Problem 29 - Chemistry An Atoms First Approach

Question

The following equilibrium pressures at a certain temperature were observed for the reaction $$\begin{aligned}2 \mathrm{NO}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \\\P_{\mathrm{NO}_{2}} &=0.55 \mathrm{atm} \\\P_{\mathrm{NO}} &=6.5 \times 10^{-5} \mathrm{atm} \\\P_{\mathrm{O}_{2}} &=4.5 \times 10^{-5} \mathrm{atm}\end{aligned}$$. Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature.

Step-by-Step Solution

Answer

The equilibrium constant \(K_p\) for the given reaction at this temperature is \(6.28 \times 10^{-12}\).

Step 1: Write down the equilibrium constant formula for this reaction

The equilibrium constant formula, based on the partial pressures, is given by: \[K_p = \frac{[NO]^2 \times [O_2]}{[NO_2]^2}\] Here, the expressions in brackets denote the partial pressures of the corresponding species at equilibrium.

Step 2: Substitute the given pressures into the formula

Substitute the given pressures of \(NO\), \(O_2\), and \(NO_2\) into the equilibrium constant formula: \[K_p = \frac{(6.5 \times 10^{-5})^2 \times (4.5 \times 10^{-5})}{(0.55)^2}\]

Step 3: Calculate the value of K_p

Now, we will calculate the numerical value for \(K_p\): \[K_p = \frac{(6.5 \times 10^{-5})^2 \times (4.5 \times 10^{-5})}{(0.55)^2} = \frac{(4.225 \times 10^{-9}) \times (4.5 \times 10^{-5})}{0.3025}\] \[K_p = \frac{1.90125 \times 10^{-12}}{0.3025} = 6.28 \times 10^{-12}\] Therefore, the equilibrium constant \(K_p\) for this reaction at the given temperature is \(6.28 \times 10^{-12}\).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products. At equilibrium, the chemical system is said to be balanced and stable, with the amounts of each substance remaining constant over time. This concept is crucial in understanding how reactions proceed and eventually reach a point where they appear to have 'stopped' though, in reality, they continue on a microscopic scale.

Partial Pressures

Partial pressures refer to the pressure exerted by a single gas in a mixture of gases. It's based on Dalton's Law, which states that the total pressure of a mixture is equal to the sum of the partial pressures of the constituent gases. In the context of chemical equilibrium involving gases, each gas has a partial pressure proportional to its mole fraction. Understanding how to measure and calculate these pressures is key to solving equilibrium problems, especially when analyzing gaseous equilibria. Their accurate determination allows us to use them in the equilibrium constant formula effectively.

Equilibrium Constant Formula

The equilibrium constant formula represents the relationship between the concentrations (or partial pressures) of the products and reactants at equilibrium. For a generic reaction where reactants A and B form products C and D, the equilibrium constant is expressed as \( K_c = \frac{[C]^c \times [D]^d}{[A]^a \times [B]^b} \) for concentrations, or \( K_p = \frac{\text{{('partial pressure of C')}}^c \times \text{{('partial pressure of D')}}^d}{\text{{('partial pressure of A')}}^a \times \text{{('partial pressure of B')}}^b} \) for partial pressures, where a, b, c, and d are the stoichiometric coefficients. This ratio is essential as it quantifies the position of equilibrium and can predict the direction a reaction will proceed under changing conditions.

Reaction Quotient

The reaction quotient (Q) is a measure that tells us how far a system is from reaching equilibrium. It is calculated in the same way as the equilibrium constant—using either concentrations or partial pressures—but for any moment in time rather than solely at equilibrium. Comparing Q to the equilibrium constant (K) indicates the direction in which the reaction needs to proceed to establish equilibrium. If \( Q < K \) the reaction will proceed forward, turning more reactants into products. Conversely, if \( Q > K \) the reaction will shift in reverse, forming more reactants from products. Finally, if \( Q = K \) the system is at equilibrium.