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barxatty [35]
3 years ago
9

The equilibrium constant, Kp, for the following reaction is 0.497 at 500K.PCl5(g) PCl3(g) + Cl2(g)If an equilibrium mixture of t

he three gases in a 18.4 L container at 500K contains PCl5 at a pressure of 0.471 atm and PCl3 at a pressure of 0.651 atm, the equilibrium partial pressure of Cl2 is atm.
Chemistry
1 answer:
Anestetic [448]3 years ago
5 0

<u>Answer:</u> The equilibrium partial pressure of chlorine gas is 0.360 atm

<u>Explanation:</u>

For the given chemical equation:

PCl_5(g)\rightleftharpoons PCl_3(g)+Cl2(g)

The expression of K_p for above reaction follows:

K_p=\frac{p_{Cl_2}\times p_{PCl_3}}{p_{PCl_5}}

We are given:

K_p=0.497\\p_{PCl_3}=0.651atm\\p_{PCl_5}=0.471atm

Putting values in above equation, we get:

0.497=\frac{p_{Cl_2}\times 0.651}{0.471}\\\\p_{Cl_2}=0.360atm

Hence, the equilibrium partial pressure of chlorine gas is 0.360 atm

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