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Consider the following system at equilibrium where AH° = 16.1 kJ, and K. = 6.50x103, at 298 K. 2NOBr(g) =2NO(g) + Br2(g) When 0.15 moles of NO(g) are added to the equilibrium system at constant temperature: The value of K. The value of Qc. V 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 Br, will

Consider the following system at equilibrium where AH° = 16.1 kJ, and K. = 6.50x103, at 298 K. 2NOBr(g) =2NO(g) + Br2(g) When 0.15 moles of NO(g) are added to the equilibrium system at constant temperature: The value of K. The value of Qc. V 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 Br, will

Chemistry for Engineering Students
Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 12.18PAE: The reaction, 3 H2(g) + N2(g) (g), has the fol lowing equilibrium constants at the temperatures...
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help me asap pls for number 6 i'll give you an upvote later

Consider the following system at equilibrium where AH° = 16.1 kJ, and K. = 6.50x103, at 298 K. 2NOBr(g) =2NO(g) + Br2(g) When 0.15 moles of NO(g) are added to the equilibrium system at constant temperature: The value of K. The value of Qc. V 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 Br, will
Transcribed Image Text:Consider the following system at equilibrium where AH° = 16.1 kJ, and K. = 6.50x103, at 298 K. 2NOBr(g) =2NO(g) + Br2(g) When 0.15 moles of NO(g) are added to the equilibrium system at constant temperature: The value of K. The value of Qc. V 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 Br, will
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