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At 100.0 C, the equilibrium constant for the reaction: CO (g) + Cl2 (g) <--> COCl2...

At 100.0 C, the equilibrium constant for the reaction: CO (g) + Cl2 (g) <--> COCl2 (g) has a value of 4.6 x 109. If 0.40 mol of COCl2 is placed into a 10.0 L flask at 100.0 C, what will be the equilibrium concentration of all species? (A simplifying approximation that will make the solution of the resulting equation easier is to note that x is much less than 0.040mol/L. This means that 0.040 -x is approximately 0.040.)

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Expert Solution

                          CO                Cl2                  COCl2

Initial                  0                        0                 0.4

equilibrium         X                       X                 0.4-X

[CO2] = X/10 = 0.1X= [Cl2] , [COCl2] = (0.4-X) /10 = 0.4/10 = 0.04   ( we approximate 0.4-X as 0.4)

Now K = [COCl2] /[CO] [Cl2]

4.6 x 10^9 = (0.04) / ( 0.1X) ( 0.1X)

4.6 x 10^7 X^2 = 0.04

X = 2.95 x 10^-5 M

hece [CO] = [Cl2] = 0.1 x 2.95 x 10^-5 = 2.95 x 10^-6 M

[COCl2] = 0.04 M    ( approximate value since X is very small)

queen_honey_blossom answered 2 years ago
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