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 mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) 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

In conclusion, the equilibrium concentrations for the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) with \(K=4.0 \times 10^{-7}\) and 1.0 mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) initially placed in a 10.0 L vessel are 0.0999 M for \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(2.0 \times 10^{-4}\) M for \(\mathrm{NO}_{2}\).

Step-by-step solution

Write an expression for the equilibrium constant K

According to the given reaction and equilibrium constant value, we can write the expression for K as follows: \[K = \frac{[NO_2]^2}{[N_{2} O_{4}]}\] Where [NO2] and [N2O4] are the equilibrium concentrations of NO2 and N2O4, respectively.

Set up an ICE table

We can set up an ICE (initial-change-equilibrium) table to find out the changes in the concentrations of reactants and products (in terms of moles per liter): | | [N2O4] | [NO2] | |------|-------------|----------------| | I | \(\frac{1.0}{10.0}\)M | 0M | | C | -x | +2x | | E | \(\frac{1.0}{10.0}\) - x | 2x | We need to find the values of x.

Solve for the equilibrium concentration of N2O4

According to the ICE table, the equilibrium concentrations can be written in terms of x: \[[NO_2] = 2x\] \[[N_{2} O_{4}] = 0.1 - x\] Now plug in these concentrations in the expression for K that we obtained in step 1: \[K = \frac{(2x)^2}{(0.1 - x)} = 4.0 \times 10^{-7}\] Solve the equation for x: \(4x^2 = 4.0 \times 10^{-7} (0.1 - x)\) Now, assuming that x is very small compared with 0.1 mole, we can simplify the equation: \(4x^2 = 4.0 \times 10^{-8}\) \(x^2 = 1.0 \times 10^{-8}\) \(x = \sqrt{1.0 \times 10^{-8}}\) Hence: \[x = 1.0 \times 10^{-4}\] So the equilibrium concentration of N2O4 is: \[[N_{2} O_{4}] = 0.1 - 1.0 \times 10^{-4} = 0.1 - 0.0001 = 0.0999 M\]

Solve for the equilibrium concentration of NO2

Using the value of x found in step 3, we can find the equilibrium concentration of NO2: \[[NO_2] = 2x = 2(1.0 \times 10^{-4}) = 2.0 \times 10^{-4} M\] In conclusion, the equilibrium concentrations of N2O4 and NO2 are 0.0999 M and 2.0 x 10^{-4} M, respectively.

Key concepts

Equilibrium Constant Expression

Understanding the equilibrium constant expression is vital for predicting the extent of a chemical reaction at equilibrium. The equilibrium constant, represented by K, is a measure of the ratio of concentrations of products to reactants at a specific temperature.

For the reaction \(N_2O_4(g) \rightleftharpoons 2 NO_2(g)\), the equilibrium constant expression is derived from the balanced chemical equation and is given by:\[K = \frac{[NO_2]^2}{[N_{2}O_{4}]}\]where \([NO_2]\) and \([N_2O_4]\) are the molar concentrations of nitrogen dioxide and dinitrogen tetroxide, respectively, at equilibrium. It's important to note that solid and liquid concentrations are not included in the K expression as their concentrations don't change appreciably with the progression of the reaction. Gas and aqueous solution concentrations do change and hence are included.

In a practical sense, knowing the equilibrium constant allows chemists to predict whether the reactants or products will be favored in the reaction at equilibrium. A large K value indicates a product-favored reaction, while a small K value reveals a reactant-favored reaction.

ICE Table

When calculating equilibrium concentrations, the ICE table (Initial, Change, Equilibrium) is an indispensable tool. It's a systematic way to organize what we know and what we need to find out for each species in a reaction.

The 'I' stands for the initial concentrations (or pressures) before the reaction has started to proceed. 'C' represents the change that occurs as the reaction moves towards equilibrium, and 'E' denotes the equilibrium concentrations of reactants and products. The table helps us express these changes in terms of a variable, typically denoted by \(x\), which reflects the amount of reactant converted to product or vice versa.

For our reaction, the ICE table is structured as follows: the initial concentration of \(N_2O_4\) is calculated by dividing the amount in moles by the volume of the reaction vessel. As the reaction proceeds towards equilibrium, \(N_2O_4\) decreases by \(x\) while \(NO_2\) increases by \(2x\), since two moles of \(NO_2\) are formed per mole of \(N_2O_4\) consumed. The equilibrium concentrations can then be expressed in terms of \(x\). This approach transforms the abstract concept of chemical change into a quantitative exercise.

Equilibrium Concentration Calculations

Equilibrium concentration calculations often require careful determination of the extent of reaction and the resulting concentrations of all species at equilibrium. By using the equilibrium constant and the ICE table, we can mathematically describe the state of equilibrium.

The process involves setting up an equation where the product of the concentrations of the products, each raised to the power of their stoichiometric coefficients, is divided by the same for the reactants. The calculated changes in concentrations during the reaction are related back to the equilibrium constant through this mathematical relationship.

In the given exercise, once we have the equilibrium constant expression and the ICE table, we proceed by substituting the equilibrium concentrations in terms of \(x\) into the K expression. From there, we solve for \(x\), which represents the change in concentration from initial to equilibrium. Several assumptions can facilitate this calculation, such as the notion that \(x\) is very small compared to the initial concentration, allowing us to simplify the problem as in the exercise provided.

Accurate knowledge of equilibrium concentrations is crucial for predicting how different conditions can affect the chemical equilibrium, such as changes in temperature or pressure, and for designing experiments and processes in chemical engineering and research laboratories.