Question

$A\rightleftharpoons B(K_c=3),B\rightleftharpoons C(K_c=5),$
$C\rightleftharpoons D(K_c=2),$
From the given values of equilibrium constant for the above reactions, the value of equilibrium constant for $D\rightleftharpoons A$ will be.

• A. 15
• B. 0.3
• C. 30
• D. 0.03

Answer 1

The correct option is D 0.03

For equlibrium, $A\rightleftharpoons B{K}_c=\frac{\left[B\right]}{\left[A\right]}$
$\Rightarrow \left[B\right]=3\left[A\right]$ ...(i)

For equlibrium, $B\rightleftharpoons C{K}_c=\frac{\left[C\right]}{\left[B\right]}$
$\Rightarrow 5=\frac{\left[C\right]}{\left[B\right]}$
from eq. (i) substituting for [B]$\Rightarrow 5=\frac{\left[C\right]}{3\left[A\right]}$
$\Rightarrow \left[A\right]=\frac{\left[C\right]}{15},\left[C\right]=15\left[A\right]$ ...(ii)

For equlibrium, $C\rightleftharpoons D$
$K_c=\frac{\left[D\right]}{\left[C\right]}$
$\Rightarrow \left[D\right]=2\left[C\right]=30\left[A\right]$ ...(iii)

For equlibrium, $D\rightleftharpoons A$
$K_c=\frac{\left[A\right]}{\left[D\right]}=\frac{\left[A\right]}{30\left[A\right]}=\frac{1}{30}=0.03$

Alternate solution: Sum of the three equilibriums result in an equilibrium
$A\rightleftharpoons D$ , whose equilibrium constant will be the multiple of the three equilibrium equilibrium constants
$=3\times 5\times 2=30.$
Equilibrium constant of the reverse equilibrium, $D\rightleftharpoons A$ will be the reciprocal $=\frac{1}{30}=0.03$