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The equilibrium constant of the reaction...

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?

A

`1.86`

B

` 0.93`

C

`2.32`

D

`0.46`

Text Solution

Verified by Experts

The correct Answer is:
A
`A_(2) (g) + B_(2) (g) hArr 2 AB (g)`
`{:(1,2,0,"inital 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 of `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 vaue of x obtined by solving may be neglected as it is greater than 1)
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