If the rate of decomposition of N 2 ​ O 5 ​ during a certain time internal is `2.4×10^−4` molL −1 min −1 . N 2 ​ O 5 ​ →2NO 2 ​ + 2 1 ​ O 2 ​ What is the rate of formation of NO 2 in ​ molL −1 min −1 ?

Explanation

To solve the problem, we need to determine the rate of formation of NO2 from the given rate of decomposition of N2O5. The balanced chemical equation for the reaction is:

\[ \text{N}_2\text{O}_5 \rightarrow 2\text{NO}_2 + \frac{1}{2}\text{O}_2 \]

### Step-by-Step Solution:

1. **Identify the Rate of Decomposition**:
The rate of decomposition of N2O5 is given as:
\[ \text{Rate of decomposition} = -\frac{d[\text{N}_2\text{O}_5]}{dt} = 2.4 \times 10^{-4} \, \text{mol L}^{-1} \text{min}^{-1} \]

2. **Write the Rate Expression**:
From the balanced equation, we can express the rates of the reactants and products in terms of their stoichiometric coefficients. The rate of formation of NO2 can be related to the rate of decomposition of N2O5 as follows:
\[ -\frac{1}{1} \frac{d[\text{N}_2\text{O}_5]}{dt} = \frac{2}{1} \frac{d[\text{NO}_2]}{dt} \]

3. **Relate the Rates**:
Rearranging the expression gives:
\[ \frac{d[\text{NO}_2]}{dt} = 2 \left(-\frac{d[\text{N}_2\text{O}_5]}{dt}\right) \]

4. **Substitute the Given Rate**:
Substituting the given rate of decomposition into the equation:
\[ \frac{d[\text{NO}_2]}{dt} = 2 \times 2.4 \times 10^{-4} \, \text{mol L}^{-1} \text{min}^{-1} \]

5. **Calculate the Rate of Formation of NO2**:
Performing the multiplication:
\[ \frac{d[\text{NO}_2]}{dt} = 4.8 \times 10^{-4} \, \text{mol L}^{-1} \text{min}^{-1} \]

### Final Answer:
The rate of formation of NO2 is:
\[ \frac{d[\text{NO}_2]}{dt} = 4.8 \times 10^{-4} \, \text{mol L}^{-1} \text{min}^{-1} \]