### Problem 5: Equilibrium Constant CalculationConsider the following reaction:\[ \text{N}_2(g) + 3\text{H}_2(g) \leftrightharpoons 2\text{NH}_3(g) \]At equilibrium, the pressure values of the substances were:- \( \text{N}_2 \): 0.62 atm- \( \text{H}_2 \): 0.50 atm- \( \text{NH}_3 \): 0.24 atm**Question:** What is \( K_{eq} \) for the reaction?### Multiple Choice Answers:a. 0.74 \b. 2.7 \c. 1.3 \d. 0.60 \e. 0.37**Explanation:**To find the equilibrium constant \( K_{eq} \) for the reaction, we use the following formula for a gas-phase reaction, where pressures are given:\[ K_{eq} = \frac{P_{\text{NH}_3}^2}{P_{\text{N}_2} \cdot P_{\text{H}_2}^3} \]Given that:- \( P_{\text{N}_2} = 0.62 \) atm- \( P_{\text{H}_2} = 0.50 \) atm- \( P_{\text{NH}_3} = 0.24 \) atmSubstitute these values into the equilibrium expression:\[ K_{eq} = \frac{(0.24)^2}{(0.62) \cdot (0.50)^3} \]Calculate the numerator:\[ (0.24)^2 = 0.0576 \]Calculate the denominator:\[ (0.62) \cdot (0.50)^3 = 0.62 \cdot 0.125 = 0.0775 \]Now, divide the numerator by the denominator:\[ K_{eq} = \frac{0.0576}{0.0775} \approx 0.74 \]Thus, the correct answer is:**a. 0.74**

Answered: Image /qna-images/answer/fa5215d2-5ae1-4c61-aa66-9d25a5f92724.jpg

5. Consider the following reaction: N2(g) + 3 H₂(g) 2 NH3(g) wAt equilibrium, the pressure values of the substances were 0.62 atm N₂, 0.50 atm H2, and 0.24 atm NH3. What is Keq for the reaction? a. 0.74 b. 2.7 c. 1.3 d. 0.60 e. 0.37

5. Consider the following reaction: N2(g) + 3 H₂(g) 2 NH3(g) wAt equilibrium, the pressure values of the substances were 0.62 atm N₂, 0.50 atm H2, and 0.24 atm NH3. What is Keq for the reaction? a. 0.74 b. 2.7 c. 1.3 d. 0.60 e. 0.37

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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