For `A+BhArrC`, the equilibrium concentrations of A and B at a temperature are `15 "mol" L^(-1)`. When volume is doubled the reaction has equilibrium concentration of A is `10 "mol" L^(-1)`. Calculate
a. `K_(c)`
b concentration of C in original equilibrium.

Explanation

`underset(15)(A)+underset(15)(B)hArrunderset(X "equilibrium concentration")(C)`
`K_(C)=x/((15)^(2))`
If volume is doubled backward reaction is favoured since the number of moles of reactions are more.
`underset (7.5+a)underset(7.5)(A)+underset(7.5+a)underset(7.5)(B)hArrunderset(x/2-a)(x/s)(C)` initial concentration equilibrium concentration
`7.5+a=10`
`a=2.5`
`:.K_(c)=(x/2-2.5)/(10xx10)`
`(x/2-2.5)/100=x/(15xx15)`
`:.x=45M`
`K_(c)=0.2`