**Understanding Equilibrium Constants in Chemical Reactions**In the reaction of ammonia with water, the equilibrium constant is given as \(1.8 \times 10^{-5}\). The reaction is represented as follows:\[ \text{NH}_3 \, (\text{aq}) + \text{H}_2\text{O} \, (\text{l}) \rightleftharpoons \text{NH}_4^+ \, (\text{aq}) + \text{OH}^- \, (\text{aq}) \]**Question:** What is the value of the equilibrium constant for the reverse reaction?**Options:**- a. \(5.6 \times 10^{4}\)- b. \(5.6 \times 10^{-6}\)- c. \(-1.8 \times 10^{-5}\)- d. \(5.6\)- e. \(1.8 \times 10^{-5}\)**Explanation:**When reversing a chemical reaction, the equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the forward reaction. Therefore, if the equilibrium constant for the forward reaction is \(1.8 \times 10^{-5}\), the equilibrium constant for the reverse reaction is calculated as:\[K_{\text{reverse}} = \frac{1}{K_{\text{forward}}} = \frac{1}{1.8 \times 10^{-5}}\]Select one of the given options based on this calculation.

The equilibrium constant for a reaction written in the reverse direction is the reciprocal of the…

The equilibrium constant is 1.8 x 10-5 for the reaction of ammonia with water. NH3 (aq) + H20 (1) S NH4+1 (aq) + OH1(aq) What is the value of the equilibrium constant for the reverse reaction? Select one: а. 5.6 х 104 b. 5.6 x 10-6 О с. -1.8 х 10-5 O d. 5.6 Ое. 1.8 х 10-5

The equilibrium constant is 1.8 x 10-5 for the reaction of ammonia with water. NH3 (aq) + H20 (1) S NH4+1 (aq) + OH1(aq) What is the value of the equilibrium constant for the reverse reaction? Select one: а. 5.6 х 104 b. 5.6 x 10-6 О с. -1.8 х 10-5 O d. 5.6 Ое. 1.8 х 10-5

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

**Understanding Equilibrium Constants in Chemical Reactions**

In the reaction of ammonia with water, the equilibrium constant is given as \(1.8 \times 10^{-5}\). The reaction is represented as follows:

\[ \text{NH}_3 \, (\text{aq}) + \text{H}_2\text{O} \, (\text{l}) \rightleftharpoons \text{NH}_4^+ \, (\text{aq}) + \text{OH}^- \, (\text{aq}) \]

**Question:** What is the value of the equilibrium constant for the reverse reaction?

**Options:**

- a. \(5.6 \times 10^{4}\)
- b. \(5.6 \times 10^{-6}\)
- c. \(-1.8 \times 10^{-5}\)
- d. \(5.6\)
- e. \(1.8 \times 10^{-5}\)

**Explanation:**
When reversing a chemical reaction, the equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the forward reaction. Therefore, if the equilibrium constant for the forward reaction is \(1.8 \times 10^{-5}\), the equilibrium constant for the reverse reaction is calculated as:

\[
K_{\text{reverse}} = \frac{1}{K_{\text{forward}}} = \frac{1}{1.8 \times 10^{-5}}
\]

Select one of the given options based on this calculation.

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