Consider the reaction, `2N_(2) O_(5) to 4NO_(2) + O_(2)` In the reaction `NO_(2)` is being formed at the rate of 0.0125 mol `L^(-1) s^(-1)`. What is the rate of reaction at this time?

Options

A. 0.0018 mol `L^(-1) s^(-1)`
B. 0.0031 mol `L^(-1) s^(-1)`
C. 0.0041 mol `L^(-1) s^(-1)`
D. 0.050 mol `L^(-1) s^(-1)`

Explanation

To find the rate of reaction for the given chemical reaction \(2N_2O_5 \rightarrow 4NO_2 + O_2\), we will use the relationship between the rate of formation of products and the rate of reaction.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The rate of formation of \(NO_2\) is given as \(0.0125 \, \text{mol L}^{-1} \text{s}^{-1}\).

2. **Write the Rate Expression:**
- The general rate expression for the reaction can be written as:
\[ \text{Rate of reaction} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{1}{1} \frac{d[O_2]}{dt} \]
- Here, the negative sign indicates the disappearance of the reactant \(N_2O_5\), while the positive signs indicate the formation of products \(NO_2\) and \(O_2\).

3. **Relate the Rate of Reaction to the Rate of Formation of \(NO_2\):**
- Since we are given the rate of formation of \(NO_2\), we can express the rate of reaction in terms of this rate:
\[ \text{Rate of reaction} = \frac{1}{4} \frac{d[NO_2]}{dt} \]

4. **Substitute the Given Rate of Formation:**
- Substitute \(\frac{d[NO_2]}{dt} = 0.0125 \, \text{mol L}^{-1} \text{s}^{-1}\) into the rate expression:
\[ \text{Rate of reaction} = \frac{1}{4} \times 0.0125 \]

5. **Calculate the Rate of Reaction:**
- Performing the calculation:
\[ \text{Rate of reaction} = \frac{0.0125}{4} = 0.003125 \, \text{mol L}^{-1} \text{s}^{-1} \]

### Final Answer:
The rate of the reaction at this time is \(0.003125 \, \text{mol L}^{-1} \text{s}^{-1}\).

---