Question

Following reaction is pseudo-unimolecular w.r.t. $C _{6} H _{5} N _{2} Cl ( A )$
$\underset{(A)}{C _{6} H _{5} N _{2} Cl} + H _{2} O\xrightarrow{\Delta} C _{6} H _{5} OH + N _{2}+ HCl$
$50 \,mL$ of $1 \,M$ benzene diazonium chloride $( A )$ is taken. After $1 \,h , 1.226\, L$ of $N _{2}$ gas at $1\,atm$ and $300 \,K$ 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)

$C _{6} H _{5} N _{2} Cl + H _{2} O \xrightarrow{\Delta} C _{6} H _{5} OH + N _{2}+ HCl$

$\begin{matrix}Initial&a&0\\ At time t &\left(a-x\right)&x\end{matrix}$

$50\, mL$ of $1\, M A =50 \times 1$ millimol

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

$\therefore a=0.05 \,mol$

At time $t,\,\,\,\,\,\, N _{2}$ formed $= x =\frac{p V}{R T}=\frac{1 \times 1.226}{0.0821 \times 300}$

$=0.0498\,mol$

$ \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 $

$=7.53 \,min$