In the reaction `2NO_(2) + O_(2) to 2NO_(2)`, if the rate of disappearance of `O_(2)` is 16gm.min 1, then the rate of appearance of `NO_(2)` is

Options

A. `90 gm. Min^(-1)`
B. `46gm."min"^(-1)`
C. `28 gm."min"^(-1)`
D. `32 gm."min"^(-1)`

Explanation

To solve the problem, we need to analyze the given reaction and the relationship between the rates of disappearance and appearance of the reactants and products.

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
The reaction is given as:
\[ 2NO + O_2 \rightarrow 2NO_2 \]

2. **Identify the rates of disappearance and appearance**:
According to the stoichiometry of the reaction, the rates of disappearance and appearance can be expressed as:
\[ -\frac{1}{2} \frac{d[NO]}{dt} = -\frac{1}{1} \frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt} \]

3. **Given information**:
We know that the rate of disappearance of \(O_2\) is given as:
\[ \frac{d[O_2]}{dt} = -16 \, \text{gm/min} \]

4. **Relate the rate of disappearance of \(O_2\) to the rate of appearance of \(NO_2\)**:
From the stoichiometric coefficients in the balanced equation, we can relate the rates:
\[ -\frac{1}{1} \frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt} \]
Rearranging gives:
\[ \frac{d[NO_2]}{dt} = -2 \frac{d[O_2]}{dt} \]

5. **Substituting the known rate**:
Substituting the value of \(\frac{d[O_2]}{dt}\):
\[ \frac{d[NO_2]}{dt} = -2 \times (-16) \, \text{gm/min} \]
\[ \frac{d[NO_2]}{dt} = 32 \, \text{gm/min} \]

6. **Conclusion**:
Therefore, the rate of appearance of \(NO_2\) is:
\[ \frac{d[NO_2]}{dt} = 32 \, \text{gm/min} \]

### Final Answer:
The rate of appearance of \(NO_2\) is **32 gm/min**.

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