When 2.00mol each of $H_{2} (g)$ and $I_{2} (g)$ are reacted in a 1.00 L container at a certain temperature. 3.50mol of HI is present at equilibrium. Calculate the value of the equilibrium constant, $K_{c}$ for the reaction:
$H_{2} + I_{2} ⇌ 2 HI$

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$2.0 \times 10^{2}$

3.7

answer is D.

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Detailed Solution

The reaction can be given as,

$\underset{2 - x}{H_{2}} + \underset{2 - x}{I_{2}} ⇌ \underset{2 x = 3.50}{2 HI}$

For this the equilibrium constant,

$K_{C} = \frac{(3.5)^{2}}{(0.25)^{2}} K_{C} = 196$

So, the value of the equilibrium constant, $K_{c}$ for the reaction is $2.0 \times 10^{2}$ .

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