Question

Phosphorus pentachloride decomposes according to the chemical equation PCl5(g)↽−−⇀PCl3(g)+Cl2(g)????c=1.80 at 250 ∘C A 0.349 mol sample of PCl5(g) is injected into an empty 4.00 L reaction vessel held at 250 ∘C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

Answer 1

Answer:

Concentration of $PCl_{5}$ at equilibrium is 0.05385 M and concentration of $PCl_{3}$ at equilibrium is 0.0334 M

Explanation:

Construct an ICE table to calculate change in concentration at equilibrium.

Initial concentration of $PCl_{5}$ = $\frac{0.349}{4.00}M = 0.08725 M$

             $PCl_{5}\rightleftharpoons PCl_{3}+Cl_{2}$

       I:           0.08725                               0             0

       C:            -x                                      +x           +x

       E:        0.08725-x                              x              x

species inside third bracket represent equilibrium concentrations

$\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}=K_{c}$

or,$\frac{x^{2}}{0.08725-x}=1.80$

or,$x^{2}+1.8x-0.15705=0$

or,$x=\frac{-1.8+\sqrt{(1.8)^{2}+(4\times 0.15705)}}{2}$

So, $x=0.0334 M$

So, $[PCl_{3}]=x=0.0334 M$, $[PCl_{5}]= 0.08725-x = 0.05385 M$    

Answer 2

Answer : The concentration of $PCl_3$ and $PCl_5$ at equilibrium are, 0.0834 M and 0.00385 M

Explanation :

First we have to calculate the concentration of $PCl_5$.

$\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of }PCl_5}=\frac{0.349mole}{4.00L}=0.08725mole/L$

The given equilibrium reaction is,

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

Initial conc.         0.08725         0           0

At equilibrium    (0.08725-x)     x           x

The expression for equilibrium constant will be,

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

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

$1.80=\frac{(x)\times (x)}{(0.08725-x)}$

By solving the term 'x', we get:

$x=0.0834M$

The concentration of $PCl_3$ at equilibrium = x = 0.0834 M

The concentration of $PCl_5$ at equilibrium = (0.08725-x) = (0.08725-0.0834) = 0.00385 M

Therefore, the concentration of $PCl_3$ and $PCl_5$ at equilibrium are, 0.0834 M and 0.00385 M