The rate of the reaction
`2N_(2)O_(5) to 4NO_(2) + O_(2)`
can be written in three ways:
`(-d[N_(2)O_(5)])/(dt)=k[N_(2)O_(5)]`
`(d[NO_(2)])/(dt)=k'[N_(2)O_(5)]`
`(d[O_(2)])/(dt)=k''[N_(2)O_(5)]`
The relationship between k and k′ and between k and k′′ are-

Options

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

Explanation

To solve the problem, we need to analyze the given reaction and the relationships between the rate constants \( k \), \( k' \), and \( k'' \). The reaction is:

\[ 2N_{2}O_{5} \rightarrow 4NO_{2} + O_{2} \]

### Step 1: Write the Rate of Reaction
The rate of reaction can be expressed in terms of the change in concentration of reactants and products. For the given reaction, we can express the rate as:

\[ -\frac{1}{2} \frac{d[N_{2}O_{5}]}{dt} = \frac{1}{4} \frac{d[NO_{2}]}{dt} = \frac{1}{1} \frac{d[O_{2}]}{dt} \]

### Step 2: Relate the Rates to the Rate Constants
From the problem, we have the following relationships based on the rate expressions:

1. \(-\frac{d[N_{2}O_{5}]}{dt} = k[N_{2}O_{5}]\)
2. \(\frac{d[NO_{2}]}{dt} = k'[N_{2}O_{5}]\)
3. \(\frac{d[O_{2}]}{dt} = k''[N_{2}O_{5}]\)

### Step 3: Express the Rate Constants
From the rate expressions, we can express the rate constants in terms of the overall rate:

- For \(N_{2}O_{5}\):
\[ -\frac{d[N_{2}O_{5}]}{dt} = k[N_{2}O_{5}] \implies k = -\frac{1}{2} \frac{d[N_{2}O_{5}]}{dt} \]

- For \(NO_{2}\):
\[ \frac{d[NO_{2}]}{dt} = k'[N_{2}O_{5}] \implies k' = \frac{4}{2} k = 2k \]

- For \(O_{2}\):
\[ \frac{d[O_{2}]}{dt} = k''[N_{2}O_{5}] \implies k'' = \frac{1}{2} k \]

### Step 4: Summarize the Relationships
From the above derivations, we can summarize the relationships between the rate constants as follows:

1. \( k' = 2k \)
2. \( k'' = \frac{1}{2}k \)

### Final Answer
Thus, the relationships between the rate constants are:

- \( k' = 2k \)
- \( k'' = \frac{1}{2}k \)