Consider a system containing `NO_(2)` and `SO_(2)` in which `NO_(2)` is consumed in the following two parallel reactions.
`2NO_(2) overset(K_(1)) to N_(2)O_(4), NO_(2) + SO_(2) overset(K_(2)) to NO + SO_(3)`
The rate of disappearance of `NO_(2)` will be equal to

Options

A. `K_(1)[NO_(2)]^(2) + K_(2)[NO_(2)]`
B. `K_(1)[NO_(2)]^(2) + K_(2)[NO_(2)[SO_(2)]`
C. `2K_(1)[NO_(2)]^(2)`
D. `2K_(1)[NO_(2)]^(2) + K_(2)[NO_(2)][SO_(2)]`

Explanation

To find the rate of disappearance of \( NO_2 \) in the given parallel reactions, we can analyze each reaction separately and then combine their rates.

### Step 1: Write the reactions and their rates
The two reactions are:
1. \( 2NO_2 \overset{K_1}{\rightarrow} N_2O_4 \)
2. \( NO_2 + SO_2 \overset{K_2}{\rightarrow} NO + SO_3 \)

### Step 2: Write the rate expressions for each reaction
For the first reaction, the rate of disappearance of \( NO_2 \) can be expressed as:
\[ -\frac{d[NO_2]}{dt} = K_1 [NO_2]^2 \]
This is because the stoichiometry of \( NO_2 \) is 2, so we divide the rate by 2.

For the second reaction, the rate of disappearance of \( NO_2 \) is:
\[ -\frac{d[NO_2]}{dt} = K_2 [NO_2][SO_2] \]
Here, the stoichiometry of \( NO_2 \) is 1, so we do not need to divide by any factor.

### Step 3: Combine the rates
Since both reactions are parallel and contribute to the disappearance of \( NO_2 \), we can combine their rates:
\[ -\frac{d[NO_2]}{dt} = K_1 [NO_2]^2 + K_2 [NO_2][SO_2] \]

### Final Expression
Thus, the rate of disappearance of \( NO_2 \) is given by:
\[ -\frac{d[NO_2]}{dt} = K_1 [NO_2]^2 + K_2 [NO_2][SO_2] \]