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Ch.15 - Chemical Equilibrium

Chapter 15, Problem 41

Consider the interconversion of A molecules (red spheres) and B molecules (blue spheres) according to the reaction A ∆ B. Each of the series of pictures at the right represents a separate experiment in which time increases from left to right: (b) What is the value of the equilibrium constant Kc for the reaction A ∆ B?

Series of images showing the interconversion of red A and blue B molecules over time.

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Hi everyone. This problem reads the reaction. P yields Q represents the inter conversion of P molecules which are orange and Q molecules which are purple. The diagrams below depict four separate experiments showing the progress of the reaction with time. So we have our four separate experiments A B C and D calculate the value K. C. For the reaction. So the goal that we want to accomplish here is calculating the value of K. C. Which is the equilibrium constant. So let's just recall that equilibrium is the rate of the forward rate reaction is equal to the rate of the reverse reaction. So what this yields is no net change in the amount of reactant and products. So at equilibrium this yields B and D. Alright, so let's just write that down. So equilibrium. So we said that the rate of the forward reaction is equal to the rate of the reverse reaction. So there is no net change in the amount of reactant and products. So at equilibrium this is B and D. Alright. So for the reaction we have P. Is that equilibrium with Q. So we have orange yields purple. Alright. So let's go ahead and define what Casey is, which is our equilibrium constant. So K. C. Is going to equal the concentration of products over the concentration of reactant since. Okay, so this is going to be our products is Q. And our reactant is P. So are looking at what yields the correct thing. We're going to go with option D. Here. Okay, orange yields purple. Alright. And so looking at option D will see that the k C is going to equal, so we're going to have nine over three. Okay. And so when it comes to the answer choices given the D is going to be the correct one, and the value of K C is 9/3. Okay, So let's go ahead. And like this. Clearly, the value of K C based off of the option D depiction Is 9/3. That is it for this problem? I hope this was helpful.
Related Practice
Textbook Question

The reaction

2 PH31g2 + As21g2 ∆ 2 AsH31g2 + P21g2

has Kp = 2.9 * 10-5 at 873 K. At the same temperature,

what is Kp for each of the following reactions?

(a) 2 AsH31g2 + P21g2 ∆ 2 PH31g2 + As21g2

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Textbook Question

Calculate the value of the equilibrium constant at 427 °C for the reaction

Na O1s2 + 1>2 O 1g2 ∆ Na O 1s2

given the following equilibrium constants at 427 °C.

Na2O1s2 ∆ 2 Na1l2 + 1>2 O21g2 Kc = 2 * 10-25 Na O 1s2 ∆ 2 Na1l2 + O 1g2 K = 5 * 10-29

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Textbook Question

Calculate the value of the equilibrium constant for the reaction 4 NH31g2 + 3 O21g2 ∆ 2 N21g2 + 6 H2O1g2

given the following equilibrium constants at a certain temperature.

2 H21g2 + O21g2 ∆ 2 H2O1g2 Kc = 3.2 * 1081

N21g2 + 3 H21g2 ∆ 2 NH31g2 Kc = 3.5 * 108

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Textbook Question
The reaction A2 + B ∆ A + AB has an equilibrium con-stant Kc = 2. The following pictures represent reaction mix- tures that contain A atoms (red), B atoms (blue), and A2 and AB molecules.

(a) Which reaction mixture is at equilibrium?
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Textbook Question

The following pictures represent the initial state and the equilibrium state for the reaction of A2 molecules (red) with B atoms (blue) to give AB molecules. (a) Write a balanced chemical equation for the reaction.

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Textbook Question
Consider the reaction A + B ∆ AB. The vessel on the right contains an equilibrium mixture of A molecules (red spheres), B molecules (blue spheres), and AB molecules. If the stopcock is opened and the contents of the two vessels are allowed to mix, will the reaction go in the forward or reverse direction? Explain.

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