Question

Question #6fe0b

Answer 1

`4.5` `"mol PCl"_3`

Explanation:

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

`"PCl"_5(g) rightleftharpoons "PCl"_3(g) + "Cl"_2(g)`

And so the equilibrium constant expression is

`K_c = (["PCl"_3]["Cl"_2])/(["PCl"_5])`

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:

• `"PCl"_5`: `3` `"mol"`

• `"PCl"_3`: `3` `"mol"`

• `"Cl"_2`: `2` `"mol"`

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

Change:

• `"PCl"_5`: `-x`

• `"PCl"_3`: `+x`

• `"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 `"PCl"_5` once equilibrium is established.)]

Since at equilibrium, there are `1.5` `"mol PCl"_5`, the change in concentration `x` must be

`3` `"mol"` `- 1.5` `"mol"` `= 1.5` `"mol"`

Which means the final concentrations are

• `"PCl"_5`: `3` `"mol"` `- 1.5` `"mol"` `= 1.5` `"mol"`

• `"PCl"_3`: `3` `"mol"` `+ 1.5` `"mol"` `= color(red)(4.5` `color(red)("mol"`

• `"Cl"_2`: `2` `"mol"` `+ 1.5` `"mol"` `= 3.5` `"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.)