An equilibrium mixture of the reaction, $2{\rm{NO}}\left( {\rm{g}} \right) + {{\rm{O}}_2}\left( {\rm{g}} \right)\;\leftrightharpoons\;2{\rm{N}}{{\rm{O}}_2}\left( {\rm{g}} \right)$ contains 0.120 mole of ${\rm{N}}{{\rm{O}}_2}$, 0.080 mole of ${\rm{NO}}$ and 0.640 mole of ${{\rm{O}}_2}$ in a 4 litre flask at constant temperature. The value of ${K_c}$ for the reaction at this temperature is:

JEE CHEMISTRY MCQ: Equilibrium

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An equilibrium mixture of the reaction,

2\mathrm{NO}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right)\;\leftrightharpoons\;2\mathrm{N}{\mathrm{O}_2}\left( \mathrm{g} \right)

contains 0.120 mole of

\mathrm{N}{\mathrm{O}_2}

, 0.080 mole of

\mathrm{NO}

and 0.640 mole of

{\mathrm{O}_2}

in a 4 litre flask at constant temperature. The value of

{K_c}

for the reaction at this temperature is:

View answer and explanation

Correct option: A

Option A (14) is correct because the equilibrium constant Kc is calculated by dividing the square of the concentration of NO2 by the product of the square of the concentration of NO and the concentration of O2.

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.