For the gaseous reaction, ${{\rm{C}}_2}{{\rm{H}}_4} + {{\rm{H}}_2}\leftrightharpoons{{\rm{C}}_2}{{\rm{H}}_6},\;\Delta H = - 130\;{\rm{kJ}}\;{\rm{ mo}}{{\rm{l}}^{ - 1}}$ carried in a closed vessel, the equilibrium concentration of the ${{\rm{C}}_2}{{\rm{H}}_6}$ can definitely be increased by

Increasing temperature and decreasing pressure

Increasing temperature and pressure both

Decreasing temperature and pressure only

NEET UG CHEMISTRY MCQ: Equilibrium

1 Easy

1 of 1

Application +4 | -1

For the gaseous reaction,

{\mathrm{C}_2}{\mathrm{H}_4} + {\mathrm{H}_2}\leftrightharpoons{\mathrm{C}_2}{\mathrm{H}_6},\;\Delta H = - 130\;\mathrm{kJ}\;\mathrm{ mo}{\mathrm{l}^{ - 1}}

carried in a closed vessel, the equilibrium concentration of the

{\mathrm{C}_2}{\mathrm{H}_6}

can definitely be increased by

View answer and explanation

Correct option: B

Decreasing temperature and increasing pressure

Option B (Decreasing temperature and increasing pressure) is correct because the reaction is exothermic (

\Delta H < 0

) and involves a decrease in the number of moles of gas, both of which favor the formation of C2H6.

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}