Question

In the reaction, $N{H}_{4}N{O}_2(aq.)\to N_2\left(g\right)+2H_{2}O\left(l\right)$, the volume of $N_2$ after $20\text{min}$ and after a long time is $40\text{mL}$ and $70\text{mL}$ respectively. The value of rate constant is:
(Given: Order of reaction = First order)

• A. $\left(\frac{1}{20}\right)\log \left(\frac{7}{4}\right)mi{n}^{-1}$
• B. $\left(\frac{2.303}{1200}\right)\log \left(\frac{7}{3}\right){\sec }^{-1}$
• C. $\left(\frac{1}{20}\right)\log \left(\frac{7}{3}\right){min}^{-1}$
• D. $\left(\frac{2.303}{20}\right)\log \left(\frac{11}{7}\right){min}^{-1}$

Answer 1

The correct option is B $\left(\frac{2.303}{1200}\right)\log \left(\frac{7}{3}\right){\sec }^{-1}$

Given,
$t=20\text{min}=20\times 60=1200\text{sec}N{H}_{4}N{O}_2\left(aq\right)\to N_2\left(g\right)+2H_{2}O\left(l\right)K=\frac{2.303}{t}\log \left(\frac{V_{\infty }+V_0}{V_{\infty }-V_t}\right)\Rightarrow V_0=0\Rightarrow K=\frac{2.303}{1200}\log \left(\frac{70-0}{70-40}\right)=\frac{2.303}{1200}\log \left(\frac{70}{30}\right){\sec }^{-1}\Rightarrow K=\frac{2.303}{1200}\log \frac{7}{3}{\sec }^{-1}$