`2N_2O_5 rarr 4 NO_2 +O_2`
If `-(D[N_2O_5])/(dt) =k_1[N_2O_5]`
`(d[NO_2])/(dt) =k_2[N_2O_5]`
` ([O_2])/(dt) =k_3[N_2O_5]`
What is the relation between ` k_1, k_2` and ` k_3 ?`.

Options

A. `2k_1=3k_2=1/2 k_3`
B. `2k_1=k_2=4k_3`
C. `k_1=2k_2=4k_3`
D. `k_1=2k_2=1/2k_3`

Answer

To find the relationship between the rate constants \( k_1 \), \( k_2 \), and \( k_3 \) for the reaction:

\[ 2N_2O_5 \rightarrow 4NO_2 + O_2 \]

we start by expressing the rates of the reaction in terms of the stoichiometry of the reactants and products.

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

- The rate of disappearance of \( N_2O_5 \):
\[ -\frac{d[N_2O_5]}{dt} = k_1 [N_2O_5] \]

- The rate of appearance of \( NO_2 \):
\[ \frac{d[NO_2]}{dt} = k_2 [N_2O_5] \]

- The rate of appearance of \( O_2 \):
\[ \frac{d[O_2]}{dt} = k_3 [N_2O_5] \]

### Step 2: Relate the rates to stoichiometry
From the stoichiometry of the reaction, we know that:

- For every 2 moles of \( N_2O_5 \) that react, 4 moles of \( NO_2 \) are produced and 1 mole of \( O_2 \) is produced.

Thus, we can express the rates in terms of the stoichiometric coefficients:

\[ -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{1}{1} \frac{d[O_2]}{dt} \]

### Step 3: Substitute the rate expressions
Substituting the rate expressions into the stoichiometric relationships:

1. From \( -\frac{1}{2} \frac{d[N_2O_5]}{dt} \):
\[ -\frac{1}{2} (k_1 [N_2O_5]) = \frac{1}{4} (k_2 [N_2O_5]) \]

2. From \( -\frac{1}{2} \frac{d[N_2O_5]}{dt} \):
\[ -\frac{1}{2} (k_1 [N_2O_5]) = (k_3 [N_2O_5]) \]

### Step 4: Cancel out \( [N_2O_5] \)
Since \( [N_2O_5] \) is common in all expressions, we can cancel it out (assuming \( [N_2O_5] \neq 0 \)):

1. From the first equation:
\[ -\frac{k_1}{2} = \frac{k_2}{4} \]
Multiplying through by 4 gives:
\[ -2k_1 = k_2 \]

2. From the second equation:
\[ -\frac{k_1}{2} = k_3 \]
Multiplying through by 2 gives:
\[ -k_1 = 2k_3 \]

### Step 5: Combine the relationships
Now we have two relationships:
1. \( k_2 = -2k_1 \)
2. \( k_3 = -\frac{k_1}{2} \)

To express all in terms of \( k_1 \), we can write:
\[ 2k_1 = k_2 \quad \text{and} \quad k_3 = \frac{k_2}{4} \]

### Final Relationship
Thus, the final relationship between \( k_1 \), \( k_2 \), and \( k_3 \) can be summarized as:
\[ 2k_1 = k_2 = 4k_3 \]