Question

Calculate the equilibrium concentrations of reactant and products when 0.331 moles of PCl3 and 0.331 moles of Cl2 are introduced into a 1.00 L vessel at 500 K?

The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K.

PCl3(g) + Cl2(g) PCl5(g)

Calculate the equilibrium concentrations of reactant and products when 0.331 moles of PCl3 and 0.331 moles of Cl2 are introduced into a 1.00 L vessel at 500 K.

PCl3=?M
Cl2=?M
PCl5=?M

Answer 1

`["PCl"_3]_(eq) = ["Cl"_2]_(eq) = "0.0573 M"`

`["PCl"_5]_(eq) = "0.2737 M"`

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Construct the ICE table:

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

`"I"" "0.331" "" "0.331" "" "" "" "0`
`"C"" "-x" "" "-x" "" "" "" "+x`
`"E"" "0.331-x" "0.331-x" "" "x`

So then, the mass action expression becomes:

`K_P = x/(0.331-x)^2 = 83.3`

`x = 83.3(0.109561 - 0.662x + x^2)`

`0 = 83.3x^2 - 56.1446x + 9.12643`

Solve the quadratic. You should get

`x = "0.4003 M", "0.2737 M"`

Which one doesn't make physical sense? (i.e. which one gives you NEGATIVE concentration?)

So you should get

`["PCl"_3]_(eq) = ["Cl"_2]_(eq) = "0.0573 M"`

`["PCl"_5]_(eq) = "0.2737 M"`