$A ⇌ C + D K_{c} = 5 \times 10^{- 6} B + C ⇌ E K_{c} = 5 \times 10^{- 6} M^{- 1}$
2 moles of A and 5 moles of B are taken in a 2 L vessel. The equilibrium concentration of C.

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Solution

$A ⇌ C + D$ $K_{c} = 5 \times 10^{- 6} = \frac{[C] [D]}{A}$
$B + C ⇌ E$ $K_{c} = 5 \times 10^{- 6} = \frac{[E]}{[B] [C]}$
$K_{c}^{2} ⟹ 25 \times 10^{- 12} = \frac{[D] [E]}{[A] [B]}$ $= \frac{[D] [E]}{[\frac{2}{2}] [\frac{5}{2}]}$
$⟹ [D] [E] = \frac{25}{2} \times 10^{- 12}$
Conc. of C $⟹ 5 \times 10^{- 6}$ = $\frac{[C]}{1}$
Hence $(C)$ is $5 \times 10^{- 6}$

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A⇌C+DKc=5×10−6B+C⇌EKc=5×10−6M−12 moles of A and 5 moles of B are taken in a 2 L vessel. The equilibrium concentration of C.

A⇌C+D Kc=5×10−6=[C][D]AB+C⇌E Kc=5×10−6=[E][B][C]K2c⟹25×10−12=[D][E][A][B] =[D][E][22][52] ⟹[D][E]=252×10−12Conc. of C ⟹5×10−6 =[C]1Hence (C) is 5×10−6

$A ⇌ C + D K_{c} = 5 \times 10^{- 6} B + C ⇌ E K_{c} = 5 \times 10^{- 6} M^{- 1}$
2 moles of A and 5 moles of B are taken in a 2 L vessel. The equilibrium concentration of C.