MCQ
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- A

- ✓

- C

- D







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$NO(g) + Br_2 (g) \rightleftharpoons NOBr_2 (g)$
$NOBr_2(g)+ NO(g) \longrightarrow 2NOBr(g)$
If the second step is the rate determining step, the order of the reaction with respect to $NO(g)$ is

Reason : Oxidation state of $Cl$ in $HClO_4$ is $+7$ and in $HClO_3$, it is $+ 5$.
$\mathop {C{H_2} = CH - Br}\limits_{(I)} $ $\mathop {{\text{C}}{{\text{H}}_3} - Br}\limits_{\left( {II} \right)} $ $\mathop {\begin{array}{*{20}{c}}
{C{H_3} - CH - Br} \\
{|\,\,} \\
{\,\,\,C{H_3}}
\end{array}}\limits_{\left( {III} \right)} $ $\mathop {C{H_3}C{H_2}Br}\limits_{\left( {IV} \right)} $
