Question

Determine \(K_{c}\) for the reaction $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NOBr}(\mathrm{g})$$ from the following information (at \(298 \mathrm{K}\) ). $$\begin{aligned} 2 \mathrm{NO}(\mathrm{g}) & \rightleftharpoons \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) K_{\mathrm{c}}=2.1 \times 10^{30} \\ \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{g}) & \rightleftharpoons \mathrm{NOBr}(\mathrm{g}) \quad K_{\mathrm{c}}=1.4 \end{aligned}$$

Short answer

Using the provided equilibrium constants and corresponding reactions, the equilibrium constant \(K_c\) for the given reaction is \(K_c = K_{c1} * K_{c2} = \sqrt{2.1 * 10^{30}} * 1.4\).

Step-by-step solution

Analysis of given reactions

First, let's examine the given reactions. The first reaction produces N2 and O2 from 2NO, and the second reaction produces NOBr from NO and Br2. Both reactions have known \(K_c\) values. Our goal is to manipulate these two reactions in a way that it will resemble our target reaction: \(\frac{1}{2}N2(g) + \frac{1}{2}O2(g) + \frac{1}{2}Br2(g) \rightleftharpoons NOBr(g)\).

Manipulate the first reaction

In the first reaction, we the N2 and O2 being produced from 2NO. However, in the our target reaction, we want only half of these amounts. To match the desired reaction, therefore, we need to divide the first reaction by 2, which gives: \(NO(g) \rightleftharpoons \frac{1}{2}N2(g) + \frac{1}{2}O2(g)\). According to the rules of chemical equilibrium, when the reaction is divided by a number (in this case, 2), we should also take the square root of the equilibrium constant to get the new \(K_c\). Therefore, the new \(K_c\) for this reaction is \(K_{c1} = \sqrt{2.1 * 10^{30}}\).

Combine the first reaction with the second reaction

The second reaction: \(NO(g) + \frac{1}{2}Br2(g) \rightleftharpoons NOBr(g)\) already matches the remainder of our target reaction, and it does not require any manipulation. So, now we can add the manipulated first reaction with the second reaction to achieve our target equation. When adding the two reactions, we also multiply the corresponding equilibrium constants. Therefore, the \(K_c\) for the desired reaction will be \(K_{c1} * K_{c2}\), which is \(\sqrt{2.1 * 10^{30}} * 1.4\).

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_c\), is a vital concept in chemical equilibrium. It tells us the ratio of the concentrations of products to reactants at equilibrium for a given reaction. The formula is expressed as:\[K_c = \frac{[products]}{[reactants]}\]where the concentrations are raised to the power of their stoichiometric coefficients.
The equilibrium constant provides insight into the reaction's position at equilibrium. A large \(K_c\) (greater than 1) indicates that products are favored at equilibrium, while a small \(K_c\) (less than 1) suggests that reactants are favored.
In the given problem, we need to find \(K_c\) for a target reaction by combining known reactions with their respective \(K_c\) values. This involves understanding how changes in reaction conditions, like scaling or reversing, can affect \(K_c\). The equilibrium constant, therefore, serves as a guide in predicting chemical behavior under equilibrium conditions.

Reaction Manipulation

Manipulating reactions is a key skill in solving problems involving chemical equilibrium. When given multiple reactions, it might be necessary to alter them to match a desired target reaction. This process involves either reversing the direction of a reaction, adjusting the coefficients, or a combination of both.
When you alter a reaction, it's important to also update its equilibrium constant accordingly. Here are some basic rules:

• If a reaction is reversed, the new \(K_c\) becomes the reciprocal of the original \(K_c\).
• If the coefficients in a reaction are multiplied by a factor, \(K_c\) is raised to the power of that factor.
• If divided, \(K_c\) is taken as the root corresponding to that factor.

In our exercise, the first reaction was divided by 2. Consequently, the new equilibrium constant is obtained by taking the square root of the original \(K_c\). This ensures that the altered reaction maintains the correct equilibrium constant reflective of its new form.

Gaseous Reactions

Gaseous reactions occur in the gas phase, characterized by their involvement with gaseous reactants and/or products. In equilibrium exercises, these reactions frequently employ equilibrium constants expressed in terms of concentration \([ ]\) or partial pressure \((P)\).
For gaseous reactions like the ones in this problem, assume containers where gases are free to expand and occupy space. Adjustments in pressure or volume can impact gaseous equilibria significantly. Still, the fundamental principle remains unchanged: the system will strive to reestablish equilibrium according to Le Chatelier's Principle.
In the given reaction scenario, we manipulate the reactions involving gases such as \(NO\), \(N_2\), \(O_2\), and \(Br_2\) to find the equilibrium state of a target gaseous reaction. By understanding gaseous properties, concentration, and pressure relationships, you can predict and control the behavior of gaseous reactions at equilibrium.