Question

For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 4.0 at 228°C. If pure PCl5 is placed in a 1.00-L container and allowed to come to equilibrium, and the equilibrium concentration of PCl5(g) is 0.26 M, what is the equilibrium concentration of PCl3?

Answer 1

Answer : The equilibrium concentration of $PCl_3$ is, 1.0 M

Explanation : Given,

Equilibrium concentration of $PCl_5$ = 0.26 M

Volume of solution = 1.00 L

Equilibrium constant $(K_c)$ = 4.0

The balanced equilibrium reaction will be,

$PCl_5\rightleftharpoons PCl_3+Cl_2$

The expression of equilibrium constant for the reaction will be:

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

From the reaction we conclude that the concentration of $PCl_3$ and $Cl_2$ are equal.

Let the concentration of $PCl_3$ be 'X'.

So, concentration of $Cl_2$ = X

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

$4.0=\frac{(X)\times (X)}{0.26}$

$4.0=\frac{(X)^2}{0.26}$

$X=1.0M$

Thus,

The concentration of $PCl_3$ at equilibrium = X = 1.0 M

The concentration of $Cl_2$ at equilibrium = X = 1.0 M