A quantity of ${\rm{PC}}{{\rm{l}}_5}$ was heated in a 10 litre vessel at $250^\circ C$ to show ${\rm{PC}}{{\rm{l}}_5}\left( {\rm{g}} \right)\leftrightharpoons{\rm{PC}}{{\rm{l}}_3}\left( {\rm{g}} \right) + {\rm{C}}{{\rm{l}}_2}\left( {\rm{g}} \right)$. At equilibrium the vessel contains 0.1 mole of ${\rm{PC}}{{\rm{l}}_5}$, 0.20 mole of ${\rm{PC}}{{\rm{l}}_3}$ and 0.20 mole of ${\rm{C}}{{\rm{l}}_2}$. The equilibrium constant of the reaction is :

JEE CHEMISTRY MCQ: Equilibrium

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A quantity of

\mathrm{PC}{\mathrm{l}_5}

was heated in a 10 litre vessel at

250^\circ C

to show

\mathrm{PC}{\mathrm{l}_5}\left( \mathrm{g} \right)\leftrightharpoons\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)

. At equilibrium the vessel contains 0.1 mole of

\mathrm{PC}{\mathrm{l}_5}

, 0.20 mole of

\mathrm{PC}{\mathrm{l}_3}

and 0.20 mole of

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

. The equilibrium constant of the reaction is :

View answer and explanation

Correct option: C

0.04

Option C (0.04) is correct because the equilibrium constant is calculated by dividing the product of the concentrations of the products by the concentration of the reactant.

More Questions from Equilibrium

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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.