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 0.1846 mol sample of PCl 5 ( g ) is injected into an empty 2.55 L reaction vessel held at 250 ∘ C. Calculate the concentrations of PCl 5 ( g ) and PCl 3 ( g ) at equilibrium.

Answer 1

Answer:

$[PCl_3]_{eq}=0.0697M$

$[PCl_5]_{eq}=0.00269M$

Explanation:

Hello, in this case, considering the undergoing gaseous state chemical reaction:

$PCl_5 \rightleftharpoons PCl_3+Cl_2$

And the given mole, volume and temperature, one could compute the initial concentration of phosphorus pentachloride as follows:

$[PCl_5]_0=\frac{0.1846mol}{2.55L}=0.0724M$

Next, one writes the law of mass action as follows:

$Kc=\frac{[Cl_2]_{eq}[PCl_3]_{eq}}{[PCl_5]_{eq}}$

Thus, by introducing the change $x$ due to the reaction extent, it becomes:

$Kc=\frac{x^2}{0.0724-x}=1.80$

In such a way, solving for $x$ with solver or the quadratic equation, one gets:

$x_1=-1.87M\\x_2=0.0697M$

With which it is clear the solution is 0.0697M, therefore, the equilibrium concentrations result:

$[PCl_3]_{eq}=x=0.0697M$

$[PCl_5]_{eq}=0.0724M-x=0.0724M-0.0697M=0.00269M$

Best regards.