For the reaction system : `2NO(g) + O_2(g) ------>2NO_2(g)`,volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to `O_2` and second order with respect to NO, the rate of reaction will

Options

A. Diminish to one-fourth of its initial value
B. Diminish to one-eighth of its initial value
C. Increase to eight times of its initial value
D. Increase to four times of its initial value.

Explanation

To solve the problem, we need to analyze how the reduction of volume affects the concentrations of the reactants and subsequently the rate of the reaction.

### Step-by-Step Solution:

1. **Identify the Reaction and Order of Reaction**:
The given reaction is:
\[ 2NO(g) + O_2(g) \rightarrow 2NO_2(g) \]
The reaction is first order with respect to \( O_2 \) and second order with respect to \( NO \).

2. **Write the Rate Law Expression**:
The rate law expression for the reaction can be written as:
\[ \text{Rate} = k [NO]^2 [O_2]^1 \]
where \( k \) is the rate constant.

3. **Understand the Effect of Volume Change**:
When the volume is suddenly reduced to half, the concentration of the gases will change. Concentration is defined as:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]
Therefore, if the volume is halved, the concentration will double:
\[ [NO]_{\text{new}} = 2[NO]_{\text{initial}} \]
\[ [O_2]_{\text{new}} = 2[O_2]_{\text{initial}} \]

4. **Substitute New Concentrations into the Rate Law**:
Now, substituting the new concentrations into the rate law:
\[ \text{Rate}_{\text{new}} = k (2[NO])^2 (2[O_2])^1 \]

5. **Calculate the New Rate**:
Expanding the expression:
\[ \text{Rate}_{\text{new}} = k (4[NO]^2) (2[O_2]) = 8k [NO]^2 [O_2] \]
Thus, we can express the new rate in terms of the initial rate:
\[ \text{Rate}_{\text{new}} = 8 \times \text{Rate}_{\text{initial}} \]

6. **Conclusion**:
Therefore, the rate of the reaction after the volume is reduced to half will be **8 times the initial rate**.