For the reaction system `2NO(g) + O_(2)(g) rarr 2NO(g)` volume is suddenly produced 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 eighth times of its initial value
D. ( d) Increase to four times of its initial value

Explanation

To solve the problem, we need to analyze how the rate of the reaction changes when the volume is suddenly halved, which effectively doubles the concentrations of the reactants.

### Step-by-Step Solution:

1. **Write the Rate Law**:
The rate of the reaction can be expressed using the rate law:
\[ \text{Rate} (R) = k [NO]^2 [O_2]^1 \]
where \( k \) is the rate constant, \([NO]\) is the concentration of nitrogen monoxide, and \([O_2]\) is the concentration of oxygen.

2. **Initial Concentrations**:
Let’s denote the initial concentrations of \( NO \) and \( O_2 \) as \([NO]_0\) and \([O_2]_0\) respectively. The initial rate of the reaction can be written as:
\[ R_0 = k [NO]_0^2 [O_2]_0 \]

3. **Effect of Halving the Volume**:
When the volume is suddenly halved, the concentrations of the gases will double:
\[ [NO] = 2[NO]_0 \quad \text{and} \quad [O_2] = 2[O_2]_0 \]

4. **New Rate of Reaction**:
Substitute the new concentrations into the rate law:
\[ R' = k (2[NO]_0)^2 (2[O_2]_0)^1 \]
Simplifying this gives:
\[ R' = k \cdot 4[NO]_0^2 \cdot 2[O_2]_0 = 8k [NO]_0^2 [O_2]_0 \]

5. **Relate New Rate to Initial Rate**:
We can express the new rate \( R' \) in terms of the initial rate \( R_0 \):
\[ R' = 8R_0 \]
This means the rate of the reaction increases to 8 times its initial value.

### Conclusion:
The rate of the reaction will increase to 8 times its initial value when the volume is suddenly halved.

### Final Answer:
The correct option is C: The rate of reaction will increase to 8 times its initial value.

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