OCl2(g) ⇄ CO(g) + Cl2(g)COCl2(g) decomposes according to the equation above. When pure COCl2(g) is injected into a rigid, previouslyevacuated flask at 690 K, the pressure in the flask is initially 1.0 atm. After the reaction reaches equilibrium at 690K, the total pressure in the flask is 1.2 atm. What is the value of Kp for the reaction at 690 K
Question
OCl2(g) ⇄ CO(g) + Cl2(g)COCl2(g) decomposes according to the equation above. When pure COCl2(g) is injected into a rigid, previouslyevacuated flask at 690 K, the pressure in the flask is initially 1.0 atm. After the reaction reaches equilibrium at 690K, the total pressure in the flask is 1.2 atm. What is the value of Kp for the reaction at 690 K
Solution
To find the value of Kp for the reaction at 690 K, we can use the equation:
Kp = (P(CO) * P(Cl2)) / P(COCl2)
Given that the initial pressure of COCl2 is 1.0 atm and the total pressure at equilibrium is 1.2 atm, we can assume that the pressure of CO and Cl2 is negligible compared to COCl2.
Therefore, we can rewrite the equation as:
Kp = P(COCl2) / P(COCl2)
Since the pressure of COCl2 at equilibrium is 1.2 atm, we can substitute the values into the equation:
Kp = 1.2 atm / 1.0 atm
Kp = 1.2
Therefore, the value of Kp for the reaction at 690 K is 1.2.
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