Question

If $A\rightleftharpoons B\text{ (}{{\text{K}}_{c}}\text{=3)}$,$B\rightleftharpoons C\text{ (}{{\text{K}}_{c}}\text{=5)}$,$C\rightleftharpoons D\text{ (}{{\text{K}}_{c}}\text{=2)}$. The value of equilibrium constant for the above reaction are given, the value of equilibrium constant for $D\rightleftharpoons A$ will be:-
(a) 15
(b) 0.3
(c) 30
(d) 0.03

Answer 1

At equilibrium the rate of forward reaction is always equal to rate of backward reaction and ${{K}_{c}}$ is the equilibrium constant. We have been given the equilibrium constants of the reactions as $A\rightleftharpoons B\text{ (}{{\text{K}}_{c}}\text{=3)}$,$B\rightleftharpoons C\text{ (}{{\text{K}}_{c}}\text{=5)}$,$C\rightleftharpoons D\text{ (}{{\text{K}}_{c}}\text{=2)}$,from this we can find the
equilibrium constant of the reaction $A\rightleftharpoons D\text{ (}{{\text{K}}_{4}}\text{)}$and then, we can easily find the equilibrium constant of the reaction $D\rightleftharpoons A$. Now solve it.

Complete answer:
Fuser of all, let's discuss what is ${{K}_{P}}$and ${{K}_{c}}$. By the term ${{K}_{c}}$, we mean the equilibrium constant. When a reaction is at equilibrium , the rate of the forward reaction and the rate of backward reaction is always equal to each other through the reactions even at the equilibrium and the equilibrium constant tells us about the relationship between reactants and the products.
Consider general reaction at equilibrium as;
$aA+bB\rightleftharpoons cC+dD$
Then, the equilibrium constant of the reaction is as;
${{K}_{c}}=\dfrac{{{C}^{c}}{{D}^{d}}}{{{A}^{a}}{{B}^{b}}}$
Here, C and D are the products , A and B are the reactants and a, b , c and d are the number of moles of reactants and products respectively and${{K}_{c}}$is the equilibrium constant.
Now considering the statement;
We know that;
$A\rightleftharpoons B\text{ (}{{\text{K}}_{1}}\text{=3)}$(given)
$B\rightleftharpoons C\text{ (}{{\text{K}}_{2}}\text{=5)}$(given)
$C\rightleftharpoons D\text{ (}{{\text{K}}_{3}}\text{=2)}$(given)
Then;
$A\rightleftharpoons D\text{ (}{{\text{K}}_{4}}\text{)}$
Now, ${{K}_{4}}$ is the product of all the three equilibrium constants i.e. ${{K}_{1}}$,${{K}_{2}}$and ${{K}_{3}}$ as;
${{K}_{4}}$=${{K}_{1}}$ ${{K}_{2}}$ ${{K}_{3}}$
$\begin{aligned} & =3\times 5\times 2 \\\ & =30 \\\ \end{aligned}$
From this we can find the equilibrium constant for the reaction;
$D\rightleftharpoons A$
Suppose, the equilibrium constant for the reaction is; ${{K}_{5}}$
Then,
$\begin{aligned} & {{K}_{5}}=\dfrac{1}{{{K}_{4}}} \\\ & \text{ =}\dfrac{1}{30} \\\ & \text{ = 0}\text{.03} \\\ \end{aligned}$

Hence, option (d) is correct.

Note:
If equilibrium constant (${{K}_{c}}$) is greater than one, then the equilibrium shifts towards the products and if the equilibrium constant is smaller than 1 , then the equilibrium constant shifts towards the reactants.