Problem 45 - Chemistry and Chemical Reactivity

Question

The equilibrium constant, \(K_{\mathrm{p}},\) for \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{g})\) is 0.15 at \(25^{\circ} \mathrm{C} .\) If the pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at equilibrium is \(0.85 \mathrm{atm},\) what is the total pressure of the gas mixture \(\left(\mathrm{N}_{2} \mathrm{O}_{4}+\mathrm{NO}_{2}\right)\) at equilibrium?

Step-by-Step Solution

Answer

The total pressure at equilibrium is 1.207 atm.

Step 1: Write the Equilibrium Expression

First, identify the equilibrium expression for the reaction \(\mathrm{N}_{2}\mathrm{O}_{4}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{g})\). The equilibrium constant expression in terms of pressure is given by: \[K_{\mathrm{p}} = \frac{P_{\mathrm{NO}_2}^2}{P_{\mathrm{N}_2\mathrm{O}_4}}\] where \(P_{\mathrm{NO}_2}\) and \(P_{\mathrm{N}_2\mathrm{O}_4}\) are the equilibrium partial pressures of \(\mathrm{NO}_2\) and \(\mathrm{N}_2\mathrm{O}_4\), respectively.

Step 2: Substitute Known Values into the Equilibrium Expression

According to the problem, \(K_{\mathrm{p}} = 0.15\) and \(P_{\mathrm{N}_2\mathrm{O}_4} = 0.85\, \mathrm{atm}\). Substitute these values into the expression:\[0.15 = \frac{P_{\mathrm{NO}_2}^2}{0.85}\]

Step 3: Solve for \(P_{\mathrm{NO}_2}\)

Rearrange the equation to solve for \(P_{\mathrm{NO}_2}\):\[P_{\mathrm{NO}_2}^2 = 0.15 \times 0.85\]Calculate the right side:\[P_{\mathrm{NO}_2}^2 = 0.1275\]Take the square root of both sides to find \(P_{\mathrm{NO}_2}\):\[P_{\mathrm{NO}_2} = \sqrt{0.1275} \approx 0.357 \mathrm{atm}\]

Step 4: Find the Total Pressure at Equilibrium

The total pressure at equilibrium is the sum of the partial pressures of \(\mathrm{N}_2\mathrm{O}_4\) and \(\mathrm{NO}_2\):\[P_{\text{total}} = P_{\mathrm{N}_2\mathrm{O}_4} + P_{\mathrm{NO}_2}\]Substitute the values:\[P_{\text{total}} = 0.85 + 0.357 = 1.207 \, \mathrm{atm}\]

Key Concepts

Understanding Partial Pressures

In a gaseous mixture, each gas component contributes to the total pressure based on its own pressure if it were the only gas present in the container. This is known as the partial pressure of a gas. In the context of chemical equilibrium, partial pressure becomes crucial because it directly influences the equilibrium state of a reaction.

For the reaction \[\mathrm{N}_2\mathrm{O}_4(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_2(\mathrm{g})\], we look at the individual pressures exerted by \(\mathrm{N}_2\mathrm{O}_4\) and \(\mathrm{NO}_2\). At equilibrium, these pressures are not arbitrary but specific, determined by the conditions of the reaction and the equilibrium constant, \(K_p\).

To find the equilibrium partial pressure of each component, we use the equilibrium constant expression. Once we know one partial pressure, like the given \(P_{\mathrm{N}_2\mathrm{O}_4} = 0.85\, \mathrm{atm}\), it can help us find the other using the equilibrium constant formula.

Equilibrium Expression Simplified

An equilibrium expression links the equilibrium constants to the concentrations or pressures of the reactants and products in a chemistry reaction. For a gaseous reaction like \[\mathrm{N}_2\mathrm{O}_4(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_2(\mathrm{g})\], we use the equilibrium constant \(K_p\), which relates the pressures of the gases.

The equilibrium expression for this reaction is: \[K_{\mathrm{p}} = \frac{P_{\mathrm{NO}_2}^2}{P_{\mathrm{N}_2\mathrm{O}_4}}\] This equation shows that the equilibrium constant \(K_p\) is directly dependent on the squares of the partial pressure of \(\mathrm{NO}_2\) and inversely on the pressure of \(\mathrm{N}_2\mathrm{O}_4\). When a value for \(K_p\) and one partial pressure is known, we can rearrange the equilibrium expression to solve for the unknown partial pressure. This approach is very useful for predicting how a system will behave when it reaches equilibrium.

Total Pressure Calculation at Equilibrium

With both partial pressures determined at equilibrium, the total pressure is simply their sum. This follows the principle that the total pressure in a mixture of non-reactive gases is the sum of the partial pressures of the individual gases. This is known as Dalton's Law of Partial Pressures.

In our example, the equation for total pressure is: \[ P_{\text{total}} = P_{\mathrm{N}_2\mathrm{O}_4} + P_{\mathrm{NO}_2} \] Here, we substitute the known values for the partial pressures: \(P_{\mathrm{N}_2\mathrm{O}_4} = 0.85\, \mathrm{atm}\) and \(P_{\mathrm{NO}_2} = 0.357\, \mathrm{atm}\). Thus, the total pressure at equilibrium becomes: \[ P_{\text{total}} = 0.85 + 0.357 = 1.207 \, \mathrm{atm} \]

By utilizing both the understanding of equilibrium concepts and the mathematical calculation, students can predict how changes to the system may alter the equilibrium's total pressure.