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 solve the problem, we need to determine the rate of the overall reaction based on the rate of formation of one of the products, \( NO_2 \). Here’s a step-by-step breakdown of the solution:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2N_2O_5 \rightarrow 4NO_2 + O_2 \]

### Step 2: Identify the rate of formation of \( NO_2 \)
We are given that the rate of formation of \( NO_2 \) is:
\[ \frac{d[NO_2]}{dt} = 0.0125 \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 3: Relate the rate of reaction to the rate of formation of \( NO_2 \)
In a chemical reaction, the rate of reaction can be expressed in terms of the rate of formation of products or the rate of consumption of reactants. For the given reaction, the stoichiometry indicates that for every 4 moles of \( NO_2 \) formed, 1 mole of reaction occurs. Therefore, the rate of reaction can be expressed as:
\[ \text{Rate of reaction} = \frac{1}{4} \frac{d[NO_2]}{dt} \]

### Step 4: Substitute the known value into the rate equation
Now, substitute the given rate of formation of \( NO_2 \) into the equation:
\[ \text{Rate of reaction} = \frac{1}{4} \times 0.0125 \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 5: Perform the calculation
Calculating the above expression:
\[ \text{Rate of reaction} = \frac{0.0125}{4} = 0.003125 \, \text{mol L}^{-1} \text{s}^{-1} \]

### Final Answer
The rate of reaction at this time is:
\[ \text{Rate of reaction} = 0.003125 \, \text{mol L}^{-1} \text{s}^{-1} \]