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

Upon mixing CO(g), Cl2(g), and COCl2(g) each with an initial partial pressure of 1 atm at 600 K, COCl2(g) will be produced, Kp will remain unchanged, CO will not be produced, Q will initially be less than K, and the system is not initially at equilibrium.

Step-by-step explanation:

When CO(g), Cl2(g), and COCl2(g) are mixed in a rigid container at 600 K with an initial partial pressure of 1 atm each, several things will happen concerning the equilibrium of the reaction CO(g) + Cl2(g) ⇌ COCl2(g) with a given Kp (equilibrium constant in terms of pressure) of 1.57.

• (T) A reaction will occur in which COCl2(g) is produced since reactants are present.
• (F) Kp will not decrease as it is a constant value at a given temperature (600 K in this case).
• (F) CO will not be produced as it is a reactant in the given forward reaction.
• (T) Q (reaction quotient) is less than K as initially, the pressure of products is zero, indicating the system will shift towards products to reach equilibrium.
• (F) The reaction is not at equilibrium initially since the reaction needs to occur for the system to reach equilibrium