Question

In: Chemistry

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 0.315 M COCl2, 6.38×10-2 M CO and 6.38×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.31×10-2 mol of Cl2(g) is added to the flask?

[COCl2] =____ M [CO] =____ M [Cl2] = ____M

Solutions

Expert Solution

CoCl2      CO + Cl2

initial concentration: a 0 0

first equilibrium concentration: a-x x x

.315 moles 6.38*10-2 6.38*10-2 (given in question) K=1.29*10-2

after adding n moles = 3.31*10-2moles of Cl2 equilibrim shifts to right , whereas equilibrium constant remains same

new equilibrium concentration a-x+n x-n x+n

therefore,

[COCl2] = a-x+n= .315+3.31*10-2= .3481 moles

[CO]= x-n = 6.38*10-2- 3.31*10-2= 6.07*10-2

[Cl2] = x+n = 6.38*10-2+3.31*10-2 = 9.69*10-2


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