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

For the reaction
`CO(g) + H_(2)O(g) hArr CO_(2)(g)+H_(2)(g)`
Initial concentration :
`0.1 M , 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)//(0.1-x)^(2) = 4.24`
`x^(2) = 4.24 (0.01 + x^(2)-0.2x)`
`x^(2) = 0.0424 + 4.24x^(2) - 0.484 x`
`3.24x^(2) - 0.848x + 0.0424 = 0`
`a = 3.24, b = -0.848, c = 0.0424`
(for quadratic equation `z^(2) + bx + x = 0`)
`x=((-b+-sqrt(b^(2)-4ac)))/(2a)`
`x = 0.848+-sqrt((0.848)^(2))-4(3.24)(0.0424)//(3.24xx2)`
`x = (0.842+-0.4118)//6.48`
`x_(1) = (0.484-0.4118)//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= 0.067 M`
`[CO] = [H_(2)O]= 0.1-0.067 = 0.033 M`