If $0.1\;{\rm{mole}}$ of ${{\rm{I}}_2}$ is introduced into $1.0\;{\rm{litre}}$ flask at $1000\;{\rm{K}},$ at equilibrium ${K_c} = {10^{ - 6}}$, which one is correct?

$\left[ {{{\rm{I}}_2}\left( {\rm{g}} \right)} \right] > \left[ {I\left( g \right)} \right]$

$\left[ {{{\rm{I}}_2}\left( {\rm{g}} \right)} \right] < \left[ {I\left( g \right)} \right]$

$\left[ {{{\rm{I}}_2}\left( {\rm{g}} \right)} \right] = \left[ {{\rm{I}}\left( {\rm{g}} \right)} \right]$

$\left[ {{{\rm{I}}_2}\left( {\rm{g}} \right)} \right] = \frac{1}{2}\left[ {{\rm{I}}\left( {\rm{g}} \right)} \right]$

NEET UG CHEMISTRY MCQ: Equilibrium

1 Easy

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Analyzing +4 | -1

0.1\;\mathrm{mole}

{\mathrm{I}_2}

is introduced into

1.0\;\mathrm{litre}

flask at

1000\;\mathrm{K},

at equilibrium

{K_c} = {10^{ - 6}}

, which one is correct?

View answer and explanation

Correct option: A

\left[ {{\mathrm{I}_2}\left( \mathrm{g} \right)} \right] > \left[ {I\left( g \right)} \right]

Option A ([I2(g)] > [I(g)]) is correct because the equilibrium constant is very small, indicating that the dissociation of I2 is minimal.

Practice All Chapter MCQs →

Mg{(HC{O_3})_2}

MgC{O_3}

Mg{(OH)_2}

MgC{O_3}

K_{\mathrm{P}}

\mathrm{CaCO}_3

4.5 \times 10^{-2}

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}),

K_{\mathrm{P}}

2.0 \mathrm{~atm}.

\mathrm{CaO}

\mathrm{CaCO}_3

+

\leftrightharpoons

\mathrm{PC}{\mathrm{l}_5}\leftrightharpoons\mathrm{PC}{\mathrm{l}_3} + \mathrm{C}{\mathrm{l}_2}

{K_{\mathrm{eq}}} = 0.41

\mathrm{p}{K_a}