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In an experiment, 0.35 mol of co and 0.40 mol of h2o were placed in a 1.00-l reaction vessel. at equilibrium, there were 0.22 mol of co remaining. keq at the temperature of the experiment is ________.

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MissPhiladelphia
We need to set up an I.C.E chart as follows:

       CO                 H20                CO2             H2 
I      0.35               0.40                  0                  0 
C   .35-x               0.40-x               x                   x 
E     0.19              0.40-x               x                   x 

We have 0.35 mol of CO at the start and 0.19 remaining. This means that x=0.35-0.19 >> 0.16 
Now substitute in 0.16 in your Ke equation. 
Just in case you're having trouble with that: 
Kb= [Products]/[Reactants] = ([CO][H2O])/([CO2][H2]) 

Therefore, the constant of equilibrium would be equal to 0.56.

Answer:

The equilibrium constant of reaction at given temperature is 0.28.

Explanation:

     [tex]CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)[/tex]

Initially  

0.35 mol        0.40 mol  

At equilibrium

(0.35 mol-x)       (0.40 mol-x)       x           x

Moles of CO left at equilibrium = 0.22 mol

0.22 mol = (0.35 mol-x)

x = 0.35 mol - 0.22 mol = 0.13 mol

Moles of [tex]H_2O[/tex]left at equilibrium = 0.40 mol - 0.13 mol = 0.27 mol

Concentration of CO at equilibrium = [tex][CO]=\frac{0.22 mol}{1 L}[/tex]

Concentration of [tex]H_2O[/tex] at equilibrium = [tex][H_2O]=\frac{0.27 mol}{1 L}[/tex]

Concentration of [tex]CO_2[/tex] at equilibrium = [tex][CO_2]=\frac{0.13 mol}{1 L}[/tex]

Concentration of [tex]H_2[/tex] at equilibrium = [tex][H_2]=\frac{0.13 mol}{1 L}[/tex]

[tex]K=\frac{[CO_2][H_2]}{[CO][H_2]}[/tex]

[tex]=\frac{\frac{0.13 mol}{1 L}\times \frac{0.13 mol}{1 L}}{\frac{0.22 mol}{1 L}\times \frac{0.27 mol}{1 L}}=0.28[/tex]

The equilibrium constant of reaction at given temperature is 0.28.

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