Question

Sulfide ion in alkaline solution reacts with solid sulfur to form polysulfide ions having formulae \[{\text{S}}_2^{2 - },{\text{S}}_3^{2 - },{\text{S}}_4^{2 - }\]and so on. The equilibrium constant for the formation of \[{\text{S}}_2^{2 - }\]​ is 12 \[\left( {{{\text{K}}_1}} \right)\] & for the formation of \[{\text{S}}_3^{2 - }\]is 132 \[\left( {{{\text{K}}_1}} \right)\], both from \[{\text{S and }}{{\text{S}}^{2 - }}\] .What is the equilibrium constant for the formation of from \[{\text{S}}_3^{2 - }\]​ from \[{\text{S and S}}_2^{2 - }\]?
A.11
B.12
C.132
D.None of these

Answer 1

Hint: We need to write the equilibrium constant for the formation of the respective ions whose equilibrium constant is given to us. By subtracting the two equations we will get the value of the final equation whose equilibrium constant we have to calculate.

Complete step by step answer:
Lastly, first write the equilibrium reaction for the formation of \[{\text{S}}_2^{2 - }{\text{ and S}}_3^{2 - }\]respectively.
The formation of \[{\text{S}}_2^{2 - }\] ions with the value of equilibrium constant \[{\text{K}} = 12\left( {{{\text{K}}_1}} \right)\]
\[{\text{S }} + {\text{ }}{{\text{S}}^{2 - }} \rightleftharpoons {\text{S}}_2^{2 - }{\text{ }}\left( {\text{i}} \right)\]
The formation of \[{\text{S}}_3^{2 - }\] ions with the value of equilibrium constant
\[{\text{2S }} + {\text{ }}{{\text{S}}^{2 - }} \rightleftharpoons {\text{S}}_3^{2 - }{\text{ }}\left( {{\text{ii}}} \right)\]
If we subtract equation 2 from equation 1 we will get the equation as follow:
\[{\text{S }} \rightleftharpoons {\text{ S}}_3^{2 - }{\text{ }} - {\text{ S}}_2^{2 - }{\text{ }}\]
Rearranging the above equation we will get the final equation of formation of \[{\text{S}}_3^{2 - }\] as follow:
\[{\text{S }} + {\text{ S}}_2^{2 - }{\text{ }} \rightleftharpoons {\text{ S}}_3^{2 - }\]
This is the equation that we have been asked in the question.
The equilibrium constant for the reaction formed by subtracting two reactions is the ratio of the equilibrium constant of two the reactions which it is formed. Let the equilibrium constant of the final reaction is \[{{\text{K}}^{''}}\]. So according to the above property of equilibrium constant we will get:
\[ \Rightarrow {{\text{K}}^{''}} = \dfrac{{132{\text{ }}{{\text{K}}_1}}}{{12{\text{ }}{{\text{K}}_1}}} = 11{\text{ }}{{\text{K}}_1}\]

Hence the correct option is A.

Additional information:
Some of the other properties of equilibrium constant are if two reactions are added up then their equilibrium constant gets multiplied to each other. If the reaction is reversed then the equilibrium constant is also reversed. When we multiply reaction by any constant then the equilibrium constant is raised to the power equals to that constant.

Note:
Equilibrium constant is a constant which depends upon the nature and the temperature of reaction. For a fixed reaction if temperature changes then the rate constant also changes.