Question

$\text{NO}_2 \text{ required for a reaction is produced by decomposition of } \text{N}_2\text{O}_5 \text{ in } \text{CCl}_4 \text{ as by equation}$
\[ 2\text{N}_2\text{O}_{5(g)} \rightarrow 4\text{NO}_{2(g)} + \text{O}_{2(g)} \]
$\text{The initial concentration of } \text{N}_2\text{O}_5 \text{ is } 3 \text{ mol L}^{-1} \text{ and it is } 2.75 \text{ mol L}^{-1} \text{ after 30 minutes.}$
$\text{The rate of formation of } \text{NO}_2 \text{ is } x \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1} , \text{ value of } x \text{ is \_\_\_\_\_\_.}$

Answer 1

Correct Answer: 17

Solution: To find the rate of formation of \( NO_2 \), we first need to determine the change in concentration of \( N_2O_4 \) over the given time period.

Initial and Final Concentration:

Initial concentration of \( N_2O_4 \): \([N_2O_4]_0 = 3 \, \text{mol L}^{-1}\)

Final concentration after 30 minutes: \([N_2O_4] = 2.75 \, \text{mol L}^{-1}\)

Change in Concentration:

The change in concentration of \( N_2O_4 \) over 30 minutes is:

\[ \Delta [N_2O_4] = [N_2O_4]_0 - [N_2O_4] = 3 - 2.75 = 0.25 \, \text{mol L}^{-1} \]

Stoichiometry of the Reaction:

According to the reaction:

\[ 2N_2O_4 \rightarrow 4NO_2 \]

For every 2 moles of \( N_2O_4 \) that decompose, 4 moles of \( NO_2 \) are formed, so the ratio is:

\[ \frac{4 \, \text{mol} \, NO_2}{2 \, \text{mol} \, N_2O_4} = 2 \]

Rate of Formation of \( NO_2 \):

The change in concentration of \( NO_2 \) formed is:

\[ \Delta [NO_2] = 2 \times \Delta [N_2O_4] = 2 \times 0.25 = 0.50 \, \text{mol L}^{-1} \]

The rate of formation of \( NO_2 \) over 30 minutes is:

\[ \text{Rate} = \frac{\Delta [NO_2]}{\Delta t} = \frac{0.50 \, \text{mol L}^{-1}}{30 \, \text{min}} = \frac{0.50}{30} \, \text{mol L}^{-1} \text{min}^{-1} = \frac{1}{60} \, \text{mol L}^{-1} \text{min}^{-1} \]

Convert to \( x \):

Given \( x \times 10^{-3} = \frac{1}{60} \), we can find \( x \):

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

Thus, the value of \( x \) is: 17