In an equilibrium A+B=C+D A & B are mized in a vessle at tempT. the initial conxntration of A was twice the initial concntration of B. After equillibrium has established ,concentrationof c was thrice the equilibrium concentration of B . calculate Kc??

dear priyanka dubey

according to the given information the equationbecomes as

2A+B=3C+D

FROM THIS Kc EQUALS TO [D][C]^3 DIVIDED BY [B][A]^2

APPROVE IF CORRECT PLEASE

CONSIDER THE INTIAL MOLES OF ''A'' BE ''a''  AND MOLES OF ''B'' BE ''b'' AND MOLES LOST BY ''A'' AND ''B'' BE X

                                                                     A  +  B ---------------->  C  +  D

 INITIAL MOLES:                                                    a         b                                       0        0

 EQUILIBRIUM MOLES :                                          a-x       b-x                                   0+x       0+x

   ACCORDING TO STICHIOMETERY AND SINCE THE COEFFICIENTS ARE 1 IF X MOLES ARE LOST BY ''B'' SO ''A'' WILL ALSO LOSE X MOLES

    IT DOESNOT MATTER IF YOU WORK IN MOLES OR CONCENTRATION BECAUSE IT IS SAMETHING  

   ACCORDING TO THE INFORMATION    a=2b  AND (0+x)=3(b-x)  SO FROM THIS EQUATION YOU GET THE VALUE OF x WHICH IS (3b)/4

  NOW K_C= [C]¹ [D]¹  DIVIDE BY [A]¹ [B]^{1          [    ] SIGN INDICATES MOLAR CONCENTRATION}

       K_{C= x * x divide by (a-x) (b-x)} 

     i didnt write the concentration which is moles upon volume beacuse volume term gets cancelled so now you have ''a'' in terms ''b'' and ''x'' in terms of ''b'' substitue those values in the equation and ''b'' term wil get cancelled and you will get your answer

      PLZ DO APPROVE THE ANSWER IF YOU ARE SATISFIED