Question

If ${K_1}$ and ${K_2}$ are equilibrium constants of the equilibriums (A) and (B) respectively, then what is the relationship between the two constants?
$(A)\,\,S{O_2}(g)\, + \,\dfrac{1}{2}{O_2}(g)\, \rightleftharpoons \,S{O_3}\,(g),\,{K_1}$
$(B)\,\,2S{O_3}(g)\,\, \rightleftharpoons \,2S{O_2}\,(g)\, + \,{O_{2\,}}(g),\,\,{K_2}$
A) ${K_1} = \,{K_2}$
B) ${K_1} = \,\dfrac{1}{{{K_2}}}$
C) ${K_2} = \,K_1^2$
D) $K_1^2 = \,\dfrac{1}{{{K_2}}}$

Answer 1

Hint: Here we have two equilibrium reactions having equilibrium constant as ${K_1}$ and ${K_2}$ now these reactions may be connected to each other. We can establish a relation between the two constants by changing the equation too. By reversing the reaction the constant gets reciprocal of the original one, on multiplying the equation by any number the constant gets it same in power.

Complete step-by-step answer:
Let’s start with the first reaction in which sulphur dioxide reacts with another half of oxygen to form sulphur trioxide see the stoichiometry of reaction and check whether the reaction is balanced or not.
Yes, we have both the equations balanced now we take first equal and reverse it, we will try to make any one equation just similar to another either by multiplication or dividing with any number.
$S{O_2}(g)\, + \,\dfrac{1}{2}{O_2}(g)\, \rightleftharpoons \,S{O_3}\,(g),\,{K_1} - - - - - - - - - - (1)$
If we reverse it, the equilibrium constant get reciprocal $S{O_3}\,(g) \rightleftharpoons \,S{O_2}(g)\, + \,\dfrac{1}{2}{O_2}(g),\,\dfrac{1}{{{K_1}}} - - - - - - - - - - (2)$
Now if we compare the equation no. $(2)\,with\,(B)$ The reactants of both the equations are the same and so as the product. We have to make the stoichiometric coefficient equal this can be achieved by multiplying the equation $(2)$ by $2$ , the equation changes as such the constant also changes take it no. $(3)$
$2S{O_3}\,(g) \rightleftharpoons \,2S{O_2}(g)\, + \,{O_2}(g),\,\,{\left( {\dfrac{1}{{{K_1}}}} \right)^2} - - - - - - - - - - (3)$
This equation is now similar to the equation (B) so, see the relation of equilibrium constant which is ${K_2} = \left( {\dfrac{1}{{{K_1}^2}}} \right)$
This is not any of the option let’s check by solving it properly, we get $K_1^2 = \,\dfrac{1}{{{K_2}}}$

Hence the correct answer is opinion ‘D’.

Note: While solving the two equilibrium equations and for finding out the relation between them, you just have to make any one of the equations by multiplying, dividing etc. Then the equation looks similar to the other equation, simultaneously you will get your required relation. Keep in mind that if the reaction gets reversed, the rate constant gets reciprocal. If you are multiplying the equation, the same changes you will see in the power of equilibrium constant.