Question

Consider the following reaction where kp = 1.57 at 600 k: co(g) + Cl₂(g) → coCl₂(g). If the three gases are mixed in a rigid container at 600 k so that the partial pressure of each gas is initially one atm, what will happen? Indicate true (t) or false (f) for each of the following: 1. A reaction will occur in which coCl₂(g) is produced. 2. Kp will decrease. 3. A reaction will occur in which co is produced. 4. Q is less than K. 5. The reaction is at equiLiBrium. No further reaction will occur.

Answer 1

In this reaction, coCl₂(g) is formed by the reaction of CO(g) and Cl₂(g). The reaction will proceed in the forward direction to form coCl₂(g). The value of Q will be less than K as the reaction is still proceeding in the forward direction.

Step-by-step explanation:

In this reaction, coCl₂(g) is formed by the reaction of CO(g) and Cl₂(g). Since the partial pressure of each gas is initially one atm, the reaction will proceed in the forward direction to form coCl₂(g). Therefore, statement 1 is true.

Since the reaction is proceeding in the forward direction, the value of Kp will not decrease. Therefore, statement 2 is false.

A reaction to produce co will not occur in this system as there are no conditions mentioned for it to occur. Therefore, statement 3 is false.

Since the reaction is proceeding in the forward direction, the value of Q will be less than K. Therefore, statement 4 is true.

The reaction is not at equilibrium as it is still proceeding in the forward direction. Therefore, statement 5 is false.