Question

A sample of pure \(\mathrm{NO}_{2}\) gas heated to \(1000 \mathrm{~K}\) decomposes: $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The equilibrium constant \(K_{P}\) is \(158 .\) Analysis shows that the partial pressure of \(\mathrm{O}_{2}\) is 0.25 atm at equilibrium. Calculate the pressure of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) in the mixture.

Short answer

\nThe pressures of \(NO_{2}\) and \(NO\) in the mixture are both approximately 0.02 atm.

Step-by-step solution

Understand the relationship through the reaction formula

The balanced chemical equation is: \(2NO_{2} \rightleftharpoons 2NO + O_{2}\). For each 2 moles of \(NO_{2}\) that decompose, 2 moles of \(NO\) and one mole of \(O_{2}\) are produced.

Set up the equilibrium expression

According to the reaction, the equilibrium constant expression for the reaction in terms of partial pressures (This is why it's called \(K_{P}\)) would be: \( K_{P} = \frac{P_{NO}^{2} \cdot P_{O_{2}}}{P_{NO_{2}}^{2}} \). Here, \(P_{NO}\), \(P_{O_{2}}\) and \(P_{NO_{2}}\) represent the partial pressures of \(NO\), \(O_{2}\) and \(NO_{2}\) respectively.

Express \(P_{NO}\) in terms of \(P_{NO_{2}}\)

Given that the partial pressure of \(O_{2}\) is 0.25 atm, you can plug this into the equilibrium expression. Furthermore, when equilibrium is reached, the pressure of \(NO_{2}\) that has decomposed will be equal to the pressure of \(NO\) that has been produced. This means we can express \(P_{NO}\) in terms of \(P_{NO_{2}}\). From the balanced chemical reaction, we get the same amount of \(NO_{2}\) and \(NO\), so we can say \(P_{NO} = P_{NO_{2}}\). After substitution, the equation becomes: \( K_{P} = \frac{P_{NO_{2}}^{2} \cdot P_{O_{2}}}{P_{NO_{2}}^{2}} = P_{O_{2}} \)

Solve the equation to get \(P_{NO_{2}}\) and \(P_{NO}\)

Now, we solve the equation for \(P_{NO_{2}}\), and given that \(K_P = 158\) and \(P_{O_{2}} = 0.25\) atm, we get: \(P_{NO_{2}} = \sqrt{\frac{P_{O_{2}}}{K_P}} = \sqrt{\frac{0.25}{158}}\). Both \(P_{NO_{2}}\) and \(P_{NO}\) have the same value.

Key concepts

Equilibrium Constant

The equilibrium constant, often represented as \( K \), plays a key role in understanding chemical equilibrium. It is a value that helps us determine the ratio of the concentrations (or partial pressures in gas reactions) of the products to reactants when a system is at equilibrium.
At equilibrium, the rates of the forward and reverse reactions are equal, and there is no net change in the concentrations of reactants and products. This balance is quantified by the equilibrium constant. In the context of the decomposition reaction, the equilibrium constant \( K_P \) measures this balance in terms of partial pressures, since the species involved are gases.
For the reaction \( 2\mathrm{NO}_2 (g) \rightleftharpoons 2\mathrm{NO} (g) + \mathrm{O}_2 (g) \), the expression for \( K_P \) is derived from the partial pressures of the gases involved:

• \( K_P = \frac{P_{NO}^2 \cdot P_{O_2}}{P_{NO_2}^2} \)

The equilibrium constant provides insight into the proportions of products and reactants at equilibrium and can be used to predict the extent of a reaction. A high \( K \) value indicates a reaction that favors the formation of products, while a low \( K \) value suggests that reactants are favored.

Partial Pressure

Partial pressure is a concept used to describe the pressure exerted by individual gas components in a mixture. It is the pressure that each gas would exert if it alone occupied the entire volume available.
The concept of partial pressure is essential when dealing with gaseous reactions, especially in understanding equilibrium for gas-phase reactions where equations like the one given for \( K_P \) are used.
In the given chemical equation, the partial pressures of \( \mathrm{NO}_2 \), \( \mathrm{NO} \), and \( \mathrm{O}_2 \) determine the position of the equilibrium. The given partial pressure for \( \mathrm{O}_2 \) is 0.25 atm, which is used alongside the equilibrium constant to solve for other unknown pressures.
Understanding how to utilize partial pressures is crucial since real-life reactions don’t occur in isolation, and these values help predict gas behavior and reaction shifts under different conditions. In experiments, knowing the partial pressures allows chemists to calculate how much of each gas is present at equilibrium, aiding in both research and industrial applications.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction in which a single compound breaks down into two or more simpler substances. This is often driven by heating, as seen in the decomposition of \( \mathrm{NO}_2 \) in this exercise. Decomposition reactions are crucial in both synthetic and analytical chemistry because they transform substances into simpler, often more useful, products.

• In the given reaction \( 2 \mathrm{NO}_2 (g) \rightleftharpoons 2 \mathrm{NO} (g) + \mathrm{O}_2 (g) \), the \( \mathrm{NO}_2 \) molecules break into \( \mathrm{NO} \) and \( \mathrm{O}_2 \).
• This reaction is reversible, meaning it can re-form the original compound under sufficient conditions (indicated by the double arrows).
• The decomposition reaction of \( \mathrm{NO}_2 \) into \( \mathrm{NO} \) and \( \mathrm{O}_2 \) shifts towards products or reactants based on the conditions such as temperature and pressure.

In this system, analyzing how changes to conditions affect the reaction equilibrium helps in understanding more complex chemical systems involving decomposition processes. This knowledge can apply to fields such as environmental science, where understanding gas decomposition assists in modeling atmospheric processes.