For a reaction A+BhArr2C the equilibrium concentration of (A) and (B) are 20 mole//Lwhen volume is doubled the new equilibrium concentration of (A) was found to be 15 mol//L then

For a reaction A+BhArr2C the equilibrium concentration of (A) and (B)

1.	For a reaction A+BhArr2C the equilibrium concentration of (A) and (B) are 20 mole//Lwhen volume is doubled the new equilibrium concentration of (A) was found to be 15 mol//L then
Answer» Ratio of CONCENTRATION of `A "and" B` at new EQUILIBRIUM is 3//4 Value of equilibrium constant for both cases are remain same Concentration of `C` at new equilibrium become half Equilibrium concentration of `C` at new equilibrium `(10sqrt20)/(sqrt20-sqrt15)` Solution :`{:(,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.