Question

At \(373 \mathrm{~K}, K_{p}=0.416\) for the equilibrium $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressures of \(\mathrm{NOBr}(g)\) and \(\mathrm{NO}(g)\) are equal, what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?

Short answer

The equilibrium pressure of Br₂(g) is \(0.416\) atm.

Step-by-step solution

Write the expression for \(K_{p}\)

Using the balanced reaction, we can write the equilibrium expression for this reaction in terms of the partial pressures (\(P\)) of the gas components. $$K_p = \frac{P_{NO}^2 \cdot P_{Br_2}}{P_{NOBr}^2}$$

Apply the given information

We are given that \(K_p = 0.416\), and the pressure of NOBr and NO are equal at equilibrium. Let's represent the equilibrium pressures of NOBr and NO as \(x\). Therefore, \(P_{NOBr} = P_{NO} = x\). We can now substitute the given information into the equilibrium expression. $$0.416 = \frac{x^2 \cdot P_{Br_2}}{x^2}$$

Solve for the equilibrium pressure of Br₂(g)

Only the equilibrium pressure of \(\mathrm{Br}_2\) is left to be found. We can now simplify and solve the equation for \(P_{Br_2}\): $$P_{Br_2} = 0.416 \cdot x^2 / x^2$$ $$P_{Br_2} = 0.416$$ The equilibrium pressure of Br₂(g) is \(0.416\) atm.

Key concepts

Understanding Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rates of the forward and reverse reactions are equal, leading to no net change in the amount of reactants and products over time. It's important to realize that this doesn't mean the reactants and products are in equal amounts, but rather that their concentrations have stabilized in a particular ratio that doesn't change with time.

When a system reaches equilibrium, it is said to be 'dynamic' because individual molecules continue to react, but their overall concentrations remain constant. The concept of equilibrium in chemistry is central to understanding how reactions proceed and how they can be controlled or shifted by changes in conditions such as temperature, pressure, or concentration.

For students, it's essential to grasp that equilibrium is about balance and not necessarily equality. When it comes to calculations involving chemical equilibrium, the understanding of equilibrium constants and partial pressures, as seen in the problem, is crucial for accurate computations.

Partial Pressure and its Role in Equilibrium

Partial pressure is a term used to describe the pressure exerted by a single gas component in a mixture of gases. Each gas in a mixture exerts pressure independently as if it were alone in the container; this pressure is the gas's partial pressure. It is directly proportional to its concentration in the gas mixture.

In the context of chemical equilibrium for reactions involving gases, partial pressures are used instead of concentrations to express the position of equilibrium. It's an indispensable concept when you're dealing with gas-phase reactions, such as the one provided in the exercise.

Understanding the relationship between the partial pressure of a gas and the total pressure of the mixture is fundamental for solving equilibrium problems. In calculations, like the one we've outlined in the problem, recognizing the significance of equal partial pressures can simplify the process and lead to quicker solutions.

The Equilibrium Constant Kp and its Calculation

The equilibrium constant, represented as 'Kp' when dealing with partial pressures, is a crucial figure in chemistry that indicates the ratio of the concentrations of products to reactants at equilibrium. For gas-phase reactions, 'Kp' provides a useful way to express this ratio in terms of partial pressures.

The equation for the equilibrium constant in terms of partial pressures is derived from the balanced chemical equation and can look different depending on the complexity of the reaction. To perform equilibrium pressure calculations, you substitute the partial pressures of the reactants and products into the equilibrium expression and solve for the unknown.

When partial pressures in the reaction are equal or when the coefficients of the reactants and products are similar, it can greatly simplify the calculation. As in our problem, such simplifications make it possible to directly relate the equilibrium constant to the unknown pressure, allowing us to solve for it without the need for additional information. Recognizing these patterns is an important skill in chemistry that facilitates problem-solving and understanding complex chemical systems.