`2N_2O_5(g)rarr4NO_2(g)+O_2(g)` what is the ratio of the rate of decomposition of `N_2O_5` to rate of formation of `O_2` ?

Options

A. `1:2`
B. `2:1`
C. `1:4`
D. `4:1`

Answer

To find the ratio of the rate of decomposition of \( N_2O_5 \) to the rate of formation of \( O_2 \) from the reaction:

\[ 2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g) \]

we can follow these steps:

### Step 1: Write the Rate Expressions
For the decomposition of \( N_2O_5 \), the rate of decomposition can be expressed as:

\[ \text{Rate of decomposition of } N_2O_5 = -\frac{1}{2} \frac{d[N_2O_5]}{dt} \]

The negative sign indicates that the concentration of \( N_2O_5 \) is decreasing.

For the formation of \( O_2 \), the rate of formation can be expressed as:

\[ \text{Rate of formation of } O_2 = \frac{1}{1} \frac{d[O_2]}{dt} \]

### Step 2: Set Up the Ratio
We need to find the ratio of the rate of decomposition of \( N_2O_5 \) to the rate of formation of \( O_2 \):

\[ \text{Ratio} = \frac{\text{Rate of decomposition of } N_2O_5}{\text{Rate of formation of } O_2} \]

Substituting the expressions from Step 1:

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

### Step 3: Substitute Coefficients
From the balanced equation, we see that:

- For every 2 moles of \( N_2O_5 \) that decompose, 1 mole of \( O_2 \) is formed.

This means that:

\[ \frac{d[O_2]}{dt} = \frac{1}{2} \left(-\frac{d[N_2O_5]}{dt}\right) \]

### Step 4: Simplify the Ratio
Now, substituting this relationship into our ratio:

\[ \text{Ratio} = \frac{-\frac{1}{2} \frac{d[N_2O_5]}{dt}}{\frac{1}{2} \left(-\frac{d[N_2O_5]}{dt}\right)} \]

The negative signs cancel out, and we simplify:

\[ \text{Ratio} = \frac{\frac{1}{2}}{\frac{1}{2}} = 2 \]

### Final Step: Write the Final Ratio
Thus, the ratio of the rate of decomposition of \( N_2O_5 \) to the rate of formation of \( O_2 \) is:

\[ \text{Ratio} = 2:1 \]