Question

Calculate equilibrium constant of co2,h2,co and h20. if only co and h20 are present initially at 10 m conc.

Answer 1

CO+H20------CO2+H2

0.1 0.1 0 0

INITIAL CONCENTRATIONS-AS REACTANTA ARE NOW IN THEIR INITIAL STAGE HAVE JUST STARTED REACTING SO CONC. OF PRODUCTS IS ZERO.NOW,FINAL CONCENTRATIONS.---0.1(1-x) 0.1(1-x) x xconc.reactants conc. productsnow,here we have done 1-x as assuming that from 1 mole of the reactant x amount is converted into products.and since in this case 1 mole of CO2 reactants with 1 mole ofH2O.so equal amounts of both the reactants will decompose to form products.in case of products,co and h2 are in the ratio of 1:1 mole so both will be formed by equal amounts.given Kc=4.24.so,4.24=x*x/(1-x)2hope you can do the remaining calculation......if you like my answer do give a thumbs up.....
For the reaction,

;;;;;;^2nd rule ;;;;;;;;;
 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