Question

At certain temperature, kc for the reaction, PCl5 ------> PCl3 + Cl2, is equal to 3.33. After .20 mole of PCL5 and 1.0 mole of PCL3 are introduced iinto a 2.00 L evacuated chamber, calculate the equilibrium concentration of PCl5.

Answer 1

Answer : The equilibrium concentration of
`PCl_5` is, 0.16 M

Explanation :

First we have to calculate the concentration of
`PCl_5\text{ and }PCl_3`

`\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of solution}}=(0.20mol)/(2.00L)=0.4M`

and,

`\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}=(1.0mol)/(2.00L)=2.0M`

The given chemical reaction is:

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

Initial conc. 0.4 2.0 0

At eqm. (0.4-x) (2.0+x) x

The expression for equilibrium constant is:

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

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

`3.33=((2.0+x)* (x))/((0.4-x))`

x = -5.57 and x = 0.24

We are neglecting the value of x = -5.57 because equilibrium concentration can not be more than initial concentration.

Thus, the value of x = 0.24

The equilibrium concentration of
`PCl_5` = (0.4-x) = (0.4-0.24) = 0.16 M

Therefore, the equilibrium concentration of
`PCl_5` is, 0.16 M