The value of `K_c` = 4.24 at 800K for the reaction, `CO_((g)) + H_2O_((g)) hArr CO_(2(g)) + H_(2(g))` Calculate equilibrium concentrations of `CO_(2), H_2, CO` and `H_2O` at 800 K, if only CO and `H_2O` are present initially at concentrations of 0.10M each.

Explanation

`{:("Reaction :", CO_((g))+, H_2O_((g)) hArr , CO_(2(g)) + , H_(2(g))),("Initial conc. :",0.1 M,0.1 M, 0 M, 0 M),("Reaction change :",-xM,-x M,+x M,+x M),("Concentration of equilibrium:",(0.1-x)M,(0.1-x)M,x M, x M):}`
where , x= amount of CO consume in reaction = amount of `H_2O`
x=made of product = amount of `CO_2` = amount `H_2`
`K_c=([CO_2][H_2])/([CO][H_2O])=((x)(x))/((0.1-x)(0.1-x))=x^2/(0.1 -x)^2`
`therefore x^2/(0.1-x)^2=4.24`
`therefore x^2=4.24(0.01-0.2x - x^2)`
`therefore x^2=0.0424-0.848x+4.24x^2`
`therefore underseta(3.24x^2)-undersetb(0.848x)+undersetc(0.0424)=0`
This is quadratic equation in which , a=3.24 , b=-0.848 and c=0.0424
This equation of quadratic , `ax^2+bx+c`
`therefore x=(-bpmsqrt(b^2-4ac))/(2a)`
`therefore x=(-(-0.848)pmsqrt((0.848)^2-4(3.24)(0.0424)))/(2(3.24))`
`therefore x=(+0.848 pm sqrt(0.7191-0.5495))/6.48`
`=(0.848pmsqrt(0.1696))/6.48 =(0.848 pm 0.4118)/6.48`
`=1.2598/6.48`
`0.4362/6.48`
=0.1944 OR 0.0673
The value 0.1994 should be neglected because it will give concentration of the reactant which is more than initial conc.
So, x=0.0673 M
`therefore` Product of equilibrium `[CO_2]=[H_2]`=0.0673 M
Reaction of equili. [CO]=`[H_2O]`= (0.1-x)M
=(0.1-0.0673)M = 0.0327 M