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 (KCat 440 $^{\circ}$ C is 0.02)

3.38%

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4.38%

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5.38%

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6.38%

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Solution

The correct option is C
5.38%

$\begin{matrix} H_{2} + & I_{2} ⇌ & 2 H I I n i t i a l m o l e s & 2 & 3 & 0 A t E q b^{'} & \frac{2 - x}{V} & \frac{3 - x}{V} & \frac{2 X}{V} \end{matrix}$

$K_{C} = \frac{4 x^{2}}{(2 - x) (3 - x)} = 0.02$

$199 x^{2} + 5 x - 6 = 0$

x = 0.1615

Out of 3 moles , 0.1615 moles of $I_{2}$ is converted into HI\)

Percentage of $I_{2}$ converted to

$H I = \frac{0.1615 \times 100}{3} = 5.38$

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