# Question fd11c

Aug 23, 2017

${K}_{\textrm{c}} = 0.27$

#### Explanation:

The chemical equation is

${\text{PCl"_5 ⇌ "PCl"_3 + "Cl}}_{2}$

$\text{Amount of PCl"_5color(white)(l) "dissociated" = "0.40 × 5 mol" = "2.0 mol}$

∴ The equilibrium moles of each component are:

$\text{PCl"_5 = "5 mol - 2.0 mol" = "3.0 mol}$
$\text{PCl"_3 = "2.0 mol}$
$\text{Cl"_2color(white)(ll) = "2.0 mol}$

The volume of the container is $\text{5 L}$, so the equilibrium concentrations of each component are

["PCl"_5] = "3.0 mol"/"5 L" = "0.60 mol/L"

["PCl"_3] = "2.0 mol"/"5 L" = "0.40 mol/L"

["Cl"_2] = "2.0 mol"/"5 L" = "0.40 mol/L"

We can set up part of an ICE table to solve this problem.

$\textcolor{w h i t e}{m m m m m l l} {\text{PCl"_5 ⇌ "PCl"_3+ "Cl}}_{2}$
$\text{E/mol·L"^"-1": color(white)(l)} 0.60 \textcolor{w h i t e}{m m} 0.40 \textcolor{w h i t e}{m l l} 0.40$

The equilibrium constant expression is

${K}_{\textrm{c}} = \left(\left[{\text{PCl"_3]["Cl"_2])/(["PCl}}_{5}\right]\right)$

K_text(c) = (0.40 × 0.40)/0.60 = 0.27#

Note: The answer can have only one significant figure, because that is all you gave for the moles of ${\text{PCl}}_{5}$ and for the volume of the container.

However, I did the calculation to two significant figures for you.