Question

2. Consider the following system at equilibrium where H° = 108 kJ/mol, and K_c = 1.29×10^-2, at 600 K.

COCl₂(g) CO (g) + Cl₂(g)

When 0.19 moles of COCl₂(g) are removed from the equilibrium system at constant temperature:

The value of K_c _________increases.decreases.remains the same.

The value of Q_c _________is greater than is equal to 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 CO will _________increase.decrease.remain the same.

Answer 1

The value of Kc is constant at a fixed temperature. For this reaction the temperature is kept constant at 600 K. So The value of Kc remains unchanged.

So, When 0.19 moles of COCl2(g) are removed from the equilibrium system at constant temperature ---

The value of Kc remains the same as Kc depends only on temperature. As the temperature is fixed so Kc is also same.

As the concentration of COCl2 decreases, the reaction will proceed towards reverse direction in order to increases the concentration of COCl2. So, the concentration of products decreases. Again, Qc is the ratio of the concentration of products to the concentration of reactants. As the product concentration decreases the value of Qc is less than Kc.

The reaction must --------- run in the reverse direction to reestablish equilibrium.   The reason is discussed in the above paragraph.

As the reaction goes to the reverse direction so the concentration of CO will decreases.