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`?

Answer

`1.868`

Explanation

`A_(2)(g)+B_(2)(g)hArr(g) K_(c)=50`
`(1-x)/(3) (2-x)/(3) (2x)/(3)`
`50=((2x)/(3)(2x)/(3))/((1-x)/(3)(2-x)/(3))=(4x_(2))/((1-x)(2-x))rArr 100-150x+50x^(2)=4x^(2)`
`therefore "no. of mol of" AB=(2x)/(3)=1.868`.