- ✓

- B

- C

- D






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$N{H_3} + {O_2}\xrightarrow[{Pt}]{\Delta }x + {H_2}O$
${x} + {O_2} \to y$
$y + {H_2}O{\text{(excess)}} \to {\text{x + z}}$
The incorrect option is
In the above first order reaction the initial concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ is $2.40 \times 10^{-2}\, \mathrm{~mol} \,\mathrm{~L}^{-1}$ at $318 \,K.$ The concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ after $1\, hour$ was $1.60 \times 10^{-2}\, \mathrm{~mol} \,\mathrm{~L}^{-1}$, The rate constant of the reaction at $318\, \mathrm{~K}$ is $.....\,\times 10^{-3} \mathrm{~min}^{-1}$. (Nearest integer)
[Given: $\log 3=0.477, \log 5=0.699$ ]
$\begin{array}{*{20}{c}}
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{{C_6}{H_5} - CH - C{H_2} - C - C{H_3}}
\end{array}\mathop {\xrightarrow{{(i)\,NaOBr}}}\limits_{(ii)\,{H_2}O/{H^ + }\,(iii)\,\Delta } $ product
product will be :