Question

Phosphorus pentachloride decomposes according to the chemical equation PCl 5 ( g ) − ⇀ ↽ − PCl 3 ( g ) + Cl 2 ( g ) K c = 1.80 at 250 ∘ C A 0.166 mol sample of PCl 5 ( g ) is injected into an empty 2.15 L reaction vessel held at 250 ∘ C. Calculate the concentrations of PCl 5 ( g ) and PCl 3 ( g ) at equilibrium.

Answer 1

Answer : The concentration of $PCl_5$ and $PCl_3$ at equilibrium is, 0.0031 M and 0.0741 M respectively.

Explanation :  Given,

Moles of $PCl_5$ = 0.166 mol

Volume of solution = 2.15 L

First we have to calculate the concentration of $PCl_5$

$\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of solution (in L)}}$

$\text{Concentration of }PCl_5=\frac{0.166mol}{2.15L}=0.0772M$

Now we have to calculate the concentration of $PCl_5$ and $PCl_3$ at equilibrium.

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

Initial conc.    0.0772           0           0

At eqm.         0.0772-x         x            x

The expression for equilibrium constant is:

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

$1.80=\frac{(x)\times (x)}{(0.0772-x)}$

By solving the term, we get the value of 'x'.

x = 0.0741

Thus,

The concentration of $PCl_5$ at equilibrium = (0.0772-x) = (0.0772-0.0741) = 0.0031 M

The concentration of $PCl_3$ at equilibrium = x = 0.0741 M