# Question 6fe0b

Jul 19, 2017

$4.5$ ${\text{mol PCl}}_{3}$

#### Explanation:

Let's first write the chemical equation for this reaction:

${\text{PCl"_5(g) rightleftharpoons "PCl"_3(g) + "Cl}}_{2} \left(g\right)$

And so the equilibrium constant expression is

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

Let's make a list of our initial concentrations (no volume is given, but this doesn't matter as we're dealing just with quantity):

Initial:

• ${\text{PCl}}_{5}$: $3$ $\text{mol}$

• ${\text{PCl}}_{3}$: $3$ $\text{mol}$

• ${\text{Cl}}_{2}$: $2$ $\text{mol}$

Using the coefficients of the chemical equation (which are all $1 : 1$ molar ratios), we can determine the change in concentration:

Change:

• ${\text{PCl}}_{5}$: $- x$

• ${\text{PCl}}_{3}$: $+ x$

• ${\text{Cl}}_{2}$: $+ x$

[I'll make the assumption that the reaction quotient ${Q}_{c} < {K}_{c}$; i.e. it proceeds to the right (which it does, because there's less ${\text{PCl}}_{5}$ once equilibrium is established.)]

Since at equilibrium, there are $1.5$ ${\text{mol PCl}}_{5}$, the change in concentration $x$ must be

$3$ $\text{mol}$ $- 1.5$ $\text{mol}$ $= 1.5$ $\text{mol}$

Which means the final concentrations are

• ${\text{PCl}}_{5}$: $3$ $\text{mol}$ $- 1.5$ $\text{mol}$ $= 1.5$ $\text{mol}$

• ${\text{PCl}}_{3}$: $3$ $\text{mol}$ $+ 1.5$ $\text{mol}$ = color(red)(4.5 color(red)("mol"

• ${\text{Cl}}_{2}$: $2$ $\text{mol}$ $+ 1.5$ $\text{mol}$ $= 3.5$ $\text{mol}$

Thus, at equilibrium, there will be color(red)(4.5 color(red)("mol PCl"_3#.

(You can even experiment the situation with different volumes to prove that you'll get this same value every time, no matter the volume.)