Question

$N _{2}$ is obtained by the decomposition of benzene diazonium chloride at constant temperature. It is a first-order reaction the various volumes obtained at different time intervals are as

Time $(\min)$	$0$	$20$	$50$	$50$	$\infty$
$N_2(mL)$	$0$	$10$	$25$	$32$	$162$

The rate constant of the reaction is

• A. $3.54 \times 10^{-2} \min ^{-1}$
• B. $3.2 \times 10^{-3} \min ^{-1}$
• C. $3.89 \times 10^{-3} \min ^{-1}$
• D. $3.1 \times 10^{-5} \min ^{-1}$

Answer 1

Answer: (B)

Decomposition of benzene diazonium chloride occur the following manner.
$C _{6} H _{5} N = NCl \longrightarrow C _{6} H _{5} Cl + N _{2}$
The volume of nitrogen obtained shows the extent of decomposition of benzene diazonium chloride so
$a \propto V_{\infty}=162\, mL$
$(a-x) \propto\left(V_{\infty}-V_{t}\right) mL$
For first order reaction,
$k=\frac{2.303}{t} \log _{10} \frac{V_{\infty}-V_{0}}{V_{\infty}-V_{t}}$
or $k=\frac{2.303}{t} \log _{10} \frac{V_{\infty}}{V_{\infty}-V_{t}}$
$\left(\because V_{0}=0\right)$
At $20\, \min$
$k =\frac{2.303}{20} \log _{10} \frac{162}{162-10}$
$=\frac{2.303}{20} \log _{10} \frac{162}{152}$
$=3.18 \times 10^{-3} \min ^{-1}$
$\approx 3.2 \times 10^{-3} \min ^{-1}$