Question

the value of Kc = 4.24 at 800K for the reaction CO(g) + H2O(g) CO2(g) + H2(g) Calculate the equilibrium concentration of CO2 ,H2O ,CO , H2 at 800K ,if only CO and H2O are pesent initially at concentration of 0.10M each. Why didn't we take 0.194 wala value for the calculation

Answer 1

                      CO(g)     +      H2O(g)        ⇌         CO2(g)        +       H2(g)
initial
concentration=  0.1M             0.1M                        0                           0

concentration
in equilibrium = 0.1-x             0.1-x                        x                            x

Hence, equilibrium constant can be written as,
 Kc = x^2 /(0.1-x)^2 = 4.24

 x^2 = 4.24(0.01 + x^2 -0.2x)
 x^2 = 0.0424 + 4.24x^2 - 0.848x

3.24x^2 – 0.848x + 0.0424 = 0

a = 3.24, b = – 0.848, c = 0.0424
for quadratic equation ax + bx + c = 0,
Thus solving we get two values of x

x1 = 0.067 x2 = 0.194
 Neglecting x2 = 0.194 because x could not be more than initial concentration.

 Hence the equilibrium concentrations are, [CO ] = [H ] = x = 0.067 M 

[CO2] = [H2] = x = 0.067 M

[CO] = [H2O] = 0.1 - x = 0.1 - 0.067 =0.033 M


i hope it will help you
regards

Answer 2

For the reaction,
 CO (g) + H2O (g) ⇔ CO2 (g) + H2 (g)
Initial concentration:
0.1M0.1 M00
Let x mole per litre of each of the productbe formed.
At equilibrium:
0.1 - x M0.1 - x Mx Mx M
where  x is the amount of CO2 and H2 at equilibrium. .
Hence, equilibrium constant can be written as,
Kc= x2/(0.1-x)2 = 4.24
x2 = 4.24(0.01 + x2-0.2x)
x2 = 0.0424 + 4.24x2-0.848x
3.24x2 – 0.848x + 0.0424 = 0
a = 3.24, b = – 0.848, c = 0.0424
for quadratic equation ax2 + bx + c = 0,
Thus solving we get two values of x 
x1= 0.067  x2=  0.194 
Neglecting x2=  0.194  becuase x could not be more than initial concentration.
Hence the equilibrium concentrations are,
[CO2] = [H2] = x = 0.067 M
[CO] = [H2O] = 0.1 – 0.067 = 0.033 M