Question

Chemical Weapon ​Phosgene, COCl₂, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine:

CO(g) + Cl₂(g) <=> COCl₂(g)

           The value of K_c for this reaction is 5.0 at 600 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which P_CO = P_Cl2 = 0.265 atm and P_COCl2 = 0.000 atm?

Answer 1

Answer – Given, P_CO = P_Cl2 = 0.265 atm and P_COCl2 = 0.000 atm

Reaction - CO(g) + Cl₂(g) <-------> COCl₂(g) , Kc = 5.0 at 600 K

We know

We know, Kp = Kc*(RT)^Δn

Δn = 1-2 = -1

      Kp = 5.0 * (0.0821 L.atm. mole^-1.K^-1*600 K)^-1

            = 0.102

Now we need to put ICE chart -

    CO(g) + Cl₂(g) <-------> COCl₂(g)
I   0.265     0.265                  0.0

C     -x           -x                     +x

E 0.265-x    0.65-x                +x

Kp = P COCl₂ / P Cl₂ * P CO

0.102 = x / (0.265-x)*(0.265-x)

0.102 (0.265-x)(0.265-x) = +x

0.102(x²-0.53x+0.0702) = x

0.102x² – 0.0541x + 0.00716 = x

0.102x² –1.0541x+0.00716 =0

By solving quadratic equation

a = 0.102 , b = -1.054 , c = 0.00716

x = -b+/- √(b²-4ac / 2a

x = 0.00680

at equilibrium pressure

P (CO) = 0.265 -x

              = 0.265*-0.00680

           = 0.258 atm

    P (Cl₂) = 0.265 -x

                = 0.265*-0.00680

                 = 0.258 atm     

P (COCl₂) = x = 0.00680 atm