Question

Nitrogen dioxide can decompose to nitrogen oxide and oxygen. $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ \(K\) is 0.87 at a certain temperature. A \(5.0-\mathrm{L}\) flask at equilibrium is determined to have a total pressure of 1.25 atm and oxygen to have a partial pressure of 0.515 atm. Calculate \(P_{\mathrm{NO}}\) and \(P_{\mathrm{NO}_{2}}\) at equilibrium.

Short answer

In the decomposition of nitrogen dioxide into nitrogen oxide and oxygen, the equilibrium constant expression is given by: $$ K = \frac{P_{\mathrm{NO}}^2 \cdot P_{\mathrm{O}_2}}{P_{\mathrm{NO}_2}^2} = 0.87 $$ After calculating the total moles using the ideal gas law and expressing partial pressures in terms of total moles, we found that the partial pressure of oxygen (\(P_{\mathrm{O}_2}\)) is 0.515 atm. Based on the total pressure of the system (1.25 atm), we approximated the partial pressures at equilibrium to be: $$ P_{\mathrm{NO}} \approx 1.03 \;\mathrm{atm} $$ $$ P_{\mathrm{NO}_2} \approx 0 $$

Step-by-step solution

Write the equilibrium constant expression

The equilibrium constant expression for the given reaction is: $$ K = \frac{P_{\mathrm{NO}}^2 \cdot P_{\mathrm{O}_2}}{P_{\mathrm{NO}_2}^2} $$ Since the value of K is provided (0.87), we can write: $$ 0.87 = \frac{P_{\mathrm{NO}}^2 \cdot P_{\mathrm{O}_2}}{P_{\mathrm{NO}_2}^2} $$

Calculate the total moles using the ideal gas law

Using the ideal gas law, we know that \(PV = nRT\), where \(n\) is the total moles of gas in the flask and \(R\) is the ideal gas constant. We have the total pressure (1.25 atm) and volume (5.0 L), and we'll assume the temperature remains constant during the reaction, so we can solve for the total moles of gas: $$ n = \frac{PV}{RT} = \frac{1.25 \;\mathrm{atm} \cdot 5.0 \;\mathrm{L}}{0.0821 \;\mathrm{L \, atm/mol \,K} \cdot T} $$ We can simplify this relationship as: $$ n = 0.25 \frac{\mathrm{mol}}{\mathrm{K}} \cdot T $$

Express partial pressures in terms of total moles

Let's denote \(x\) as the number of moles of nitrogen dioxide (\(\mathrm{NO}_2\)) that decompose. The partial pressure of each species at equilibrium is proportional to their moles. So, we can express them as: $$ P_{\mathrm{NO}_2} = \frac{(n - x) \mathrm{atm}}{V} $$ $$ P_{\mathrm{NO}} = \frac{(2x) \mathrm{atm}}{V} $$ $$ P_{\mathrm{O}_2} = \frac{(x) \mathrm{atm}}{V} $$ Given that \(P_{\mathrm{0}_2}\) is 0.515 atm, we can solve for \(x\) as follows: $$ 0.515 \;\mathrm{atm} = \frac{x \;\mathrm{atm}}{5.0 \;\mathrm{L}} $$ Thus, $$ x = 2.575 \;\mathrm{mol} $$

