Question

Consider the following equilibrium system:

COCl2(g) <----> CO(g) + Cl2(g)

A 10.00 L evacuated flask is filled with 0.4121 mol COCl2(g) at 297.9 K. The temperature is then raised to 606.9 K, where the decomposition of COCl2 gas takes place to an appreciable extent. When equilibrium is established, the total pressure in the flask is 3.000 atm.

What is the value of the equilbrium constant in terms of concentrations, Kc, at 606.9 K?

Kc =_______

Answer 1

        COCl2(g) <----> CO(g) + Cl2(g)

initial 2.05 atm         0      0

change    - x            +x      +x

equil    2.05-x           x       x

Kp = pCl2*pCO/pCOCL2

initial pressure of COCl2 = nRT/V

                          = 0.4121*0.0821*606.9/10

              = 2.05 atm

Finally,total pressure = 3 atm

       3 = 2.05-x+x+x

x = 0.95 atm

so that,

Kp = 0.95*0.95/(2.05-0.95) = 0.82

Kp = Kc*(RT)^Dn

Dn = 2-1 = 1

0.82 = Kc*(0.0821*606.9)^1

Kc = 0.0164