Problem 91 - Chemistry: An Atoms-Focused Approach

Question

A flask containing pure \(\mathrm{NO}_{2}\) is heated to \(1000 \mathrm{K},\) a temperature at which \(K_{p}=158\) for the decomposition of \(\mathrm{NO}_{2}\) $$ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$ The partial pressure of \(\mathrm{O}_{2}\) at equilibrium is 0.136 atm. a. Calculate the partial pressures of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\). b. Calculate the total pressure in the flask at equilibrium.

Step-by-Step Solution

Answer

The partial pressures for NO and NO2 are approximately 0.225 atm and 0.1805 atm, respectively. The total pressure in the flask at equilibrium is approximately 0.5415 atm.

Step 1: Set up the expression for the equilibrium constant

Write down the expression for the equilibrium constant \(K_{p}\), based on the stoichiometry of the given reaction: $$ K_{p}=\frac{P_{\mathrm{NO}}^2 P_{\mathrm{O}_{2}}}{P_{\mathrm{NO}_{2}}^2} $$ where \(P_{\mathrm{NO}}\) represents the partial pressure of \(\mathrm{NO}\), \(P_{\mathrm{O}_{2}}\) represents the partial pressure of \(\mathrm{O}_{2}\), and \(P_{\mathrm{NO}_{2}}\) represents the partial pressure of \(\mathrm{NO}_{2}\) at equilibrium.

Step 2: Plug in known values

We are given that the equilibrium constant \(K_{p}=158\) and the partial pressure of \(\mathrm{O}_{2}\) at equilibrium is 0.136 atm. Substitute these values into the expression for \(K_{p}\). $$ 158 = \frac{P_{\mathrm{NO}}^2 \cdot 0.136 }{P_{\mathrm{NO}_{2}}^2} $$

Step 3: Solve for the ratio of partial pressures

Rearrange the equation to solve for the ratio of \(\frac{P_{\mathrm{NO}}^2}{P_{\mathrm{NO}_{2}}^2}\). $$ \frac{P_{\mathrm{NO}}^2}{P_{\mathrm{NO}_{2}}^2} = \frac{158}{0.136} $$

Step 4: Use the stoichiometry of the reaction

Since the stoichiometry of the reaction is such that 2 moles of \(\mathrm{NO_{2}}\) form 2 moles of \(\mathrm{NO}\) and 1 mole of \(\mathrm{O_{2}}\), we can write: $$ 2P_{\mathrm{NO}_{2}} - P_{\mathrm{O}_{2}} = 2P_{\mathrm{NO}} $$ Substitute the value of \(P_{\mathrm{O}_{2}} = 0.136\) atm and rearrange to find the relationship between \(P_{\mathrm{NO}_{2}}\) and \(P_{\mathrm{NO}}\) $$ 2P_{\mathrm{NO}_{2}} - 0.136 = 2P_{\mathrm{NO}} $$

Step 5: Combine the expressions and solve for partial pressures

Using the relationship between \(P_{\mathrm{NO}_{2}}\) and \(P_{\mathrm{NO}}\) and the expression for the ratio of partial pressures, we can set up and solve the equations: $$ \frac{P_{\mathrm{NO}}^2}{P_{\mathrm{NO}_{2}}^2} = \frac{158}{0.136} $$ $$ 2P_{\mathrm{NO}_{2}} - 0.136 = 2P_{\mathrm{NO}} $$ Solve the above system of equations to find \(P_{\mathrm{NO}} = 0.225\) atm and \(P_{\mathrm{NO}_{2}} = 0.1805\) atm.

Step 6: Calculate total pressure

To find the total pressure in the flask, simply add the partial pressures of \(\mathrm{NO}\), \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\). $$ P_{\mathrm{total}} = P_{\mathrm{NO}} + P_{\mathrm{NO}_{2}} + P_{\mathrm{O}_{2}} = 0.225 + 0.1805 + 0.136 = 0.5415 \,\mathrm{atm} $$ The partial pressure of \(\mathrm{NO}\) is 0.225 atm, the partial pressure of \(\mathrm{NO}_{2}\) is 0.1805 atm, and the total pressure in the flask at equilibrium is 0.5415 atm.

Key Concepts

Partial Pressure

When gases react, the total pressure in the system is divided among the individual gases present. This division is known as partial pressure. Each gas in a mixture has its own partial pressure, which contributes to the total pressure of the mixture.
In chemical equilibrium, understanding partial pressure helps determine how gases behave and interact.

• Partial pressure is directly influenced by the amount of gas present and its temperature.
• In our equilibrium calculation, the partial pressures of \(\text{NO}_2\), \(\text{NO}\), and \(\text{O}_2\) were compared to analyze the reaction progress.
• To find partial pressures, we use the formula \(P = \frac{nRT}{V}\), where \(n\) is the number of moles, \(R\) is the gas constant, \(T\) is temperature, and \(V\) is volume. However, stoichiometry simplifies this in our setup.

Equilibrium Constant

The equilibrium constant \(K_p\) is a significant value that determines the relationship between the concentrations of reactants and products at equilibrium.
For gases, \(K_p\) is expressed using partial pressures and is crucial for predicting the direction and extent of a reaction.

• In the given reaction, \(K_p = 158\) at 1000 K indicates that products are favored at equilibrium.
• The formula \(K_p = \frac{P_{\text{NO}}^2 P_{\text{O}_2}}{P_{\text{NO}_2}^2}\) is derived from the balanced chemical equation and accounts for the stoichiometry.
• A high \(K_p\) value suggests a higher concentration of products compared to reactants at equilibrium, highlighting which side of the reaction is more stable under given conditions.

Reversible Reactions

Reversible reactions are those that can proceed in both forward and backward directions until equilibrium is reached. At equilibrium, the rates of the forward and reverse reactions are equal, resulting in no net change in concentration.

• The reaction \(2\, \text{NO}_2(g) \rightleftharpoons 2\, \text{NO}(g) + \text{O}_2(g)\) is reversible and reaches a dynamic equilibrium.
• Equilibrium doesn't always mean equal concentrations of reactants and products, but rather a balance in their rates.
• This concept is vital for industrial processes, like the Haber process, where maximizing yield requires controlling reaction conditions.