The decomposition of NOCl to form NO and Cl₂, 2 NOCl(g) ⇄ 2 NO(g) + CL₂ (g), has a K_p value of 1.6 *10^-5 at some temperature.If 0.500 atm of NOCl is placed in a closed vessel and allowed to come to equilibrium, what is its approximate equilibrium partial pressure?

Question

Quizplus · Accepted Answer

Given the equilibrium constant (Kp) is 1.6 x 10^-5 and the initial pressure of NOCl is 0.500 atm, we can set up an ICE table to solve for the equilibrium pressures. Let x be the change in pressure for NOCl, then the change for NO and Cl2 will also be x (since they are produced in a 1:1 ratio). At equilibrium, the pressure of NOCl will be 0.500 - x, and the pressures of NO and Cl2 will each be x. The Kp expression is Kp = (P_NO)^2 * (P_Cl2) / (P_NOCl)^2. Substituting the equilibrium pressures into the Kp expression and solving for x gives a small value, indicating that the decomposition of NOCl is minimal. Therefore, the equilibrium pressure of NOCl is approximately its initial pressure, 0.500 atm, minus a small amount, which is closest to 0.48 atm.

The Decomposition of NOCl to Form NO and Cl2, 2

The Decomposition of NOCl to Form NO and Cl₂, 2