Question

In the thermal decomposition of $N_{2}O$ at 830 K, the time required to decompose half of the reactant was 263 sec at the initial pressure 290 mm. It takes 212 sec to decompose half of the reactant if initial pressure was 360 mm. What is the order of the reaction? Also calculate $t_{1/2}$ for $N_{2}O$ decomposition if initial pressure of $N_{2}O$ is 1 atm.

• A. $O.R=2$, Half life $=100.2sec$
• B. $O.R=1$, Half life $=10.02sec$
• C. $O.R=4$, Half life $=1.002sec$
• D. None of these

Answer 1

Answer: (A)

$O.R=2$, Half life $=100.2sec$

$\frac{t_{1/2,1}}{t_{1/2,2}}=\frac{263}{212}=1.24$

$\frac{P_2}{P_1}=\frac{360}{290}=1.24$

Thus $\frac{t_{1/2,1}}{t_{1/2,2}}=\frac{P_2}{P_1}$

$t_{1/2}\propto \frac{1}{P_i}$

Thus, the half life period is inversely proportional to the initial pressure. This is characteristic of second order reaction.

$\frac{t_{1/2,}}{263}=\frac{290}{760}$ (because 1 atm = 760 mm Hg)

$t_{1/2}=100.2sec$