Question

In: Chemistry

Be sure to answer all parts. Phosgene (COCl2) is a toxic substance that forms readily from...

Be sure to answer all parts. Phosgene (COCl2) is a toxic substance that forms readily from carbon monoxide and chlorine at elevated temperatures: CO(g) + Cl2(g) ⇌ COCl2(g) If 0.470 mol of each reactant is placed in a 0.500−L flask at 600 K, what are the concentrations of all three substances at equilibrium (Kc = 4.95 at this temperature)?

Solutions

Expert Solution

Initial concentration of CO = mol of CO / volume

= 0.470 mol / 0.500 L

= 0.940 M

Initial concentration of Cl2 = mol of Cl2 / volume

= 0.470 mol / 0.500 L

= 0.940 M

ICE Table:

Equilibrium constant expression is

Kc = [COCl2]/[CO]*[Cl2]

4.95 = (1*x)/((0.94-1*x)(0.94-1*x))

4.95 = (1*x)/(0.8836-1.88*x + 1*x^2)

4.374-9.306*x + 4.95*x^2 = 1*x

4.374-10.31*x + 4.95*x^2 = 0

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

a = 4.95

b = -10.31

c = 4.374

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 = 19.61

roots are :

x = 1.488 and x = 0.5937

x can't be 1.488 as this will make the concentration negative.so,

x = 0.5937

At equilibrium:

[CO] = 0.94-1x = 0.94-1*0.5937 = 0.346 M

[Cl2] = 0.94-1x = 0.94-1*0.5937 = 0.346 M

[COCl2] = +1x = +1*0.5937 = 0.594 M

[CO] = 0.346 M

[Cl2] = 0.346 M

[COCl2] = 0.594 M


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