# An equilibrium mixture of 2 moles each of PCl_5, PCl_3 and Cl_2 in V ltrs. in T temp. has a total pressure of 3 atm. Cl_2 is added at the same temp. and pressure till the volume is doubled. What is the number of moles of Cl_2 added? Thank you:)

Oct 2, 2017

$\frac{20}{3}$, or $6.7$ moles of $C {l}_{2}$ must be added.

#### Explanation:

The chemical equation is PCl_5 ⇄PCl_3 + Cl_2.

First, we have to find the equilibrium constant $K$:
$K = \frac{\left[P C {l}_{5}\right]}{\left[P C {l}_{3}\right] \left[C {l}_{2}\right]} = \frac{\frac{2}{V}}{\left(\frac{2}{V}\right) \left(\frac{2}{V}\right)} = \frac{V}{2} \left(\text{L/mol}\right)$

Let $x \left(m o l\right)$ to be the amount of $C {l}_{2}$ you need to add.
Adding $C {l}_{2}$ will move the equilibrium to the left($\left[P C {l}_{5}\right]$ increases. $\left[P C {l}_{3}\right]$ and $\left[C {l}_{2}\right]$ decease.)
Let $y \left(m o l\right)$ to be the amount of $C {l}_{2}$ used to reach $\textcolor{red}{\text{the new equilibrium}}$ after adding Cl_2".

At the new equilibrium point, amount of $P C {l}_{5} , P C {l}_{3}$ and $C {l}_{2}$ will be $2 + y$, $2 - y$ , and $2 + x - y$ moles, respectively.

Here you obtain two equations:
(1) Since the volume is doubled at the constant pressure and temparature, the total amount of substance must be doubled too.
$\left(2 + y\right) + \left(2 - y\right) + \left(2 + x - y\right) = 6 \times 2 = 12$
$x - y = 6$ ・・・(A)

(2) The equilibrium constant doesn't change since the temparature is the same .Though, the volume is now $2 V$.
$K = \frac{\left[P C {l}_{5}\right]}{\left[P C {l}_{3}\right] \left[C {l}_{2}\right]} = \frac{\frac{2 + y}{2 V}}{\left(\frac{2 - y}{2 V}\right) \left(\frac{2 + x - y}{2 V}\right)}$
$= \frac{\frac{2 + y}{2 V}}{\left(\frac{2 - y}{2 V}\right) \cdot \frac{8}{2 V}}$ (using (A))

=(V(2+y))/(4(2-y)

$K$ must be the same as before:
$\frac{V \left(2 + y\right)}{4 \left(2 - y\right)} = \frac{V}{2}$ ・・・(B)
Multiply equation (B) by $\frac{4 \left(2 - y\right)}{V}$ leads to:
$2 + y = 2 \left(2 - y\right)$ and $y = \frac{2}{3}$ is obtained.

Substitute $y = \frac{2}{3}$ to (A) and you did it!
x=20/3=6.66… (mole)