For the reaction, `N_2O_5(g) rarr 2NO_2(g)+1//2O_2(g)`, the value of rate of disappearance of `N_2O_5` is given as `6.25xx10^(-3) "mol L"^(-1)s^(-1)`. The rate of formation of `NO_2 and O_2` is given respectively as

Options

A. `6.25xx10^(-3)"mol L"^(-1)& 6.25xx10^(-3)"mol L"^(-1)s^(-1)`
B. `1.25xx10^(-2)"mol L"^(-1)& 3.125xx10^(-3)"mol L"^(-1)s^(-1)`
C. `6.25xx10^(-3)"mol L"^(-1) & 3.125xx10^(-3)"mol L"^(-1)s^(-1)`
D. `1.25xx10^(-2)"mol L"^(-1)& 6.25xx10^(-3)"mol L"^(-1)s^(-1)`

Answer

To solve the problem, we need to determine the rates of formation of \( NO_2 \) and \( O_2 \) from the given rate of disappearance of \( N_2O_5 \). The balanced reaction is:

\[ N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2}O_2(g) \]

### Step 1: Write the Rate Expressions
For the reaction, the rate of disappearance of \( N_2O_5 \) and the rates of formation of \( NO_2 \) and \( O_2 \) can be expressed as:

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

### Step 2: Substitute the Given Rate
We know the rate of disappearance of \( N_2O_5 \):

\[ -\frac{d[N_2O_5]}{dt} = 6.25 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 3: Calculate the Rate of Formation of \( NO_2 \)
From the rate expression, we can find the rate of formation of \( NO_2 \):

\[ \frac{d[NO_2]}{dt} = -2 \left(-\frac{d[N_2O_5]}{dt}\right) = 2 \times 6.25 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]

Calculating this gives:

\[ \frac{d[NO_2]}{dt} = 12.5 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} = 1.25 \times 10^{-2} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 4: Calculate the Rate of Formation of \( O_2 \)
Next, we calculate the rate of formation of \( O_2 \):

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

Since the stoichiometric coefficient of \( O_2 \) is \( \frac{1}{2} \):

\[ \frac{d[O_2]}{dt} = \frac{1}{2} \left(-\frac{d[N_2O_5]}{dt}\right) = \frac{1}{2} \times 6.25 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]

Calculating this gives:

\[ \frac{d[O_2]}{dt} = 3.125 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Final Answers
Thus, the rates of formation are:

\[ \frac{d[NO_2]}{dt} = 1.25 \times 10^{-2} \, \text{mol L}^{-1} \text{s}^{-1} \]
\[ \frac{d[O_2]}{dt} = 3.125 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]