Consider the reactions and their respective equilibrium constants: NO(g) + 1/2 Br (g) ⇌ NOBr(g) K p = 5.3 2 NO(g) ⇌ N 2 (g) + O 2 (g) K p = 2.1⨉10 30 Use these reactions and their equilibrium constants to predict the equilibrium constant for the following reaction: N 2 (g) + O 2 (g) + Br 2 (g) ⇌ 2 NOBr(g)

All right. Hello everyone. So this question is asking us to calculate the equilibrium constant for the reaction of one mole of a two as a gas with one mole of B, two as a gas and one mole of C two as a gas to produce two moles of gaseous ABC given the reactions below and their respective equilibrium constants. So in addition to the two reactions and two equilibrium constants, we have four different answer choices labeled A through D proposing different values for the new equilibrium constant. Now because we don't have the equilibrium constant for the reaction provided in the text of the question, we have to somehow modify the two that we are given. Now, after those modifications are done through the two reactions that we are given, we can calculate the equilibrium constant for our target reaction by multiplying our two equilibrium constants together after all needed modifications. So let's start with the first reaction that were provided here, we have one mole of A B with one half of a mole of C two to produce only one mole of ABC. Now, we can see here that ABC is listed as the product However, the intended reaction is supposed to have two moles of ABC, whereas the one that we are provided produces only one mole of ABC. This means that to modify the first reaction that we provided, we have to multiply all of our coefficients by two. And because we have to multiply all coefficients by two, we're going to take our first equilibrium constant and we're going to raise that by the appropriate power. So because we're multiplying the exponents of the equation by two, we're going to race the first equilibrium constant to the second power. And so the new equilibrium constant, which I'm going to label KP one is equal to 5.8 multiplied by 10 to the first power. And just to go ahead and make sure that I'm keeping track of all of my changes on the right side of the screen, I'm going to go ahead and write the new equation after multiplying each coefficient in the first equation by two. So now that's two moles of A B and one mole of C two producing two moles of ABC. So now let's consider our second equation. We have two moles of gaseous ab producing one mole of A two as a gas and one mole of B two as a gas. So here we can see two th going back to our target equation, which is the, the reaction that we're trying to find the equilibrium constant of A two and B two alongside C two are the three reactants in this equation. However, in the second reaction that were provided a two and B two are actually products. So what that means is that we're going to have to reverse this reaction so that the products are the reactants and vice versa. This means that the new equilibrium constant, let's say KP two is going to be the inverse of the given equilibrium constant. So if KP or the second reaction here is equal to 3.3 multiplied by 10 to the 25th power, KP two, which is what I get after I reversed, this reaction is equal to one divided by 3.3 multiplied by 10 to the 25th power. And this is going to equal now 3.0 multiplied by 10 to the negative 26th power. And once again, on the right side of the screen here, I'm going to go ahead and write the new reaction after reversing the second one. So now we have a two and B two on the react inside, which means that the two moles of A B are going to be on the product side. But before we can go ahead and calculate our intended equilibrium constant, we have to make sure that the modifications of the two reactions actually give us the overall reaction that we're looking for. So let's go ahead and add these together. Now, when it comes to adding reactions together, all reactants are placed on one side and all products are placed on the other side. But notice the fact that both reactions after modification have two moles of A B on one side. For example, the first reaction has two moles of A B on the left side of the arrow. And the second reaction has two moles of A B on the right side. This means that the two moles of A B are going to cancel out. So I'm going to cross them out. But then after crossing out anything that's noted on both sides of the equations here, then the overall reaction is just like what we're looking for. So now let's go ahead and calculate the overall equilibrium constant or KP. So KP overall is going to be equal to KP one multiplied by KP two. So that's 5.8 multiplied by 10 to the first power multiplied by 3.0 multiplied by 10 to the negative 26 power. After evaluating this expression, our final answer becomes 1.7 multiplied by 10 to the negative 24th power which corresponds to option D in the multiple choice. And so there you have it if you watch this video all the way through. Thank you so very much. And I hope you found this helpful.

Ch.15 - Chemical Equilibrium

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Tro 4th Edition

Ch.15 - Chemical Equilibrium

Problem 29

Chapter 15, Problem 29

Consider the reactions and their respective equilibrium
constants:
NO(g) + 1/2 Br (g) ⇌ NOBr(g) K_p = 5.3
2 NO(g) ⇌ N₂(g) + O₂(g) K_p = 2.1⨉10³⁰
Use these reactions and their equilibrium constants to predict
the equilibrium constant for the following reaction: N₂(g) + O₂(g) + Br₂(g) ⇌ 2 NOBr(g)

Verified step by step guidance

Identify the target reaction: \( \text{N}_2(g) + \text{O}_2(g) + \text{Br}_2(g) \rightleftharpoons 2\text{NOBr}(g) \).

Recognize that the target reaction can be obtained by manipulating the given reactions.

Reverse the second reaction: \( \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g) \), which changes its equilibrium constant to \( K' = \frac{1}{K_p} = \frac{1}{2.1 \times 10^{30}} \).

Multiply the first reaction by 2: \( 2\text{NO}(g) + \text{Br}_2(g) \rightleftharpoons 2\text{NOBr}(g) \), which squares its equilibrium constant to \( K'' = (5.3)^2 \).

Combine the modified reactions to match the target reaction and multiply their equilibrium constants: \( K_{\text{target}} = K' \times K'' \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium Constant (K)

The equilibrium constant (K) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given chemical reaction. It provides insight into the extent to which a reaction proceeds; a large K indicates a reaction that favors products, while a small K suggests a preference for reactants. Understanding K is essential for predicting the behavior of chemical systems at equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust to counteract the change and restore a new equilibrium. This principle helps predict how changes in concentration, pressure, or temperature will affect the position of equilibrium, which is crucial when manipulating reactions to achieve desired outcomes.

Manipulating Equilibrium Reactions

To find the equilibrium constant for a new reaction based on known reactions, one can manipulate the given reactions by reversing them or multiplying them to derive the desired equation. The equilibrium constants for these manipulated reactions can then be combined using multiplication or inversion rules. This concept is vital for predicting the equilibrium constant of complex reactions from simpler, known reactions.

Consider the reactions and their respective equilibriumconstants:NO(g) + 1/2 Br (g) ⇌ NOBr(g) Kp = 5.32 NO(g) ⇌ N2(g) + O2(g) Kp = 2.1⨉1030Use these reactions and their equilibrium constants to predictthe equilibrium constant for the following reaction: N2(g) + O2(g) + Br2(g) ⇌ 2 NOBr(g)

Verified Solution

Key Concepts

Equilibrium Constant (K)

Le Chatelier's Principle

Manipulating Equilibrium Reactions