Two reactions and their equilibrium constants are given. A+2B 2C K = 2.81 2C D K2 5 = 0.248 Calculate the value of the equilibrium constant for the reaction D=A+2B. K =
Two reactions and their equilibrium constants are given. A+2B 2C K = 2.81 2C D K2 5 = 0.248 Calculate the value of the equilibrium constant for the reaction D=A+2B. K =
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![The image presents two chemical reactions with their respective equilibrium constants:
1. \( \text{A} + 2\text{B} \rightleftharpoons 2\text{C} \) with equilibrium constant \( K_1 = 2.81 \).
2. \( 2\text{C} \rightleftharpoons \text{D} \) with equilibrium constant \( K_2 = 0.248 \).
The task is to calculate the value of the equilibrium constant for the reverse reaction:
\[ \text{D} \rightleftharpoons \text{A} + 2\text{B} \]
The result of the calculation should be entered in the provided space labeled:
\[ K = \]
**Explanation for Calculation:**
To find the equilibrium constant for the reverse reaction \( \text{D} \rightleftharpoons \text{A} + 2\text{B} \), the following relationship is used:
\[ K_{\text{reverse}} = \frac{1}{K_{\text{forward}}} \]
Thus, we first need to determine \( K_{\text{forward}} \) for the reaction \( \text{D} \rightleftharpoons \text{A} + 2\text{B} \), which can be found by combining the given reactions.
1. Reverse the second reaction to get \( \text{C} \rightleftharpoons \frac{1}{2}\text{D} \), where the equilibrium constant becomes \( \frac{1}{K_2} = \frac{1}{0.248} \).
2. The overall reaction is obtained by combining these adjusted reactions to yield \( \text{D} \rightleftharpoons \text{A} + 2\text{B} \).
3. The equilibrium constant for the overall reaction is calculated by multiplying the modified constants \( K_1 \) and \( \frac{1}{K_2} \).
Finally, take the reciprocal of this equilibrium constant to find \( K \) for the reverse reaction.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe9fe32c-b492-41de-b627-5e5dec5bc081%2F4145ff06-72f0-4afa-9b63-cca723d8876b%2Fmyh7og_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image presents two chemical reactions with their respective equilibrium constants:
1. \( \text{A} + 2\text{B} \rightleftharpoons 2\text{C} \) with equilibrium constant \( K_1 = 2.81 \).
2. \( 2\text{C} \rightleftharpoons \text{D} \) with equilibrium constant \( K_2 = 0.248 \).
The task is to calculate the value of the equilibrium constant for the reverse reaction:
\[ \text{D} \rightleftharpoons \text{A} + 2\text{B} \]
The result of the calculation should be entered in the provided space labeled:
\[ K = \]
**Explanation for Calculation:**
To find the equilibrium constant for the reverse reaction \( \text{D} \rightleftharpoons \text{A} + 2\text{B} \), the following relationship is used:
\[ K_{\text{reverse}} = \frac{1}{K_{\text{forward}}} \]
Thus, we first need to determine \( K_{\text{forward}} \) for the reaction \( \text{D} \rightleftharpoons \text{A} + 2\text{B} \), which can be found by combining the given reactions.
1. Reverse the second reaction to get \( \text{C} \rightleftharpoons \frac{1}{2}\text{D} \), where the equilibrium constant becomes \( \frac{1}{K_2} = \frac{1}{0.248} \).
2. The overall reaction is obtained by combining these adjusted reactions to yield \( \text{D} \rightleftharpoons \text{A} + 2\text{B} \).
3. The equilibrium constant for the overall reaction is calculated by multiplying the modified constants \( K_1 \) and \( \frac{1}{K_2} \).
Finally, take the reciprocal of this equilibrium constant to find \( K \) for the reverse reaction.
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