Consider the following system at equilibrium where delta H degree = -16.1 kJ/mol
ID: 486598 • Letter: C
Question
Consider the following system at equilibrium where delta H degree = -16.1 kJ/mol, and K_c = 154, at 298 K. 2 NO(g) + Br_2(g) 2NOBr(g) When 0.34 moles of NOBr(g) are removed from the equilibrium system at constant temperature: the value K_c increases. decreases. remains the same. the value of Q_c is greater than K_c. is equal to K_c is less than K_c. the reaction must: run in the forward direction to reestablish equilibrium. run in the reverse direction to reestablish equilibrium. remain the same. It is already at equilibrium. the concentration of Br_2 will: increase. decrease. remain the same.Explanation / Answer
KC= [NOBr]2/ [NO]2 [Br2]
NOBr when removed, [NOBr] will be less accordinly Q = [NOBr]2/ [NO]2 [Br2] will be less than KC
To keep KC constant NO and Br2 wiill have to decrease to keep KC constant . So the reaction molved towards products side.
1. Kc remains the same and is a function of temperature
2. Q<KC
3. Reaction proceeds in the forward direction
4. Concentration of Br2 decreases.