The value of `K_(c)=4.24` at `800K` for the reaction, `CO(g)+H_(2)O(g)hArrCO_(2)(g)+H_(2)(g)`
Calcualte equilibrium concentrations of `CO_(2)`, `H_(2)`, `CO` and `H_(2)O` at `800K`, if only `CO` and `H_(2)O` are present initially at concentrations of `0.1M` each.

Explanation

For the reaction
`CO(g)+H_(2)(O)(g) hArr CO_(2)(g)+H_(2)(g)`
Initial concentration :
`0.1M " " 0.1 M " " 0 " "0)`
Let x mole per litre of each of the product be formed.
At equilibrium :
`{:((0.1-x)M*, (0.1-x)M,xM,xM):}`
Where x is the amount of `CO_(2) and H_(2)` at equilibrium.
Hence, equilibrium constant can be written as,
`K_(C)=x^(2)(0.1x)^(2)=4.24`
`x^(2)=4.24(0.01+x^(2)-0.2x)`
`x^(2)=(0.0424+4.4x^(2)-0.848x`
`a=3.24,b=-0.484,c=0.0424`
(for quadratic equation `ax^(2)+bx+c=0)`
`x=((-b+- sqrt(b^(2)-4ac)))/(2a)`
`x=0.484+-sqrt((0.848)^(2))-4(3.24(0.424)//(3.24xx2)`
`x=(0.848+-0.4118).//6.48`
`x_(1)=(0.848-0.4118)//6.48//6.48=0.067`
`x_(2)=(0.848+0.4118)//6.48=0.194`
the value 0.19 shoule be neglected because it will given concentration of the reactant which is more than initial concentration.
Hence the equlibirum concentrations, are,
`[CO_(2)]=[H_(2)]=x=0.067M`
`[CO]=[H_(2)O]=0.1-0.067=0.033M`