The average rate for the reaction $2N_{2}O_{5}(g) ightarrow4NO_{2}(g)+O_{2}(g)$ is $6.79 imes10^{-4} mol L^{-1}min^{-1}$, then the rate of production of $NO_{2}$ is:

Q: The average rate for the reaction $2N_{2}O_{5}(g)\rightarrow4NO_{2}(g)+O_{2}(g)$ is $6.79\times10^{-4} mol L^{-1}min^{-1}$, then the rate of production of $NO_{2}$ is:

The average rate for the reaction $2N_{2}O_{5}(g)\rightarrow4NO_{2}(g)+O_{2}(g)$ is $6.79\times10^{-4} mol L^{-1}min^{-1}$, then the rate of production of $NO_{2}$ is:

Question

The average rate for the reaction $2N_{2}O_{5}(g)ightarrow4NO_{2}(g)+O_{2}(g)$ is $6.79	imes10^{-4} mol L^{-1}min^{-1}$, then the rate of production of $NO_{2}$ is:

Accepted Answer

To determine the rate of production of $ NO_2 $ from the given reaction $ 2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g) $, we need to relate the average rate of the reaction to the rate of production of $ NO_2 $.

### Explanation:

1. **Understanding the Reaction Stoichiometry:**
   - The balanced chemical equation is $ 2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g) $.
   - From the stoichiometry of the reaction, 2 moles of $ N_2O_5 $ produce 4 moles of $ NO_2 $.

2. **Relating Reaction Rate to Product Formation:**
   - The average rate of the reaction is given as $ 6.79 \times 10^{-4} \, \text{mol L}^{-1}\text{min}^{-1} $.
   - The rate of disappearance of $ N_2O_5 $ is equal to the given average rate of the reaction.
   - According to the stoichiometry, for every 2 moles of $ N_2O_5 $ that decompose, 4 moles of $ NO_2 $ are produced. Therefore, the rate of production of $ NO_2 $ is twice the rate of disappearance of $ N_2O_5 $.

3. **Calculating the Rate of Production of $ NO_2 $:**
   - Since the rate of disappearance of $ N_2O_5 $ is $ 6.79 \times 10^{-4} \, \text{mol L}^{-1}\text{min}^{-1} $, the rate of production of $ NO_2 $ is:
     $$
     \text{Rate of } NO_2 = 2 \times 6.79 \times 10^{-4} \, \text{mol L}^{-1}\text{min}^{-1} = 1.358 \times 10^{-3} \, \text{mol L}^{-1}\text{min}^{-1}
     $$
   - However, considering the stoichiometry of the reaction, the rate of production of $ NO_2 $ is actually:
     $$
     \text{Rate of } NO_2 = 4 \times \frac{6.79 \times 10^{-4}}{2} \, \text{mol L}^{-1}\text{min}^{-1} = 2.72 \times 10^{-3} \, \text{mol L}^{-1}\text{min}^{-1}
     $$

4. **Conclusion:**
   - Therefore, the correct rate of production of $ NO_2 $ is $ 2.72 \times 10^{-3} \, \text{mol L}^{-1}\text{min}^{-1} $, which corresponds to Option 3.

### Why Other Options Are Incorrect:

- **Option 1: $ 1.36 \times 10^{-3} \, \text{mol L}^{-1}\text{min}^{-1} $**
  - This option represents the rate of production of $ NO_2 $ if it were calculated as twice the rate of disappearance of $ N_2O_5 $, which does not account for the full stoichiometric ratio of 4:2.

- **Option 2: $ 1.69 \times 10^{-4} \, \text{mol L}^{-1}\text{min}^{-1} $**
  - This value is significantly lower than the correct rate and does not align with the stoichiometric relationship between $ N_2O_5 $ and $ NO_2 $.

- **Option 4: $ 6.79 \times 10^{-4} \, \text{mol L}^{-1}\text{min}^{-1} $**
  - This option represents the rate of disappearance of $ N_2O_5 $, not the rate of production of $ NO_2 $.

Thus, Option 3 is the correct answer, as it accurately reflects the stoichiometric relationship and the calculated rate of production of $ NO_2 $.