Problem 62 - Chemistry and Chemical Reactivity

Question

The equilibrium constant, \(K_{\mathrm{p}}\), is 0.15 at \(25^{\circ} \mathrm{C}\) for the following reaction: $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{g}) $$ If the total pressure of the gas mixture is 2.5 atm at equilibrium, what is the partial pressure of each gas?

Step-by-Step Solution

Answer

\( P_{\mathrm{N}_2\mathrm{O}_4} = 1.96 \) atm, \( P_{\mathrm{NO}_2} = 0.54 \) atm.

Step 1: Define Variables and Expression

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, the expression for the equilibrium constant in terms of partial pressures is: \[ K_{\mathrm{p}} = \frac{(P_{\mathrm{NO}_2})^2}{P_{\mathrm{N}_2\mathrm{O}_4}} \] Given that \( K_{\mathrm{p}} = 0.15 \).

Step 2: Setup Expression with Total Pressure

The total pressure at equilibrium is given as 2.5 atm. Hence, \[ P_{\mathrm{N}_2\mathrm{O}_4} + P_{\mathrm{NO}_2} = 2.5 \] We will solve simultaneously using the expression for \( K_{\mathrm{p}} \).

Step 3: Relate Partial Pressures

Let \( P_{\mathrm{N}_2\mathrm{O}_4} = x \). Thus, the change would be such that \( P_{\mathrm{NO}_2} = 2.5 - x \). Substituting gives: \[ K_{\mathrm{p}} = \frac{(2.5 - x)^2}{x} \]

Step 4: Solve Quadratic Equation

Substitute \( K_{\mathrm{p}} = 0.15 \) into the equation: \[ 0.15 = \frac{(2.5 - x)^2}{x} \] \[ 0.15x = (2.5 - x)^2 \] Expand it: \[ 0.15x = 6.25 - 5x + x^2 \] Re-arrange: \[ x^2 - 5.15x + 6.25 = 0 \] Solve the quadratic equation for \( x \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -5.15 \), \( c = 6.25 \).

Step 5: Calculate Values

Compute the roots: \[ x = \frac{-(-5.15) \pm \sqrt{(-5.15)^2 - 4 \times 1 \times 6.25}}{2 \times 1} \] \[ x = \frac{5.15 \pm \sqrt{26.5225 - 25}}{2} \] \[ x = \frac{5.15 \pm \sqrt{1.5225}}{2} \] \[ x \approx \frac{5.15 \pm 1.233}{2} \] Calculate the two potential values and consider physical significance (only positive values): \( x_1 = 3.1915 \) and \( x_2 = 1.9585 \). Choose \( x_2 = 1.9585 \) (valid as \( x \) cannot exceed total pressure).

Step 6: Determine Partial Pressures

The partial pressure calculated above for \( \mathrm{N}_{2} \mathrm{O}_{4} \) is \( x = 1.9585 \) atm. The partial pressure for \( \mathrm{NO}_{2} \) is \( 2.5 - 1.9585 = 0.5415 \) atm.

Key Concepts

Partial Pressure Explained

Partial pressure is the pressure exerted by a single type of gas within a mixture of gases. In the context of our problem, we are dealing with a gaseous reaction at equilibrium, and we need to determine the partial pressures of the components involved: nitrogen tetroxide \((\text{N}_2\text{O}_4)\) and nitrogen dioxide \((\text{NO}_2)\).
The total pressure of the gas mixture is the sum of the partial pressures of the individual gases, as described by Dalton’s Law of Partial Pressures:

• If a mixture consists of several gases, the total pressure is the sum of the pressures each gas would exert if it alone occupied the whole volume.
• Mathematically, it can be expressed as: \( P_{\text{total}} = P_{\text{N}_2\text{O}_4} + P_{\text{NO}_2} \).

In our exercise, the total equilibrium pressure is given as 2.5 atm. Thus, by representing the partial pressure of \(\text{N}_2\text{O}_4\) as \(x\) and using the relationship \(P_{\text{NO}_2} = 2.5 - x\), we calculate the values required for chemical equilibrium.

Understanding Chemical Equilibrium

Chemical equilibrium occurs when the forward and reverse reactions happen at the same rate, meaning the concentration of reactants and products remains constant over time.
In our example, the reaction between nitrogen tetroxide \((\text{N}_2\text{O}_4)\) and nitrogen dioxide \((\text{NO}_2)\) reaches equilibrium. Here’s how:

• The equilibrium constant \(K_{\text{p}}\) quantifies this state. For the reaction \(\text{N}_2\text{O}_4 \rightleftarrows 2\text{NO}_2\), \(K_{\text{p}}\) is defined based on the partial pressures:
• \( K_{\text{p}} = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}} \) is given as 0.15.

By substituting \(2.5 - x\) for \(P_{\text{NO}_2}\) and \(x\) for \(P_{\text{N}_2\text{O}_4}\) into this equation, we find the value of \(x\) that satisfies the equilibrium condition. This step is crucial because it allows us to describe the equilibrium state using measurable pressures, central to solving the problem.

Gas Mixtures and Their Role

Gas mixtures like the one in this exercise are common in chemical reactions, particularly those involving gases at equilibrium.
Understanding how gas mixtures behave and how to manipulate their properties is essential to mastering gas equilibria.

• When gases mix, they do not chemically interact in the way solids or liquids might; instead, they share kinetic energy which determines the mixture's temperature and pressure.
• The behavior of gas mixtures can be predicted and analyzed using the ideal gas laws and principles of partial pressures.

In reaction scenarios like \(\text{N}_2\text{O}_4 \rightleftarrows 2\text{NO}_2\), understanding the composition of gas mixtures helps us track the progress of reaction and establish equilibrium state. This understanding allows us to calculate individual gas pressures, crucial for finding partial pressures contributing to the total pressure of the mixture.