The equilibrium constant of the reaction `A _(2) (g) + B_2 (g) hArr 2 AB (g) at 100^(@)C is 50.` If a one-litre flask containing one mole of `A_(2)` is connected to a two litre flask conatining two moles of `B_(2)` how many moles of AB will be formed at 373 K?

Options

A. 1.86
B. 0.93
C. 2.32
D. 0.46

Answer

A

Explanation

`{:(A_(2)(g),+,B_(2)(g),hArr,2AB(g),),(1,,2,,0,"initial moles"),((1-x),,(2-x),,2x,"moles at equilibrium"),(((1-x))/(3),,((2-x))/(3),,(2x)/(3),"molar concentration at equi."):}`
Where x is moles `A_(2)` converted to AB at equilibrium and the total volume of the container
`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 `K_(p) = 50` at 373 K) , x = 0.93
(The other value of x obtained by solbing may be neglected as it is greater than 1 )
Moles of `AB = 2 xx 0.93 = 1.86`