The equilibrium constant of the reaction `A_(2)(g)+B_(2)(g)hArr2AB(g)` at `100^(@)C` is 50. If a one litre flask containing one mole of `A_(2)` is connected to a two litre flask containing two moles of `B_(2)`, how many moles of `AB` will be formed at `373 K`?

Explanation

`overset(1)(A_(2)(g))+overset(2)(B_(2)(g)) Leftrightarrow overset(0)(2AB(g))" Initial moles"`
`(1-x) " (2-x) 2x Moles at equilibrium"`
`((1-x))/3" "((2-x))/(3)" "(2x)/3" Molar concentration at eqb"`
where x is moles of `A_2` converted to AB at equilibrium and the total volume of the container=1+2=3 litres.
`K_(c)=([AB]^2)/([A_2][B_2])=((2x)/3)^(2)/((1-x)) 3.((2-x))/(3)`
`4x^(2)/((1-x)(2-x))=50 ("given that K"_p="50 at 373K")`
x=0.93
(the other value of x may be neglected as it is greater than 1) Moles of `AB=2 xx 0.93 =1.86`