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)hArrCO_(2)(g)+H_(2)(g)`
Initial concentration
`{:(0.1M,0.1M,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)//(0.1-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^(2)+bx+c=0`,
`x=((-b+-sqrt(b^(2)-4ac)))/(2a)`
`x=0.848+-(0.848)^(2)-4(3.24)(0.0424)//(3.24xx2)`
`x=(0.848+-0.4118)//6.48`
`x_(1)=(0.848-0.4118)//6.48=0.067`
`x_(2)=(0.848+0.4118)//6.48=0.194`
the value `0.194` should be neglected because it will give concentration of the reactant which is more than initial concentration .
Hence the equilibrium concentrations are,
`[CO_(2)]=[H_(2)]=x=0.067M`
`[CO]=[H_(2)O]=0.1-0.067=0.33M`