How to Calculate Kp for a Sealed Vessel at Equilibrium?

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To calculate Kp for the reaction I2 (g) + Cl2 (g) <-> 2 ICl (g) at equilibrium, the initial partial pressures are 0.3 atm for I2 and Cl2, and 0.5 atm for ICl. At equilibrium, the partial pressure of ICl increases by 52.8%, resulting in 0.764 atm. The equilibrium expression Kp can be calculated using the formula Kp = (P_ICl^2) / (P_I2 * P_Cl2), where the changes in pressures must be accounted for correctly. The confusion arises from the need to adjust all partial pressures based on the stoichiometry of the reaction, as all pressures change during the process. Understanding the relationship between the changes in pressures and the total pressure is crucial for solving the problem accurately.
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Homework Statement



Given a sealed vessel at 2000K, initially contains I2 (g) + Cl2 (g) <- -> 2 ICl (g) with partial pressures .3 atm, .3 atm and .5 atm respectively. At equilibrium the partial pressure of ICl has increased by 52.8% (so .764 atm)... calculate Kp. But that is one part that confuses me. Does only one pressure change?

Anyway.. I just did a Kp calculation for a previous question with no change and got the proper answer (it's given)... but using an ICE table in that question allowed me to solve for x to figure out the diffferent pressures... I tried to do that here but Got:

I2 Cl2 ICl
(.3-x) (.3-x) (.5+2x)

but if I add them together and equal to the total pressure (1.1 atm) ... everything cancels out. I admit I don't have a great understanding of pressures yet... any help would be greatly appreciated.
 
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All partial pressures change in this case.
 

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