At $25^\circ C$, the equilibrium ${K_1},{K_2}$ and ${K_3}$ of three reactions are : ${{\rm{N}}_2} + 3{{\rm{H}}_2}\;\leftrightharpoons\;2{\rm{N}}{{\rm{H}}_3};\;{K_1}$ ${{\rm{N}}_2} + {{\rm{O}}_2}\;\leftrightharpoons\;2{\rm{NO}};\;{K_2}$ ${{\rm{H}}_2} + \frac{1}{2}{{\rm{O}}_2}\;\leftrightharpoons\;{{\rm{H}}_2}{\rm{O}};\;{K_3}$ The equilibrium constants for the oxidation of ${\rm{N}}{{\rm{H}}_3}$ by oxygen to give ${\rm{NO}}$ is :

JEE CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1

25^\circ C

, the equilibrium

{K_1},{K_2}

and

{K_3}

of three reactions are :

{\mathrm{N}_2} + 3{\mathrm{H}_2}\;\leftrightharpoons\;2\mathrm{N}{\mathrm{H}_3};\;{K_1}

{\mathrm{N}_2} + {\mathrm{O}_2}\;\leftrightharpoons\;2\mathrm{NO};\;{K_2}

{\mathrm{H}_2} + \frac{1}{2}{\mathrm{O}_2}\;\leftrightharpoons\;{\mathrm{H}_2}\mathrm{O};\;{K_3}

The equilibrium constants for the oxidation of

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

by oxygen to give

\mathrm{NO}

is :

View answer and explanation

Correct option: D

{K_2}K_3^3/{K_1}

Option D (

{K_2}K_3^3/{K_1}

) is correct because it represents the equilibrium constant for the oxidation of ammonia to nitric oxide, derived from the given equilibrium constants.

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.