Question

sulphide ion reacts with solid sulphur S^2- + S--> S2^2- ,K1= 10 and S^2- + 2S --> S3^2- ,K2=130 the equilibrium constant for the formation of S3^2- from S2^2- and sulphur is ?

Answer 1

The chemical equilibria representing the reaction of sulfide ion with solid sulfur is,

Equilibrium -1 :$S^{2-}(aq)+S(s)<===> S_{2}^{2-}(aq)       K_{1}=10$

Equilibrium -2 :$S^{2-}(aq)+2S(s)<==> S_{3}^{2-}   K_{2} = 130$

The above equilibria can be reaaranged to give the new equilibrium reaction for the formation of $S_{3}^{2-}$ from $S_{2}^{2-}$ and sulfur,

Reversing equilibrium -1: $S_{2}^{2-}<==> S^{2-}+ S     K = \frac{1}{K_{1} } =\frac{1}{10} = 0.1$

Adding equilbrium 2: $S^{2-} + 2S<==> S_{3}^{2-}     K2 = 130$

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Net equilibrium : $S_{2}^{2-} + S <=>S_{3}^{2-}    ==> K_{net} = \frac{1}{K_{1} }  * (K_{2} )$

Therefore the equilibrium constant for the given equilibrium = $\frac{1}{10}*130=13$

Answer 2

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