osphorus pentachloride decomposes according to the chemical

osphorus pentachloride decomposes according to the chemical equation PC15(g) PC,(g) + C12 e) K,=1.80 at 250°C A 0.230 mol sample of PCIs(g) is injected into an empty 3.70 L reaction vessel held at 250 Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium. Number PC151 = Number PCI,-

Solution

initial concentration of PCl5 = number of mol of PCl5 / volume
= 0.230 mol / 3.70 L
= 0.0622 M

ICE Table:

                    [PCl5]              [PCl3]              [Cl2]             

initial             0.0622              0                   0                 

change              -1x                 +1x                 +1x               

equilibrium         0.0622-1x           +1x                 +1x               

Equilibrium constant expression is
Kc = [PCl3]*[Cl2]/[PCl5]
1.8 = (1*x)(1*x)/((0.0622-1*x))
1.8 = (1*x^2)/(0.0622-1*x)
0.11196-1.8*x = 1*x^2
0.11196-1.8*x-1*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = -1
b = -1.8
c = 0.112

Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 3.688

roots are :
x = -1.86 and x = 6.019*10^-2

since x can\'t be negative, the possible value of x is
x = 6.019*10^-2

At equilibrium:
[PCl5] = 0.0622-1x = 0.0622-1*0.06019 = 0.00201 M
[PCl3] = +1x = +1*0.06019 = 0.06019 M

Answer:
[PCl5] = 0.00201 M
[PCl3] = 0.0602 M