Question

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 ________.

Answer 1

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 2

Answer:

The equilibrium constant of reaction at given temperature is 0.28.

Explanation:

     $CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

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 $H_2O$left at equilibrium = 0.40 mol - 0.13 mol = 0.27 mol

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

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

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

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

$K=\frac{[CO_2][H_2]}{[CO][H_2]}$

$=\frac{\frac{0.13 mol}{1 L}\times \frac{0.13 mol}{1 L}}{\frac{0.22 mol}{1 L}\times \frac{0.27 mol}{1 L}}=0.28$

The equilibrium constant of reaction at given temperature is 0.28.