Question

Given the following data for the decomposition of ammonium nitrite in aqueos solution which follows first order kinetics. Calculate the value of rate constant (k) of the reaction.

Time (minutes)	10	15	20	25	$\infty$
Volume of $N_2(c.c)$	6.25	9.00	11.40	13.65	35.05

• A. $k=8.2mi{n}^{-1}$
• B. $k=0.9mi{n}^{-1}$
• C. $k=0.201mi{n}^{-1}$
• D. $k=0.019mi{n}^{-1}$

Answer 1

The correct option is D $k=0.019mi{n}^{-1}$

$N{H}_{4}N{O}_2(aq.)\to N_2\left(g\right)+2H_{2}O\left(l\right)$

Consider,
$V_t$ is the volume of $N_2$ collected at time 't'
$V_{\infty }$ is the volume of $N_2$ collected at the send of the reaction

$1^{st}$ order integrated rate law,
$kt=ln\frac{a}{a-x}$

$a\text{(initial conceentration)}\propto V_{\infty }$
$x\propto V_t$
$a-x\propto V_{\infty }-V_t$

$\therefore$
$k=\frac{1}{t}ln\frac{V_{\infty }}{V\infty -V_t};V_{\infty }=35.05$

The values of $k$ at different times are obtained as follows:

Time	$V_{\infty }-V_t$	$\frac{1}{t}ln\frac{V_{\infty }}{V_{\infty }-V_t}=k_1$
10 min	$35.05-6.25=28.80$	$\frac{1}{10}ln\frac{35.05}{28.80}=0.01976mi{n}^{-1}$
15 min	$35.05-9.00=26.05$	$\frac{1}{15}ln\frac{35.05}{26.05}=0.01976mi{n}^{-1}$
20 min	$35.05-11.40=23.65$	$\frac{1}{20}ln\frac{35.05}{23.65}=0.01964mi{n}^{-1}$
25 min	$35.05-13.65=21.40$	$\frac{1}{25}ln\frac{35.05}{21.40}=0.01971mi{n}^{-1}$

A constant value of $k$ shows that the reaction is of the first order.