Q

# Please,please help me Sulphide ions in alkaline solution react with solid sulphur to form polyvalent sulphide ions. The equikibrium constant for the formation of Cand S32-from S and S2-ions are 1.7 and 5.3 respectively. Equilibrium constant for the f

Sulphide ions in alkaline solution react with solid sulphur to form polyvalent sulphide ions. The equikibrium constant for the formation of C and S32- from S and S2- ions are 1.7 and 5.3 respectively. Equilibrium constant for the formation of S32-   from  S32-  and S is

• Option 1)

1.33

• Option 2)

3.11

• Option 3)

4.21

• Option 4)

1.63

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As we learnt in

Law of Chemical equilibrium -

At a given temperature, the product of concentration of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentration of the reactants raised to their individual stoichiometric coefficients has a constant value.

- wherein

$aA+bB\rightleftharpoons cC+dD$

$K_{c}=\frac{[C]^{c\:[D]^{d}}}{[A]^{a}\:[B]^{b}}$

$[A],\:[B],\:[C]\:[D]$

are equilibrium concentration

Given is

$S+S^{2-}\leftrightharpoons S{_2}^{2-}, K_{1}=1.7$

$2S+S^{2-}\leftrightharpoons S_{3}^{2-}, K_{2}=5.3$

So, the desired reaction is

$S_{2}^{2-}+S\leftrightharpoons S_{3}^{2-}, K_{3}$

$K_{1}=\frac{S_{2}^{2-}}{S^{2-}}$$K_{2}= \frac{S_{3}^{2-}}{S^{2-}}$

Clearly $K_{3}=\frac{K_{2}}{K_{1}}=\frac{[S_{3}^{2-}]}{[S_{2}^{2-}]}=\frac{53}{1.7}$

= 3.11

Option 1)

1.33

This option is incorrect

Option 2)

3.11

This option is correct

Option 3)

4.21

This option is incorrect

Option 4)

1.63

This option is incorrect

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