For the reaction, `N_2O_5 to 2NO_2 + O_2` ,
Given `(-d)/(dt)[N_2O_5]=K_1[N_2O_5]`
`d/(dt)[NO_2]=K_2[N_2O_5]`
`d/(dt)[O_2]=K_3[N_2O_5]`, the relation in between of `K_1K_2K_3` is

Options

A. `2K_1=K_2=2K_3`
B. `K_1=K_2=K_3`
C. `2K_1=4K_2=K_3`
D. `2K_1=2K_2=3K_3`

Answer

To derive the relationship between the rate constants \( K_1 \), \( K_2 \), and \( K_3 \) for the reaction

\[ N_2O_5 \rightarrow 2NO_2 + O_2, \]

we start with the rate expressions given in the problem:

1. The rate of disappearance of \( N_2O_5 \):
\[ -\frac{d[N_2O_5]}{dt} = K_1 [N_2O_5] \]

2. The rate of formation of \( NO_2 \):
\[ \frac{d[NO_2]}{dt} = K_2 [N_2O_5] \]

3. The rate of formation of \( O_2 \):
\[ \frac{d[O_2]}{dt} = K_3 [N_2O_5] \]

### Step 1: Write the rate expression based on stoichiometry

From the balanced chemical equation, we can express the rates in terms of the stoichiometric coefficients:

\[ -\frac{d[N_2O_5]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt} = \frac{1}{2} \frac{d[O_2]}{dt} \]

### Step 2: Substitute the rate expressions into the stoichiometric relationship

Substituting the expressions for the rates into the stoichiometric relationship gives us:

\[ -K_1 [N_2O_5] = \frac{1}{2} K_2 [N_2O_5] = \frac{1}{2} K_3 [N_2O_5] \]

### Step 3: Cancel out the common term

Since \( [N_2O_5] \) is common in all terms, we can cancel it out (assuming \( [N_2O_5] \neq 0 \)):

\[ -K_1 = \frac{1}{2} K_2 = \frac{1}{2} K_3 \]

### Step 4: Express relationships between \( K_1 \), \( K_2 \), and \( K_3 \)

From \( -K_1 = \frac{1}{2} K_2 \), we can express \( K_2 \) in terms of \( K_1 \):

\[ K_2 = -2K_1 \]

From \( -K_1 = \frac{1}{2} K_3 \), we can express \( K_3 \) in terms of \( K_1 \):

\[ K_3 = -2K_1 \]

### Step 5: Final relationships

Thus, we can summarize the relationships as follows:

\[ K_1 = K_2/2 = K_3/2 \]

or equivalently:

\[ 2K_1 = K_2 \quad \text{and} \quad 2K_1 = K_3 \]

### Conclusion

The final relationship between the rate constants is:

\[ K_2 = 2K_1 \quad \text{and} \quad K_3 = 2K_1 \]