For the reaction `2N_(2)O_(4)iff 4NO_(2)`, given that `(-d[N_(2)O_(4)])/(dt)=K " and "(d[NO_(2)])/(dt)=K`, then

Options

A. `2k' = k`
B. `k' = 2k`
C. `k' = k`
D. `k = (1)/(4) k'`

Explanation

To solve the problem, we need to analyze the given reaction and the relationships between the rates of change of the concentrations of the reactants and products.

### Step-by-Step Solution:

1. **Write the Reaction**:
The reaction given is:
\[ 2N_{2}O_{4} \rightleftharpoons 4NO_{2} \]

2. **Define the Rates**:
We are given that:
\[ -\frac{d[N_{2}O_{4}]}{dt} = K \]
\[ \frac{d[NO_{2}]}{dt} = K' \]
Here, \( K \) is the rate of decrease of the reactant \( N_{2}O_{4} \) and \( K' \) is the rate of increase of the product \( NO_{2} \).

3. **Relate the Rates to Stoichiometry**:
From the stoichiometry of the reaction, we can express the rates in terms of the change in concentration:
\[ -\frac{1}{2} \frac{d[N_{2}O_{4}]}{dt} = \frac{1}{4} \frac{d[NO_{2}]}{dt} \]
This means that for every 2 moles of \( N_{2}O_{4} \) that react, 4 moles of \( NO_{2} \) are produced.

4. **Substitute the Given Rates**:
Substitute the expressions for the rates into the stoichiometric relationship:
\[ -\frac{1}{2} K = \frac{1}{4} K' \]

5. **Solve for the Relationship Between \( K \) and \( K' \)**:
Rearranging the equation gives:
\[ K' = -2K \]
However, since \( K' \) represents a rate of formation, we consider the absolute values and write:
\[ K = 2K' \]

6. **Conclusion**:
The correct relationship between \( K \) and \( K' \) is:
\[ K = 2K' \]
Thus, the correct option is that \( K \) is twice \( K' \).