JEE CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1

For the reaction

2 {H}_2+2 {NO} \rightleftharpoons {N}_2+2 {H}_2 {O}

the following mechanism has been suggested :

\begin{aligned} & 2 {NO} \rightleftharpoons {N}_2 {O}_2 \text { [equilibrium constant }=K_1 \text { (fast)] } \\ & {N}_2 {O}_2+{H}_2 \xrightarrow{{K}_2} {~N}_2 {O}+{H}_2 {O} \text { (slow) } \\ & {N}_2 {O}+{H}_2 \xrightarrow{{K}_3} {~N}_2+{H}_2 {O} \text { (fast) } \end{aligned}

According to following mechanism is correctly matched.

View answer and explanation

Correct option: D

-\frac{1}{2} \frac{d[{NO}]}{d t}=K[N O]^2\left[H_2\right]

where

K=K_2 \times K_1

Option D is correct because the rate law derived from the slow step, considering the equilibrium of the fast step, matches the given rate expression.

More Questions from Equilibrium

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\mathrm{K}_1

\mathrm{K}_2

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

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

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

\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}

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

\mathrm{pH}

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

\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

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

\text{O}_2