Question

Consider the following system at equilibrium where H° = -111 kJ, and Kc = 0.159, at 723 K.

N2(g) + 3H2(g) --> 2NH3(g) When 0.32 moles of H2(g) are removed from the equilibrium system at constant temperature:

1- the value of Kc

A. increases.

B. decreases.

C. remains the same.

2- the value of Qc

A. is greater than Kc.

B. is equal to Kc.

C. is less than Kc. t

3- the reaction must:

A. run in the forward direction to reestablish equilibrium.

B. run in the reverse direction to reestablish equilibrium.

C. remain the same. It is already at equilibrium.

4-the concentration of N2 will:

A. increase.

B. decrease.

C. remain the same.

Answer 1

The given reaction is at equilibrium, therefore, any disturbance will move the reaction in the direction which will nullify the change according to the Le Chateliers principle. Hence, when 0.32 moles of H2 are removed from the equilibrium the reaction will move backward to form H2 and N2 to compensate the loss of H2 from the system. This will lead to a decrease in product NH3.

1) Kc remains the same. Since there is no temperature change. Kc changes when the temperature changes.

2) Since the reaction will shift to the left to nullify the effect of H2 removed. Therefore, the reaction coefficient Qc is greater than Kc.

3) The reaction must run in the reverse to reestablish the equilibrium.

4) Since the reaction will shift to the left more amount of N2 will be formed, hence, the concentration of N2 will increase.