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.

Consider the given equilibrium reaction is as follows; A + 2B ⇌ 2C K1=2.81…

Answered: Two reactions and their equilibrium…

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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Question

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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