For `A+B hArr C`, the equilibrium concentration of A and B at a temperature are `15 mol L^(-1)`. When volume is doubled the reaction has equilibrium concentration of A as `10 mol L^(-1)`, calculate
a. `K_(c)`

Options

A. Ratio of concentration of `A "and" B` at new equilibrium is 3/4
B. Value of equilibrium constant for both cases are same
C. Concentration of `C` at new equilibrium become half
D. Equilibrium concentration of `C` at new equilibrium `(10sqrt20)/(sqrt20-sqrt15)`

Answer

A, B, D

Explanation

`{:(,A,+,2B,hArr,2C),("at vol=v lit",20,,20,,a),("at vol=2v lit",(20)/(2),,(20)/(2),,(a)/(2)):}`
at new equilibrium
`((20)/(2)+(x)/(2))((20)/(2)+x)((a)/(2)-x)`
given that `(A)_(new)=15=((20)/(2)+(x)/(2)rArrx=(15-10)xxL`
`x=101`
for Ist equilibrium `K_(C)=((a)/(2))^(2)/(((20)/(2)(20)/(2))^(2))=((a)/(2))^(2)/(10+100)=(((a)/(2))^(2)/(1000))`
for IInd equilibrium `K_(C)=((a)/(2)-10)^(2)/((20/(2)+(x)/(2))((20)/(2)-x)^(2))=((a)/(2)-10)^(2)/(15xx25)=((a)/(2)-10)^(2)/(275)`
`(K_(C) "for" I=K_(C) "for II")`
`((a)/(2))^(2)/(1000)=((a)/(2)-10)^(2)/(375)`
`((a)/(2))/(31.62)=((a)/(2)-10)/(19.365)`
`19.365xx(a)/(2)=31.62xx(a)/(2)-31.62xx10`
`9.6825xx0=15.81xx0-316.2`
`6.1275xxa=316.2`
`a=51.60M`
`K_(C)=(C)^(2)/((20)(20)^(2))=(51.6xx51.6)/(20xx20xx20)=332.82xx10^(-3)`
`K_(C)=0.333` approxly.