Question

What volume flask should be used to get 0.37 moles `I` present for every mole `I_2` at equilibrium?

For the dissociation of `I_2"(g)"` at 1200°C, `K_c=0.011`. What volume flask should we use if we want 0.37 moles of `I` to be present for every mole of `I_2` present at equilibrium?

`I_2"(g)"\rightleftharpoons2I"(g)"`

---

My work:

(can't see? it sets equilibrium pressures as stated below:)
`P_I=\frac{0.37*0.0821(1200+273)}{V}=\frac{0.37*121}{V}" atm"`
`P_(I_2)=\frac{1.00*0.0821(1200+273)}{V}=\frac{1.00*121}{V}=121/V" atm"`

after using those pressures for `K_p` in the formula `\color(red)(K_c=K_p(RT)^(\Deltan))`, I got `V=0.0307/0.011\approx0.279" L"`

Answer 1

Are we missing information? We need to be able to figure out the exact number of mols of `"I"` or `"I"_2`... no ratios, an actual number.

---

We begin with the ICE table:

`"I"_2(g) " "rightleftharpoons" " 2"I"(g)`

`"I"" "["I"_2]_i" "" "" "" "0`
`"C"" "-x" "" "" "+2x`
`"E"" "["I"_2]_i-x" "" "2x`

So we write the mass action expression to get:

`K_c = 0.011 = (["I"]^2)/(["I"_2])`

`= (2x)^2/(["I"_2]_i - x)`

What we apparently want is

`(2x)/(["I"_2]_i - x) = 0.37 = (["I"])/(["I"_2]) = ("mols I"//cancel"L")/("mols I"_2//cancel"L")`

We now have a system of equations:

`(2x)^2/(["I"_2]_i - x) = 0.011` `" "" "bb((1))`
(Note that `0.011` is in implied units of `"M"`.)

`(2x)/(["I"_2]_i - x) = 0.37` `" "" "bb((2))`
(Note that `0.37` is unitless.)

By inspection, we have:

`(2x)^2/(["I"_2]_i - x) = 0.011 = 0.37(2x) = 0.74x`

So,

`x = ("0.011 M")/0.74 = ul"0.0149 M"`

As a result, take `(2)` to get

`(2 cdot "0.0149 M")/(["I"_2]_i - "0.0149 M") = 0.37`

`2 cdot "0.0149 M" = 0.37(["I"_2]_i - "0.0149 M")`

`"0.0297 M" = 0.37["I"_2]_i - "0.0055 M"`

Therefore, the initial concentration of `"I"_2` is:

`color(blue)(["I"_2]_i) = ("0.0352 M")/(0.37) = color(blue)("0.0952 M")`

and the fraction of dissociation is

`alpha = x/(["I"_2]_i) = 0.1565`

Now we just need to know the ACTUAL mols of either `"I"` or `"I"_2`... the ratio desired can be ANY NUMBER of combinations of quantities...

The fraction of dissociation `alpha` and the concentration lost `x` also vary with concentration `["I"_2]_i`, so we cannot use those to determine the volume.