The equilibrium constant of the reaction
`A_(2)(g)+B_(2)(g) hArr 2AB(g)`
at `100^(@)C` is `50`. 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

`A_(2)(g) + B_(2)(g) rarr 2AB(g) Delta n = 0`
`K = [AB]^(2)/([A_(2)][B_(2)]) = (n_(AB))^(2)/(n_(A_(2)).n_(B_(2))) = (2x)^(2)/((1-x)(2-x))`
`implies 5-0 = 4x^(2)/x^(2)+3x+2 implies 23x^(2) -7x +50 = 0`
`implies x = (75 +- sqrt(75^(2)-4xx23xx50))/46 = 0.93,2.32`
2.32 is not acceptable because x cannot be greater than 1, Mole of AB = 2x = 2 xx 0.93 = 1.86