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

Given
`A+B+ hArr C+D`
`K_(C)`
`K_(C)=([C])/([A][B])`
[A]= 10 moles/lit
[B]= 10 moles/lit
[C]= 10 moles/lit
`=(20)/(10xx10)=0.2`