Question

The two most abundant atmospheric gases react to a tiny extent at \(298 \mathrm{~K}\) in the presence of a catalyst: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad K_{\mathrm{p}}=4.35 \times 10^{-31} $$ (a) What are the equilibrium pressures of the three gases when the atmospheric partial pressures of \(\mathrm{O}_{2}(0.210 \mathrm{~atm})\) and of \(\mathrm{N}_{2}(0.780 \mathrm{~atm})\) are put into an evacuated \(1.00-\mathrm{L}\) flask at \(298 \mathrm{~K}\) with the catalyst? (b) What is \(P_{\text {total }}\) in the container? (c) Find \(K_{\text {c }}\) at \(298 \mathrm{~K}\).

Short answer

The equilibrium pressures are \(P_{\mathrm{N}_{2}}\approx 0.780 \mathrm{~atm}, \;P_{\mathrm{O}_{2}} \approx 0.210 \mathrm{~atm}, \;P_{\mathrm{NO}} \approx 8.44 \times 10^{-16} \mathrm{~atm}\).\;The total pressure is approximately \( P_{\text {total}}=0.990 \mathrm{~atm}\). \(K_{\text{c}} \text{ is }4.35 \times 10^{-31}\).

Step-by-step solution

Understand the Equilibrium Reaction

The chemical equilibrium reaction given is \ \ \ \ \ \ \ \ \( \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \). \ \ \ \ \( K_{p} \) is provided as \(4.35 \times 10^{-31}\) at \(298\mathrm{~K} \).

Initial Pressures

The initial pressures are given: \(P_{ \mathrm{N}_{2}} = 0.780 \mathrm{~atm}\) and \(P_{ O_{2}} = 0.210 ~\mathrm{atm} \).

Set Up the ICE Table

Define the change in pressure using \(x\): Initial: \ \(P_{\mathrm{N}_{2, Initial}} = 0.780 \mathrm{~atm}\) \ \(P_{\mathrm{O}_{2, Initial}} = 0.210 \mathrm{~atm} \) \ \(P_{\mathrm{NO, Initial}} = 0 \mathrm{~atm}\)\ \( \;\) Change: \ \(P_{\mathrm{N}_{2, Change}} = -x \mathrm{~atm}\) \ \(P_{\mathrm{O}_{2, Change}} = -x \mathrm{~atm}\) \ \(P_{\mathrm{NO, Change}} = +2x \mathrm{~atm}\)\ \( \;\) Equilibrium: \ \(P_{\mathrm{N}_{2, Equilibrium}} = 0.780 - x \mathrm{~atm}\) \ \(P_{\mathrm{O}_{2, Equilibrium}} = 0.210 - x \mathrm{~atm} \) \ \(P_{\mathrm{NO, Equilibrium}} = 2x \mathrm{~atm}\)

Expression for Equilibrium Constant

Set up the expression for the equilibrium constant: \ \( K_p = \dfrac{P_{ \mathrm{NO}}^2}{P_{ \mathrm{N}}_{2} \cdot P_{ \mathrm{O}}_{2}} = 4.35 \times 10^{-31} \) \ Substitute the equilibrium pressures into the equation: \ \(4.35 \times 10^{-31} = \dfrac{(2x)^2}{(0.780 - x)(0.210 - x)} \)

Simplify and Solve for x

Simplify the equation \(-Combine Like Terms \ and \; -Solve \ for \; x: \) \ \(4.35 \times 10^{-31} = \dfrac{4x^2}{(0.780 - x)(0.210 - x)} \) \ \(4.35 \times 10^{-31} = \dfrac{4x^2}{0.1638 - 0.990x + x^2} \) \ Given the extremely small value of \( K_p \), assume \( x \approx 0 \), thus: \ \(4x^2 = 4.35 \times 10^{-31} \cdot 0.1638 \) \ Therefore, \(4x^2 \approx 7.125 \times 10^{-32} \) \ \( x^2 \approx \dfrac{7.125 \times 10^{-32}}{4} \approx 1.78125 \times 10^{-32} \) \ \( x \approx \sqrt{1.78125 \times 10^{-32}} \approx 4.22 \times 10^{-16} \text{atm} \)

