Question

Diazonium salt decomposes as $C_6{H}_5{N}_2^{+}C{l}^{-}\to C_6{H}_{5}Cl+N_2$. At $0^{\circ }C$ the evolution of$N_2$ becomes two times faster when the initial concentration of the salt is doubled. Therefore, it is

• A. A first order reaction
• B. A second order reaction
• C. Independent of the initial concentration of the salt
• D. A zero order reaction

Answer 1

Answer: (A)

For reaction $C_6{H}_5{N}_2^{+}C{l}^{-}\to C_6{H}_{5}Cl+N_2$
Let the rate law be $r=\frac{d\left[N_2\right]}{dt}=k{\left[C_6{H}_5{N}_2^{+}C{l}^{-}\right]}^x$
$=k{\left[Salt\right]}^x....\left(i\right)$
From given, if $2r=\frac{2d\left[N_2\right]}{dt}=k{\left(2\left[salt\right]\right)}^x...\left(ii\right)$
Then, on dividing$\left(i\right)$ by $\left(ii\right)$ we get
$\frac{1}{2}={\left(\frac{1}{2}\right)}^x\Rightarrow x=1$
$\therefore$ Reaction is of 1st order