**Chemical Equilibrium Analysis**Consider the following reaction where \( K_c = 6.30 \) at 723 K.\[ 2\text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\text{H}_2(g) \]A reaction mixture was found to contain \( 6.21 \times 10^{-4} \) moles of \(\text{NH}_3(g)\), \( 4.61 \times 10^{-2} \) moles of \(\text{N}_2(g)\), and \( 3.99 \times 10^{-2} \) moles of \(\text{H}_2(g)\) in a 1.00 liter container.**Questions to Consider:**1. Is the reaction at equilibrium?2. If not, in what direction must it run in order to reach equilibrium?**Given Data:**- Reaction quotient, \( Q_c \): [Input required for calculation] **Options for Analysis:**- **The reaction:** - A. must run in the forward direction to reach equilibrium. - B. must run in the reverse direction to reach equilibrium. - C. is at equilibrium.**Instructions:**Calculate \( Q_c \) using the concentrations provided and compare it with \( K_c \) to determine the direction in which the reaction must proceed to reach equilibrium. Insert the required value of \( Q_c \) in the designated box and select the correct conclusion from the options given.

Given: The reaction is as follows, 2NH3(g) ⇋ N2(g) + 3H2(g) Kc = 6.30Moles of NH3 = 6.21×10-4 molMoles of N2 = 4.61×10-2 molMoles of H2 = 3.99×10-2 molTo find the molarity of each species as follows, Volume of solution = 1.00 LMoles of NH3 = 6.21×10-4 molMoles of N2 = 4.61×10-2 molMoles of H2 = 3.99×10-2 mol Substituting the values in following equation, Molarity = No. of molesVolumeMolarity of NH3 = 6.21×10-4 mol1.00 L = 6.21×10-4 MMolarity of N2 = 4.61×10-2 mol1.00 L = 4.61×10-2 MMolarity of H2 = 3.99×10-2 mol1.00 L = 3.99×10-2 MThe given reaction is as follows, 2NH3(g) ⇋ N2(g) + 3H2(g) To find the reaction quotient as follows, Molarity of NH3 = 6.21×10-4 MMolarity of N2 = 4.61×10-2 MMolarity of H2 = 3.99×10-2 M Substituting the values in reaction quotient expression, Qc = N2 H23NH32= 4.61×10-2 3.99×10-236.21×10-42Qc = 7.593The value of Qc is greater than Kc i.e. Qc > Kc. Therefore, the reaction is not at equilibrium and proceeds in backward direction. Therefore, the value of Qc is 7.593. The reaction should proceed in reverse direction to reach equilibrium. The correct option is B.

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Chemistry

Consider the following reaction where K. = 6.30 at 723 K. 2NH3(g)N2(g) + 3H2(g) A reaction mixture was found to contain 6.21×104 moles of NH3(g), 4.61×10-² moles of N2(g), and 3.99x10² moles of H2(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qc. equals | The reaction A. must run in the forward direction to reach equilibrium. B. must run in the reverse direction to reach equilibrium. C. is at equilibrium.

Consider the following reaction where K. = 6.30 at 723 K. 2NH3(g)N2(g) + 3H2(g) A reaction mixture was found to contain 6.21×104 moles of NH3(g), 4.61×10-² moles of N2(g), and 3.99x10² moles of H2(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qc. equals | The reaction A. must run in the forward direction to reach equilibrium. B. must run in the reverse direction to reach equilibrium. C. is at equilibrium.

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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