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.

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Alleei Alleei · Answer 1 · 2020-02-12T09:09:49+01:00

Answer : The concentration of [tex]PCl_5[/tex] and [tex]PCl_3[/tex] at equilibrium is, 0.0031 M and 0.0741 M respectively.

Explanation : Given,

Moles of [tex]PCl_5[/tex] = 0.166 mol

Volume of solution = 2.15 L

First we have to calculate the concentration of [tex]PCl_5[/tex]

[tex]\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of solution (in L)}}[/tex]

[tex]\text{Concentration of }PCl_5=\frac{0.166mol}{2.15L}=0.0772M[/tex]

Now we have to calculate the concentration of [tex]PCl_5[/tex] and [tex]PCl_3[/tex] at equilibrium.

[tex]PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)[/tex]

Initial conc. 0.0772 0 0

At eqm. 0.0772-x x x

The expression for equilibrium constant is:

[tex]K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]

[tex]1.80=\frac{(x)\times (x)}{(0.0772-x)}[/tex]

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

x = 0.0741

Thus,

The concentration of [tex]PCl_5[/tex] at equilibrium = (0.0772-x) = (0.0772-0.0741) = 0.0031 M

The concentration of [tex]PCl_3[/tex] at equilibrium = x = 0.0741 M