A mixture of $H_{2}$ and $I_{2}$ in molecular proportion of 2 : 3 was heated at $440^{\circ} C$ till the reaction $H_{2} + I_{2}$ $⇌$ 2HI reached equilibrium state. Calculate the percentage of iodine converted into HI $K_{c} a t 440$ $^{\circ}$ C is 0.02)

Q. 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

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Equivalent amounts of H2 and I2 are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen can be converted to HI the KC at this temperature is:

The correct option is B 64 H2+I2 ⇌ 2HIAt initial 1 1 0At equal (1−0.8) (1−0.8) 2×0.8 =0.2 =0.2 =1.6∵ Kc=[HI]2[H2][I2]∴ Kc=1.6×1.60.2×0.2⇒Kc=64.

Equivalent amounts of $H_{2}$ and $I_{2}$ are heated in a closed vessel till equilibrium is obtained. If $80 %$ of the hydrogen can be converted to HI the $K_{C}$ at this temperature is:

The correct option is B $64$
$H_{2} + I_{2} ⇌ 2 H I$
At initial $1$ $1$ $0$
At equal $(1 - 0.8)$ $(1 - 0.8)$ $2 \times 0.8$
$= 0.2$ $= 0.2$ $= 1.6$

$∵$ $K_{c} = \frac{[H I]^{2}}{[H_{2}] [I_{2}]}$

$∴$ $K_{c} = \frac{1.6 \times 1.6}{0.2 \times 0.2}$

$\Rightarrow K_{c} = 64$ .