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Question: Sulphide ion in alkaline solution react with solid sulfur to form polysulfide ions having formula \(...

Sulphide ion in alkaline solution react with solid sulfur to form polysulfide ions having formula S22−,S32−,S42−{S_2}^{2 - }, {S_3}^{2 - }, {S_4}^{2 - } and so on. The equilibrium constant for the formation of S22−{S_2}^{2-} is 12(K1)12({K_1}) and for the formation of S32−{S_3}^{2 - } is 132(K1)132({K_1}) , both from S and S2−{S^{2 - }} . What is the equilibrium constant for the formation of S32−{S_3}^{2 - } from S22−{S_2}^{2 - } and S?
A.11
B.12
C.132
D.None of these

Explanation

Solution

At first we will know what an equilibrium constant means. Once we define that we will look into the balanced equation of these and we will sort out the data given in the question. We will write the equations and then add the two equations to get the final equation. Now we will calculate the final equilibrium constant by multiplying the constant of those two equations.

Complete Step by step solutions:
Step1. Equilibrium constant is the value of its reaction quotient at the point where reaction can neither go backward or forward after ample time has elapsed or we can say that there is no possibility of further change. That point is called equilibrium. In general it is denoted by K.
Step2. Given in the question that S22−{S_2}^{2 - } is formed by S and S2−{S^{2 - }} . And the equilibrium constant is 12(K1)12({K_1}) .
S+S2−⇌S22−,K1=12S + {S^{2 - }} \rightleftharpoons {S_2}^{2 - },{K_1} = 12
Here the K1{K_1} is the equilibrium constant.
The second equation is formation of S32−{S_3}^{2 - } from S and S2−{S^{2 - }} , and the equilibrium constant is 132(K1)132({K_1})
2S+S2−⇌S23−,K2=1322S + {S^{2 - }} \rightleftharpoons {S_2}^{3 - },{K_2} = 132
The K2{K_2} is the equilibrium constant.
We need to find the equilibrium constant of
S+S22−⇌S32−,K=?S + {S_2}^{2 - } \rightleftharpoons {S_3}^{2 - },K = ?
Step3.
At first reverse the first equation. At reversing the equilibrium constant becomes the reciprocal of itself. Let it callK3{K_3}. And then the second equation is written as it is
S22−⇌S+S2−,K3=112{S_2}^{2 - } \rightleftharpoons S + {S^{2 - }},{K_3} = \dfrac{1}{{12}}
2S+S2−⇌S23−,K2=1322S + {S^{2 - }} \rightleftharpoons {S_2}^{3 - },{K_2} = 132
On addition of these reactions the S2−{S^{2 - }} ion cancels it and the final equation is
S+S22−⇌S32−,K=13212S + {S_2}^{2 - } \rightleftharpoons {S_3}^{2 - },K = \dfrac{{132}}{{12}}
Since the equations were added then their equilibrium constant gets multiplied.
Hence we get K=11K = 11 .

Note: The position of equilibrium constant is changed if we change the concentration of something present in the mixture. According to Le Chatelier’s Principle , The equilibrium will move in the direction which will try to undo the change made in the reaction.