Question

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.

Answer 1

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