Question

At \(900^{\circ} \mathrm{C}, K_{p}=51.2\) for the equilibrium $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressure of \(\mathrm{NO}(g)\) is half the pressure of \(\mathrm{NOBr}(g)\), what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?

Short answer

The equilibrium pressure of \(Br_2(g)\) is 204.8 atm.

Step-by-step solution

Write the expression for Kp

Since the equilibrium constant is given in terms of pressure, we will write the expression for Kp in terms of the partial pressures of the reactants and products. For the reaction $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g),$$ the expression for Kp is given by: \[K_p = \frac{(\mathrm{P_{NO}})^2 \times (\mathrm{P_{Br_2}})}{(\mathrm{P_{NOBr}})^2}\]

Set up the relationship between the pressures

We are given that the pressure of NO(g) is half the pressure of NOBr(g). Let's denote the equilibrium pressure of NOBr(g) as x. Therefore, the pressure of NO(g) will be equal to 0.5x. Now, we can rewrite the pressures in our equation in terms of x.

Substitute the pressures into the Kp expression

Using the expression for Kp, we substitute the pressures of NO(g) and NOBr(g) as functions of x and plug in the given value for Kp: \[51.2 = \frac{(0.5x)^2 \times (\mathrm{P_{Br_2}})}{(x)^2}\]

Solve for the Br2 pressure

Now, we solve the equation for the pressure of Br2(g) using algebraic steps: \[51.2 = \frac{0.25x^2 \times (\mathrm{P_{Br_2}})}{(x)^2}\] \[(51.2)(x^2) = 0.25x^2 \times (\mathrm{P_{Br_2}})\] Divide both sides by (0.25x^2): \[\frac{(51.2)(x^2)}{0.25x^2} = \mathrm{P_{Br_2}}\] Simplify: \[204.8 = \mathrm{P_{Br_2}}\] So, the equilibrium pressure of Br2(g) is 204.8 atm.

Key concepts

Equilibrium Constant

The equilibrium constant is a crucial concept in chemical equilibrium that characterizes the balance between reactants and products in a reversible chemical reaction. It is denoted as either \(K_c\) or \(K_p\), depending on whether concentrations or partial pressures are used in the expression. In the given problem, \(K_p\) is used because we are dealing with gases and measuring pressure. At equilibrium, the forward and backward reaction rates are equal, and the ratio of the products to reactants remains constant. For the reaction \(2\ \mathrm{NOBr}(g) \rightleftharpoons 2\ \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\), the equilibrium expression in terms of partial pressures is written as:

• \(K_p = \frac{(\mathrm{P_{NO}})^2 \times (\mathrm{P_{Br_2}})}{(\mathrm{P_{NOBr}})^2}\)

This formula tells us how the pressures of each gaseous component relate to each other at equilibrium. By knowing \(K_p\), we can predict the behavior of the reaction under various conditions, such as changes in pressure or temperature.

Partial Pressure

Partial pressure refers to the pressure exerted by an individual gas in a mixture of gases. It is an important concept when dealing with reactions involving gases and is key to understanding chemical equilibria in gaseous systems. In the exercise, it is given that the pressure of \(\mathrm{NO}(g)\) is half that of \(\mathrm{NOBr}(g)\). This relationship is important for substituting into the equilibrium constant expression:

• \(\mathrm{P_{NO}} = 0.5x\)
• \(\mathrm{P_{NOBr}} = x\)

Given this relationship, knowing one component's pressure allows us to find others, shaping our understanding of the system. Partial pressures are used in the formula for \(K_p\) rather than total pressures, because they more accurately reflect the contribution of each gas to the equilibrium state. This concept helps us analyze and solve problems where different gases are evolving to reach equilibrium at the same time.

Reaction Dynamics

Reaction dynamics is the study of how reactions occur, focusing on the path from reactants to products and the speed at which this transformation happens. In terms of chemical equilibrium, it involves analyzing how changes in conditions like concentration, temperature, or pressure affect the rates and balance of forward and reverse reactions. This dynamic balance is the reason why reactions can "settle" at equilibrium points, where the rates of forward reactions (forming products) and reverse reactions (reforming reactants) are equal.
Factors such as temperature changes can shift the equilibrium position, altering the equilibrium constant \(K_p\). Understanding reaction dynamics allows us to manipulate these conditions to favor the formation of products or reactants based on the desired outcome. For instance, altering pressure conditions at high temperatures (like at \(900^{\circ}\mathrm{C}\), as in the problem) can significantly impact how the reaction proceeds and the concentrations of gases at equilibrium.