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The value of K(c) for the reaction: A(...

The value of `K_(c)` for the reaction:
`A_(2)(g)+B_(2)(g) hArr 2AB(g)`
at `100^(@)C` is `49`. If `1.0 L` flask containing one mole of `A_(2)` is connected with a `2.0 L` flask containing one mole of `B_(2)`, how many moles of AB will be formed at `100^(@)C`?

Text Solution

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`A_(2)(g)+B_(2)(g) hArr 2AB(g)`
As the two vessels are connected, the final volume is now `3.0 L`. Let `x` mol each of `A_(2)` and `B_(2)` react to form `2x` moles of `AB_(2)` (from stoichiometry of reaction).

`K_(c )=([AB]^(2))/([A_(2)][B_(2)])=49`
Concentration species at equilibrium are
`[A_(2)]=(1-x)//3, [B_(2)]=(1-x)//3, [AB]=2x//3`
`K_(c )=(((2x)/3)^(2))/(((1-x)/3)((1-x)/3))=(4x^(2))/((1-x)^(2))=49`
Taking square root on both sides:
`rArr (2x)/(1-x)=7rArr x=0.78`
rArr "moles" of `AB(g)` formed at equilibrium `=2x=1.56`
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