Problem 43 - Chemistry The Science in Context

Question

Calculate the value of the equilibrium constant \(K_{\mathrm{p}}\) at \(298 \mathrm{K}\) for the reaction $$ \mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ from the following \(K_{\mathrm{p}}\) values at \(298 \mathrm{K}\) $$\begin{aligned} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g) & & K_{\mathrm{p}}=4.4 \times 10^{-31} \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}_{2}(g) & & K_{\mathrm{p}}=2.4 \times 10^{12} \end{aligned}$$

Step-by-Step Solution

Answer

Question: Calculate the equilibrium constant (\(K_{\mathrm{p}}\)) for the reaction \(\mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) at \(298 \mathrm{K}\), given that the equilibrium constants for the reactions \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\) and \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) are \(4.4 \times 10^{-31}\) and \(2.4 \times 10^{12}\), respectively. Answer: The equilibrium constant (\(K_{\mathrm{p}}\)) for the reaction \(\mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) at \(298 \mathrm{K}\) is \(1.056 \times 10^{-18}\).

Step 1: Analyze the given reactions

Observe the reactions provided and see how they can be combined to form the desired reaction. Notice that if we add both reactions together, we can obtain the desired reaction after cancelling out \(\mathrm{NO}(g)\) molecules from both sides: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) (1) $$ $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) (2) $$

Step 2: Combine the reactions

Add the two reactions (1) and (2) together to obtain the desired reaction and cancel out the \(\mathrm{NO}(g)\) molecules: $$ (\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g))+ (2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)) \rightleftharpoons (2 \mathrm{NO}(g)) + (2 \mathrm{NO}_{2}(g)) $$ Simplify this to obtain the desired reaction: $$ \mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \hspace{15pt} (3) $$

Step 3: Evaluate the \(K_{\mathrm{p}}\) value for the combined reaction

Since we have added the two reactions, we will multiply the respective \(K_{\mathrm{p}}\) values to find the combined reaction's \(K_{\mathrm{p}}\). Therefore, we have: $$ K_{\mathrm{p, combined}} = K_{\mathrm{p1}} \times K_{\mathrm{p2}} $$ Substitute the given values: $$ K_{\mathrm{p, combined}} = (4.4 \times 10^{-31}) \times (2.4 \times 10^{12}) $$

Step 4: Calculate the equilibrium constant

Perform the multiplication to find the equilibrium constant \(K_{\mathrm{p}}\) for the desired reaction (3): $$ K_{\mathrm{p, combined}} = 1.056 \times 10^{-18} $$ Thus, the value of the equilibrium constant \(K_{\mathrm{p}}\) at \(298 \mathrm{K}\) for the reaction \(\mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) is \(1.056 \times 10^{-18}\).

Key Concepts

Chemical Equilibrium

Understanding chemical equilibrium is key for students tackling reactions like \[\[\begin{align*}\mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\end{align*}\]\]. In simple terms, chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, meaning that the concentrations of reactants and products remain constant over time. This state does not imply that the reactants and products are present in equal amounts, but rather that their ratios are stable.

At equilibrium, the rate at which \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) react to form \(\mathrm{NO}_{2}(g)\) is precisely matched by the rate \(\mathrm{NO}_{2}(g)\) decomposes back into \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\). This dynamic balance is crucial in calculating the equilibrium constant (\(K_{\mathrm{p}}\)), which we obtained in our exercise as \(1.056 \times 10^{-18}\).

It's important to note that reaching equilibrium does not depend on the initial amounts of reactants and products but on their inherent properties and the conditions under which the reaction occurs, such as temperature and pressure. The equilibrium constant offers a numerical value illustrating the ratio of product concentrations to reactant concentrations at this point, each raised to the power of their coefficients in the balanced equation.

Reaction Quotient (Q)

The reaction quotient (\(Q\)) plays a pivotal role in determining the direction in which a reaction mixture will proceed to achieve equilibrium. It's calculated using the same formula as the equilibrium constant but with the initial concentrations or partial pressures of reactants and products, not at equilibrium.

For our exercise, if we had initial partial pressures and needed to predict the reaction progression, we'd use the formula for \(Q\) to compare it with the known \(K_{\mathrm{p}}\). If \(Q > K_{\mathrm{p}}\), the reaction would shift left, converting products back into reactants. Conversely, if \(Q < K_{\mathrm{p}}\), the reaction would shift right, favoring the production of more products. Only when \(Q = K_{\mathrm{p}}\) is the system at equilibrium, and there is no net change in concentrations of reactants and products.

Understanding the relationship between \(Q\) and \(K_{\mathrm{p}}\) is thus integral for predicting the reactions' behavior and comprehending the ongoing dynamic within a chemical system not yet at equilibrium.

Le Chatelier's Principle

Le Chatelier's principle provides insight into how a system at equilibrium responds to external changes. It states that if a dynamic equilibrium system undergoes a change in concentration, temperature, volume, or pressure, the system adjusts itself to counteract the imposed change and a new equilibrium is established.

In the context of our exercise, if the concentration of \(\mathrm{O}_{2}(g)\) were to increase, Le Chatelier's principle predicts that the reaction would shift towards the formation of more \(\mathrm{NO}_{2}(g)\) to reduce the excess \(\mathrm{O}_{2}(g)\). Similarly, a change in temperature or pressure would also cause the system to shift in a direction that partially undoes the change—this could result in producing more reactants or more products, depending on the specifics of the shift.

Le Chatelier's principle is essential for understanding how equilibrium can be manipulated in chemical processes, making it possible to increase the yield of a desired product or to shift a reaction toward a state that is more favorable under different conditions.

Gibbs Free Energy

Gibbs free energy (\(G\)) is a thermodynamic function representing the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. It links the concepts of spontaneity and equilibrium in chemical reactions.

A reaction at equilibrium has a \(\Delta G = 0\), indicating that it can do no work. The sign of \(\Delta G\) helps us predict whether the reaction will proceed spontaneously. If \(\Delta G < 0\), the reaction is spontaneous in the forward direction; if \(\Delta G > 0\), the reaction is non-spontaneous under standard conditions. In relation to equilibrium, \(\Delta G = \Delta G^{\circ} + RT \ln Q\), where \(\Delta G^{\circ}\) is the standard Gibbs free energy change, \(R\) is the gas constant, \(T\) is the temperature, and \(Q\) is the reaction quotient.

In our exercise, knowing the Gibbs free energy change for the reactions would provide further context to the favorability and spontaneity of the formation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) under the given conditions. A deep understanding of Gibbs free energy not only helps in explaining equilibrium but also serves as a bridge to the energetics of chemical reactions and their feasibility.