The rate law of a chemical reaction given below:
`2NO+O_(2)rarr 2NO_(2)`
is given as rate `=K[NO]^(2)[O_(2)]`. How will the rate of reaction change if the volume of reaction vessel is reduced to `1//4th` of its original valur?

Answer

To determine how the rate of the reaction changes when the volume of the reaction vessel is reduced to one-fourth of its original value, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Rate Law**: The rate law for the reaction \( 2NO + O_2 \rightarrow 2NO_2 \) is given as:
\[ \text{Rate} = k[NO]^2[O_2] \]

2. **Identify Initial Concentrations**: Let the initial concentrations of \( [NO] \) and \( [O_2] \) be \( C \) and \( D \) respectively.

3. **Effect of Volume Change on Concentration**: When the volume of the reaction vessel is reduced to one-fourth, the concentration of each reactant increases. The concentration is inversely proportional to the volume. Thus, if the volume is reduced to \( \frac{1}{4} \), the new concentrations will be:
\[ [NO]_{new} = 4C \]
\[ [O_2]_{new} = 4D \]

4. **Calculate the New Rate**: Substitute the new concentrations into the rate law:
\[ \text{Rate}_{new} = k[NO]_{new}^2[O_2]_{new} = k(4C)^2(4D) \]
\[ = k \cdot 16C^2 \cdot 4D = 64kC^2D \]

5. **Relate New Rate to Original Rate**: The original rate can be expressed as:
\[ \text{Rate}_{original} = kC^2D \]
Therefore, the new rate is:
\[ \text{Rate}_{new} = 64 \cdot \text{Rate}_{original} \]

6. **Conclusion**: The rate of the reaction increases by a factor of 64 when the volume of the reaction vessel is reduced to one-fourth of its original value.

### Final Answer:
The rate of reaction will change to \( 64 \) times the original rate.