The equilibrium constant, K, for the following reaction is 1.29×10-2 at 600 K. COCl2(g) CO(g) +...

The equilibrium constant, K, for the following reaction is 1.29×10-2 at 600 K. COCl2(g) CO(g) + Cl2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 600 K contains 0.294 M COCl2, 6.16×10-2 M CO and 6.16×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 4.50×10-2 mol of CO(g) is added to the flask? [COCl2] = M

[CO] = M

[Cl2] = M

Solutions

Expert Solution

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

I 0.294 6.16*10^-2 + 4.5*10^-2 6.16*10^-2

C +x -x -x

E 0.294+x 10.66*10^-2 -x 6.16*10^-2 -x

kc = [CO][Cl2]/[COCl2]

1.29*10^-2 = (10.66*10^-2-x)(6.16*10^-2 -x) /0.294+x

1.29*10^-2 *(0.294+x) = (10.66*10^-2-x)(6.16*10^-2 -x)

x = 0.0169

[CoCl2] = 0.294 + x = 0.294+0.0169 = 0.311M

[CO] = 10.66*10^-2-x = 10.66*10^-2 -0.0169 = 0.0897M

[Cl2] = 6.16*10^-2 -x = 6.16*10^-2 -0.0169 = 0.0447M

queen_honey_blossom answered 1 year ago

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