Question

Following reaction is pseudo-unimolecular w.r.t. $C_6{H}_5{N}_{2}Cl(A)$
$\underset{(A)}{C_6{H}_5{N}_{2}Cl}+H_{2}O\stackrel{\Delta }{\to }{C}_6{H}_{5}OH+N_2+HCl$
$50mL$ of $1M$ benzene diazonium chloride $(A)$ is taken. After $1h,1.226L$ of $N_2$ gas at $1atm$ and $300K$ is obtained. Thus, half-life of the reaction is $(\log 250=2.40)$

• A. 0.092 min
• B. 7.53 min
• C. 10.86 min
• D. None of these

Answer 1

Answer: (B)

$C_6{H}_5{N}_{2}Cl+H_{2}O\stackrel{\Delta }{\to }{C}_6{H}_{5}OH+N_2+HCl$

$\begin{array}{ccc}Initial& a& 0\\ Attimet& \left(a-x\right)& x\end{array}$

$50mL$ of $1MA=50\times 1$ millimol

$=\frac{50\times 1}{1000}mol=0.05mol$

$\therefore a=0.05mol$

At time $t,N_2$ formed $=x=\frac{pV}{RT}=\frac{1\times 1.226}{0.0821\times 300}$

$=0.0498mol$

$\therefore (a-x)=0.050-0.0498$

$=2\times 10^{-4}mol$

$k=\frac{2.303}{t}\log \left(\frac{a}{a-x}\right)$

$=\frac{2.303}{60\min }\log \frac{0.05}{2\times 10^{-4}}=\frac{2.303}{60}\log 250$

$=\frac{2.303}{60}\times 2.40$

Half-life $\left(t_{50}\right)=\frac{2.303\log 2}{k}=\frac{2.303\times 0.3010}{2.303\times 2.4}\times 60$

$=7.53min$