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 $PCl_5\text{ and }PCl_3$

$\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of solution}}=\frac{0.20mol}{2.00L}=0.4M$

and,

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}=\frac{1.0mol}{2.00L}=2.0M$

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:

$K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

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

$3.33=\frac{(2.0+x)\times (x)}{(0.4-x)}$

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