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

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