Consider the following system at equilibrium where AH° = 16.1 kJ, and Kc = 6.50×10-3, at 298 К. 2NOBr(g)=2NO(g) + Br2(g) When 0.26 moles of NO(g) are removed from the equilibrium system at constant temperature: The value of K The value of Qc Kc. The reaction must O run in the forward direction to restablish equilibrium. O run in the reverse direction to restablish equilibrium. O remain the same. It is already at equilibrium. The concentration of Br2 will
Consider the following system at equilibrium where AH° = 16.1 kJ, and Kc = 6.50×10-3, at 298 К. 2NOBr(g)=2NO(g) + Br2(g) When 0.26 moles of NO(g) are removed from the equilibrium system at constant temperature: The value of K The value of Qc Kc. The reaction must O run in the forward direction to restablish equilibrium. O run in the reverse direction to restablish equilibrium. O remain the same. It is already at equilibrium. The concentration of Br2 will
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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![Consider the following system at equilibrium where AH° = 16.1 kJ, and Kc = 6.50×10-3, at 298
К.
2NOBr(g) =2NO(g) + Br2(g)
When 0.26 moles of NO(g) are removed from the equilibrium system at constant temperature:
The value of Kc
The value of Qc
The reaction must
O run in the forward direction to restablish equilibrium.
O run in the reverse direction to restablish equilibrium.
O remain the same. It is already at equilibrium.
The concentration of Br2 will](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faac9dbc5-af90-4a50-8279-c1633a9e4832%2F1c37bdfc-5317-4870-8067-294ebcc9ecd1%2F8aijoxh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the following system at equilibrium where AH° = 16.1 kJ, and Kc = 6.50×10-3, at 298
К.
2NOBr(g) =2NO(g) + Br2(g)
When 0.26 moles of NO(g) are removed from the equilibrium system at constant temperature:
The value of Kc
The value of Qc
The reaction must
O run in the forward direction to restablish equilibrium.
O run in the reverse direction to restablish equilibrium.
O remain the same. It is already at equilibrium.
The concentration of Br2 will
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