Problem 77 - Chemistry: Principles and Reactions

Question

Consider the following reaction at a certain temperature: $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ A reaction mixture contains \(0.70 \mathrm{~atm}\) of \(\mathrm{O}_{2}\) and \(0.81\) atm of NO. When equilibrium is established, the total pressure in the reaction vessel is \(1.20 \mathrm{~atm}\). Find \(K\)

Step-by-Step Solution

Answer

Answer: The equilibrium constant (K) for the given reaction is approximately \(10.97\).

Step 1: Write the equilibrium constant expression

For the reaction: $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ The equilibrium constant expression is given by: $$K = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{NO}]^2[\mathrm{O}_{2}]}$$ Here, we can replace concentrations with partial pressures because of the ideal gas law, which states that the concentration is directly proportional to the partial pressure. So the expression becomes: $$K = \frac{P_{\mathrm{NO}_{2}}^2}{P_{\mathrm{NO}}^2 P_{\mathrm{O}_{2}}}$$

Step 2: Set up the system of equations using given information

Given initial pressures of O₂ and NO: $$P_{\mathrm{O}_{2}, \text{initial}} = 0.70 \ \mathrm{atm}$$ $$P_{\mathrm{NO}, \text{initial}} = 0.81 \ \mathrm{atm}$$ Let \(x\) be the change in pressure for NO and O₂ and \(2x\) be the change in pressure for NO₂ as the reaction proceeds and reaches equilibrium. We have: $$P_{\mathrm{NO}, \text{equilibrium}} = P_{\mathrm{NO}, \text{initial}} - 2x = 0.81 - 2x$$ $$P_{\mathrm{O}_{2}, \text{equilibrium}} = P_{\mathrm{O}_{2}, \text{initial}} - x = 0.70 - x$$ $$P_{\mathrm{NO}_{2}, \text{equilibrium}} = 2x$$ We are also given the total pressure at equilibrium as \(1.20 \ \mathrm{atm}\). So, the sum of the equilibrium pressures equals the total pressure: $$P_{\mathrm{NO}, \text{equilibrium}} + P_{\mathrm{O}_{2}, \text{equilibrium}} + P_{\mathrm{NO}_{2}, \text{equilibrium}} = 1.20 \ \mathrm{atm}$$ Substituting the expressions for the equilibrium pressures, we get: $$(0.81 - 2x) + (0.70 - x) + 2x = 1.20$$

Step 3: Solve the system of equations for x and the equilibrium pressures

Solving the equation for \(x\): $$0.81 - 2x + 0.70 - x + 2x = 1.20$$ → \(x = 0.31\) Now substitute the value of \(x\) to find the equilibrium pressures: $$P_{\mathrm{NO}, \text{equilibrium}} = 0.81 - 2(0.31) = 0.19 \ \mathrm{atm}$$ $$P_{\mathrm{O}_{2}, \text{equilibrium}} = 0.70 - 0.31 = 0.39 \ \mathrm{atm}$$ $$P_{\mathrm{NO}_{2}, \text{equilibrium}} = 2(0.31) = 0.62 \ \mathrm{atm}$$

Step 4: Calculate K using the equilibrium pressures

Now, we can substitute the equilibrium pressures into the K expression and solve for K: $$K = \frac{P_{\mathrm{NO}_{2}}^2}{P_{\mathrm{NO}}^2 P_{\mathrm{O}_{2}}}$$ $$K = \frac{(0.62)^2}{(0.19)^2(0.39)} = 10.97$$ Hence, the equilibrium constant, \(K\), for this reaction at the given temperature is approximately \(10.97\).

Key Concepts

Chemical Equilibrium

The concept of chemical equilibrium plays a critical role in understanding how chemical reactions occur. It refers to the state in a reversible reaction where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. This does not mean the reactants and products are in equal concentrations but that their concentrations remain constant.

For the reaction in the problem, \(2 \text{{NO}}(g) + \text{{O}}_{2}(g) \rightleftharpoons 2 \text{{NO}}_{2}(g)\), achieving chemical equilibrium implies that as much NO and \(O_{2}\) react to form \(NO_{2}\) at the same rate as \(NO_{2}\) decomposes back into NO and \(O_{2}\). This dynamic balance is essential when calculating concentrations or partial pressures at equilibrium in order to use them in the equilibrium constant expression.

Partial Pressures in Chemistry

In the realm of chemistry, partial pressures are of special interest when dealing with gases. The partial pressure of a specific gas in a mixture is the pressure it would exert if it alone occupied the entire volume. It is directly proportional to its mole fraction in a mixture according to Dalton's Law of Partial Pressures. This proportionality allows chemists to treat gas pressures as concentrations, which simplifies calculations for gas-phase reactions under the assumptions of ideal behavior.

For example, if you know the total pressure in a system and the mole fraction of each gas, you can calculate each gas's partial pressure. This is important because, as demonstrated in the exercise, equilibrium constants for gas-phase reactions are often expressed in terms of partial pressures. Understanding this relationship is fundamental when setting up the equilibrium constant expression for reactions involving gases.

Equilibrium Constant Expression

The equilibrium constant expression, often symbolized by \(K\), quantifies the balance between products and reactants at equilibrium for a particular reaction and at a specific temperature. For the chemical reaction involving gases, the equilibrium constant can be expressed in terms of partial pressures. The general form of the equilibrium constant expression for the reaction \(aA + bB \rightleftharpoons cC + dD\) is given by:

\[K = \frac{{(P_C)^c (P_D)^d}}{{(P_A)^a (P_B)^b}}\]
Where \(P_X\) represents the partial pressure of substance X and the exponents represent the stoichiometric coefficients from the balanced equation. In the problem's context, the expression \(K = \frac{{(P_{\text{{NO}}_{2}})^2}}{{(P_{\text{{NO}}})^2 (P_{\text{{O}}_{2}})}}\) relates the equilibrium pressures of nitrogen dioxide (\(NO_{2}\)), nitric oxide (NO), and oxygen (\(O_{2}\)). Understanding how to derive and calculate this expression is crucial for students studying chemical equilibrium in gaseous systems.