Q. In a $10litre$ box $2.5mole$ hydroiodic acid is taken. After equilibrium $2HI\rightleftharpoonsH_{2}+I_{2}$ the concentration of $HI$ is found to be $0.1 \, \text{mol} \, \text{L}^{- 1}$ . The concentration of $\left[H_{2}\right]$ at equilibrium in $molL^{- 1}$ , is

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A

$2.4$

B

$0.15$

C

$1.5$

D

$7.5\times 10^{- 2}$

Solution:

$0.25-2$ at equilibrium
$0.25-2x=0.1$
$x=7.5\times 10^{- 2}$

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5. In a $10litre$ box $2.5mole$ hydroiodic acid is taken. After equilibrium $2HI\rightleftharpoonsH_{2}+I_{2}$ the concentration of $HI$ is found to be $0.1 \, \text{mol} \, \text{L}^{- 1}$ . The concentration of $\left[H_{2}\right]$ at equilibrium in $molL^{- 1}$ , is

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