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Introduction

The equilibri...

Question

The equilibrium constant for the reaction:

$H_{2} (g) + I_{2} (g) \Leftrightarrow 2 H I (g)$ is 64, at certain temperature.The equilibrium concentrations of $H_{2}$ and HI are 2 and 16 $m o l l i t^{- 1}$ respectively. The equilibrium concentration (in $m o l L i t^{- 1}$ ) of $I_{2}$ is ?

A

16

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B

4

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C

8

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D

2

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Solution

The correct option is C 2
The equilibrium reaction is $H_{2} (g) + I_{2} (g) \Leftrightarrow 2 H I (g)$ .
The expression for the equilibrium constant is $K = \frac{[H I]^{2}}{[H_{2}] [I_{2}]}$ .
But $K = 64, [H I] = 16 M, [H_{2}] = 2 M$ .
Substitute values in the above expression.
$64 = \frac{(16)^{2}}{2 \times [I_{2}]}$ .
$[I_{2}] = 2 M$ .

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