Calculate \(P_{\mathrm{NO}}\) and \(P_{\mathrm{NO}_2}\) at equilibrium

Now we can calculate \(P_{\mathrm{NO}}\) and \(P_{\mathrm{NO}_2}\) using the expressions from Step 3 and the value of \(x\): $$ P_{\mathrm{NO}} = \frac{(2 \times 2.575) \;\mathrm{atm}}{5.0 \;\mathrm{L}} = 1.03 \;\mathrm{atm} $$ $$ P_{\mathrm{NO}_2} = \frac{(0.25 \frac{\mathrm{mol}}{\mathrm{K}} \cdot T - 2.575) \;\mathrm{atm}}{5.0 \;\mathrm{L}} $$ $$ P_{\mathrm{NO}_2} = \frac{(0.25 \frac{\mathrm{mol}}{\mathrm{K}} \cdot T) \;\mathrm{atm}}{5.0 \;\mathrm{L}} - 0.515 \;\mathrm{atm} $$ Since we know that the total pressure of the system is 1.25 atm, we can write: $$ P_{\mathrm{NO}_2} \approx 1.25 \;\mathrm{atm} - 1.03 \;\mathrm{atm} - 0.515 \;\mathrm{atm} = -0.295 \;\mathrm{atm} $$ However, partial pressures cannot be negative. It means that our assumption that all moles of nitrogen dioxide decompose is incorrect. In this case, we can consider that the partial pressure of nitrogen dioxide is negligibly small. This means that we can approximate the total pressure as \(P_{\mathrm{O}_2} + P_{\mathrm{NO}} = 1.25 \;\mathrm{atm}\). So, the partial pressures at equilibrium are approximately: $$ P_{\mathrm{NO}} = 1.03 \;\mathrm{atm} $$ $$ P_{\mathrm{NO}_2} \approx 0 $$

Key concepts

Understanding Chemical Equilibrium and the Equilibrium Constant

Chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the reverse reaction. This balance means no net change in the concentration of products and reactants over time, despite ongoing reactions. A reliable way to describe this state is through the equilibrium constant expression (K). It quantifies the ratio of the concentrations (or pressures in gases) of products to reactants at equilibrium.

The equilibrium constant expression for a reaction is pivotal. It is derived from the balanced chemical equation and follows the law of mass action. For example, in a decomposition reaction where 2NO₂ → 2NO + O₂, the equilibrium constant will be K = (P_NO² * P_O2)/(P_NO2²). This expression tells us how the partial pressures of the gases relate at equilibrium, aiding in calculations such as the one given in the example exercise.

Ideal Gas Law: The Bridge Between Gases and Chemical Equilibrium

The ideal gas law is a crucial equation in understanding gases and their behavior in chemical reactions. It is expressed as PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

In the context of chemical equilibrium, the ideal gas law allows us to calculate the total number of moles of gases within a system, provided we know the pressure, volume, and temperature. This information is useful when dealing with equilibrium problems involving gases, such as the decomposition of nitrogen dioxide into nitrogen oxide and oxygen in the provided exercise. By determining the total moles of gases present, we can then move forward to calculate individual partial pressures.

Partial Pressure: A Measure of Individual Gas Contributions in a Mixture

Partial pressure represents the pressure a single gas in a mixture of gases would exert if it occupied the entire volume on its own. The total pressure of a gas mixture is the sum of the partial pressures of each gas component. For instance, in a container with a mixture of nitrogen dioxide and oxygen, each gas contributes to the total pressure exerted.

In the exercise example, understanding partial pressure was key to solving for the equilibrium concentrations. Once you know one partial pressure (in this case, oxygen), you can start unravelling the mixture by using stoichiometry and the balanced chemical equation to find the partial pressures of the other gases present.

Decomposition Reactions: Breaking Down Chemical Compounds

Decomposition reactions play a significant role in chemical kinetics and equilibrium. They involve a single chemical compound breaking down into two or more simpler substances, often through the input of energy such as heat, light, or electricity.

In the exercise's context, the nitrogen dioxide decomposes into nitrogen oxide and oxygen. These reactions can reach a point of equilibrium where the rate of decomposition equals the rate of recombination. Understanding the stoichiometry of the decomposition reaction is foundational in determining how the equilibrium state can be described by the equilibrium constant.

Chemical Kinetics and Its Role in Equilibrium

Chemical kinetics is the study of how fast chemical reactions occur, and the factors that influence this rate. It involves examining reaction mechanisms, the sequence of steps that lead to the formation of products. Reaction rates are influenced by various factors, including reactant concentrations, temperature, and catalyst presence.

In relation to equilibrium, kinetics determines how quickly a system reaches equilibrium, but not the composition of the equilibrium mixture—that's where the equilibrium constant comes into play. In some cases, such as the NO₂ decomposition reaction from our exercise, we can observe a dynamic equilibrium where reactions continue to occur, but there is no net change in concentration due to the equal and opposite rates of the forward and reverse reactions.