JEE Chemistry 2024 Question with Solution

Given: Reaction is

$$2N_2O_4 \rightleftharpoons 4NO_2 + O_2$$

Initial concentration of $N_2O_4$ is $3 \, \text{mol L}^{-1}$, final concentration after $30 \, \text{min}$ is $2.75 \, \text{mol L}^{-1}$.

Find: The value of $x$ in $x \times 10^{-3} \, \text{mol L}^{-1} \text{ min}^{-1}$ for the rate of formation of $NO_2$.

First, calculate the decrease in concentration of $N_2O_4$:

$$\Delta [N_2O_4] = 3 - 2.75 = 0.25 \, \text{mol L}^{-1}$$

From the stoichiometry

$$2N_2O_4 \rightarrow 4NO_2$$

for every $2$ moles of $N_2O_4$ decomposed, $4$ moles of $NO_2$ are formed. Hence, the concentration of $NO_2$ formed is:

$$\Delta [NO_2] = 2 \times 0.25 = 0.50 \, \text{mol L}^{-1}$$

Therefore, the average rate of formation of $NO_2$ over $30 \, \text{min}$ is:

$$\text{Rate} = \frac{0.50}{30} \, \text{mol L}^{-1} \text{ min}^{-1} = \frac{1}{60} \, \text{mol L}^{-1} \text{ min}^{-1}$$

Now compare with the given form:

$$x \times 10^{-3} = \frac{1}{60}$$

So,

$$x = \frac{1}{60} \times 1000 = 16.67 \approx 17$$

Therefore, the correct option is B.

Given:

$$2N_2O_4 \rightleftharpoons 4NO_2 + O_2$$

$[N_2O_4]_0 = 3 \, \text{mol L}^{-1}$, $[N_2O_4] = 2.75 \, \text{mol L}^{-1}$ after $30 \, \text{min}$.

Find: Rate of formation of $NO_2$.

Average rate of decomposition of $N_2O_4$ is:

$$-\frac{\Delta [N_2O_4]}{\Delta t} = -\frac{2.75 - 3.00}{30} = \frac{0.25}{30} = 8.333 \times 10^{-3} \, \text{mol L}^{-1} \text{ min}^{-1}$$

Using stoichiometry,

$$\frac{\text{Rate of formation of } NO_2}{\text{Rate of decomposition of } N_2O_4} = \frac{4}{2} = 2$$

Hence,

$$\text{Rate of formation of } NO_2 = 2 \times 8.333 \times 10^{-3} = 16.666 \times 10^{-3} \, \text{mol L}^{-1} \text{ min}^{-1}$$

Thus,

$$x \approx 16.7 \approx 17$$

Therefore, the value of $x$ is 17, so the correct option is B.

Note: The second extracted approach in the source solution mentions $N_2O_5$, which is inconsistent with the question. The numerical working still uses the same stoichiometric ratio and gives the same final answer.