Question

At a certain temperature, `"PCl"_5(g)` decomposes. What are the concentrations of `"PCl"_5` and `"PCl"_3` in a `"3.70-L"` container that begins with `"0.287 mols"` `"PCl"_5(g)` that dissociates? `K_c = 1.80` at this temperature.

Answer 1

`"0.00308 M"`
`"0.0745 M"`

I'll let you decide which is which, by reading the answer down below. ;)

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Construct an ICE table and mass action expression in terms of concentration:

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

`"I"" "0.287/3.70" "" "" "0" "" "" "0`

`"C"" "-x" "" "" "+x" "" "+x`

`"E"" "0.287/3.70 - x" "x" "" "" "x`

And thus,

`K_c = x^2/(0.287/3.70 - x)`

`K_c` is not small enough for the small `x` approximation, so this requires the full quadratic formula.

`0.287/3.70K_c - K_cx = x^2`

`=> x^2 + K_cx - 0.287/3.70K_c`

`= x^2 + 1.80x - 0.1396 = 0`

Solve to obtain a physical value of `x = 0.07449` `"M"`. Therefore:

`[PCl_5] = 0.287/3.70 - 0.0745 = ul"0.00308 M"`

`[PCl_3] = ul"0.0745 M"`

Why would the small `x` approximation fail? What is the percent dissociation here?