InterviewSolution
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InterviewSolution
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For a reaction `A+BhArr2C` the equilibrium concentration of `(A)` and `(B)` are `20` mole//L when volume is doubled the new equilibrium concentration of `(A)` was found to be `15` mol//L thenA. Ratio of concentration of `A "and" B` at new equilibrium is 3//4B. Value of equilibrium constant for both cases are remain sameC. Concentration of `C` at new equilibrium become halfD. Equilibrium concentration of `C` at new equilibrium `(10sqrt20)/(sqrt20-sqrt15)` |
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Answer» Correct Answer - A::B::D `{:(,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. |
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