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.

Jun 26, 2017

$\text{0.00308 M}$
$\text{0.0745 M}$

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

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

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

$\text{I"" "0.287/3.70" "" "" "0" "" "" } 0$

$\text{C"" "-x" "" "" "+x" "" } + x$

$\text{E"" "0.287/3.70 - x" "x" "" "" } x$

And thus,

${K}_{c} = {x}^{2} / \left(\frac{0.287}{3.70} - x\right)$

${K}_{c}$ is not small enough for the small $x$ approximation, so this requires the full quadratic formula.

$\frac{0.287}{3.70} {K}_{c} - {K}_{c} x = {x}^{2}$

$\implies {x}^{2} + {K}_{c} x - \frac{0.287}{3.70} {K}_{c}$

$= {x}^{2} + 1.80 x - 0.1396 = 0$

Solve to obtain a physical value of $x = 0.07449$ $\text{M}$. Therefore:

$\left[P C {l}_{5}\right] = \frac{0.287}{3.70} - 0.0745 = \underline{\text{0.00308 M}}$

$\left[P C {l}_{3}\right] = \underline{\text{0.0745 M}}$

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