Question

The dissociation of chlorine into atoms has been determiend to be 0.570 at 2000k.

Cl2 (g) <--> 2Cl (g)

A- If 0.1 mol of chlorine are present in a volume of 5L at that temperature. Converted Kp into Kc (leave in mol/dm^3). What is the degree of dissociation for the molecule? (the fraction of chlorine atoms present)

B- The volume is expanded to 20L what will the degree of dissociation be then?

C- Will this result agree with Le Chatliers Principle?

Answer 1

Find Kp:

Kp = Kc*(RT)^dn

Kp = (0.57)(0.082*2000)^1

Kp = 93.48

[Cl2] = 0.1/5 = 0.02 M

for

Kp = Cl^2 / Cl2

93.48 = (2x)^2 / (0.02-x)

93.48 *0.02 - 93.48 x = 4x^2

4x^2 + 93.48x - 1.8696 = 0

x = 0.01999

then

ratio

[Cl-] = [Cl2]

[Cl-] = 2*0.01999 = 0.03998

[Cl2] = 0.02 -0.01999 = 10^-5

B)

IF V goes from 5 L to 20, the ratio is 4x

so concentratio ngoes by 4

0.1/(20) = 0.005 M of Cl2

Kp = 93.48

Kp =  Cl^2 / Cl2

93.48 = (0+2*x)^2/(0.005-x)

93.48 *0.005-93.48 x = 4x^2

4x^2 +93.48 x - 0.4674 = 0

x = 0.00499

[Cl[= 2*0.00499 = 0.009

[Cl2] = 0.005-0.00499 = 10^-5

c)

This agrees

since

Lechatelier states that an increase in volume makes mor space for multiple mol creation, that is, the atomic Chlorine atoms (products)