Question

The container of 2 litre contains 4 mole of $N_2{O}_5$. On heating to ${100}^{o}C,N_2{O}_5$ undergoes complete dissociation to $N{O}_2$ and $O_2$. Select the correct answers if rate constant for decomposition of $N_2{O}_5$ is $6.2\times {10}^{-4}se{c}^{-1}$.
1. The mole ratio before and after dissociation is 4:2.
2. The half-life of $N_2{O}_5$ is 1117.7 sec and it is independent of temperature.
3. The time required to complete 40% of reaction is 824 sec.
4. If the volume of the container is doubled, the rate of decomposition becomes half of the initial rate

• A. 1, 3, 4
• B. 1, 2, 3, 4
  
• C. 3, 4
  
• D. None of these

Answer 1

The correct option is C 3, 4

$N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$
Initial no. of mole $4$ $0$ $0$
Mole after dissociation $0$ $8$ $2$
$\therefore$ Molar ratio $=\frac{4}{10}=2:5$

$t_{1/2}=\frac{0.693}{K}=\frac{0.693}{6.2\times {10}^{-4}}=1117.7sec$

It depends upon temperature as K also depends upon temperature.

$t_{40}$% $=\frac{2.303}{6.2\times {10}^{-4}}log\frac{100}{60}=824sec$

$Rate=K\left[N_2{O}_5\right]$

Thus, $r_1=K\left[N_2{O}_5\right]$

If V is doubled, the concentration becomes half.

$\therefore r_2=K\times \frac{1}{2}\left[N_2{O}_5\right]$

$\therefore \frac{r_1}{r_2}=\frac{2}{1}$