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

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Answer: The equilibrium partial pressure of chlorine gas is 0.360 atm

Step-by-step explanation:

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=(p_(Cl_2)* 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=(p_(Cl_2)* 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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