Question

A container of $2litre$ capacity contains 4 moles of $N_2{O}_5$. On heating to $100{}^{\circ }C,N_2{O}_5$ undergoes dissociation to give $N{O}_2$ and $O_2$. Select the correct answer among the following if the rate constant for decomposition of $N_2{O}_5$ is $6.3\times {10}^{-4}se{c}^{-1}.(ln2=0.693)$

• A. The mole ratio before and after complete dissociation (of total gaseous moles) is 4 : 2
• B. Half life of $N_2{O}_5$ is 1100 sec and it is independent of temperature
• C. Time required to complete 87.5% of reaction is 11 min
• D. If the volume of container is doubled at constant temperature, the rate of decomposition of $N_2{O}_5$ becomes half of the initial rate

Answer 1

The correct option is D If the volume of container is doubled at constant temperature, the rate of decomposition of $N_2{O}_5$ becomes half of the initial rate

$$\begin{array}{cccccc}& N_2{O}_5& \to & 2N{O}_2& +& \frac{1}{2}{O}_2\\ \text{initial mole}& 4& & 0& & 0\\ \text{Moles after diss.}& 0& & 8& & 2\end{array}$$
$\therefore$ Mole ratio $=\frac{4}{10}=\frac{2}{5}$
$\frac{t_1}{2}=\frac{0.693}{K}=\frac{0.693}{6.3\times {10}^{-4}}=1100$ sec but it depends upon temperature as K also depends upon temperature
$t_{87.5\%}=\frac{1}{63\times {10}^{-4}}ln\frac{100}{100-87.5}=330sec=5.5min$
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
$\therefore r_2=K\frac{1}{2}\left[N_2{O}_5\right]\therefore \frac{r_1}{r_2}=\frac{2}{1}$