The equilibrium, ${{\rm{P}}_4}\left( s \right) + 6{\rm{C}}{{\rm{l}}_2}\left( {\rm{g}} \right)\;\leftrightharpoons\;4{\rm{PC}}{{\rm{l}}_3}\left( {\rm{g}} \right)$ attained by mixing equal moles of ${{\rm{P}}_4}$ and ${\rm{C}}{{\rm{l}}_2}$ in a evacuated vessel. Then, at equilibrium,

$\left[ {{{\rm{P}}_4}} \right] > \left[ {{\rm{C}}{{\rm{l}}_2}} \right]$

$\left[ {{\rm{C}}{{\rm{l}}_2}} \right] > \left[ {{\rm{PC}}{{\rm{l}}_3}} \right]$

$\left[ {{\rm{C}}{{\rm{l}}_2}} \right] > \left[ {{{\rm{P}}_4}} \right]$

$\left[ {{\rm{PC}}{{\rm{l}}_3}} \right] > \left[ {{{\rm{P}}_4}} \right]$

Back to Equilibrium

JEE CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1

The equilibrium,

{\mathrm{P}_4}\left( s \right) + 6\mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)\;\leftrightharpoons\;4\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right)

attained by mixing equal moles of

{\mathrm{P}_4}

and

\mathrm{C}{\mathrm{l}_2}

in a evacuated vessel. Then, at equilibrium,

View answer and explanation

Correct option: C

\left[ {{\mathrm{P}_4}} \right] > \left[ {\mathrm{C}{\mathrm{l}_2}} \right]

The correct answer is C ([P₄] > [Cl₂]) because, based on the stoichiometry of the reaction and the initial equal moles, more P₄ remains at equilibrium than Cl₂.

More Questions from Equilibrium

Practice All Chapter MCQs →

The indicator used in titrating oxalic acid with caustic soda solution is:

\mathrm{K}_1

and

\mathrm{K}_2

are equilibrium constant for reactions (i) and (ii)

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

(i)

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

(ii)

Then,

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

are the respective ionisation constants for the following reactions

\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}

The correct relationship between

K{a_1},\,\,K{a_2}\,,K{a_3}

is The

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

solution is (For

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

) Reaction

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

In equilibrium condition pressure of

\text{O}_2

depends on: A : At equilibrium concentration of the reactant and product does not change with time for a chemical reaction.
R: The rate of reaction is zero at equilibrium.