The equilibrium constant for the disproportionation of HgCl2 into HgCl+ and HgCl3- is Given textHgCl+ + textCl- leftharpoons textHgCl2 ; textK1 = 3 × 106 HgCl2+Cl-leftharpoons HgCl3-;K2=9.0

Question

The equilibrium constant for the disproportionation of HgCl2 into HgCl+ and HgCl3- is Given textHgCl+ + textCl- leftharpoons textHgCl2 ; textK1 = 3 × 106 HgCl2+Cl-leftharpoons HgCl3-;K2=9.0 (A) 27× 106 (B) 3.3× 10- 7 (C) 3.3× 10- 6 (D) 3× 10- 6

Tardigrade Admin · Accepted Answer

K1=([. H g C l2 ]./[. H g C l+ ]. [. C l- ].);K2=([. H g C l3- ]./[. H g C l2 ]. [. C l- ].) So K3=([. H g C l+ ]. [. H g C l3- ]./[. H g C l2 ].2)=(K2/K1)=(9/3 × 1 06)=3× 10- 6

Q. The equilibrium constant for the disproportionation of $H g C l_{2}$ into $H g C l^{+}$ and $H g C l_{3}^{-}$ is

Given $HgCl^{+} + Cl^{-} ⇌ HgCl_{2}; K_{1} = 3 \times 1 0^{6}$

$H g C l_{2} + C l^{-} ⇌ H g C l_{3}^{-}; K_{2} = 9.0$

$27 \times 1 0^{6}$

$3.3 \times 1 0^{- 7}$

$3.3 \times 1 0^{- 6}$

$3 \times 1 0^{- 6}$

$K_{1} = \frac{[ H g C l _{2} ]}{[ H g C l ^{+} ] [ C l ^{-} ]}; K_{2} = \frac{[ H g C l _{3}^{-} ]}{[ H g C l _{2} ] [ C l ^{-} ]}$

So $K_{3} = \frac{[ H g C l ^{+} ] [ H g C l _{3}^{-} ]}{[ H g C l _{2} ] ^{2}} = \frac{K _{2}}{K _{1}} = \frac{9}{3 \times 1 0 ^{6}} = 3 \times 1 0^{- 6}$

Q. The equilibrium constant for the disproportionation of HgCl2​ into HgCl+ and HgCl3−​ is Given HgCl++Cl−⇌HgCl2​;K1​=3×106 HgCl2​+Cl−⇌HgCl3−​;K2​=9.0

27×106

3.3×10−7

3.3×10−6

3×10−6