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Q. For the reactions and their equilibrium constants given below,
$CuCl _{4}^{2-}+ Br ^{-} \rightleftharpoons CuCl _{3} Br ^{2-}+ Cl ^{-} ; K_{1} $
$CuCl _{3} Br ^{2-}+ Br ^{-} \rightleftharpoons CuCl _{2} Br _{2}^{2-}+ Cl ^{-} ; K_{2} $
$CuCl _{2} Br _{2}^{2-}+ Br ^{-} \rightleftharpoons CuClBr _{3}^{2-}+ Cl ^{-} ; K_{3} $
$CuClBr _{3}^{2-}+ Br ^{-} \rightleftharpoons CuBr _{4}^{2-}+ Cl ^{-} ; K_{4}$
The equilibrium constant, $K$ for the reaction
$CuCl _{4}^{2-}+3 Br ^{-} \rightleftharpoons CuClBr _{3}^{2-}+3 Cl ^{-}$, is

KVPYKVPY 2009Equilibrium

Solution:

For the reactions,
$CuCl _{4}^{2-}+ Br ^{-} \rightleftharpoons CuCl _{3} Br ^{2-}+ Cl ^{-} ; K_{1}\,\,\, \ldots(i)$
$CuCl _{3} Br ^{2-}+ Br ^{-} \rightleftharpoons CuCl _{2} Br ^{2-}+ Cl ^{-} ; K_{2}\,\,\,...(ii)$
$CuCl _{2} Br _{2}^{2-}+ Br ^{-} \rightleftharpoons CuClBr _{3}^{2-}+ Cl ^{-} K_{3}\,\,\,\,...(iii)$
$CuClBr _{3}^{2-}+ Br ^{-} \rightleftharpoons CuBr _{4}^{2-}+ Cl ^{-} ; K_{4} \,\,\,\ldots (iv)$
To obtain reaction,
$CuCl _{4}^{2-}+3 Br ^{-} \rightleftharpoons CuClBr _{3}^{2-}+3 Cl ^{-}$
Add equations (i), (ii) and (iii)
$\therefore K=K_{1} \times K_{2} \times K_{3}$