1. Given reaction

$2\,\mathrm{N_2O_5} \rightarrow 4\,\mathrm{NO_2} + \mathrm{O_2}$

2. Change in concentration of $\mathrm{N_2O_5}$ in 30 min

Initial concentration: $[\mathrm{N_2O_5}]_0 = 3.00\,\mathrm{mol\,L^{-1}}$

After 30 min: $[\mathrm{N_2O_5}] = 2.75\,\mathrm{mol\,L^{-1}}$

Decrease in concentration: $\Delta [\mathrm{N_2O_5}] = 3.00 - 2.75 = 0.25\,\mathrm{mol\,L^{-1}}$

So, average rate of disappearance of $\mathrm{N_2O_5}$ is $\text{Rate of disappearance of } \mathrm{N_2O_5} = \frac{0.25}{30} = 8.33\times 10^{-3}\,\mathrm{mol\,L^{-1}\,min^{-1}}$

3. Use stoichiometry to find rate of formation of $\mathrm{NO_2}$

From the balanced equation: $2\,\mathrm{N_2O_5} \rightarrow 4\,\mathrm{NO_2}$

This means:

• for every $2$ moles of $\mathrm{N_2O_5}$ consumed,
• $4$ moles of $\mathrm{NO_2}$ are formed.

Hence, $\frac{\text{rate of formation of }\mathrm{NO_2}}{\text{rate of disappearance of }\mathrm{N_2O_5}} = \frac{4}{2} = 2$

Therefore, $\text{Rate of formation of }\mathrm{NO_2} = 2\times 8.33\times 10^{-3}$ $= 16.67\times 10^{-3}\,\mathrm{mol\,L^{-1}\,min^{-1}}$

So, $x = 16.67$

Nearest integer: $x \approx 17$

4. Comparison with stored answer

Stored correct answer = $17$

Our derived answer also = $17$.

Hence, the answer agrees with the stored correct answer.