Question

For the reaction PCl5(g) 4 PCl3(g) 1 Cl2(g) Kp 5 23.6 at 500 K a. Calculate the equilibrium partial pressures of the reactants and products at 500 K if the initial pressures are PPCl5 5 0.560 atm and PPCl3 5 0.500 atm.

Answer 1

Answer:
`Cl_2` = 42.9 atm,
`PCl_3` = 93.4 atm and
`PCl_5` = 7.66 atm

Explanation: The given balanced equation is:

`PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)`

`K_p=523.6`

Initial pressure of
`PCl_5` = 50.560 atm

initial pressure of
`PCl_3` = 50.500 atm

Let's say the change in pressure is p. Then:

equilibrium partial pressure of
`PCl_5` = (50.560 - p) atm

equilibrium partial pressure of
`PCl_3` = (50.500 + p) atm

equilibrium partial pressure of
`Cl_2` = p atm

`K_p=((PCl_3)(Cl_2))/(PCl_5)`

Let's plug in the values in it:

`523.6=((50.500+p)(p))/(50.560-p)`

on cross multiply:

`26473.216-523.6p=50.500p+p^2`

on rearranging the above equation:

`p^2+574.1p-26473.216=0`

It's a quadratic equation. On solving this equation:

p = 42.9

So, the equalibrium partial pressure of
`Cl_2` = 42.9 atm

equilibrium partial pressure of
`PCl_3` = 50.500 + 42.9 = 93.4 atm

equilibrium partial pressure of
`PCl_5` = 50.560 - 42.9 = 7.66 atm