For the gaseous phase reaction, $2NO\leftrightharpoons{N_2} + {\rm{ }}{O_2}$, $\Delta {H^ \circ } = {\rm{ }} - {\rm{ }}43.5{\rm{ }}\;k\;cal{\rm{ }}\;mo{l^{ - 1}},$ which statement is correct for, ${N_2}\left( g \right){\rm{ }} + {\rm{ }}{O_2}\left( g \right){\rm{ }}\leftrightharpoons{\rm{ }}2NO\left( g \right)?$

$K$ is independent of temperature

$K$ increases as temperature decreases

$K$ varies with addition of $NO$

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JEE CHEMISTRY MCQ: Equilibrium

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For the gaseous phase reaction,

2NO\leftrightharpoons{N_2} + \mathrm{ }{O_2}

\Delta {H^ \circ } = \mathrm{ } - \mathrm{ }43.5\mathrm{ }\;k\;cal\mathrm{ }\;mo{l^{ - 1}},

  which statement is correct for,

{N_2}\left( g \right)\mathrm{ } + \mathrm{ }{O_2}\left( g \right)\mathrm{ }\leftrightharpoons\mathrm{ }2NO\left( g \right)?

View answer and explanation

Correct option: C

K

decreases as temperature decreases

Option C (

K

decreases as temperature decreases) is correct because the reverse reaction (formation of NO) is endothermic, so decreasing the temperature will shift the equilibrium to the left, decreasing K.

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.