Question

For the equilibrium reaction: PCis(g) = PC13(g) + Cl2(g); Kc = 0.00183 a. Write the concentration equilibrium equation for the reaction. [1 mark] b. If 2.0 gram of PCis(g) is introduced into a 1.5 L flask what will be the equilibrium concentrations of PCls, PC13 and Cl2? [4marks]

Answer 1

a)

Kc is defined as concentration of product by concentration of reactant with each concentration term raised to power that is equal to its stoichiometric coefficient in balanced equation

So,

Kc = [PCl3][Cl2]/[PCl5]

b)

Molar mass of PCl5,

MM = 1*MM(P) + 5*MM(Cl)

= 1*30.97 + 5*35.45

= 208.22 g/mol

mass(PCl5)= 2.0 g

use:

number of mol of PCl5,

n = mass of PCl5/molar mass of PCl5

=(2 g)/(2.082*10^2 g/mol)

= 9.605*10^-3 mol

volume , V = 1.5 L

use:

Molarity,

M = number of mol / volume in L

= 9.605*10^-3/1.5

= 6.403*10^-3 M

ICE Table:

[РC151 [РCС13] [C12] initial 0.006403 change -1x +1x +1x equilibrium 0.006403-1x +1x +1x

Equilibrium constant expression is

Kc = [PCl3]*[Cl2]/[PCl5]

0.00183 = (1*x)(1*x)/((6.403*10^-3-1*x))

0.00183 = (1*x^2)/(6.403*10^-3-1*x)

1.172*10^-5-1.83*10^-3*x = 1*x^2

1.172*10^-5-1.83*10^-3*x-1*x^2 = 0

This is quadratic equation (ax^2+bx+c=0)

a = -1

b = -1.83*10^-3

c = 1.172*10^-5

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 = 5.022*10^-5

roots are :

x = -4.458*10^-3 and x = 2.628*10^-3

since x can't be negative, the possible value of x is

x = 2.628*10^-3

At equilibrium:

[PCl5] = 0.006403-1x = 0.006403-1*0.002628 = 0.00377 M

[PCl3] = +1x = +1*0.002628 = 0.00263 M

[Cl2] = +1x = +1*0.002628 = 0.00263 M

Answer:

[PCl5] = 0.00377 M

[PCl3] = 0.00263 M

[Cl2] = 0.00263 M