Sulphide ion (S2-) reacts with solid sulphur forming S2-2 and S2-3 with formation constants 12 and 132 respectively. Formation constant of S2-3 from sulphur and S2-2 is :

Question

Sulphide ion (S2-) reacts with solid sulphur forming S2-2 and S2-3 with formation constants 12 and 132 respectively. Formation constant of S2-3 from sulphur and S2-2 is : (A) 12 (B) 132 (C) 132 x 12 (D) 11

Tardigrade Admin · Accepted Answer

(l) S 2-+ S leftharpoons S 22- ; K 1=12 (II) S 2-+ S leftharpoons S 32- ; K 1=132 (III) S 22-+ S leftharpoons S 32- ; K = ? To get eqn. (III), subtract (I) from (II) K =(K2/K1)=(132/12)=11

Q. Sulphide ion $(S^{2 -})$ reacts with solid sulphur forming $S_{2}^{2 -}$ and $S_{3}^{2 -}$ with formation constants 12 and 132 respectively. Formation constant of $S_{3}^{2 -}$ from sulphur and $S_{2}^{2 -}$ is :

12

132

132 x 12

11

(l) $S^{2 -} + S ⇌ S_{2}^{2 -}; K_{1} = 12$
(II) $S^{2 -} + S ⇌ S_{3}^{2 -}; K_{1} = 132$
(III) $S_{2}^{2 -} + S ⇌ S_{3}^{2 -}; K =$ ?
To get eqn. (III), subtract (I) from (II)
$K = \frac{K _{2}}{K _{1}} = \frac{132}{12} = 11$

Q. Sulphide ion (S2−) reacts with solid sulphur forming S22−​ and S32−​ with formation constants 12 and 132 respectively. Formation constant of S32−​ from sulphur and S22−​ is :

12

132

132 x 12

11

(l) S2−+S⇌S22−​;K1​=12 (II) S2−+S⇌S32−​;K1​=132 (III) S22−​+S⇌S32−​;K= ? To get eqn. (III), subtract (I) from (II) K=K1​K2​​=12132​=11

Q. Sulphide ion $(S^{2 -})$ reacts with solid sulphur forming $S_{2}^{2 -}$ and $S_{3}^{2 -}$ with formation constants 12 and 132 respectively. Formation constant of $S_{3}^{2 -}$ from sulphur and $S_{2}^{2 -}$ is :

(l) $S^{2 -} + S ⇌ S_{2}^{2 -}; K_{1} = 12$
(II) $S^{2 -} + S ⇌ S_{3}^{2 -}; K_{1} = 132$
(III) $S_{2}^{2 -} + S ⇌ S_{3}^{2 -}; K =$ ?
To get eqn. (III), subtract (I) from (II)
$K = \frac{K _{2}}{K _{1}} = \frac{132}{12} = 11$