Question:
If $1.0$ mole of $I_{2}$ is introduced into $1.0$ litre flask at $1000\, K$, at equilibrium $\left(K_{c}=10^{-6}\right)$, which one is correct?
If $1.0$ mole of $I_{2}$ is introduced into $1.0$ litre flask at $1000\, K$, at equilibrium $\left(K_{c}=10^{-6}\right)$, which one is correct?
Updated On: Dec 15, 2025
- $\left[ I _{2}( g )\right]>\left[ I ^{-}( g )\right]$
- $\left[ I _{2}( g )\right]
- $\left[ I _{2}( g )\right]=\left[ I ^{-}( g )\right]$
- $\left[ I _{2}( g )\right]=\frac{1}{2}\left[ I ^{-}( g )\right]$
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The Correct Option is B
Solution and Explanation
${ $\underset{\text{ 1 -x}}{{I_{2} }}$
<=> $\underset{\text{ 2x }}{{ 2 I^{-} }}$
}$
$K_c = \frac{(2x)^2}{(1-x)} = 10^{-6}$
It shows that (1 - x) < 2x
<=> $\underset{\text{ 2x }}{{ 2 I^{-} }}$
}$
$K_c = \frac{(2x)^2}{(1-x)} = 10^{-6}$
It shows that (1 - x) < 2x
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