Question

Sulphide ion react with solid Sulphur
${ S }_{ (aq) }^{ 2- }+{ S }_{ (s) }\rightleftharpoons { S }_{ 2(aq) }^{ 2- }\quad ;\quad { K }_{ 1 }=10$
$S_{(aq)}^{2-}+2{{S}_{(s)}}\rightleftharpoons S_{3(aq)}^{2-}\quad ;\quad K_{1}^{\prime }=130$
The equilibrium constant for the formation of ${ S }_{ 3(aq) }^{ 2- }$from ${ S }_{ 2(aq) }^{ 2- }$ and sulphur is
(A) 10
(B) 13
(C) 130
(D) 1300

Answer 1

Hint: For the given equations we can formulate the equilibrium constant by dividing the concentration of products by concentration of reactants. After formulating equilibrium constant for the first two reactions, we can compare these equations to get the equilibrium constant of the desired product.

Complete step by step answer:
   - Let's start with the idea of equilibrium constant. In a chemical reaction at equilibrium the concentration of each reactant and product is related. Their concentration depends on each other.
  - Equilibrium constant ( ${{K}_{eq}}$) can be formulated as amounts of products divided by amounts of reactants and each amount (either pressure or concentration) is raised to the power of its coefficient in the balanced chemical equation.


The first equation is given as
       ${ S }_{ (aq) }^{ 2- }+{ S }_{ (s) }\rightleftharpoons { S }_{ 2(aq) }^{ 2- }\quad ;\quad { K }_{ 1 }=10$
 From this we can formulate the equilibrium constant as follows
${{K}_{1}}=\dfrac{\left[ S_{2}^{2-} \right]}{\left[ {{S}^{2-}} \right]\left[ S \right]}=10$ →Equation (1)
The second equation is given as
$S_{(aq)}^{2-}+2{{S}_{(s)}}\rightleftharpoons S_{3(aq)}^{2-}\quad ;\quad K_{1}^{\prime }=130$

 We can find the equilibrium constant for this reaction as follows
  $K_{1}^{\prime }=\dfrac{\left[ S_{3}^{2-} \right]}{\left[ {{S}^{2-}} \right]{{\left[ S \right]}^{2}}}=130$ → Equation (2)
 We are asked to find the equilibrium constant for the reaction in which ${ S }_{ 3(aq) }^{ 2- }$ is formed from the reaction between Sulphur and ${ S }_{ 2(aq) }^{ 2- }$. Hence, we can write this reaction as follows
 ${ S }_{ 2 }^{ 2- }+S\rightleftharpoons { S }_{ 3 }^{ 2- }$

From this, we can formulate the equilibrium constant. Let's take this value as ${ K }_{ 2 }$. ${ K }_{ 2 }$ can be written as
  ${ K }_{ 2 }=\cfrac { \left[ { S }_{ 3 }^{ 2- } \right] }{ \left[ S \right] \left[ { S }_{ 2 }^{ 2- } \right] }$ → Equation (3)
 From comparing equation (3), (2) and (1), we can write the equilibrium constant ${ K }_{ 2 }$ as
  ${ K }_{ 2 }=\cfrac { { K }_{ 1 }^{ \prime } }{ { K }_{ 1 } }$
         = $\dfrac { 130 }{ 10 }$
         =13
So, the correct answer is “Option B”.

Note: The value of the equilibrium constant will allow us to determine the direction a reaction will proceed to achieve equilibrium and the ratios of the concentrations of reactants and products when equilibrium is reached.
If the equilibrium constant is a large number, it means that the equilibrium concentration of the products is large and, in this case, the reaction as written will proceed to the right.
If equilibrium constant is a small number, it means that the equilibrium concentration of the reactants is large and, in this case, the reaction is written as it will proceed to the left.