Question

Phosphorous pentachloride decomposes according to the reaction

PCl5(g)=PCl3(g)+Cl2(g)


A 12.3 g sample of PCl5 is added to a sealed 1.50 L flask and the reaction is allowed to come to equilibrium at a constant temperature. At equilibrium, 31.8% of the PCl5 remains. What is the equilibrium constant, Kc, for the reaction?


Question is asking for Kc

Answer 1

Answer: The equilibrium constant,
`K_c`, for the reaction is 0.061.

Step-by-step explanation:

Initial concentration of
`PCl_5` =
`\frac{\text {given mass}}{\text {Molar mass}* Volume in L}}=(12.3g)/(208.2g/mol* 1.50L)=0.039M`

Equilibrium concentration of
`PCl_5` =
`(31.8)/(100)* 0.039=0.012M`

The given balanced equilibrium reaction is,

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

Initial conc. 0.039 M 0 M 0 M

At eqm. conc. (0.039-x) M (x) M (x) M

Given : (0.039-x) = 0.012

x = 0.027

The expression for equilibrium constant for this reaction will be,

`K_c=([Cl_2]* [PCl_3])/([PCl_5])`

Now put all the given values in this expression, we get :

`K_c=(0.027* 0.027)/(0.012)=0.061`

The equilibrium constant,
`K_c`, for the reaction is 0.061.