Problem 48 - Chemistry and Chemical Reactivity

Question

The total pressure for a mixture of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) is 1.5 atm. If \(K_{p}=6.75\) (at \(25^{\circ} \mathrm{C}\) ), calculate the partial pressure of each gas in the mixture. $$ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) $$

Step-by-Step Solution

Answer

Partial pressures: \( P_{\mathrm{NO}_{2}} = 0.70 \, \text{atm} \) and \( P_{\mathrm{N}_{2}\mathrm{O}_{4}} = 0.80 \, \text{atm} \).

Step 1: Express the Equilibrium Reaction

Write down the equilibrium reaction given in the problem:\[ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{N}_{2}\mathrm{O}_{4}(\mathrm{g}) \]This reaction shows the conversion of \( \mathrm{NO}_{2} \) gas to \( \mathrm{N}_{2}\mathrm{O}_{4} \) gas and vice versa.

Step 2: Define Partial Pressures at Equilibrium

Let the partial pressure of \( \mathrm{N}_{2}\mathrm{O}_{4} \) be \( P_{\mathrm{N}_{2}\mathrm{O}_{4}} \) and the partial pressure of \( \mathrm{NO}_{2} \) be \( P_{\mathrm{NO}_{2}} \). At equilibrium, these pressures will change due to the reaction, so define initial pressures as uniformly contributing to the total pressure:\[ P_{\mathrm{N}_{2}\mathrm{O}_{4}} + P_{\mathrm{NO}_{2}} = 1.5 \, \text{atm} \]

Step 3: Use the Equilibrium Constant Expression

Since \( K_p = 6.75 \) at \( 25^{\circ} \mathrm{C} \), use the expression for the reaction:\[ K_p = \frac{P_{\mathrm{N}_{2}\mathrm{O}_{4}}}{(P_{\mathrm{NO}_{2}})^2} \]Substitute the known \( K_p \) value:\[ 6.75 = \frac{P_{\mathrm{N}_{2}\mathrm{O}_{4}}}{(P_{\mathrm{NO}_{2}})^2} \]

Step 4: Solve for Partial Pressures

Substitute \( P_{\mathrm{N}_{2}\mathrm{O}_{4}} = 1.5 - x \) and \( P_{\mathrm{NO}_{2}} = x \) into the equation from Step 3 and solve for \( x \):\[ 6.75 = \frac{1.5 - x}{x^2} \]Rearrange and solve the quadratic equation to find \( x \), then use \( x \) to find both \( P_{\mathrm{N}_{2}\mathrm{O}_{4}} \) and \( P_{\mathrm{NO}_{2}} \).

Step 5: Calculate Partial Pressures

From solving the equation, if we consider the one positive and meaningful value of \( x \),\[ x = 0.70 \, \text{atm} \]Then:\[ P_{\mathrm{NO}_{2}} = x = 0.70 \, \text{atm} \]\[ P_{\mathrm{N}_{2}\mathrm{O}_{4}} = 1.5 - x = 0.80 \, \text{atm} \]

Key Concepts

Equilibrium Constant

In a chemical reaction, the equilibrium constant is a crucial concept that defines the extent of the reaction at equilibrium. This value, denoted as \(K_p\) in reactions involving gases, measures the ratio of the concentrations/products of gases when the reaction has reached a steady state.
For the given exercise, where the reaction is \( 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{N}_{2}\mathrm{O}_{4}(\mathrm{g}) \), the equilibrium constant \(K_p\) is 6.75 at 25°C. This means that at equilibrium, the pressure of \( \mathrm{N}_{2}\mathrm{O}_{4}\) relative to the pressure of \( \mathrm{NO}_{2}\) is standardized to this ratio.
When interpreting \(K_p\):

• If \(K_p > 1\), the reaction favors products (in this case \(\mathrm{N}_{2}\mathrm{O}_{4}\)).
• If \(K_p < 1\), the reactants (here \(\mathrm{NO}_{2}\)) are favored at equilibrium.

The relationship provided by \(K_p\) helps us calculate the individual partial pressures necessary to balance the reaction at the given temperature.

Partial Pressure

Partial pressure is the pressure exerted by a single gas in a mixture of gases. It contributes to the total pressure in proportion to its abundance.
In this particular chemical equilibrium exercise, the total pressure of the system is 1.5 atm, which is the sum of the partial pressures of \(\mathrm{N}_{2}\mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\). Defining them as \(P_{\mathrm{N}_{2}\mathrm{O}_{4}}\) and \(P_{\mathrm{NO}_{2}}\) respectively, they together satisfy the equation:

• \(P_{\mathrm{N}_{2}\mathrm{O}_{4}} + P_{\mathrm{NO}_{2}} = 1.5\, \text{atm}\)

Each gas exerts pressure independently, regardless of what other gases are present. This means that the behavior of gases can be treated independently in calculations.
Using the equilibrium constant and the relationship between the partial pressures, we can solve for each gas's contributions to the total pressure. Partial pressures give a detailed insight into the dynamics of gas-phase reactions and are essential for understanding chemical equilibria involving gases.

Gaseous Reaction

Gaseous reactions are transformations of substances that involve gases as either reactants or products. They often occur at a significant velocity, especially at elevated temperatures and pressures.
In this example, the reaction \( 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{N}_{2}\mathrm{O}_{4}(\mathrm{g}) \) is a dynamic equilibrium where both forward and backward reactions occur simultaneously at the same rate. This dual characteristic is a hallmark of gaseous reactions in equilibrium.
To reach equilibrium, both gases adjust their concentrations (or partial pressures in this case) until no net change is observed, though a perceptible fluctuation continues at the microscopic level. Understanding "gaseous reactions" requires

• Recognizing the role of conditions such as temperature and pressure,
• Employing the ideal gas law for quantifying pressure-volume relations,
• Interpreting these balances using equilibrium constants.

Gaseous reactions like this one demonstrate how chemical compositions adjust dynamically, ensuring that the reaction's equilibrium state is attained, consistent with the reaction's \(K_p\).