Question

The reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ has \(K_{\mathrm{p}}=109\) at \(25^{\circ} \mathrm{C}\) . If the equilibrium partial pressure of \(\mathrm{Br}_{2}\) is 0.0159 atm and the equilibrium partial pressure of NOBr is 0.0768 atm, calculate the partial pressure of \(\mathrm{NO}\) at equilibrium.

Short answer

The partial pressure of NO at equilibrium is approximately 0.0302 atm.

Step-by-step solution

Write the equilibrium expression

The equilibrium expression for this reaction is \[K_p = \frac{(P_{\mathrm{NOBr}})^2}{(P_{\mathrm{NO}})^2 \times (P_{\mathrm{Br}_{2}})}\] where \(K_p\) is the equilibrium constant, and \(P_{\mathrm{NO}}\), \(P_{\mathrm{Br}_{2}}\), and \(P_{\mathrm{NOBr}}\) represent the partial pressures of NO, Br2, and NOBr, respectively, at equilibrium.

Plug in the given values

We are given \(K_p = 109\), \(P_{\mathrm{Br}_{2}} = 0.0159\), and \(P_{\mathrm{NOBr}} = 0.0768\). Plugging these values into the equilibrium expression, we get: \[109 = \frac{(0.0768)^2}{(P_{\mathrm{NO}})^2 \times (0.0159)}\]

Solve for the partial pressure of NO

To solve for \(P_{\mathrm{NO}}\), we will first multiply both sides of the equation by \((P_{\mathrm{NO}})^2 \times (0.0159)\). Then, we will take the square root of both sides to solve for \(P_{\mathrm{NO}}\). \[(P_{\mathrm{NO}})^2 \times (0.0159) \times 109 = (0.0768)^2\] \[(P_{\mathrm{NO}})^2 = \frac{(0.0768)^2}{(0.0159) \times 109}\] \[P_{\mathrm{NO}} = \sqrt{\frac{(0.0768)^2}{(0.0159) \times 109}}\] Calculating the value, we get: \[P_{\mathrm{NO}} \approx 0.0302\] So, the partial pressure of NO at equilibrium is approximately 0.0302 atm.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry, referring to a state in a chemical reaction where the concentrations or pressures of reactants and products remain constant over time. This occurs when the forward and reverse reactions occur at equal rates, an indication that the system is balanced and stable. Understanding this concept is essential when studying reactions, as it helps predict and manipulate outcomes. When a reaction reaches equilibrium, it doesn’t mean the reactants and products stop reacting; instead, their changes occur at the same rate, maintaining stable conditions. In our exercise, the reaction described \[2 \text{NO}(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{NOBr}(g) \]reaches equilibrium at a constant temperature, thereby allowing us to use equilibrium principles to find unknown quantities like pressure.

Partial Pressure

In chemical reactions involving gases, partial pressure is crucial for understanding how equilibrium is achieved. Partial pressure denotes the pressure contributed by a single gas in a mixture of gases. It's the hypothetical pressure of that gas if it occupied the entire volume by itself, at the same temperature.
Understanding partial pressure is key, as it affects reaction rates and equilibrium conditions significantly. In our reaction, each gaseous component—NO, Br $_{2}$ , and NOBr—has its partial pressure at equilibrium. Given values, like that of Br $_{2}$ (0.0159 atm) and NOBr (0.0768 atm), allow us to solve for unknown quantities, such as NO, using the equilibrium constant and expression.

Reaction Quotient

The reaction quotient, often represented as \(Q\), helps determine the direction in which a reaction needs to shift to reach equilibrium. It is calculated similarly to the equilibrium constant but can be assessed at any stage of the reaction, not just at equilibrium.
For a reaction to proceed towards equilibrium, comparing \(Q\) with the equilibrium constant \(K\) provides insightful information:

• If \(Q < K\), the forward reaction needs to occur more to reach equilibrium.
• If \(Q > K\), the reverse reaction is favored until equilibrium is achieved.

In our problem, since we are already given equilibrium pressures and the equilibrium constant, we solved for a missing partial pressure to ensure the system adheres to its equilibrium constraints defined by \(K_p\).

Equilibrium Expression

The equilibrium expression is a mathematical relation that ties together the concentrations or pressures of reactants and products at equilibrium through the equilibrium constant. In the given equation, the equilibrium constant is represented by \(K_p\), signifying partial pressures involvement.
For the reaction \[2 \text{NO}(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{NOBr}(g)\] the equilibrium expression is given by:\[K_p = \frac{(P_{\text{NOBr}})^2}{(P_{\text{NO}})^2 \times (P_{\text{Br}_2})}\] This states that the equilibrium constant \(K_p\) is equal to the ratio of the products’ partial pressures raised to their stoichiometric coefficients over the reactants’ pressures similarly adjusted.

• This expression illustrates the relationship amongst the pressures and helps solve for unknowns in gaseous reactions at equilibrium.
• Plugging in the known values allows calculation of any other partial pressure needed, such as \(P_{\text{NO}}\) in our exercise.