Calculate Equilibrium Pressures

Now find the equilibrium pressures using the value of \( x \): \ \(P_{\mathrm{N}_{2, Equilibrium}} = 0.780 - x \approx 0.780 \mathrm{~atm}\)\ \(P_{\mathrm{O}_{2, Equilibrium}} = 0.210 - x \approx 0.210 \mathrm{~atm} \)\ \(P_{\mathrm{NO, Equilibrium}} = 2x \approx 2 \times 4.22 \times 10^{-16} = 8.44 \times 10^{-16} \mathrm{~atm}\)

Find Total Pressure

Calculate the total pressure: \ \(P_{\text{total}} = P_{\mathrm{N}_{2}} + P_{\mathrm{O}_{2}} + P_{\mathrm{NO}}\)\ \(P_{\text{total}} = 0.780 + 0.210 + 8.44 \times 10^{-16} \approx 0.990 \mathrm{~atm}\)

Calculate Kc

Find \(K_c\) using the equation: \ \ \ \(K_p = K_c (RT)^{ \Delta n} \)\ Here, \( R = 0.0821 \text{ L atm K}^{-1} \text{mol}^{-1}\) and \( \Delta n = 2 - (1+1) = 0 \).\ Thus, \( K_p = K_c \) \ Therefore, \(K_c = 4.35 \times 10^{-31} \)

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_p\) or \(K_c\), is a fundamental concept in chemical equilibrium. It represents the ratio of the concentrations or partial pressures of the products to reactants at equilibrium. In the given exercise, \(K_p\) is used for gases and is defined as: \[K_p = \frac{(P_{products})}{(P_{reactants})}\] Remember that each pressure term is raised to the power of its coefficient in the balanced equation. Here, \(K_p\) for the reaction \(N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)\) is given as \(4.35 \times 10^{-31}\). A very small \(K_p\) value indicates that at equilibrium, the formation of products is extremely low compared to reactants. Thus, the forward reaction hardly proceeds.

Partial Pressure

Partial pressure is the pressure exerted by an individual gas in a mixture of gases. Each gas in a mixture behaves independently and contributes to the total pressure. In the exercise:

• Initial pressures are given as \(P_{N_{2}} = 0.780 \text{atm}\) and \(P_{O_{2}} = 0.210 \text{atm}\).
• Partial pressure of NO at equilibrium is found using the change in pressure method (ICE Table), ultimately calculated as \(2x \text{atm}\).

The total pressure of the gas mixture is the sum of the individual partial pressures, which aligns with Dalton's Law of Partial Pressures. This exercise exemplifies how minimal the change in pressures is due to the tiny equilibrium constant, reinforcing that the formation of products is negligible.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is like the equilibrium constant but for non-equilibrium conditions. It shows the ratio of the concentrations or partial pressures of products to reactants at any point in time: \[Q = \frac{(P_{products})}{(P_{reactants})}\] If \(Q = K_p\), the system is at equilibrium. If \(Q > K_p\), the reaction will shift towards reactants. If \(Q < K_p\), it will shift towards products. While solving the exercise, we implicitly use this concept by assuming initial concentrations and calculating until the equilibrium constant criteria are met. This approach verifies that our assumption holds true or must be adjusted.

Ideal Gas Law

The Ideal Gas Law is a crucial principle for understanding gas behavior: \[PV = nRT\] where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the universal gas constant (0.0821 \text{L atm K}^{-1} \text{mol}^{-1}), and \(T\) is the temperature in Kelvin. In the exercise, while calculating \(K_c\), we use the relationship between \(K_p\) and \(K_c\) given by: \[K_p = K_c (RT)^{\Delta n}\] Here, \( \Delta n\) represents the change in the number of moles of gas between reactants and products. For the exercise's reaction, \(\text{\Delta n}\ = 0\), simplifying \(K_p = K_c\). This law expresses the interrelationship between gas properties and is cornerstone content in any chemical equilibrium discussion.