

Set up the ICE table

Next, we'll set up an ICE (Initial, Change, Equilibrium) table to keep track of how the concentrations change during the reaction. We assign x as the amount of N₂O₄ that decomposes, which will result in the production of 2x NO₂. The table looks as follows: | | N₂O₄ | 2 NO₂ | |----------|------|-------| | Initial | 0.1 | 0 | | Change | -x | +2x | | Equilibrium| 0.1-x| 2x |

Write the equilibrium constant expression

The equilibrium constant expression for the reaction is: K = \(\frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]}\).

Substitute equilibrium concentrations into the K expression

Substitute the equilibrium concentrations from the ICE table into the K expression: K = \(\frac{(2x)^2}{0.1-x}\).

Solve for x

Now, we are given that K = 4.0 × 10⁻⁷. Replace K in the equation: 4.0 × 10⁻⁷ = \(\frac{(2x)^2}{0.1-x}\). Solve this equation for x: x ≈ 2.0 × 10⁻⁴

Calculate equilibrium concentrations

Finally, substitute x back into the equilibrium expressions for the concentrations of N₂O₄ and NO₂: \[ [\mathrm{N}_2\mathrm{O}_4]_{eq} = 0.1 - x \approx 0.1 - 2.0 × 10^{-4} = 0.0998 ~M \] \[ [\mathrm{NO}_2]_{eq} = 2x \approx 2(2.0 × 10^{-4}) = 4.0 × 10^{-4} ~M \] At equilibrium, the concentration of N₂O₄ is approximately 0.0998 M, and the concentration of NO₂ is approximately 4.0 × 10⁻⁴ M.

Key concepts

Understanding the ICE Table

The ICE table stands for Initial, Change, and Equilibrium - a straightforward method used to organize and calculate changes in concentration throughout a chemical reaction until equilibrium is reached.

Imagine lining up three columns for each substance involved in the reaction: one for the initial concentration (I), one for the change during the reaction (C), and one for the equilibrium concentration (E). Let's consider a situation where we have a container with a reactant and we know the volume of the container. If the initial amount of product is zero, as in the case of our textbook problem, this simplifies our initial setup.

When a reaction proceeds, the concentration of reactants decreases, meaning we subtract the change, while the concentration of products increases, indicated by adding the change. This change correlates with the stoichiometry of the reaction - for each mole of N₂O₄, two moles of NO₂ are produced. Symbolically, we often represent this change as 'x', a variable amount of moles that react, hence the change in N₂O₄ concentration is '-x', and the change for NO₂ is '+2x'.

Using this table allows students to visualize the shifts in concentration and understand how the amounts of reactants and products adjust as a reaction approaches equilibrium. It also provides a structured approach to solving for the unknown 'x' and ultimately the equilibrium concentrations.

Equilibrium Constant Expression

At the heart of understanding chemical equilibrium is mastering the equilibrium constant expression, often denoted as 'K'. This expression quantifies the ratio of product concentrations to reactant concentrations at equilibrium, raised to the power of their respective coefficients in the balanced chemical equation.

For the reaction in our exercise, \( \mathrm{N}_{2}O_{4}(g) \rightleftharpoons 2 NO_{2}(g) \), the equilibrium constant expression is given by \( K = \frac{[NO_2]^2}{[N_2O_4]} \). Doubling the coefficient for NO₂ in the reaction is reflected in squaring its concentration in the formula.

It is important for students to remember that only gaseous and aqueous species appear in the equilibrium constant expression. Solids and liquids are not included because their concentrations do not change in a reaction. By substituting equilibrium concentrations from the ICE table into the K expression and solving for 'x', we obtain the variable needed to calculate the equilibrium concentrations of the reactants and products.

Chemical Reaction Calculations

Chemical reaction calculations can span a broad range of operations, from balancing equations to computing the amounts of reactants and products.

In the context of equilibrium, such calculations determine the concentrations of substances when the system is at rest. This is done by applying the ICE table and equilibrium constant expression together, as shown in our textbook problem. Once 'x' is isolated by rearranging the K expression, it serves as the key component to unlocking the equilibrium concentrations. Remember, solving for 'x' often involves some algebraic manipulation, and depending on the complexity of the reaction, you might need to employ quadratic formulas or approximations.

For the given exercise, after finding 'x', students are asked to substitute it back into the equilibrium expressions for N₂O₄ and NO₂. This reflects a critical thinking approach in problem-solving, wherein one piece of data is used to deduce another. By the end of such a process, students will have learned not just to find numeric answers, but also have gained a deeper understanding of the dynamic nature of chemical reactions and the concept of equilibrium.