Hl was heated in a closed tube at 440∘c till equilibrium is obtaine

Question

Hl was heated in a closed tube at

440^{\circ} c

till equilibrium is obtained. At this temperature

22 %

of Hl was dissociated. The equilibrium constant for this dissociation will be

Options:

A . 0.282

B . 0.0796

C . 0.0199

D . 1.99

Answer: Option C
:
C

2 H I ⇋ H_{2} + l_{2}

\begin{matrix} I n i t i a l n o o f m o l e s - & 2 m o l e s & 0 & 0 a t e q u i l i b r i u m & 2 m o l e s - 0.44 & 0.22 & 0.22 \end{matrix}

K_{c} = \frac{[H_{2}] [l_{2}]}{[H I]^{2}} = \frac{0.22 \times 0.22}{{1.56}^{2}} = 0.0199

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