Question

Calculate the rate constant for decomposition of ammonium nitrate from the following data;
$N{H}_{4}N{O}_2\to N_2+2H_{2}O$

Time (minutes)	10	25	$\infty$
Vol.of $N_2$ ml	6.25	13.65	35.05

Given that the reaction follows first order kinectics.

• A. $0.0197mi{n}^{-1}$
• B. $1.97mi{n}^{-1}$
• C. $0.197mi{n}^{-1}$
• D. None of these

Answer 1

Answer: (A)

$0.0197mi{n}^{-1}$

The decomposition of ammonium nitrate is given by

$N{H}_{4}N{O}_2⟶N_2+2H_{2}O$

The kinetics of this reaction is studied by measuring volume of $N_2$ formed at different intervals.

Let, $V_{\infty }$ denotes volume of $H_2$ formed at $\infty$ time.

(time at which reaction is complete)

$\Rightarrow$ $V_{\infty }\propto$ amount of $N{H}_{4}N{O}_2$ present initially $\left(a\right)$

$\Rightarrow$ $V_{\infty }\propto a$ $-\left(i\right)$

Let, $Vt$ is the volume of $N_2$ at any time $t$.

$\Rightarrow$ $Vt\propto$ amount of $N{H}_{4}N{O}_2$ reacted $\left(x\right)$

$\Rightarrow$ $Vt\propto x$ $-\left(ii\right)$

Now, $(a-x)\propto (V_{\infty }-Vt)$ $-\left(iii\right)$

Now, since this reaction follows first order kinetics, so,

$K=\frac{1}{t}ln\left(\frac{a}{a-x}\right)$ $-\left(iv\right)$

where, $K$= rate constant, $t$= time, $a$= initial concentration of reactant, $x$= amount of reactant reacted in time $t$

Putting (i) & (iii) in (iv):-

$K=\frac{1}{t}ln\left(\frac{V_{\infty }}{V_{\infty }-Vt}\right)$ $-\left(v\right)$

According to question, $V_{\infty }=a=35.05$

Now, at $t=10$ min, $V_t=6.25$ ml

Putting the values in $\left(v\right)$ :-

$K=\frac{1}{10}ln\left(\frac{35.05}{35.05-6.25}\right)$

$=\frac{1}{10}ln\left(\frac{35.05}{20.00}\right)$

$=\frac{1}{10}ln\left(1.22\right)=\frac{1}{10}\times 0.1900$

$=0.01900$ $mi{n}^{-1}$