The equilibrium constant of the reaction
`A_(2)(g)+B_(2)(g) hArr 2AB(g)`
at `100^(@)C` is `49`. If a `1L` flask containing `1` mol of `A_(2)` is connected to a `2 L` flask containing `2` mol of `B_(2)`, how many moles of AB will be formed at `373 K`?

Explanation

If x is the moles of `A_2` converted to AB at equilibrium and the total volume is 1+2 =3L
`{:(,A_2+,B_2 hArr , 2AB),("Initial conc.",1,2,0),("Moles at equ.", (1-x),(2-x),2x):}`
`[A_2]=(1-x)/3, [B_2]=(2-x)/3, [AB]=(2x)/3`
`K_c=[AB]^2/([A_2][B_2])=((2x)/3)^2/(((1-x)/3)((2-x)/3))=50`
`(4x^2)/((1-x)(2-x))=50`
Solving for x, we get x =0.955 , Moles of AB formed = 2 x 0.955 =1.91 mol