Question

Phosphorus pentachloride decomposes into Phosphorous trichloride and Chlorine gas. 0.670 moles of pure Phosphorus pentachloride is placed in a 4.00 L bottle. What are the resultingconcentrations?

Answer 1

The resulting concentrations at equilibrium are:

[PCl5] = 0.168 M

[PCl3] = 0.0595 M

[Cl2] = 0.0595 M

Step-by-step explanation:

The question requires us to calculate the concentrations of PCl3 and Cl2, given that 0.670 moles of PCl5 were placed in a 4.00 L container and the equilibrium constant of the following reaction is 0.0211 M:

`PCl_(5(g))\rightleftarrows PCl_(3(g))+Cl_(2(g))`

We can solve this problem using the equilibrium constant expression, considering that the concentration of PCl3 and Cl2 at the equilibrium will be the same (note that they present the same stoichiometric coefficient).

We can write the expression for the equilibrium constant as:

`K=([PCl_3\rbrack*[Cl_2\rbrack)/([PCl_5\rbrack)`

Considering that [PCl3] = [Cl2] = x, we can rearrange this expression to calculate x:

`x^2=K*[PCl_5\rbrack\rightarrow x=\sqrt[x]{K*[PCl_5\rbrack}`

The concentration of PCl5 can be obtained from the number of moles given (0.670 moles) and the volume of the container (4.00L):

`[PCl_5\rbrack=(0.670mol)/(4.00L)=0.168mol/L=0.168M`

Now, applying the values of K (0.0211M) and [PCl5] (0.168M) to the expression we wrote above, we'll have:

`\begin{gathered} \begin{equation*} x=\sqrt[x]{K*[PCl_5\rbrack} \end{equation*} \\ \\ x=\sqrt[2]{0.0211M*0.168M}=0.0595M \end{gathered}`

Therefore, the resulting concentrations at equilibrium are:

[PCl5] = 0.168 M

[PCl3] = 0.0595 M

[Cl2] = 0.0595 M