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The molar concentration of A and B are 0...

The molar concentration of `A` and `B` are `0.80 mol//litre` each . On mixing them, the reaction starts to proceed as : `A+BhArrC+D`, and attain equilibrium. At equilibrium molar concentration of `C` is `0.60 mol//litre`. Find the value of `K_(c)` of the reaction.

Text Solution

Verified by Experts

Given, `[C]` at equilibrium `=0.6 mol//litre`
`[D]=0.6 mol//litre`
Thus, concentration of `A` and `B`, which reacts, are `0.6 mol//litre` at equilibrium.
`{:(,A,+,B,hArr,C,+,D),("Initial conc.",0.8,,0.8,,0,,0),("Equilibrium conc.",(0.8-0.6),,(0.8-0.6),,0.6,,0.6),(,=0.2,,=0.2,,0.6,,0.6):}`
`:. K_(c)=([C][D])/([A][B])=(0.6xx0.6)/(0.2xx0.2)=9`
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