40% of a mixture of 0.2 mole of ${{\rm{N}}_2}$ and 0.6 mole of ${{\rm{H}}_2}$ react to give ${\rm{N}}{{\rm{H}}_3}$ according to the equation,${{\rm{N}}_2}\left( {\rm{g}} \right) + 3{{\rm{H}}_2}\left( {\rm{g}} \right)\leftrightharpoons2{\rm{N}}{{\rm{H}}_3}\left( {\rm{g}} \right)$ at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is :

NEET UG CHEMISTRY MCQ: Equilibrium

1 Easy

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Application +4 | -1

40% of a mixture of 0.2 mole of

{\mathrm{N}_2}

and 0.6 mole of

{\mathrm{H}_2}

react to give

\mathrm{N}{\mathrm{H}_3}

according to the equation,

{\mathrm{N}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right)

at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is :

View answer and explanation

Correct option: A

4\mathrm{ }:\mathrm{ }5

Option A (

4\mathrm{ }:\mathrm{ }5

) is correct because the ratio of final to initial volume is calculated based on the stoichiometry of the reaction and the percentage conversion.

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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}