Question

The following data represents for the decomposition of $N{H}_{4}N{O}_2$ is an aqueous solution:

Time in min	10	15	20	25	$\infty$
Volume of $N_2$ (in mL)	6.25	9.0	11.40	13.64	33.05

Velocity constant is:

• A. $2\times {10}^{-3}mi{n}^{-1}$
• B. $4\times {10}^{-3}mi{n}^{-1}$
• C. $3\times {10}^{-3}mi{n}^{-1}$
• D. none of these

Answer 1

The correct option is A $2\times {10}^{-3}mi{n}^{-1}$

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

At $t=0$	$a$	$0$	$0$
At $t=t$	$a-X$	$X$	$X$

The volume of Nitrogen formed at any time is proportional to the mass of $N{H}_{4}N{O}_2$ decomposed at that time.
Therefore,
At $t=\infty ,V_{N_2}=33.05mL\therefore a\propto 33.05$
I $t=10,V_{N_2}=6.25mL\therefore X\propto 6.25$
II $t=15,V_{N_2}=9.0mL\therefore X\propto 9.0$
III $t=20,V_{N_2}=11.40mL\therefore X\propto 11.40$
IV $t=25,V_{N_2}=13.65mL\therefore X\propto 13.65$

$\therefore K=\frac{2.303}{t}log\frac{a}{(a-X)}$

$\therefore$ for I, $t=10a\propto 33.05$ and $X\propto 6.25$

$K_1=\frac{2.303}{10}log\frac{33.05}{33.05-6.25}=2\times {10}^{-3}mi{n}^{-1}$

Similarly, calculate $K$ for each case. The value of $K$ comes almost constant and thus, reaction obeys $I^{st}$ order kinetics.