Halved volume with chemical equilibrium

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SUMMARY

The discussion centers on the chemical equilibrium of the reaction PCl5 (g) <=> PCl3 (g) + Cl2, where the initial pressures are 0.14 atm, 0.36 atm, and 0.50 atm, respectively. When the volume is halved, the total pressure of the mixture is calculated to be 1.88 atm. The equilibrium constant Kp is determined to be 1.29 atm, indicating that the ideal gas law applies in this scenario, despite initial assumptions. The equilibrium shifts to counteract the change in pressure due to the volume reduction.

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Homework Statement


The reaction is PCl5 (g) <=> PCl3 (g) + Cl2
In an unknow volume, the following mixture is balanced:
0.14 atm of PCl5
0.36 atm of PCl3
0.50 atm of Cl2

What would be the the pressure of the whole mixture if the volume would be halved?

(the answer is 1.88 atm)

Homework Equations


p1+p2+p3=1 atm
Kp=p3*p2/p1

The Attempt at a Solution


Since the answer is 1.88 atm I assumed the ideal gas laws aren't valid in this question because then the pressure would be 2 atm. I calculated Kp and it's 1.29 atm. But now I'm stuck.
 
Physics news on Phys.org
1. If the ideal gas law isn't valid for this question, good luck then. Luckily, they are applicable. (Generally are, unless Van der waal's Gas is given.)

2. Since the system is disturbed, the equilibrium will shift so as to minimize the effect. Since the volume is halved or the pressure is doubled, the equilibrium will shift to decrease the pressure (Guess how?) But it can only do it to some extent.

Try gathering some relevant know-how on how to solve these types of problems (they are generally book illustrations).
 

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