### Equilibrium Constant Calculation**Problem Statement:**Consider the following reversible reaction in a 3.00 liter container, with the following amounts found in equilibrium at 400 °C:- 0.0420 mole N₂- 0.516 mole H₂- 0.0357 mole NH₃Evaluate the equilibrium constant \( K_c \).**Reaction:**\[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \]**Options:**(a) 0.202 (b) 1.99 (c) 16.0 (d) 4.94 (e) 0.503**Solution Steps:**1. **Determine Concentrations:** - N₂: \(\frac{0.0420 \text{ moles}}{3 \text{ L}} = 0.0140 \text{ M}\) - H₂: \(\frac{0.516 \text{ moles}}{3 \text{ L}} = 0.172 \text{ M}\) - NH₃: \(\frac{0.0357 \text{ moles}}{3 \text{ L}} = 0.0119 \text{ M}\)2. **Write the Expression for \( K_c \):** \[ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \]3. **Substitute Values:** \[ K_c = \frac{(0.0119)^2}{(0.0140)(0.172)^3} \]4. **Calculate \( K_c \):** Complete the calculation to solve for \( K_c \), selecting the correct answer from the options provided.

Kc=product÷reactant Molarity=mole÷volume Kc=0.88

6. Consider the following reversible reaction. In a 3.00 liter container, the following amounts are found in equilibrium at 400 °C: 0.0420 mole N2, 0.516 mole H2 and 0.0357 mole NH3. Evaluate Kc. N2(g) + 3H2(g) 2NH3(g) (a) 0.202 (b) 1.99 (c) 16.0 (d) 4.94 (e) 0.503

6. Consider the following reversible reaction. In a 3.00 liter container, the following amounts are found in equilibrium at 400 °C: 0.0420 mole N2, 0.516 mole H2 and 0.0357 mole NH3. Evaluate Kc. N2(g) + 3H2(g) 2NH3(g) (a) 0.202 (b) 1.99 (c) 16.0 (d) 4.94 (e) 0.503

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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