The value of `K_(c)=4.24` at 800 K for the reaction
`CO(g)+H_(2)O(g)hArrCO_(2)(g)+H_(2)(g)`
Calculate equlibrium concentrations of `CO_(2) H_(2),CO and H_(2)O` at 800 K, if only CO and `H_(2)O` are present initially at concentration of 0.10 M each?

Explanation

`{:(,CO(g)+,H_2O(g)hArr,CO_2(g)+,H_2(g)),("Initial conc.", 0.1,0.1,0,0),("At equi",0.1-x,0.1-x,x,x):}`
where x is the amount of `CO_2` and `H_2` formed at equilibrium
`K_c=([CO_2][H_2])/([CO][H_2O])`
`K_c=(x xx x)/((0.1-x)(0.1-x))=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`
Solving for x ,
`x=(0.848pm sqrt((0.848)^(2)-4(3.24)(0.0424)))/(3.24xx2)`
x=0.067 and 0.194
[The value x=0.194 should be neglected because it will give concentration of the reactant which is more than initial concentration ]
Hence equilibrium concentrations are :
`[CO_2]=[H_2]`=0.067 M
`[CO]=[H_2O]`=0.1-0.067 =0.033 M