For gasesous reaction `A+BhArrC`, the equilibrium concentration of `A` and `B` at a temperature are `15 mol litre^(-1)`. When volume is doubled the reaction has equilibrium concentration of `A` as `10 mol litre^(-1)`. Calculate:
(`i`) `K_(c)`
(`ii`) Concentration of `C` in originl equilibrium.

Answer

A, B, C

Explanation

`{:(,,A,+,B,hArr,C),(I,"Conc at 1",15,,15,,a),("equilibrium",,,,,,),(("Volume in VL"),,,,,,),(II,"Conc when V",,,,,),("is doubled",,,,,,),(,,15/2,,15/2,,a/2),(("Volume is 2VL"),,,,,,):}`
Since on increasing volume, pressure decreases and the reaction proceeds in the direction where it shows an increase in moles, i.e. backward reaction.
Conc at II equilibrium
`[15/2+x][15/2+x][a/2-x]`
`:' 15/2+x=10 :' x=5/2`
Now,
`K_(c)=([C])/([A][B])=([a/2-5/2])/([15/2+5/2][15/2+5/2])` ...(i) for (II equilibrium)
`K_(c)=([C])/([A][B])=a/(15xx15)` ...(ii) for (I equilibrium)
`:' K_(c)` are same.
`:. ((a-5)//2)/(10xx10)=a/(15xx15)" " :. a=45 M`
Now `K_(c)=45/(15xx15)=0.2 "mol"^(-1) L^(+1)`