Question

Phosphorus pentachloride decomposes according to the chemical equation PCI; (g) =PC12 (9) + CI, (g) Kc = 1.80 at 250 °C A 0.371 mol sample of PCI; (g) is injected into an empty 3.95 L reaction vessel held at 250 °C. Calculate the concentrations of PCI;(9) and PCI, (g) at equilibrium. [PCI,] = [PCI,] =

Answer 1

Concentration = number of mol/volume

Chemical reaction

​​PCl_{​​​​5}(g) <================> PCl_{​​​​2}(g) +. Cl_{​​​​​2}​(g)

Initially 0.371mol. 0.0mol. 0.0mol

At equilibrium (0.371 - x)mol. x mol. x mol

Concentration

At equilibrium (0.371-x)mol/3.95L. xmol/3.95L. xmol/3.95L

K_{​​​​​​C} = [PCl_​3 ][Cl_{​​​​​2} ]/[PCl_​5 ]

K_{​​​​​​C} = (x.mol/3.95L](x.mol/3.95L)/{(0.371-x)mol/3.95L]

K_{​​​​​​C} = 1.80mol/L

Now

1.80mol/L = (x^​2​​​​​mol/L)/{0.371-x)(3.95)

x^{​​​​​​2} = 2.64 - 7.11x

x^{​​​​​​2} +7.11x - 2.64 = 0

on solving this equation we get

x = .354

concentration of PCl₅ =(0.371 - 0.354)mol/3.95L =0.0043037mol/L

concentration of PCl_​5 = 4.3037×10^-3M

concentration of PCl_{​​​​3} = 0.354mol/3.95L = 0.089M