Question

If $A\rightleftharpoons B({{K}_{c}}=3),B\rightleftharpoons C({{K}_{c}}=5),C\rightleftharpoons D({{K}_{c}}=2)$
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.$0\cdot 3$
C.30
D.$0\cdot 03$

Answer 1

Hint:To find the equilibrium constant for a reaction which is sum of two or more reactions we use Multi-Equilibrium Rule which states that the equilibrium constant for sum of two or more reactions is the product of equilibrium constant for each step. So, for reaction going from A to D we can have the product of the middle reactions involved and for D to A we can reciprocal the equilibrium constant of A to D.
Formula Used:
   ${{k}_{4}}={{k}_{1}}{{k}_{2}}{{k}_{3}}$
${{k}_{4}}$ is equilibrium constant for reaction $A\rightleftharpoons D$
${{k}_{1}}$ is equilibrium constant for reaction$A\rightleftharpoons B$
${{k}_{2}}$ is equilibrium constant for reaction$B\rightleftharpoons C$
${{k}_{3}}$ is equilibrium constant for reaction $C\rightleftharpoons D$

Complete step by step answer:
Given, $A\rightleftharpoons B({{K}_{c}}=3),B\rightleftharpoons C({{K}_{c}}=5),C\rightleftharpoons D({{K}_{c}}=2)$
The equilibrium constant for reaction $A\rightleftharpoons D$ $=3\times 5\times 2$
Therefore, ${{k}_{4}}$$=30$
To find the equilibrium constant for the reaction$D\rightleftharpoons A$, reverse the equilibrium constant for
$A\rightleftharpoons D$. So, equilibrium constant for the reaction $D\rightleftharpoons A$$=\dfrac{1}{{{k}_{4}}}$
   $\begin{align}
  & =\dfrac{1}{30} \\
 & =0.03 \\
\end{align}$

So, the correct choice is (D).

Additional Information: The equilibrium constant, K, expresses the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit. At equilibrium, the rate of the forward and reverse reaction is equal, which is demonstrated by the arrows. The equilibrium constant, however, gives the ratio of the units (pressure or concentration) of the products to the reactants when the reaction is at equilibrium.
The numerical value of equilibrium constant is obtained by letting a single reaction proceed to equilibrium and then measuring the concentrations of each substance involved in that reaction. The ratio of the product concentrations to reactant concentrations is calculated. Because the concentrations are measured at equilibrium, the equilibrium constant remains the same for a given reaction independent of initial concentrations.

Note:
The final reaction can be the sum of two or more reactions but to find its equilibrium constant one has to take the products of equilibrium constants involved in individual reactions. Remember the product of equilibrium constants not the sum. The equilibrium constant can be reversed i.e. we can take its inverse if the reaction is reversed.
If $K>1$ then equilibrium favors products
If $K<1$ then equilibrium favors the reactants