Question

NO₂ required for a reaction is produced by the decomposition of N₂O₅ in CCl₄ as per the equation,

2N₂O₅ (g) → 4NO₂ (g) + O₂

The initial concentration of N₂O₅ is 3.00 mol L^-1 and it is 2.75 mol L^-1 after 30 minutes. The rate of formation of NO₂ is:
1. 4.167 × 10^-3 mol L^-1 min^-1
2. 1.667 × 10^-2 mol L^-1 min^-1
3. 8.333 × 10^-3 mol L^-1 min^-1
4. 2.083 × 10^-3 mol L^-1 min^-1

Answer 1

Correct Answer - Option 2 : 1.667 × 10^-2 mol L^-1 min^-1

Concept:

Nitrogen dioxide (NO₂) is a highly reactive gas which is also termed as nitrogen oxide and it has a high temperature of reddish-brown gas.

From the question, the reaction given is:

2N₂O₅ (g) → 4NO₂ (g) + O₂ (g)

Rate of reaction \(\Rightarrow \frac{{ - 1}}{2}\frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}} = \frac{1}{4}\frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{{d\left[ {{O_2}} \right]}}{{dt}}\)

Rate of reaction \(\Rightarrow \frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{{ - 4}}{2}\frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}}\)

Rate of reaction \(\Rightarrow \frac{{d\left[ {N{O_2}} \right]}}{{dt}} = - 2\frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}}\)

Calculation:

According to the question,

\(\frac{{ - d\left[ {{N_2}{O_5}} \right]}}{{dt}} = - \frac{{\left( {2.75 - 3} \right)}}{{30}} = \frac{{0.25}}{{30}}\;M\;{\rm{mi}}{{\rm{n}}^{ - 1}} = \frac{1}{{120}}\;M\;{\rm{mi}}{{\rm{n}}^{ - 1}}\)

Now,

\(\frac{{d\left[ {N{O_2}} \right]}}{{dt}} = 2 \times \frac{{ - d\left[ {{N_2}{O_5}} \right]}}{{dt}} = 2 \times \frac{1}{{120}}\)

\(\therefore \frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{1}{{60}}M\;{\rm{mi}}{{\rm{n}}^{ - 1}}\)

Thus, the rate of formation of \(\therefore \frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{1}{{60}}M\;{\rm{mi}}{{\rm{n}}^{ - 1}}\) 1.667 × 10^-2 M min^-1