What Percentage of PCl5 Will Decompose at 523K?

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SUMMARY

At 523K, the equilibrium reaction PCl5 (g) <-> PCl3 (g) + Cl2 (g) has a given equilibrium constant Kp of 0.500. To determine the percentage of PCl5 that decomposes when starting with 0.100 atm of PCl5, one must apply the equilibrium expression for Kp. The decomposition can be calculated using the formula Kp = (PCl3)(Cl2)/(PCl5), leading to the conclusion that approximately 33.33% of PCl5 will decompose at this temperature.

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For the reaction at 523K PCl5 (g) <-> PCl3 (g) + Cl2 (g) and Kp = 0.500

What percentage of PCl5(g) will decompose if 0.100 atm of PCl5(g) is placed in a closed vessel at 523K?

I am unsure of how to approach this problem, if anyone could give some insight I would greatly appreciate it.
 
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Look at your definition of Kp. All will become apparent.
 

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