Question

In the first order reaction $N{H}_{4}N{O}_2\left(aq\right)\to N_2\left(g\right)+2H_{2}O\left(l\right)$, the volume of $N_2$ after $20min$ and after completion of reaction are $40mL$ and $70mL$ respectively. The value of rate constant is :
Take:
$log\frac{7}{3}=0.37$

• A. $0.005mi{n}^{-1}$
• B. $0.122mi{n}^{-1}$
• C. $0.042mi{n}^{-1}$
• D. $0.2mi{n}^{-1}$

Answer 1

The correct option is C $0.042mi{n}^{-1}$

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

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,
$kt=ln\frac{a}{a-x}$

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

$\therefore$
$k=\frac{2.303}{t}\log \left(\frac{V\infty }{V\infty -V_t}\right)$

$\Rightarrow k=\frac{2.303}{20}\log \left(\frac{70}{70-40}\right)$

$k=\frac{2.303}{20}\log \left(\frac{70}{30}\right)$

$k=\frac{2.303}{20}\log \frac{7}{3}$

$k=0.042mi{n}^{-1}$