$K_{p}$ for the reaction ${N H}_{4} I (s) ⇌ {N H}_{3} (g) + H I (g)$ is $1 / 4$ at a particular temperature . Above equilibrium is established by taking 4 moles of ${N H}_{4} I$ in 100 litre container, then moles of ${N H}_{4} I (s)$ left in the container at equilibrium is: $[T a k e R = 1 / 12 L t . a t m {m o l}^{- 1} K^{- 1}]$

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Solution

According to the problem:
Let us consider the relation of Kp and Kc is,
$K_{p} = K_{c} = {(R T)}^{Δ}$
where,
$Δ n = 2$
$R = 0.082$
$T = 298$
putting the given values,
$\frac{1}{4} = K_{c} {(0.082 \times 298)}^{2}$
implies that
$K_{c} = 0.00042$
For reaction
$N H_{4} I (s) N H_{3} (g) + H I (g)$
$0.04 00$
$- x + x + x$
$0.04 - x x x$
$k_{c} = \frac{[N H_{3}] [H I]}{[N H_{4} I]}$
implies that
$0.0042 = \frac{x^{2}}{(0.04 - x)}$
$x^{2} + 0.00042 x - 0.000016 g = 0$
$x = 0.004$
At equilibrium, concentration of $0.04 - 0.004 = 0.036 M$
left= $0.04$ - $0.004$ = $0.036$ M.
Moles of $N H_{4} I$ left= $con c e n t r a t i o n \times v o l u m e$
$0.036 \times 100 = 3.6 m o l e s$
Hence
$3.6$ moles of $N H_{4} I$ are left.

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Kp for the reaction NH4I(s)⇌NH3(g)+HI(g) is 1/4 at a particular temperature . Above equilibrium is established by taking 4 moles of NH4I in 100 litre container, then moles of NH4I(s) left in the container at equilibrium is:[TakeR=1/12Lt.atmmol−1K−1]